General Chemistry Flashcards
Which compound has the bond with the smallest dipole moment?
A) H2O
B) H2S
C) H2Se
D) H2Te
D) H2Te
Dipole moment: u=qr
q=magnitude of partial charge separated across a distance r
Te is the largest, all group 16 elements, which makes H-Te the longest bond in the given compounds. However, the electronegativity difference is the smallest in the H-Te bond because electronegativity tends to decrease moving down a group (column) on the periodic table
Therefore, the H-Te bond is the longest but has the smallest dipole moment of the given compounds because the atoms in the H-Te bond have almost no difference in electronegativity, ie negligible partial charge separated across the bond
The slope of the trendline for the data shown in Figure 1 is -4911 and its y-intercept is 5.4. Which of the following is an expression for deltaG* at temperature T stated in terms of the data variables?
A) 4911R - T(5.4R)
B) 5.4R + T(4911R)
C) 4911/R -T5.4/R)
D) (5.4/R) + T(4911/R)
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A) 4911R - T(5.4R)
Given van’t Hoff equation: lnKa = (-deltaH* / R)(1/T) + (deltaS*/R)
equate to y=mx+b
-4911=(-deltaH* / R)
deltaH* = 4911R
5.4 =(deltaS/R)
deltaS = 5.4R
Would changes in the can’t Hoff plot be observed if the reaction rate were increased by adding a catalyst during the experiment?
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No, because deltaG. deltaH, and deltaS* are independent of the reaction rate
What volume of the solutions tested would contain 0.080g of ocytocin acetate?
A) 0.020 mL
B) 0.32 mL
C) 20 mL
D) 320 mL
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D) 320 mL
0.080 g oxytocin x (1000 mg/1g) x (1mL / 0.25 mg) = 320 mL
(1mL / 0.25 mg) = sample oxytocin concentration given in passage
Suppose the 10 mM MgCl2 solution used in experiment 2 is replaced with a 5 mM NaCl solution. Compared to the Cl- concentration in the 10 mM MgCl2 solution, the concentraiton of Cl- in the 5mM NaCl solution would be:
A) greater
B) the same
C) 2 times less
D) 4 times less
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D) 4 times less
NaCL solution:
(5 mmol NaCl / 1 mL solution) x (1 mmol Cl-/1mmol NaCL) = 5 mM Cl-
MgCl2 solution:
(10 mmol MgCl2 / 1 mL solution) x (2 mmol Cl- / 1 mmol MgCl2) = 20 mM Cl
Which of the following experimental changes to the initially prepared oxytocin acetate solutions will decrease the percennt of oxytocin acetate recovered in Experiment 1 after the 4-week storage period?
A) Adding more sodium citrate after the NaOH has been added
B) decreasing the initial concentration of oxytocin acetate
C) storing the solutions at 0C rather than 5C
D) increasing the citrate buffer concentration
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A) Adding more sodium citrate after the NaOH has been added
Dissociation of a weak acid: Ka = ([HA] [A-]) / [HA]
Henderson Hasselbach: pH = pKa + log ([A-] / [HA])
Adding more acid will decrease the pH, adding more base will increase the pH
In Experiment 1, the citrate buffer is made of citric acid [HA] and sodium citrate [A-]. Adding more sodium citrate to the initially prepared solutions will increase the pH of the solution.
The passage states that oxytocin is more stable at a pH of 4.5, and Figure 1 shows that at the higher pH of 6.8 (water), less oxytocin was recovered. Therefore, adding more sodium citrate will decrease the percent recovery of oxytocin acetate.
If 16 mL of the initial oxytocin acetate solution containing the 50 mM Zn2+ additive was analyzed, how many moles of oxytocin acetate (MW = 1067 g/mol) were recovered from the initial sample after 4 weeks at 50C?
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16 mL x (0.25mg / 1 mL) x (1 g / 1000 mg) x (1 mol/1067 g) x (80 mol recovered / 100 mol original ) = 3.0 x 10^-6 mol oxytocin acetate
Suppose that citric acid (H3C6H5O7) is titrated with 0.1M NaOH to form a citrate buffer solution with a pH of 4.5. What is the pH at the first equivalence point? Note: pKa1 = 3.13, pKa2 = 4.76, pKa3 = 6.40.
A) Less than 3.13
B) Between 3.13 and 4.76
C) Equal to 3.13
D) Greater than 4.76
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B) Between 3.13 and 4.76
The pH at the first equivalence point must be between the pKa of the first acidic hydrogen (pKa1) and the second acidic hydrogen (pKa2)
A radioactive atom decays by 5 alpha, 3 beta minus, and 2 gamma emissions to yield 211PO. What was the original nucleus?
