General Chemistry Flashcards

Question

If a radioactive sample with an activity of 20 counts per minute has an activity of 1.25 counts per minute after 24 hours has passed, what is the half-life of this compound? * 8 hours * 4 hours * 12 hours * 6 hours

Answer 1

6 hours; ## Footnote 20 =\> 10 =\> 5=\> 2.5 =\> 1.25 1 2 3 4 Since it was after 24 hours, and there were 4 so you get 6

Answer 2

3.13% After every half-life, only half of the sample remains. So after 5 half lives only 3.125%

Answer 3

6 hrs After every half-life, only half of the sample remains. If 87.5% has decayed then 12.5% remains (100% - 87.5% = 12.5%) which is how much would remain after 3 half-lives So 3 half-lives = 18hrs, therefore 1 half-life = 6hrs

Answer 4

1/32 After every half-life, only half of the sample remains. If t1/2 = 2.5hrs, then 12.5hrs is equal to 5 half-lives

Answer 5

Be The 3rd ionization energy is the energy required to remove the 3rd electron. Beryllium is the only answer choice that has only 2 valence electrons (or fewer). Therefore the 3rd electron to remove is a core electron for beryllium and is much more difficult to remove than a valence electron.

Answer 6

I- As long as the ions being compared all have the same charge, then the trend for ionic radius will be the same as that for atomic radius. Atomic radius increases going down a group (column) and to the left across a period (row) on the periodic table. The answer choices are all in the same group and I- is the furthest down the group and therefore has the largest ionic radius.

Answer 7

C-O Correct!

Answer 8

F ## Footnote Electron affinities increase (get more negative) going left to right across period 2, with the exception that N has a much lower electron affinity than C, therefore F has the highest (most negative) electron affinity.

Answer 9

Na * The 2nd ionization energy is the energy required to remove the 2nd electron. * Sodium is the only answer choice that has only 1 valence electron (all the others have more than 1 valence electron). * Therefore the 2nd electron to remove is a core electron for sodium and is much more difficult to remove than a valence electron.

Answer 10

Ge * Atomic radius increases going down a group (column) and to the left across a period (row) on the periodic table. * Ge is the furthest down and furthest left out of the choices listed and is the correct answer.

Answer 11

Cl * Electron affinities increase (get more negative) going left to right across a period

Answer 12

C * Electron affinities increase (get more negative) going left to right across a period but the groups (periods) containing Be and N are exceptions to the trend having either positive electron affinities or values close to zero. Therefore carbon has the highest (most negative) electron affinity rather than nitrogen.

Answer 13

N * 1st ionization energy increases going up a group (column) and to the right across a period (row) on the periodic table. * But the answer choices here involve one of the exceptions to the general trend. Had this not been an exception we might have expected oxygen to be the correct answer. * But the Group 5A elements (N, P, As) have higher 1stionization energies than the Group 6A elements (O, S, Se) in the same period: N \> O, P \> S, and As \> Se. The group 5A elements have electron configurations that end with s2p3. With the p subshell half-filled these tend to be more stable electron configurations thus explaining the higher energy required to remove an electron.

Answer 14

Te * the comparison here is challenging as phosphorus (P) is furthest to the left while tellurium (Te) is furthest down on the periodic table. The key is realizing that the increase in size in going down a group two periods is larger than going from right to left across a period one group, therefore iodine (r = 133pm) has the largest atomic radius (phosphorus: r = 110pm).

Answer 15

KCl * An ionic compound is one composed of ions and is typically composed of either a metal and a nonmetal or of polyatomic ions. KCl is the only compound in the list composed of both a metal and nonmetal and is an ionic compound. * CO2 is a molecular compound being made of 2 nonmetals. * H3AsO4 and H2S are both molecular compounds as well as acids.

Answer 16

low melting points

Answer 17

C (diamond)

Answer 18

PCl5 * PCl5 is the only molecular compound listed, being composed of 2 nonmetals. * K2S, MgCl2, and LiF are all ionic compounds, being composed of a metal and a nonmetal.

Answer 19

NH4Br ## Footnote Even though NH4Br is composed of all nonmetals it is still an ionic compound in that it contains a polyatomic ion (the ammonium ion). Ammonium salts should file away in your head as ionic compounds that are composed of all nonmetals.

Answer 20

high melting points

Answer 21

In looking for a molecule with 3 resonance structures we can probably start by looking to see which molecules/ions have 3 atoms around the central atom:\*only NO3- and PBr3 do.\* But PBr3 does not have resonance.

Answer 22

90° ## Footnote With six electron domains around the central Xe atom, the electron domain geometry (EDG) is octahedral while the molecular geometry is square planar.\* Either way the bond angles are 90°.

Answer 23

With 3 electron domains around the central S atom, it will need 3 hybrid orbitals and therefore need to combine 3 orbitals to make the hybrids.\* Therefore S will combine 1 s, and 2 p orbitals to make 3 sp2 hybrid orbitals.

Answer 24

Five Electron Domains (1 non-bonding, 4 bonding): Molecular Geometry = see-saw

Answer 25

sp3d ## Footnote With 5 electron domains around the central Xe atom, it will need 5 hybrid orbitals and therefore need to combine 5 orbitals to make the hybrids.\* Therefore Xe will combine 1 s, 3 p and 1 d orbitals to make 5 sp3d hybrid orbitals.

Answer 26

CHCl3 Correct! ## Footnote These 4 compounds are very different in molecular weight and therefore the determining factor will be molecular weight rather than polarity. CHCl3 is significantly heavier than the other 3 choices and will therefore have significantly greater London Dispersion forces than the other 3 and the greatest overall intermolecular forces.

Answer 27

NaCl ## Footnote The compound with the highest intermolecular forces will have the highest boiling point. Typically we can rank intermolecular forces as follows: ionic or network covalent \>\> hydrogen bonding \> dipole-dipole \> London dispersion forces.

Answer 28

H2Te Correct! These 3 compounds are very different in molecular weight and therefore the determining factor will be molecular weight rather than polarity. H2Te is significantly heavier than the other 2 choices and will therefore have significantly greater London Dispersion forces than the other 2 and the greatest overall intermolecular forces.

Answer 29

NaOH ## Footnote The compound with the highest intermolecular forces will have the highest boiling point. Typically we can rank intermolecular forces as follows: ionic or network covalent \>\> hydrogen bonding \> dipole-dipole \> London dispersion forces. NaOH is the only ionic compound listed among the answer choices and has a much higher boiling point than the other choices.

Answer 30

H2O Correct!

Answer 31

The highest vapor pressure will be for the compound/element with the lowest intermolecular forces. These are all noble gases and are non-polar and therefore only have London dispersion forces. Helium is the smallest (and has the smallest surface area) and therefore has the lowest London dispersion forces and the highest vapor pressure.

Answer 32

CH3CH2CH2OH Correct!

Answer 33

The heat causing the water to heat up comes from the metal, so we can say that the metal loses heat as it cools down, and the water gains that heat and gets hotter... so heat (q) is being transferred from the metal to the water. If this is a perfect, isolated system, we can assume that the heat lost by the metal = heat (q) gained by the water, which is where we get: q water = -q metal =\> qmetal +qwater = 0 (m _water) (C_water) (delta T_water)= -(m_metal)(C_metal)(delta T_metal) (50_g)(1 _{cal/ g °})(10°C- 0°C)= -(50g)(C_metal) (10°C-50°C) C_{metal = 0.25 cal/ g °C}

Answer 34

Sublimation Correct!

Answer 35

None of these Correct!

Answer 36

*q *= *mcdelta*T= (50g) (1 cal/g °C)(79.4°C-10°C)= 3470 cal Q=change heat m=mass c=specific heat T = Tf - Ti

Answer 37

Condensation Correct!

Answer 38

D Correct!

Answer 39

BC Correct!

Answer 40

B Correct!

Answer 41

D Correct!

Answer 42

It is 6 times smaller Correct! ## Footnote We can solve this by looking at the relationships we can form from the equation PV=nRT. Pressure and volume are inversely proportional, so tripling the pressure will reduce the volume to 1/3 of its original value. Volume and temperature are directly proportional but must be examined using the absolute (Kelvin) scale. The temperature is cut in half in reducing it from 546K (273oC) to 273K (0oC) which thereby cuts the volume in half. Taking into a account both the pressure and temperature changes (1/3x1/2)\*we conclude that the volume will be 1/6 of it's original value (i.e. 6 times smaller).

Answer 43

All of the above Solution: The major assumptions made for ideal gases include: 1) Gas molecules have negligible volume compared with the volume of the container. 2) There are no attractive forces between molecules (i.e. all collisions are elastic). 3) Pressure is a result of collisions between gas molecules and with the walls of the container Therefore A, B, and C are all correct.

Answer 44

H2, 4 times faster Correct!

Answer 45

Ne ## Footnote In deciding which gas behaves the most ideally we should focus on which gas satisfies the following assumption of ideal gases the best: There are no attractive forces between molecules, i.e. all collisions are elastic. Therefore the gas with the lowest intermolecular forces can be expected to behave the most ideally. H2O has hydrogen bonding and can be expected to significantly deviate from ideal behavior. Ne, CO2, and Xe are all non-polar and therefore only have London dispersion forces. But as Ne is the smallest (lowest surface area) it will have the weakest London dispersion forces (and therefore lowest overall intermolecular forces) and will be expected to behave the most like an ideal gas.

Answer 46

Ne ## Footnote As both gases are in the same balloon it can be assumed they are at the same temperature and therefore have the same average kinetic energy eliminating choices A and B. With the same average kinetic energy, the lighter Ne will have a larger rms (like an avg) velocity which is why Ne escapes faster (choice D).

Answer 47

500 ## Footnote This is testing upon the concept of Dalton’s Law of Partial Pressures. If 40% of the molecules result in a partial pressure of 200torr, then 200torr must equal 40% of the total pressure. 200torr = 0.40 (Ptotal)

Answer 48

A gas’ kinetic energy is a function of its temperature and since all these gases are at the same temperature (150 C) they will all have the same kinetic energy. Since K.E. =1/2mv^2 , the gas with the largest mass (m) will have the lowest average velocity (v) and Cl2 has the largest mass of the choices listed.

Answer 49

108 Dividing the number of moles of each reactant by its coefficient in the balanced reaction reveals that NO2 is the limiting reagent.