A) 231Pa
B) 223At
C) 231U
D) 191IR
A) 231Pa
Alpha emission
release of a helium nucleus (two protons and two neutrons). Each alpha emission reduces the atomic mass and atomic number of the atom by 4 and 2, respectively
Beta minus emission
release of an electron from the nucleus and turns a neutron into a proton; it increases the atomic number by 1. The mass of an electron is negligible compared to the nucleus, so beta emission does not alter the atomic mass
Gamma emission
the release of a high energy photon that occurs when the protons and neutrons in a nucleus change configuration. it does not alter the atomic mass or atomic number of an atom
Consider the dimerization of nitrogen dioxide at room temp:
2No2(g) –> N2O4 (g) ; Kp = 8atm^-1
If the equilibrium partial pressure of nitrogen dioxide in a container is 0.5atm, what is the partial pressure of dinitrogen tetroxide?
2 atm
Kp = (Pproducts)^m /(Preactants)^n
=PN2O4 / PNO2 x PNo2
8atm^-1 = PN2O4 / (0.5 atm)^2
PN2O4 - 2 arm
Researchers wished to mimic the conditions of the medial Golgi (ph=6.3). Which of the buffers shown in Table 1 would be best suited for this experiment?
Acetic acid : Ka = 1.7 x 10^-5
MES: Ka = 7.1 x 10^-7
HEPES : 2.8 x 10^-8
Tris: 6.3 x 10^-9
MES: Ka = 7.1 x 10^-7
this equals pka of 7.1, within +/- region of 6.3, given pH
The chemical structures of sulfur compounds H2S, SF6, S2Cl2, and S4N4 each contain only single bonds. Based on trends in the atomic radii of the elements, the compound that contains the largest bond between sulfur and an atom of another element is:
S2Cl2
The length of a sigma bond can be estimated as the sum of the atomic radii of the bonded atoms.
Because Cl has a larger atomic radius than H, N, or F, the S-Cl bond will be the longest bond with sulfur among the given compounds.
The anion SO3 2- and sulfur trioxide, SO3, have the same elemental composition. These two chemical species are:
A) the same compound, because both have the same elemental composition
B) the same compound, because both have the same number of electrons and bonding configuration
C) different compounds, because the atoms in SO3 2- are held together by ionic bonds
D) Different compounds, because SO3 2- has two more electrons and a different bonding configuration
D) Different compounds, because SO3 2- has two more electrons and a different bonding configuration
positron emission
B+ decay
Suppose that the (p-n) nuclear reaction described in the passage was attempted but resulted in the 18 O nucleus absorbing a proton within ejecting a neutron. Which of the following statements correctly describes the resulting atom that would be formed?
A) the process would form an isotope of flourine with very different chemical properties than 18F
B) the process would form an isotope of oxygen with nearly identical chemical properties
C) the process would form an isotope of flourine with nearly identical chemical properties
D) the process would form an element other than O and F with very different chemical properties
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C) the process would form an isotope of flourine with nearly identical chemical properties
If the 18 O nucleus were to absorb a proton, the atomic number would increase by 1 to a total of 9 protons. Because all atoms with 9 protons are flourine atoms, the resulting nucleus would be a flourine isotope. If no neutron was ejected following the absorption of the proton, the mass number would also increase by 1 to give a flourine isotope with a mass number of 19. The flourine cation initially formed would quickly acquire the missing valence electron from the surrounding environment.
Because 18F and 19F are isotopes of the same element with the same electron configuration, the chemical properties of these nuclei would be nearly identical.
If the initial activity from a 1-mL dose of 18F-FDG is 200 millicuries, what will be the approximate activity of the dose after 1.5 half-lives of 18F have passed?
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70 millicuries
Measured activity = (fraction remaining) x (initial activity)
fraction remaining = 35%, on graph half life = 110 minutes, therefore 1.5 half-lives = 165 minutes. At one 165 minutes, only 35% remains
Measured activity = .35 x 200 = 70
Compared to the effective nuclear charge of 18O, the effective nuclear charge of 18F is:
A) higher, because 18F and 18O have the same number of core electrons but 18F has fewer neutrons
B) Lower, because 18F has more valence electrons than 18O
C) higher, because 18F and 18O have the same number of core electrons but has 18 has more protons
D) lower, because 18F has an equal number of protons and electrons
C) higher, because 18F and 18O have the same number of core electrons but has 18 has more protons
90Sr is converted to 90Y by which of the following processes?
A) Electron capture
B) positron emission
C) gamma ray emission
D) beta minus decay
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D) beta minus decay
Electron capture
a special form of beta decay in which a nucleus absorbs an electron from an external source without emitting a particle. The absorbed electron converts a proton into a neutron, deceasing the atomic number by one
In its nuclear decay chain, 90St decays into 90Y and then to 90Zr. Which of the following elements exhibits chemical properties most similar to zirconium-90?