Answer 50

What is the maximum number of moles of CO2 that could be produced when 28 moles of O2 are reacted with excess C5H12? C5H12 + 8O2\* 5CO2 + 6H2O\* 44. 8 17. 5 3. 5 770

Answer 51

400ml (200gCaCO₃)( 1mol CaCO₃/100g CaCO₃)(2 mol HCL/1 mol CaCO₃)= 4 mol HCL 4 mol HCl/ 10 mol/L HCL= 0.4 L HCL= 400 ml

Answer 52

One of the keys to this problem is to remember that 1 mole of gas at STP occupies 22.4L. Since the molar mass of CaCO3 is approximately 100g, we have 200g present in the reaction, and therefore 2 moles of CaCO3 react with excess HCl . From the balanced equation we can see that CaCO3and CO2 are in a 1:1 ratio, so 2 moles of CaCO3reacts to produce 2 moles of CO2, or (22.4L/mole) x (2 moles) = 44.8L.

Answer 53

Decreasing the temperature and increasing the pressure Correct! Gases are more soluble in liquids at low temperatures and high pressures.

Answer 54

800 g M= mol/ V Rearranged to: (M)(V)= mol solute (2M)(10L)=20 mol NaOH

Answer 55

We’ll use the following equation for the definition of molarity: 0.1 moles ## Footnote M = moles / VL where VL is the volume in liters Rearranging to solve for moles: moles = (M)(VL) ``` M = 0.2M V = 500ml = 0.5L (the volume must be in liters) ``` ``` moles = (M)(VL) moles = (0.2M)(0.5L) = 0.1moles ```

Answer 56

AgNO3(aq), KCl(aq), and KNO3(aq) are strong electrolytes and so the ionic equation is Ag+(aq) + Cl-(aq) + K+(aq) + NO3-(aq) → AgCl(s) + K+(aq) + NO3-(aq) K+(aq) and NO3-(aq) are both spectator ions (as they appear on both sides of the reaction as free ions and therefore haven't done anything) and are eliminated to yield the net ionic equation: Ag+(aq) + Cl-(aq) → AgCl(s)

Answer 57

We’ll use the following equation for the definition of molarity: M = moles / VL where VL is the volume in liters Rearranging to solve for volume: VL = moles / M To solve we’ll first have to convert 29g KF into moles: 29g KF (1mole KF / 58g KF) = 0.5moles KF ``` VL = moles / M VL = 0.5moles / 4M = 0.125L = 125mL ```

Answer 58

We’ll use the following equation for the definition of molarity: M = moles / VL where VL is the volume in liters Rearranging to solve for moles: moles = (M)(VL) ``` M = 0.6M V = 3L ``` ``` moles = (M)(VL) moles = (0.6M)(3L) = 1.8 moles ```

Answer 59

We’ll use the following equation for the definition of molarity: M = moles / VL where VL is the volume in liters Rearranging to solve for volume: VL = moles / M To solve we’ll first have to convert 50g CaBr2 into moles: 50g CaBr2 (1mole CaBr2 / 200g CaBr2) = 0.25moles CaBr2 ``` VL = moles / M VL = 0.25moles / 0.1M = 2.5L ```

Answer 60

We’ll use the following equation for the definition of molarity: M = moles / VL where VL is the volume in liters We’ll rearrange this to solve for the moles of KF and then convert to grams. Rearranging to solve for moles: moles = (M)(VL) ``` M = 0.25M V = 400ml = 0.4L (the volume must be in liters) ``` ``` moles = (M)(VL) moles = (0.25M)(0.4L) = 0.1 moles KF ``` 0.1 moles KF (58g KF / 1 mole KF) = 5.8g KF

Answer 61

We’ll use the following equation for the definition of molarity: M = moles / VL where VL is the volume in liters We’ll rearrange this to solve for the moles of NaBr and then convert to grams. Rearranging to solve for moles: moles = (M)(VL) ``` M = 0.4M V = 1.5L ``` ``` moles = (M)(VL) moles = (0.4M)(1.5L) = 0.6 moles NaBr ``` 0.6 moles NaBr (103g NaBr / 1 mole NaBr) = 62g NaBr

Answer 62

The greater the concentration of solute, the greater the freezing pt. depression. 1m NaCl = 2m ions since NaCl dissociates into 2 ions 1. 5m CH3CH2OH=1.5m as it is a non-electrolyte and doesn’t dissociate. 0. 7m AlCl3 = 2.8m ions since AlCl3 dissociates into 4 ions (1 Al3+ and 3Cl-) 1. 2m CsNO3 = 2.4m ions since CsNO3 dissociates into 2 ions (1 Cs+ and 1NO3-)

Answer 63

i = 1 for a non-electrolyte so m=0.33m

Answer 64

change in T_f= -K_Fm= -(1.86 ºC/m) (2m)= -3.72ºC Therefore TF=0ºC - 3.72ºC = -3.72

Answer 65

5.5m CH3OH has 5.5 moles CH3OH for every 1kg H2O. (1 kg H₂O)(1000g/1kg)(1 mol H₂O/ 18g H₂O)= 55.5 mol H₂O

Answer 66

5mol CaCl2 = 15mol ions since it dissociates into 3 ions (1 Ca2+ and 2Cl-) 5mol BaSO4 = 10 mol ions since it dissociates into 2 ions (1 Ba²⁺ and 1 SO₄^2-) 5mol KCl = 10 mol ions since it dissociates into 2 ions (1 K⁺ and 1 Cl^-) 5mol CH₃OH = 5mol CH₃OH since it is a non-electrolyte and doesn't dissociate into ions CaCl2 dissociates to give the highest concentration of ions and therefore has the highest boiling point elevation.

Answer 67

The highest freezing pt. will be the one that has been depressed the least and therefore is the solution with the lowest concentration of dissolved species. NaCl dissociates into 2 ions (1 Na+ and 1 Cl-) and so 1m NaCl results in a 2m solution of dissolved species. 0. 7m AlCl3 dissociates into 4 ions (1 Al3+ and 3 Cl-) and so 0.7m AlCl3 results in a 2.8m solution of dissolved species. 1. 2m CsNO3 dissociates into 2 ions (1 Cs+ and 1 NO3-) and so 1.2m CsNO3 results in a 2.4m solution of dissolved species. CH3CH2OH is non-electrolyte and doesn't dissociate into ions and therefore it's 1.5m concentration is the lowest concentration of dissolved species amongst the answer choices and this solution will have the smallest decrease (depression) in its freezing point leaving it with the overall highest freezing point.

Answer 68

The mole to mole ratio for C3H8 to CO2 is 1:3 based upon the coefficients in the reaction listed. This means that for every 1 mole of C3H8 that is consumed, 3 moles of CO2 are produced. So if the rate of disappearance of C3H8 is 0.3 atm/min, then the rate of appearance of CO2 will be three times as much or 0.9 atm/min.

Answer 69

0.50M/min Correct! ## Footnote The mole to mole ratio between Cu2+ and Al3+ is 3:2. This means that for every 3 moles of Cu2+ that are consumed, 2 moles of Al3+are produced. So if the rate of disappearance of Cu2+ is 0.75M/min, then the rate of appearance of Al3+ must be 2/3 as much or 0.50M/min.

Answer 70

-0.50M/min Correct! ## Footnote The mole to mole ratio between Cl2 and PCl3 is 3:1. This means that for every 3 moles of Cl2 that are produced, 1 mole of PCl3 is consumed. So if the rate of appearance of Cl2 is 1.50M/min, then the rate of consumption of PCl3 is 1/3 of that or 0.50M/min. But the question didn’t ask for the rate of consumption but for the rate of change. In this case we need to indicate that the concentration is decreasing by a negative rate of change and therefore the correct answer is -0.50M/min. Had the question asked for the “rate of consumption” it would already have been implied the concentration was decreasing and the answer would have been positive rather than negative.

Answer 71

0.75M/min Correct! ## Footnote The mole to mole ratio between HCl and Cl2 is 1:1 (reduces from 3:3). This means that for every 3 moles of HCl that are consumed, 3 moles of Cl2 are produced. So if the rate of disappearance of HCl is 0.75M/min, then the rate of appearance of Cl2 must also be 0.75M/min.

Answer 72

35kJ Correct! If Ea=50kJ and H=15kJ then the activation energy of the reverse reaction will be the difference between these (see diagram).

Answer 73

4M/min Correct! ## Footnote The mole to mole ratio for N2 to NH3 is 1:2 based upon the coefficients in the reaction listed. This means that for every 1 mole of N2 that is consumed, 2 moles of NH3 are produced. So if the rate of disappearance of N2 is 2M/min, then the rate of appearance of NH3 will be twice as much or 4M/min

Answer 74

4M/min Correct!

Answer 75

Comparing trials 2 and 3 we can see that the concentration of Y remains constant while the concentration of X doubles while the rate also doubles. Since Rate [X]order 2=2(order) so the order must be 1. Comparing trials 1 and 2 we can see that the concentration of X remains constant while the concentration of Y triples while the rate increases by a factor of 9 (convert 1.8x10-4 to 18x10-5 to see this more easily). Since Rate [Y]order 9=3(order) so the order must be 2 (Students will commonly, albeit incorrectly deduce that Y is 3rd order. Remember that while 3x3=9, it is 32=9 that is relevant here. Therefore the overall rate law is 1st order with respect to X and 2nd order with respect to Y.

Answer 76

Comparing trials 1 and 2 we can see that the concentration of X remains constant while the concentration of Y doubles (0.040/0.020 = 2) and the rate also doubles. Rate [Y]order 2=2(order) so the order must be 1 From here it gets a little tricky as there aren’t two trials we can compare in which the concentration of Y is held constant. The key is remembering that Y is a 1st order reactant (as determined above). From here we can compare trials 2 and 3 or trials 1 and 3 (as long as [X]0 changes) and factor in that Y is a first order reactant. Comparing trials 2 and 3 we can see that the concentration of X doubles (0.020/0.010 = 2), the concentration of Y doubles (0.080/0.040 = 2) and the rate increases by a factor of 4 (8.0x10-4/2.0x10-4 = 4). We know that the doubling of the concentration of Y should also result in a doubling of the rate as it is first order so this accounts for an increase by a factor of 2. Again, the rate increased by a factor of 4 and so the change in the concentration of X will account for the remaining factor of 2. Since the concentration of X doubling accounts for this remaining factor of 2 we can determine that X must be first order. Rate [X]order[Y]order 4 = (2)order(2)1 so the order with respect to X must be 1 Therefore the overall rate law is 1st order with respect to both X and Y i.e. Rate=k[X][Y].

Answer 77

Rate increases by a factor of 100 Correct! ## Footnote For a 2nd order reactant the rate depends upon the reactants concentration squared (Rate = k[A]2). So if the reactant concentration is increase by a factor of 10, the rate will be increased by a factor of 102 or 100.

Answer 78

Rate=k[X]2 Correct! ## Footnote Comparing trials 1 and 2 we can see that the concentration of X remains constant while the concentration of Y doubles (0.040/0.020 = 2) and the rate doesn’t change which is an indication of a zero order reactant (one that has no effect on the rate). From here it gets a little tricky as there aren’t two trials we can compare in which the concentration of Y is held constant. The key is remembering that it is a zero order reactant (as determined above) and has no effect on the rate. Therefore we can compare trials 2 and 3 or trials 1 and 3 (as long as [X]0changes) and ignore the fact that the concentration of Y is changing. Comparing trials 2 and 3 we can see that the concentration of X doubles (0.020/0.010 = 2) while the rate quadruples (8.0x10-4/2.0x10-4 = 4). Rate [Y]order 4=2(order) so the order must be 2 Therefore the overall rate law is 2nd order with respect to X and zero order with respect to Y i.e. Rate=k[X]2[Y]0=k[X]2.

Answer 79

Rate=k[X][Y] Correct! Comparing trials 1 and 2 we can see that the concentration of Y remains constant while the concentration of X quadruples (0.80/0.20 = 4) and the rate quadruples (4.4x10-4/1.1x10-4 = 4). Rate [X]order 4=4(order) so the order must be 1. Comparing trials 1 and 3 we can see that the concentration of X remains constant while the concentration of Y doubles (1.0/0.50 = 2) while the rate also doubles (2.2x10-4/1.1x10-4 = 2). Rate [Y]order 2=2(order) so the order must be 1 Therefore the overall rate law is 1st order with respect to X and 1st order with respect to Y i.e. Rate=k[X][Y].

Answer 80

Rate=k[Y] Correct! ## Footnote Comparing trials 2 and 3 we can see that the concentration of Y remains constant while the concentration of X doubles but rate remains constant. Since Rate [X]order 1=2(order) so the order must be 0. Comparing trials 1 and 2 we can see that the concentration of X remains constant while the concentration of Y triples while the rate also triples. Since Rate\* [Y]order 3=3(order) so the order must be 1 Therefore the overall rate law is zero order with respect to X and 1st order with respect to Y.

Answer 81

Rate=k[Y] Correct! Comparing trials 1 and 2 we can see that the concentration of X remains constant while the concentration of Y doubles (0.040/0.020 = 2) and the rate also doubles. Rate [Y]order 2=2(order) so the order must be 1 From here it gets a little tricky as there aren’t two trials we can compare in which the concentration of Y is held constant. The key is remembering that Y is a 1st order reactant (as determined above). From here we can compare trials 2 and 3 or trials 1 and 3 (as long as [X]0 changes) and factor in that Y is a first order reactant. Comparing trials 2 and 3 we can see that the concentration of X doubles (0.020/0.010 = 2), the concentration of Y doubles (0.080/0.040 = 2) and the rate doubles (8.0x10-4/4.0x10-4 = 2). We know that the doubling of the concentration of Y should also result in a doubling of the rate as it is first order. As the rate simply doubles we can determine that doubling the concentration of X had no impact on the rate of the reaction and therefore must be a zero order reactant. Therefore the overall rate law is zero order with respect to X and 1st order with respect to Y i.e. Rate=k[X]0[Y]=k[Y] .

Answer 82

A and C Correct! At higher temperatures, molecules have a greater average velocity and will therefore collide more often (answer choice A) and with a greater average energy (answer choice C). The activation energy however is independent of temperature which eliminates answer choice B.

Answer 83

A correct! ## Footnote A catalyst will not appear in the overall reaction as it as both a reactant and a product (it is not consumed in a reaction). It is most easily recognized as a reactant in an earlier step and a product in a later step. Only A qualifies.

Answer 84

All of the above In looking at a rate law (Rate=k[A]x) it is easy to see that reactant concentrations (that aren’t zero order) and anything that affects k will affect the rate of the reaction.\*The dependence of the rate constant upon the activation energy (influenced by a catalyst) and the temperature is demonstrated by the Arrhenious equation:\* .\* And so we see that because both a catalyst and temperature will affect the rate constant they will also affect the rate as well so the answer is all of the above.

Answer 85

Both a catalyst and temperature Correct! This one is a little tricky. In looking at a rate law (Rate=k[A]x) it is easy to see that reactant concentrations (that aren’t zero order) and anything that affects k will affect the rate of the reaction. But while reactant concentrations affect the rate of the reaction, they do not affect k, the rate constant. The dependence of the rate constant upon the activation energy (influenced by a catalyst) and the temperature is demonstrated by the Arrhenious equation:\*

Answer 86

Increasing the temperature Correct! ## Footnote If the forward reaction is exothermic, increasing the temperature decreases the value of the equilibrium constant. If the forward reaction is endothermic, increasing the temperature increases the value of the equilibrium constant. This reaction has a positive enthalpy value so is endothermic.

Answer 87

Adding H2(g) Correct! ## Footnote Adding NH3 results in a shift left (as adding a product typically does). Removing N2 results in a shift left (as removing a reactant typically does). Increasing the volume of the container results in a shift left as increasing the volume (decreasing the pressure) favors the side of the reaction with more moles of gas. Adding H2 is the only answer that results in a shift right (as adding a reactant typically does).

Answer 88

lowering the pH Correct! ## Footnote Lowering the pH means making the solution more acidic and a more acidic solution will have a reduction in the concentration of OH-. A reduction in the concentration of OH- (a reactant) will indeed result in the shift to the left and is the correct answer.

Answer 89

addition of CO2 Correct! ## Footnote Generally the addition of a reactant or the removal of a product to a system at equilibrium will result in a shift to the right. The key is to remember that as solids and liquids don’t show up in equilibrium constant expressions, they therefore can’t result in a shift in the equilibrium. Therefore any answer involving the addition or removal of CaO(s) or CaCO3(s) won’t result in a shift at all and can be eliminated. However the addition of CO2(g), a reactant, will indeed result in a shift right.

Answer 90

Decreasing the pH Correct! ## Footnote ``` To increase the yield of H2, the reaction needs to shift right. Decreasing the pH (making the solution more acidic) increases the [H+], a reactant, therefore resulting in a shift to the right. Removing Mg(s) results in no shift as the addition/removal of solids don’t shift an equilibrium as they don’t show up in the equilibrium constant expression. Removing Mg2+ results in a shift left (as removing a reactant typically does). Increasing the pressure results in a shift left as it favors the side of the reaction with fewer moles of gas.\* Keep in mind that solids, liquids, and aqueous species are largely unaffected by a change in pressure.\* The reactants side has zero moles of gas while the products side has 1 mole of gas so the reactants side is indeed the side with fewer moles of gas. ```

Answer 91

none of these Correct! ## Footnote Generally the addition of a product or the removal of a reactant to a system at equilibrium will result in a shift to the left, whereas the addition of a reactant or the removal of a product to a system at equilibrium will result in a shift to the right. But don’t forget that as solids and liquids don’t show up in equilibrium constant expressions, they therefore can’t result in a shift in the equilibrium. Therefore the addition of CaCO3 won’t result in a shift at all and can be eliminated. The removal of Ca2+ and the removal of CO32- (both are products) will both result in a shift to the right and can be eliminated. Therefore “none of these” will result in a shift to the left which is the correct answer.

Answer 92

The reaction will proceed to the left as there are fewer moles of gas on the reactants side Correct! ## Footnote An increase in pressure (or a reduction in volume) will shift an equilibrium to the side that has the fewer number of moles of gas, whereas a decrease in pressure (or an increase in volume) will have the opposite affect. In this case it is important to note that Sb is a solid, not a gas and there are 3 moles of gaseous reactants whereas there are 5 moles of gaseous products. Increasing the pressure will therefore result in a shift to the left where there are fewer moles of gas.

Answer 93

The solubility reaction and the corresponding Ksp expression are as follows: ScF3(s) =\> Sc3+(aq) + 3F-(aq) Ksp = [Sc3+][F-]3 NaF is a strong electrolyte (all group 1 metal salts are) and dissociates completely resulting in a fluoride concentration of [F-] = 0.0001M. We can substitute this into the Ksp expression to figure out the maximum concentration of Sc3+ possible before a precipitate forms: ``` Ksp = [Sc3+][F-]3 4.2x10-18 = [Sc3+](0.0001)3 4.2x10-18 = [Sc3+](1x10-4)3 4.2x10-18 = [Sc3+](1x10-12) [Sc3+] = 4.2x10-18 / 1x10-12 [Sc3+] = 4.2x10-6M ```

Answer 94

The solubility reaction and the corresponding Ksp expression are as follows: FePO4(s) Fe3+(aq) + PO43-(aq) Ksp = [Fe3+][PO43-] If an ‘x’ molar concentration (the molar solubility) of FePO4 dissolves this will result in equilibrium concentrations of Fe3+ and PO43- of ‘x’ as well as can be seen in the following ICE table: Substituting in the equilibrium values into the Ksp expression and then substituting in the molar solubility, ‘x’ given allows us to solve for the Ksp. ``` Ksp = [Fe3+][PO43-] Ksp = (x)(x) Ksp = x2 Ksp = (1.1x10-11)2 Ksp = 1.2x10-22 ```

Answer 95

AB3(s) → A3+(aq) + 3B-(aq) Ksp=[A3+][B-]3 = (x)(3x)3 = 27x4 = 3.0x10-19 \* x4=0.11x10-19 x4=1.1x10-20 x=(1.1x10-20)1/4 x~1.0x10^-5=molar solubility

Answer 96

MX2(s) → M2+(aq) + 2X-(aq) Ksp = [M2+][X-]2 = (x)(2x)2 = 4x3 Remember that the molar solubility = x and that we’re solving for Ksp, not the other way around. ``` Ksp = 4(2.0x10-6)3 Ksp = 4(8.0x10-18) = 32x10-18 = 3.2x10-17 ```

Answer 97

BaF2(s) → Ba2+(aq) + 2F-(aq) Ksp = [Ba2+][F-]² 3.2x10^-8 = (0.032)[F-]² (1.0x10-6)1/2=[F-] [F-]= 1.0x10-3M

Answer 98

3.2x10-6M Correct! BaF2(s) → Ba²⁺(aq) + 2F-(aq) Ksp = [Ba²⁺][F-]² = (x)(0.1+2x)² But x (the molar solubility) will be small and so 2x will be negligible when added to 0.1 and so the expression reduces to the following: Ksp = (x)(0.1)²= 3.2x10^-8

Answer 99

A precipitate will not form because Qsp \< Ksp Correct! BiI3(s) → Bi3+(aq) + 3I-(aq) Ksp = [Bi3+][I-]3 (at equilibrium) Qsp = [Bi3+][I-]3 (not necessarily at equilibrium) Qsp = (0.0002)(1.0x10-5)3 Qsp = (2.0x10-4)(1.0x10-5)3 = 2.0x10-19 which is \< 8.2x10-19 Since Qsp \< Ksp, the solution is below the saturation point and will not form a precipitate.

Answer 100

A precipitate will not form because Qsp \< Ksp Correct! ## Footnote This question is tricky. Mixing 500ml of 2x10-5M AgNO3 with 500ml of 1x10-5M NaCl dilutes both concentrations in half (You can use M1V1 = M2V2 to figure this out). These final concentrations after mixing are therefore 1x10-5M AgNO3 and 0.5x10-5M NaCl or 5.0x10-6M NaCl. AgCl(s) → Ag+(aq) + Cl-(aq) Ksp = [Ag+][Cl-] (at equilibrium) ``` Qsp = [Ag+][Cl-] (not necessarily at equilibrium) Qsp = (1x10-5)(5x10-6) = 5x10-11 which is \< 1.8x10-10 ```

Answer 101

O^2- Correct! ## Footnote This one is tricky. The question asks for the conjugate base of OH- so OH- must be the conjugate acid (which is why this is tricky as you’ve probably only ever seen OH- act as a base and a strong one at that). As the Bronsted-Lowry definition of an acid is a H+ (proton) donor, simply having OH- donate an H+ leads us the the conjugate base, O-2.

Answer 102

HClO₄ Correct!

Answer 103

HF (Ka=6.8x10-4) Correct! ## Footnote The strongest acid is the one that dissociates the most which is the one with the highest Ka value. HF has the highest Ka out of the 4 answer choices. (Note that 6.8x10-4 \> 1.8x10-4)

Answer 104

HC2H3O2 (pKa=4.75) Correct! ## Footnote Acid strength increases as pKa decreases.

Answer 105

HCN (Ka=4.9x10-10) Correct! ## Footnote The stronger the acid, the weaker the conjugate base. So the weakest acid will have the strongest conjugate base. A lower Ka means a weaker acid so we're looking for the acid with the lowest Ka and HCN has the lowest Ka of the acids listed. (Note that 4.9x10-10 \< 5.6x10-10)

Answer 106

H2Te Correct! ## Footnote For binary acids, acid strength increases down the periodic table with increasing size (as a larger anion is a more stable conjugate base).

Answer 107

9.0 Correct! pH = 14 – pOH and pOH = -log [OH-] ``` pOH = -log (1.0 x 10-5) = 5 pH = 14 – 5 = 9 ```

Answer 108

[H+] = 10-^pH = 10^-3.2which will be between 1.0x10^-4M and 1.0x10^-3M The only answer choice in this range is 6.3x10-4M which therefore must be the correct answer.

Answer 109

7.9x10-10M Correct! ## Footnote If the pH = 4.9, then the pOH = 9.1 [OH-] = 10-pOH = 10-9.1 which will be between 1.0x10-10M and 1.0x10-9M which still leaves us with 2 possible answer choices (1.6x10-10M and 7.9x10-10M) so we’ll have to be more precise. You’ll have to remember not only the [OH-] values when the pOH is an integer but also for the halves (.5) too. The half pOH occurs when [OH-] = 3.16x10-whatever as demonstrated below: ``` [OH-] = 1x10-9M pOH = 9 [OH-] = 3.16x10-10M pOH = 9.5 [OH-] = 1x10-10M pOH = 10 ``` Since a pOH of 9.1 is between 9 and 9.5, the [OH-] must be between 3.16x10-10M and 1x10-9M and therefore 7.9x10-10M must be the correct answer.

Answer 110

``` pH = -log [H+] pH = -log (4.0x10-3) which should be between 2 and 3. As [H+] increases, pH decreases (they’re inversely related). Since the -log of 1x10-3 = 3, then the -log of a slightly higher [H+] will yield a slightly lower pH. ``` But this still leaves us with two possible correct answer choices (2.4 and 2.9) and so you’ll have to remember not only the [H+] values when the pH is an integer but also for the halves (.5) too. The half pH occurs when [H+] = 3.16x10-whatever as demonstrated below: ``` [H+] = 1x10-2M pH = 2 [H+] = 3.16x10-3M pH = 2.5 [H+] = 1x10-3M pH = 3 ``` Since 4.0x10-3 is between 3.16x10-2 and 1x10-2, then the corresponding pH will be between 2.5 and 2 making 2.4 the correct answer in this case.

Answer 111

A difference of 2.5 pH units is a factor of 102.5 times more acidic. Since 102 = 100 and 103 = 1000, 102.5 must be between 100 and 1000, there is only one answer choice that falls in this range, pH 1.5 is approximately 300 times more acidic than pH 4.0.

Answer 112

pH = 14 – pOH and pOH = -log [OH-] As [OH-] increases, pOH decreases (they’re inversely related). Since the -log of 1x10-10 = 10, then the -log of a slightly higher [OH-] will yield a slightly lower pOH. pOH = -log (7.1x10-10) which should therefore be between 9 and 10 and so the pH should be between 4 and 5 which still leaves us with 2 possible answer choices (4.1 and 4.8) so we’ll have to be more precise. You’ll have to remember not only the [OH-] values when the pOH is an integer but also for the halves (.5) too. The half pOH occurs when [OH-] = 3.16x10-whatever as demonstrated below: ``` [OH-] = 1x10-9M pOH = 9 [OH-] = 3.16x10-10M pOH = 9.5 [OH-] = 1x10-10M pOH = 10 ``` Since 6.0 x 10-10M is between 3.16x10-10M and 1x10-9M, then the corresponding pOH will be between 9.5 and 9. Subtracting from 14 reveals that the pH must be between 4.5 and 5 and so 4.8 must be the correct answer.

Answer 113

[H+] = 10-pH = 10-6.6 which will be between 1.0x10-7M and 1.0x10-6M which still leaves us with 2 possible answer choices (2.5x10-7M and 6.6x10-7M) so we’ll have to be more precise. You’ll have to remember not only the [H+] values when the pH is an integer but also for the halves (.5) too. The half pH occurs when [H+] = 3.16x10-whatever as demonstrated below: ``` [H+] = 1x10-6M pH = 6 [H+] = 3.16x10-7M pH = 6.5 [H+] = 1x10-7M pH = 7 ``` Since a pH of 6.6 is between 6.5 and 7, the [H+] must be between 1x10-7M and 3.16x10-7M and therefore 2.5x10-7M must be the correct answer.

Answer 114

8.9 ## Footnote pH = 14 – pOH and pOH = -log [OH-] pOH = -log (8.6x10-6) which is definitely \<6 but closer to 5…say 5.1 (As [OH-] increases, pOH decreases…they’re inversely related.) So the pH = 14 – 5.1 ~8.9

Answer 115

[H+]=1.0x10-11M Correct! ## Footnote It is easiest to compare relative acidities and basicities based upon pH as this is what most of us are most familiar with. The most basic solution is the one with the highest pH. Pure water pH = 7 pOH = 12 so pH = 2 If NH3 were a strong base it would dissociate completely yielding [OH-] = 0.001M and so pOH = -log (0.001) = 3 and pH = 11. Since it’s a weak base we know the pH will be \<11. [H+] = 1.0x10-11M so pH = -log (1.0x10-11M) = 11 which is the most basic of the choices.

Answer 116

7.2 Correct! ## Footnote ``` pH = -log [H+] pH = -log (6.4x10-8) which should be between 7 and 8. As [H+] increases, pH decreases (they’re inversely related). Since the -log of 1x10-8 = 8, then the -log of a slightly higher [H+] will yield a slightly lower pH. ``` But this still leaves us with two possible correct answer choices (7.2 and 7.8) and so you’ll have to remember not only the [H+] values when the pH is an integer but also for the halves (.5) too. The half pH occurs when [H+] = 3.16x10-whatever as demonstrated below: ``` [H+] = 1x10-7M pH = 7 [H+] = 3.16x10-8M pH = 7.5 [H+] = 1x10-8M pH = 8 ``` Since 6.4x10-8 is between 3.16x10-8 and 1x10-7, then the corresponding pH will be between 7.5 and 7 making 7.2 the correct answer in this case.

Answer 117

13.3 Correct! ## Footnote NaOH is a strong base and dissociates completely to yield [OH-] = 0.2M = 2x10-1M pOH = -log[OH-] = -log(2x10-1M) = 0.7 pH = 14 – 0.7 = 13.3 As [OH-] increases, pOH decreases (they’re inversely related). Since the -log(1x10-1) = 1, then the -log of a slightly higher [OH-] will yield a slightly lower pOH so the pOH should be between 0 and 1 and therefore the pH must be between 13 and 14 (pH = 14 – pOH) and therefore the correct answer must be 13.3 being the only answer choice in this range.

Answer 118

HCl is a strong acid and dissociates completely so [H+] = 0.08M \* \*(between 1 and 2 but closer to 1)

Answer 119

12 Correct! Ba(OH)2 is a strong base and dissociates completely to yield [OH-] = 2(0.005M) =0.01M

Answer 120

HBr is a strong acid and dissociates completely so [H+] = 0.001M = 1x10-3M pH = -log[H+] = -log(1x10-3M) = 3

Answer 121

A little less than 3 Correct! This question is a little tricky. H2SO4 is a strong acid and a diprotic (2 acidic protons) acid but only the 1st proton dissociates completely; the 2nd is only weakly acidic and therefore only dissociates to a small extent. So for 1.0x10-3M H2SO4, [H+] ~ 1.0x10-3M (just slightly higher) And the pH ~ 3 (just slightly lower) and so the best answer is a little less than 3

Answer 122

2.5 Correct! HNO3 is a strong acid and dissociates completely so [H+] = 0. 0032M = 3.2x10-3M pH = -log[H+] = -log(3.2x10-3M) = 2.5 As [H+] increases, pH decreases (they’re inversely related). Since the -log (1x10-3) = 3, then the -log of a slightly higher [H+] will yield a slightly lower pH so the pH should be between 2 and 3 and therefore 2.5 must be the correct answer being the only answer choice in this range.

Answer 123

2.6 Correct! ## Footnote pH = -log [H+] = -log(2.5x10-3) = 2.6 As [H+] increases, pH decreases (they’re inversely related). Since the -log(1x10-3) = 3, then the -log of a slightly higher [H+] will yield a slightly lower pH so the pH should be between 2 and 3 and therefore 2.6 must be the correct answer being the only answer choice in this range.

Answer 124

pOH = -log[OH-] = -log(3x10-3) = 2.5 pH = 14 – pOH = 14 – 2.5 = 11.5 As [OH-] increases, pOH decreases (they’re inversely related). Since the -log(1x10-3) = 3, then the -log of a slightly higher [OH-] will yield a slightly lower pOH so the pOH should be between 2 and 3. The pH should then be between 11 and 12 (pH = 14 – pOH) and therefore 11.5 must be the correct answer being the only answer choice in this range.

Answer 125

pH = -log [H+] = -log(8.4x10-3) = 2.1 As [H+] increases, pH decreases (they’re inversely related). Since the -log (1x10-3) = 3, then the -log of a slightly higher [H+] will yield a slightly lower pH so the pH should be between 2 and 3 and therefore 2.1 must be the correct answer being the only answer choice in this range.

Answer 126

pOH = -log[OH-] = -log(1.3x10-4) = 3.9 pH = 14 – pOH = 14 – 3.9 = 10.1 As [OH-] increases, pOH decreases (they’re inversely related). Since the -log(1x10-4) = 4, then the -log of a slightly higher [OH-] will yield a slightly lower pOH so the pOH should be between 3 and 4. The pH should then be between 10 and 11 (pH = 14 – pOH) and therefore 10.1 must be the correct answer being the only answer choice in this range.

Answer 127

pOH = -log[OH-] = -log(3.9x10-6) = 5.4 pH = 14 – pOH = 14 – 5.4 = 8.6

Answer 128

pOH=-log[OH-]=-log(1.2x10-6)~5.9 pH=14-pOH=14-5.9=8.1

Answer 129

pH = -log [H+] = -log(1.7x10-5) = 4.8 As [H+] increases, pH decreases (they’re inversely related). Since the -log(1x10-5) = 5, then the -log of a slightly higher [H+] will yield a slightly lower pH so the pH should be between 4 and 5 and therefore 4.8 must be the correct answer being the only answer choice in this range.

Answer 130

Na2SO4 Correct! ## Footnote This question test knowledge of the potential acidity/basicity of salts. Cations: Group 1 and 2 metal ions or metal ions with a +1 charge are negligible; all other cations are acidic. Anions: The general rule is the that conjugate base of a strong acid is a negligible base (Cl-, Br-, I-, NO3-, ClO4-, ClO3-) or an acid (in the case of HSO4-). But almost all other anions are basic. pH\>7 means a basic solution and it is the anions that can act as bases so we’ll focus on them. KBr \* Br- is negligible Na2SO4 \*\*\* SO42- isn’t the conjugate base of a strong acid and is therefore basic CsI \*\* I- is negligible RbNO3 \*NO3- is negligible

Answer 131

CaCl2 Correct! ## Footnote This question test knowledge of the potential acidity/basicity of salts. Cations: Group 1 and 2 metal ions or metal ions with a +1 charge are negligible; all other cations are acidic. Anions: The general rule is the that conjugate base of a strong acid is a negligible base (Cl-, Br-, I-, NO3-, ClO4-, ClO3-) or an acid (in the case of HSO4-). But almost all other anions are basic. NaHCO3 \*\*\* Na+ is negligible (Group 1) but HCO3-is a base (Overall: Basic Salt) KClO2\*\*\* K+ is negligible (Group 1) but ClO2- is a base (Overall: Basic Salt) CaCl2\*\*\* Ca2+ is negligible (Group 2) and Cl- is negligible (Overall: Neutral Salt - This solution will have a pH closest to 7) Al(NO3)3 \*\*\* Al3+ is acidic but NO3- is negligible (Overall: Acidic Salt)

Answer 132

4.0 Correct! ## Footnote For buffers every power of 10 is 1 pH unit as its a log10 based scale. When [H+] = [A-], pH = pKa 10x more acid, 1 pH unit lower (more acidic) than the pKa 100x more acid, 2 pH units lower than the pKa 10x more conjugate base, 1 pH unit higher (more basic) than the pKa 100x more conjugate base, 2 pH units higher than the pKa In this problem with almost 10x more conjugate base we could anticipate the pH to be almost 1 pH unit higher than the pKa of 3.2

Answer 133

0.1M HClO2 and 0.1M NaClO2 A buffer is the mixture of a weak acid with its conjugate base (or weak base with its conjugate acid) in as close to a 1:1 ratio as possible. A buffer can be made in 3 ways: i) ~1 eq weak acid / 1 eq of conjugate base ii) ~1 eq weak acid / eq of strong base iii) ~1 eq weak base / eq of strong acid Choice A: strong acid and conjugate base; not a buffer Choice B: strong acid and strong base; not a buffer Choice C: weak acid and conjugate base (1:1 ratio too); a buffer Choice D: weak acid and strong base (1:1 ratio); not a buffer

Answer 134

HF is a weak acid and NaF is the salt of its conjugate base and the mixture of the 2 forms a buffer. For calculations with buffers, one can use Ka, Kb, or the Henderson-Hasselback Equation which is my preference:

Answer 135

The 1L of solution containing 0.5M HCN and 0.5M NaCN is a buffer having a weak acid and the salt of its conjugate base. Buffers resist changes in pH. Therefore adding a little HCl (strong acid) will decrease the pH only a little, not a lot.

Answer 136

HC2H3O2 (pKa = 4.75) Correct! In the titration of a weak acid with a strong base, pH = pKa at the half-equivalence point which can be seen to be just below a pH of 5.Therefore HC2H3O2 (pKa = 4.75) is the correct answer

Answer 137

8mL ## Footnote To relate the volumes and molarities of an acid and a base in a neutralization reaction we can use the following equation: nAMAVA = nBMBVB where nA is the number of acidic H+ ions the acid can donate and nB is the number of H+ ions the base can accept (typically equal to the number of OH- in the base). nAMAVA = nBMBVB (2)(0.15M)(VA) = (1)(0.1M)(24mL) VB = 8mL

Answer 138

To relate the volumes and molarities of an acid and a base in a neutralization reaction we can use the following equation: nAMAVA = nBMBVB where nA is the number of acidic H+ ions the acid can donate and nB is the number of H+ ions the base can accept (typically equal to the number of OH- in the base). nAMAVA = nBMBVB (2)(0.1M)(30mL) = (1)(0.2M)(VB) VB = 30mL

Answer 139

HC2H3O2 is a weak acid and CsOH is a strong base and the pH at the equivalence point of the titration of a weak acid with a strong base will be slightly basic, a pH between 7 and 10 being typical. Only 8.4 is in this range and is the correct answer.

Answer 140

NaCN is a salt that dissociates into Na+ and CN-. CN- is a weak base and HClO4 is a strong acid and the pH at the equivalence point of the titration of a weak base with a strong acid will be slightly acidic, a pH between 4 and 7 being typical. Only 4.8 is in this range and is the correct answer.

Answer 141

NH3 is a weak base and HCl is a strong acid and the pH at the equivalence point of the titration of a weak base with a strong acid will be slightly acidic, a pH between 4 and 7 being typical. Only 5.2 is in this range and is the correct answer.

Answer 142

RbOH is a strong base and HCl is a strong acid and the pH at the equivalence point of the titration of a strong base with a strong acid will be exactly 7.0.

Answer 143

H2SeO3 Correct! ## Footnote The titration curve has two equivalence points and therefore indicates that the acid being titrated is diprotic. Of the choices listed HCl and HClO2 are monoprotic and H3PO4 is triprotic, but only H2SeO3 is diprotic and is therefore the correct answer.

Answer 144

To relate the volumes and molarities of an acid and a base in a neutralization reaction we can use the following equation: nAMAVA = nBMBVB where nA is the number of acidic H+ ions the acid can donate and nB is the number of H+ ions the base can accept (typically equal to the number of OH- in the base). nAMAVA = nBMBVB (3)(0.4M)(VA) = (2)(0.1M)(30mL) VB = 5mL

Answer 145

In the titration of a weak base with a strong acid, pOH = pKb at the half-equivalence point (and pH = pKa of the base’s conjugate acid) which can be seen to be somewhere between a pH of 9 and 10.\* This corresponds to a pOH of between 4 and 5 and the only base with a pKb in this range is ammonia (pKb = 4.7

Answer 146

To relate the volumes and molarities of an acid and a base in a neutralization reaction we can use the following equation: nAMAVA = nBMBVB where nA is the number of acidic H+ ions the acid can donate and nB is the number of H+ ions the base can accept (typically equal to the number of OH- in the base). nAMAVA = nAMBVB (1)(0.5M)(40mL) = (1)(0.1M)(VB) VB = 200mL

Answer 147

9.1 Correct! HOCl is a weak acid and NaOH is a strong base and the pH at the equivalence point of the titration of a weak acid with a strong base will be slightly basic, a pH between 7 and 10 being typical. Only 9.1 is in this range and is the correct answer.

Answer 148

In the titration of a weak acid with a strong base, pH = pKa at the half-equivalence point which can be seen to be just above a pH of 7. Therefore HOCl (pKa = 7.5) is the correct answer.

Answer 149

It is the equivalence point in a titration that can be used to calculate the concentration of the acid or base being titrated. At the equivalence point, the number of equivalents of acid are equal to the number of equivalents of base added which can be expressed by the following equation: nAMAVA = nBMBVB where nA is the number of acidic H+ ions the acid can donate and nB is the number of H+ ions the base can accept (typically equal to the number of OH- in the base). nB = 1 for NaOH nA = 1 as this is a monoprotic acid as there is only one equivalence point on the titration curve nAMAVA= nBMBVB (1)( MA)(10mL) = (1)(0.1M)(25mL) MA = 0.25M

Answer 150

HBr is a strong acid and KOH is a strong base and the pH at the equivalence point of the titration of a strong acid with a strong base will be exactly 7.0.

Answer 151

The system gains energy Correct! ## Footnote Since the gaseous system is heated (the gaseous system gains energy), q is positive. Since the gaseous system is compressed, ΔV is negative and w is positive (w=-PΔV) ΔE = q + w Since q and w are both positive, ΔE must be positive (the system gains energy)

Answer 152

ΔE = q + w = +18J – 10J = +8J Heat is transferred into the system (the system gains energy) and so q is positive. Work is done by the system (the system loses energy) and so w is negative.

Answer 153

For a spontaneous process, ΔS universe \> 0 Correct!

Answer 154

A state function is one that depends only upon the initial and final states of a system but is independent of the pathway from the initial state to the final state. You should simply remember that most of the properties that we speak of in Thermodynamics are state functions (ΔG, ΔH, ΔS, etc.) with the two big exceptions being q and w which completely depend on the pathway taken.

Answer 155

The following three phase changes all result in a decrease in disorder and are therefore incorrect: Na(l) =\> Na(s) NH3(g) =\> NH3(l) CO2(g) =\>CO2(s) Only C6H6(s) C6H6(l) reslts in an increase in disorder and will therefore have a positive enthalpy

Answer 156

fusion Correct!

Answer 157

The key here is realizing that 1271kJ is released for every 2 mol Ca reacted (as the coefficient for Ca is 2) and so multiplying this by the number of moles of Ca gives us the amount of heat released. The other little trick is that while ΔH is negative, the question asks for the amount of heat released, not ΔH. And while ΔH = -1907kJ, this means that 1907kJ of heat is released.\* We would not say that -1907kJ of heat is released; the fact that we are saying that the heat is released already indicates the sign of ΔH.\* This really is a matter of semantics but this is important.

Answer 158

deposition Correct!

Answer 159

CaO(s) + O2(g) CaCO3(s) Correct! he single largest indicator for DS is a change in the number of moles of gas.\* An increase in the number of moles of gas from reactants to products indicates an increase in entropy (DS\>0).\* A decrease in the number of moles of gas from reactants to products indicates a decrease in entropy (DS\<0). Choice A involves 0 mol gas ® 1 mol gas and thereforeDS\>0. Choice B involves 2 mol gas ® 4 mol gas and thereforeDS\>0. Choice C involves 1 mol gas ® 2 mol gas and thereforeDS\>0. Choice D involves 1 mol gas ® 0 mol gas and thereforeDS\<0.

Answer 160

The single largest indicator for S is a change in the number of moles of gas. An increase in the number of moles of gas from reactants to products indicates an increase in entropy ( S \> 0). A decrease in the number of moles of gas from reactants to products indicates a decrease in entropy ( S \< 0). Choice A involves 3 mol gas 2 mol gas and therefore\* S \< 0. Choice B involves 1 mol gas\* \*2 mol gas and therefore\* S \> 0. Choice C involves 1 mol gas\* \*1 mol gas and therefore\* S will have a small magnitude. Choice D involves 3 mol gas\* \*1 mol gas and therefore\* S \< 0.

Answer 161

-97kJ Correct! Using bond enthalpies (D), ΔHrxn = Dbroken\*- Dformed Examining the reactants we see that 1 NºN bond (941kJ) and 3 H-H (436kJ) bonds are broken.\*\* Examining the products we see that 6 N-H (391kJ) bonds are formed. ΔH = [941 + 3(436)] – [6(391)] = -97kJ

Answer 162

rxn for which ΔHrxn =ΔHf \*must be a formation rxn.\*A formation rxn forms 1 mole of a single product from individual elements (eliminates choice D as H2O is a compound not an element) in their standard states (eliminates choices A and C as the standard state of sodium is Na(s), the standard state of chlorine is Cl2(g)). Only choice B is left to satisfy the definition of a formation rxn.

Answer 163

A formation rxn forms 1 mole of a single product (eliminates choices B and C) from individual elements in their standard states (eliminates choice A as the standard state of carbon is graphite). Only choice D is left to satisfy the definition of a formation rxn.

Answer 164

Here we use Hess' Law. To make sure the Fe2O3(s) lines up as we need it in the desired reaction (3Fe2O3(s) on the reactants side) we'll need to reverse the 1st reaction provided and triple it which changes the sign and triples the\* H value. 3Fe2O3(s) ® 6Fe(s) + 9/2O2(g)\*\*\*\* ΔH°rxn =\*(-3)(-824 kJ) = 2472 kJ To make sure the Fe3O4(s)\*lines up as we need it in the desired readtion (2Fe3O4(s)\*on the products side) we’ll need to double the second reaction provided which\*also doubles the DH\*value 6Fe(s) + 4O2(g) ® 2Fe3O4(s) ΔH°rxn =\*(2)(-1120 kJ) = -2240 kJ Adding these two reactions together now gives\*the desired reaction and so adding theirDH values also gives the DH of the desired reaction. 3Fe2O3(s) ® 6Fe(s) + 9/2O2(g)\*\*\*\* ΔH°rxn =\*(-3)(-824 kJ) = 2472 kJ 6Fe(s) + 4O2(g) ® 2Fe3O4(s) ΔH°rxn =\*(2)(-1120 kJ) = -2240 kJ 3Fe2O3(s)\*® 2Fe3O4(s) + ½O2(g)\*\*\*\* ΔH°rxn =\*2472 kJ + (-2240 kJ) = 232kJ

Answer 165

Here we use Hess’ Law. To make sure the A lines up as we need it in the desired rxn (4A on the reactants\*side) we’ll need to quadruple the 1st rxn provided which increases ΔH by a factor of 4.\*This also takes care of getting 8B on the reactants side.\* To make sure that the D lines up as we need it in the desired rxn (2D on the products side) we’ll double and reverse the second rxn provided (multiply by -2).\*This also takes care of getting 6E on the products side.\* Adding the two rxns gives the desired rxn and so adding their ΔH values also gives the ΔH of the desired rxn. 4A + 8B → 4C\*\*\* ΔH° = (-100kJ)(4) = -400kJ 4C → 2D + 6E\*\* ΔH° = (150kJ)(-2)\*\* = -300kJ 4A + 8B → 2D + 6E\*ΔH° = -700kJ

Answer 166

Here we use Hess’ Law. To make sure H2O2(l) lines up as we need it in the desired reaction (H2O2(l) on the reactants side) we’ll need to reverse the 2nd reaction provided which changes the sign of the DH value. H2O2(l) ® H2O(l) + ½O2(g) ΔH°rxn = -98 kJ To make sure H2(g) lines up as we need it in the desired reaction (H2(g) on the products side) we’ll need to reverse the 1st reaction provided which changes the sign of the DH value. H2O(l) ® H2(g) + ½O2(g) ΔH°rxn = 286 kJ Adding these two reactions together now gives the desired reaction and so adding their DH values also gives theDH of the desired reaction. H2O2(l) ® H2O(l) + ½O2(g) ΔH°rxn = -98 kJ H2O(l) ® H2(g) + ½O2(g) ΔH°rxn = 286 kJ H2O2(l) + H2O(l) ® H2O(l) + ½O2(g) + H2(g) + ½O2(g) which reduces to H2O2(l) ® H2(g) + O2(g) ΔH°rxn = -98 kJ + 286 kJ = 188kJ

Answer 167

To make sure Cu2S(s) lines up as we need it in the desired reaction (Cu2S(s) on the reactants side) we'll need to reverse the 2nd reaction provided which changes the sign of the H value. Cu2S(s) + O2(g) 2Cu(s) + SO2(g) H°rxn = -217 kJ Using the above reaction as we did also assures Cu(s) lines up as we need it to in the desired reaction also (2Cu(s) on the products side). To make sure the S(s) lines up as we need it in the desired reaction (S(s) on the products side) we'll need to reverse the 1st reaction provided which changes the sign of the H value. SO2(g) S(s) + O2(g) H°rxn = 297 kJ Adding these two reactions together now gives the desired reaction and so adding their H values also gives the H of the desired reaction. Cu2S(s) + O2(g) 2Cu(s) + SO2(g) H°rxn= -217 kJ SO2(g) S(s) + O2(g) H°rxn = 297 kJ Cu2S(s) + O2(g) + S)2(g) 2Cu(s) + SO2(g) + S(s) + O2(g) which reduces to Cu2S(s) 2Cu(s) + S(s) H°rxn = -217 kJ + 297 kJ = 80 kJ

Answer 168

Here we use Hess’ Law. To make sure the CO(g) lines up as we need it in the desired rxn (2CO(g) on the reactants side) we’ll need to reverse the 1st rxn provided which changes the sign of ΔH (multiply by -1).\* The O2(g) shows up in both provided rxns so we’ll skip and save until the end.\* Finally to make sure that the CO2(g) lines up as we need it in the desired rxn (2CO2(g) on the products side) we’ll simply use the second provided rxn as is.\*Adding the two rxns gives the desired rxn and so adding their ΔH values also gives the ΔH of the desired rxn. 2CO(g) → 2C(s) + O2(g) \* ΔH°rxn = (-221 kJ)(-1) = +221kJ 2C(s) + 2O2(g) → 2CO2(g) \*\* ΔH°rxn = -787 kJ \* \_ 2CO(g) + O2(g) → 2CO2(g) ΔH°rxn\*= (+221kJ) + (-787kJ) = -566kJ

Answer 169

To make sure NO(g) lines up as we need it in the desired reaction (2NO(g) on the reactants side) we’ll need to reverse the 3rd reaction provided which changes the sign of the DH value. 2NO(g) ® N2(g) + O2(g) ΔH°rxn = -180 kJ To make sure H2O(l) lines up as we need it in the desired reaction (3H2O(l) on the reactants side) we’ll need to reverse the 2nd reaction provided and multiply by 3/2 which changes the sign of theDH value and increases it by a factor of 3/2. 3H2O(l) ® 3H2(g) + 3/2O2(g) ΔH°rxn = (3/2)(572 kJ) = 858 kJ To make sure NH3(g) lines up as we need it in the desired reaction (2NH3(g) on the products side) we’ll need to add the 1st reaction provided as it is which leaves DH value unchanged. N2(g) + 3H2(g) 2NH3(g) ΔH°rxn= -92 kJ Adding these three reactions together now gives the desired reaction and so adding their DH values also gives the DH of the desired reaction. 2NO(g) ® N2(g) + O2(g) ΔH°rxn = -180 kJ 3H2O(l) ® 3H2(g) + 3/2O2(g) ΔH°rxn = (3/2)(572 kJ) = 858 kJ N2(g) + 3H2(g) ® 2NH3(g) ΔH°rxn= -92 kJ 2NO(g) + 3H2O(l) + N2(g) + 3H2(g) ® N2(g) + O2(g) + 3H2(g) + 3/2O2(g) + 2NH3(g) which reduces to 2NO(g) + 3H2O(l) 2NH3(g) + 5/2O2(g) ΔH°rxn = -180 kJ + 858 kJ + (-92 kJ) = 586 kJ

Answer 170

The equilibrium constant will be much smaller than 1 Correct! With ΔG° \< 0, this reaction is spontaneous under standard conditions (° means standard conditions). ΔG° = -RTlnKeq shows the relationship between ΔG° and Keq. When ΔG° \< 0, Keq \> 1 and the reaction will favor the products. Since ΔG° = ΔH° – TΔS° and ΔG° \< 0, then ΔH°-TΔS° \< 0. As stated above, ΔG° = -RTlnKeq shows the relationship between ΔG° and Keq. When ΔG° \< 0, Keq \> 1 which implies that there will be more products than reactants present at equilibrium and not the other way around.

Answer 171

For a single SO32- ion, oxygen is assigned first as having a -2 oxidation state according to rule #5 provided in the outline. Sulfur is then left to balance the overall charge of a sulfite ion to -2. Sulfur must therefore be in the +4 oxidations state: S + 3(Ox) = -2 S + 3(-2) = -2 S = +4 It would be more difficult to figure out the oxidation state of sulfur in the context of the entire compound, Al2(SO3)3, but we can do it. Aluminum is in the +3 oxidation state according to rule #4 and oxygen is in the -2 oxidation state according to rule #5 as discussed above. To balance the charge of the entire compound to zero sulfur must be in the +4 oxidation state: 2(Al) + 3(S) + 9(Ox) = 0 2(+3) + 3(S) + 9(-2) = 0 3(S) = 12 S = +4 All of the rules for assigning oxidation states are provided below for reference: Oxidation States 1) Elements in their elemental form are in the zero oxidation state. 2) Group 1 metals are +1 and Group 2 metals are +2. 3) Hydrogen is +1 except when bonded to metals (when it’s –1). 4) Transition elements must be determined by anion’s charge (except Al=+3, Zn=+2, Cd=+2, Ag=+1). 5) The most electronegative elements get their typical oxidation state. 6) The last element not assigned balances the charge of the compound/ion.

Answer 172

O2 Correct! ## Footnote The oxidizing agent or oxidant is the reactant species that is being reduced. In the conversion of O2 to VO2+oxygen is changing oxidation states from zero to -2 and is being reduced (gaining electrons) and therefore O2 is the oxidant. How the oxidation states were determined is shown below. O2 is the elemental form of oxygen and is therefore in the zero oxidation state (according to rule #1). Oxygen is in the -2 oxidation state in VO2+ according to rule #5. All of the rules for assigning oxidation states are provided below for reference: Oxidation States 1) Elements in their elemental form are in the zero oxidation state. 2) Group 1 metals are +1 and Group 2 metals are +2. 3) Hydrogen is +1 except when bonded to metals (when it’s –1). 4) Transition elements must be determined by anion’s charge (except Al=+3, Zn=+2, Cd=+2, Ag=+1). 5) The most electronegative elements get their typical oxidation state. 6) The last element not assigned balances the charge of the compound/ion.

Answer 173

Cu Correct! The reductant (or reducing agent) is the species that gets oxidized (loses electrons). First of all the oxidant or reductant is always a reactant so choices C and D can be eliminated. Ag+ to Ag : +1 to 0 \* \* \* \* Ag+ is reduced and is therefore the oxidant. Cu to Cu2+ : 0 to +2 \* \* \* \* Cu is oxidized and is therefore the reductant (therefore choice B is correct)

Answer 174

Oxygen = -2 so Fe must = +3 2(Fe) + 3(Ox) = 0 2(Fe) + 3(-2) = 0 2(Fe) = +6 Fe = +3

Answer 175

The oxidizing agent (or oxidant) is the species that gets reduced (gains electrons). First of all the oxidant or reductant is always a reactant so\*Mn2+ can be eliminated. H+ to H2O : +1 to +1 H+ is not oxidized or reduced. MnO₄^- to Mn2+ : +7 to +2 MnO4- is reduced and is therefore the oxidizing agent and the correct answer choice. Fe2+ Fe3+ : +2 +3 Fe2+ is oxidized and is therefore the reducing agent rather than the oxidizing agent.

Answer 176

Cr₂O₇^2- Cr³⁺ Cr₂O₇^2-to 2Cr³⁺ balance Cr 1st since it’s the element changing oxidation states Cr₂O₇^2- to 2Cr3+ + 7H2O balance oxygen by adding H2O 14H+ + Cr₂O₇^2- to 2Cr³⁺ + 7 H₂O balance hydrogen by adding H+ Reactants = +12 charge; products = +6 charge (electrons needed on reactants side) 6e^-+14H⁺ + Cr₂O₇^2- to 2Cr³⁺ + 7H₂O add electrons to balance the charge

Answer 177

+2 Correct! ## Footnote Magnesium is the first element assigned and as a group 2 metal it will simply be in the +2 oxidation state according to rule #2 of the rules provided in the outlines. All of the rules for assigning oxidation states are provided below for reference: Oxidation States 1) Elements in their elemental form are in the zero oxidation state. 2) Group 1 metals are +1 and Group 2 metals are +2. 3) Hydrogen is +1 except when bonded to metals (when it’s –1). 4) Transition elements must be determined by anion’s charge (except Al=+3, Zn=+2, Cd=+2, Ag=+1). 5) The most electronegative elements get their typical oxidation state. 6) The last element not assigned balances the charge of the compound/ion.

Answer 178

The reductant or reducing agent is the reactant species that is oxidized. In the conversion of Zn to Zn(OH)2, zinc is changing oxidations states from zero to +2 (a loss of electrons) and is being oxidized and therefore Zn(s) is the reducing agent. How the oxidation states were determined is shown below. Zn is the elemental form of zinc and is therefore in the zero oxidation state (according to rule #1). Zinc is in the +2 oxidation state in Zn(OH)2 according to rule #4 of the rules provided in the outlines. Zinc is one of the few transition metals that always has the same oxidation state in ionic compounds. All of the rules for assigning oxidation states are provided below for reference: Oxidation States 1) Elements in their elemental form are in the zero oxidation state. 2) Group 1 metals are +1 and Group 2 metals are +2. 3) Hydrogen is +1 except when bonded to metals (when it’s –1). 4) Transition elements must be determined by anion’s charge (except Al=+3, Zn=+2, Cd=+2, Ag=+1). 5) The most electronegative elements get their typical oxidation state. 6) The last element not assigned balances the charge of the compound/ion.

Answer 179

According to the rules provided in the outline we assign potassium first as being in the +1 oxidation state according to rule #2. Oxygen must therefore be -1 according to rule #6. This is an example of a peroxide in which oxygen is in the -1 oxidation state rather than its typical -2 oxidation state. All of the rules for assigning oxidation states are provided below for reference: Oxidation States 1) Elements in their elemental form are in the zero oxidation state. 2) Group 1 metals are +1 and Group 2 metals are +2. 3) Hydrogen is +1 except when bonded to metals (when it’s –1). 4) Transition elements must be determined by anion’s charge (except Al=+3, Zn=+2, Cd=+2, Ag=+1). 5) The most electronegative elements get their typical oxidation state. 6) The last element not assigned balances the charge of the compound/ion.

Answer 180

Oxidation occurs at the cathode Correct! ## Footnote Electrolytic cells consume energy in causing a non-spontaneous rxn to occur so choices A and D are true statements and are eliminated. Electrons travel from anode to cathode in all electrochemical cells (both galvanic and electrolytic) so choice B is a true statements and is eliminated. Reduction occurs at the cathode (RED CAT) and oxidation occurs at the anode (AN OX) in all electrochemical cells so choice C is the false statement and the correct answer.

Answer 181

The cathode is negatively charged Correct! ## Footnote In a voltaic cell electrons travel spontaneously from the anode (which is negative) to the cathode (which is positive) and so choice A is the false statement. Incidentally, in an electrolytic cell electrons travel non-spontaneously from anode (which is positive)to cathode (which is negative).

Answer 182

PbO2 Correct! ## Footnote RED CAT – Reduction occurs at the cathode PbO2\*→\*\*PbSO4 : lead goes from +4\*→\*+2 therefore PbO2 is reduced and is the cathode (PbO2 is correct) Pb\*→\*PbSO4 : lead goes from zero\*→\*\*+2 therefore Pb(s) is oxidized and is the anode

Answer 183

Electrons travel from anode to cathode through the wire Correct! First, electrons travel through the wire whereas ions travel through the salt bridge and so choices A and B can be eliminated. Second, electrons always travel from anode to cathode in both galvanic and electrolytic cells and so choice C is correct and choice D is eliminated.

Answer 184

decreasing the pH Correct! ## Footnote Shifting the equilibrium to the right (toward the products) according to Le Chatelier’s Principle results in an increase in the cell potential. Increasing [Fe3+] results in a shift left which decreases the cell potential eliminating choice A. Decreasing [MnO4-] results in a shift left which decreases the cell potential eliminating choice B. Decreasing the pH is the same as increasing [H+] which results in a shift to the right and an increase in the cell potential and therefore C is the correct answer. Adding a catalyst doesn’t shift an equilibrium (it only gets a rxn to equilibrium faster) and thereby doesn’t change the cell potential eliminating choice D.

Answer 185

A balanced rxn isn’t provided in this question but to be a redox reaction, one half-rxn will be oxidation and the other reduction. And since we’re told that it’s a galvanic cell we know the rxn will be spontaneous and the emf \> 0. Fe²⁺+ 2e- =\> Fe -0.44V Ni²⁺+ 2e- =\> Ni -0.25V These reduction reactions and reduction potentials are taken out of the table provided. One of the reactions will need to be reversed (to an oxidation half-rxn) and the oxidation potential will have the opposite sign. For the overall emf to come out positive since it’s galvanic, it is the iron half-rxn that will have to be reversed. Ni²⁺ + 2e- =\> Ni -0.25V Fe =\> Fe²⁺ + 2e- +0.44V -0.25V + 0.44V = +0.19V

Answer 186

2e-+ Cu2+ → Cu\*\* Eºred =+0.34V Mn =\> Mn²⁺ + 2e-\* Eºox=+1.18V\*(This rxn is reversed and the sign changed)

Answer 187

Shifting the equilibrium to the right (toward the products) according to Le Chatelier’s Principle results in an increase in the cell potential. Increasing [Li+] results in a shift to the left and decreases the cell potential eliminating choice A. Removing I₂ doesn’t change the cell potential as solids don’t shift equilibria eliminating choice B. Increasing [IO3-] results in a shift to the right and increases the cell potential and therefore C is the correct answer choice. Increasing the size of the anode (Li is the anode) is increasing the size of a solid which don’t shift equilibria eliminating choice D.

Answer 188

Co and Cu2+ Correct! ## Footnote First of all, for a redox rxn we need a species that can be oxidized (on the products side of the table of reduction potentials) and one that can be reduced (on the reactants side of the table of reduction potentials). According to the table, both Al and Ag can be oxidized but not reduced (both only appear on the products side) and so an oxidation/reduction reaction between these two isn’t even possible. According to the table, both Cu2+ and Ni2+ can only be reduced but not oxidized (both only appear on the reactants side) and so an oxidation/reduction reaction between these two isn’t even possible. According to the table, Fe2+ can be reduced (-0.44V) and Ag can be oxidized (-0.80V) but the overall potential would be -1.24V, the negative value indicating that these two would not react spontaneously. According to the table, Co could be oxidized (+0.28V) and Cu2+ could be reduced (+0.34V). The overall cell potential would be +0.62V, the positive value indicating that these two would indeed react spontaneously.

Answer 189

Zn and Ni²⁺ Correct! ## Footnote First of all, for a redox rxn we need a species that can be oxidized (on the products side of the table of reduction potentials) and one that can be reduced (on the reactants side of the table of reduction potentials). According to the table, both Cu and H2 can be oxidized but not reduced (both only appear on the products side) and so an oxidation/reduction reaction between these two isn’t even possible. According to the table, both H+ and Ag+ can only be reduced but not oxidized (both only appear on the reactants side) and so an oxidation/reduction reaction between these two isn’t even possible. According to the table, Cr3+ can be reduced (-0.74V) and H2 can be oxidized (zero V) but the overall potential would be -0.74V, the negative value indicating that these two would not react spontaneously. According to the table, Zn could be oxidized (+0.76V) and Ni2+ could be reduced (-0.25V). The overall cell potential would be +0.51V, the positive value indicating that these two would indeed react spontaneously.

Answer 190

Al and H+ Correct! ## Footnote First of all, for a redox rxn we need a species that can be oxidized (on the products side of the table of reduction potentials) and one that can be reduced (on the reactants side of the table of reduction potentials). Co2+ can be reduced (-0.28V) and H2 can be oxidized (0.0V) but E cell=-0.28V which is \<0 and therefore non-spontaneous eliminating choice A. Co and H2 both can only be oxidized so redox isn’t possible eliminating choice B. Ag+ and Co2+ both can only be reduced based upon the table of reduction potentials provided and so redox isn’t possible eliminating choice C. Al can be oxidized (+1.66V) and H+ can be reduced (0.0V) and E cell=+1.66V which is \>0 and therefore spontaneous and so choice D is the correct answer

Answer 191

Ag+ Correct! ## Footnote The strongest oxidizing agent is the species that is most spontaneously reduced (most positive reduction potential). The most positive reduction potential listed is 0.80V and Ag+ (not Ag) is the reactant that is getting reduced and is the strongest oxidizing agent.

Answer 192

Cu2+ Correct! ## Footnote The strongest oxidizing agent is the species that is most spontaneously reduced (most positive reduction potential). The most positive reduction potential listed is 0.34V and Cu2+ (not Cu) is the reactant that is getting reduced and is the strongest oxidizing agent

Answer 193

Al Correct! ## Footnote The strongest reducing agent is the species that is most spontaneously oxidized (most positive oxidationpotential). The oxidation reactions are the reverse of those listed in the table of reduction potentials and the oxidation potentials are the negative of those listed. The most positive oxidation potential would be +1.66Vand Al (not Al3+) is the reactant that is getting oxidized and is the strongest reducing agent (choice B). Al =\> Al³⁺ + 3e^- +1.66V

Answer 194

Since reduction happens at the cathode (RED CAT) choices B and D could be eliminated as they are oxidation half reactions (electrons on the product side are lost). In the molten electrolysis of a binary salt, the cation is reduced at the cathode to its elementary form: Cd2+ + 2e- → Cd and so answer choice A is correct.

Answer 195

In the molten electrolysis of a binary salt, the cation is reduced at the cathode to its elementary form. Na+ + e-\*\* \*Na In the molten electrolysis of a binary salt, the anion is oxidized at the anode to its elementary form. 2Cl- Cl2 + 2e- So the products are Na at the cathode and Cl2 at the anode and choice C is correct.

Answer 196

The half-rxn in which Al(s) is produced is\* Al³⁺+ 3e- =\>Al.\* Therefore 3 moles of electrons are required to produce 1 mole of aluminum. 54g Al X 1 mol Al/ 27 g Al X (3 mol e- / 1 mol Al) =6 mol electrons

Answer 197

The half reaction in which Cr metal is produced is as follows: Cr3+ + 3e- Cr So it takes 3 moles of electrons to produce 1 mole of Cr. In this case we’re producing 26.0g of Cr which is 0.5 moles. Finally based on the 3:1 ratio of moles of electrons to moles Cr, it will require 3/2 moles of electrons to make 0.5 moles of Cr.

Answer 198

The product here is an ketal which results from the addition of 2eq of alcohol to a ketone. The key is realizing that the carbon bonded to both oxygens of the ketal product was the carbonyl of the reactant ketone that was attacked by the alcohols.

Answer 199

PhMgBr (a Grignard reagent) is a strong nucleophile and reacts just as if it were a phenyl anion attacking the carbonyl carbon of the aldehyde.\* H3O+\*then protonates the intermediate forming an alcohol (choice A).

Answer 200

Esters can react with 2 equivalents of a Grignard reagent.\* The first equivalent undergoes nucleophilic acyl substitution (as the ester has a leaving group) to produce a ketone.\* The 2nd equivalent undergoes nucleophilic addition with the ketone to produce an alcohol (choice D).

Answer 201

This question is a little tricky.\* A Grignard reagent shouldn’t be in any kind of protic solution as it will be protonated by anything protic in an acid-base reaction which effectively destroys the Grignard reagent as is the case here.\* In the first step, the Grignard reagent deprotonates the alcohol.\* The addition of H3O+simply reprotonates the alkoxide converting it back into an alcohol and so the product is the same as the original reactant (choice A).

Answer 202

A hemi-acetal results from the addition of an\*alcohol to an aldehyde (a hemi-ketal is similar and results from the addition\*of an alcohol to a ketone).\* It can be recognized by finding a single carbon single bonded to 2 oxygen atoms: one as an alcohol and one as an ether.

Answer 203

The product here is similar to an imine (shown below) and the only difference is that the variable group (R-group) must be a hydroxyl group explaining why choice C is correct.\* Recognizing the reactant ketone as an electrophile and then seeing the product should lead us to predict that the missing reagent is a nucleophile that has a nitrogen atom bonded to a hydroxyl group (N-OH).\* This structural element is only present in choice C.

Answer 204

This is an aldol condensation as an aldehyde and ketone are mixed under basic conditions with heat. \*Only acetone has enolizable hydrogens (a-hydrogens) and will therefore have to serve as the nucleophile. \*The abbreviated mechanism below shows the intermediate b-hydroxy ketone that, when heated, undergoes elimination (dehydration specifically) to form the conjugated enone which matches choice A. As benzaldehyde has 7 carbons and acetone has 3, the product should have 10 carbons which could have allowed us to eliminate choices B and D from the start.\* And choice C has the hydroxyl group in the wrong place.

Answer 205

This is an aldol condensation performed under basic conditions. The steps are as follows: 1) Deprotonation of the alpha carbon to form the enolate ion. 2) The enolate ion attacks another ketone molecule. 3) The resulting alkoxide is protonated. 4) Elimination to form the carbon-carbon double bond (one of the rare cases where the hydroxyl group is the leaving group).

Answer 206

This question is testing your knowledge of b-decarboxylation. \*When heated, a carboxylic acid will undergo decarboxylation forming CO2\*only if the b-carbon is a carbonyl.\* Theb-carbons (relative to the carboxyl groups) that are highlighted below in red are at a carbonyl and will result in decarboxylation of the carboxyl groups indicated with arrows. \*The b-carbon highlighted in blue is not a carbonyl and the carboxylic acid at the bottom of the molecule will be stable and not undergo decarboxylation (choice D).

Answer 207

This is an aldol condensation performed under basic conditions. The steps are as follows: 1) Deprotonation of the alpha carbon to form the enolate ion. 2) The enolate ion attacks another acetone molecule. 3) The resulting alkoxide is protonated. 4) Elimination to form the carbon-carbon double bond (one of the rare cases where the hydroxyl group is the leaving group).

Answer 208

IV \< I \< III \< II Correct!