A) 90Nb
B) 184Hf
C) 46Sc
D) 48Car
B) 184Hf
Zirconium is in the same column as hafnium HF, and therefore any isotope of Hf will have similar chemical properties to any isotope of Zr
Physical properties (color, density, atomic radius) are not easily determined by column grouping
Fission of 235U produces multuple elements. Based on figure 1, which of the following is the atomic number of the most common fission product?
A) 38
B) 55
C) 92
D) 133
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B) 55
The tallest peak indicates the greatest fission yield , 133
Of the answer choices given, only cesium CS with an atomic number of 55 has an atomic mass near 133
The uranium isotope required for Reaction 1 can be formed by the emission of an alpha particle from which of the following nuclei?
Reaction 1: 235U –> 90Sr + Z
A) 239U
B) 239Pu
235U
235Pa
B) 239Pu
A patient is injected with 20 ng of 90Y. Assuming the yttrium is not excreted, how long will it take for the initial amount of 90Y to be reduced to 2.5ng?
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192 Hours
2.5/20 = 1/8
1/2^3 = 1/8
therefore 3 half-lives
per the passage, the half-life of 90Y = 64 hours
64 x 3 = 192 hours
What is the approximate molar concentration of Na+ ions in a solution by dissolving 8.2 g of Na3PO4 (MM=163.94 g/mol) into 250 mL of water?
- 0 x 10^-1 M
- 2 g x (1 mol Na3PO4 / 163.94 g Na3PO4 ) = 0.050 mol Na3PO4
- 050 mol Na3PO4 x (3 mol Na+ / 1 mol Na3Po4) = 0.15. mol Na+
- 15 mol Na+ / 0.25 L = 0.60 mol/L
The boiling point of CH3CH2Br (bromoethane) is 38.4C whereas the boiling point of CH2CH2I (iodoethane) is 71.5C. Which of the following best explains the difference between the boiling points of the two alkyl halides?
A) the C-Br bond in bromoethane is less reactive than the C-I bond in iodoethane
B) the bromine atom in bromoethane is more polarizable than the iodine atom in iodoethane
C) The C-Br bond in bromoethane is more polar than the C-I bond in iodoethane
D) Iodoethane has greater London dispersion forces than bromoethane
D) Iodoethane has greater London dispersion forces than bromoethane
Iodine substituent is larger and more polarizable than bromine, therefore iodoethane experiences greater London dispersion forces and has a higher boiling point
Which of the following describes the standard change in entropy deltaS for the sublimation of (P4)n(s) and the overall conversion of (P4)n(s) to P4(s), respectively?
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DeltaS sublimation > 0 , DeltaSconversion > 0
Sublimation of (P4)n is the direct conversion of (P4)n (s) to (P4)n(g). This process increases the total entropy of the system, ie 1 mole of solid converts to 1 mole of gas, resulting in deltaS > 0
Conversion of (P4)n(s) to P4(s) breaks up the polymer chains of repeating P4 monomer units. This results in less restricted movement of the P4 units (ie, the number of possible arrangements increases) and yields more disorder
The phosphorous allotrope that has the highest melting point is:
A) (P4)n, because it has the strongest intermolecular forces
B) P4 because it has the stronger intermoleculer forces
C) (P4)n, because it has the stronger P-P bonds that require more energy to break
D) P4, because it has the stronger P-P bonds that are much more polar
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A) (P4)n, because it has the strongest intermolecular forces
Both P4 and (P4)n are nonpolar molecules that experience London dispersion forces. however, (P4)n comprises repeating P4 monomer units linked together, resulting in a larger molecular size and orbital surface area. Therefore, (p4)n will have a higher melting point than P4 due to its stronger London dispersion forces.
What is the specific heat capacity of solid P4?
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1.35
According to the heating curve obtained using the 5.74 g sample of P4, a total of 350 J of heat (x-axis) is required to raise the temperature (y axis) of solid P4 from 0C to its melting point of 45C.
C = q / m (delta T)
= 350 J / 5.74(45-0)
=1.35
The bonds of four salts, MgBr2, NaCl, MgCl2, and NaBr, are evalutated for ionic character based on electronegativity differences between atoms. What is the expected order of these four salts if listed according to bond character from the least ionic character to the greatest ionic character?
MgBr2 < MgCl2 < Nabr < NaCl
Comparing NaBr and MgBr2, NaBr has a higher EN difference and higher ionic character because Na is further to the left or Br on the periodic table than Mg.
Similarly, NaCl has a higher ionic character than MgCl2.
Of the two higher-ranking sodium salts, NaCl has the greatest electronegativity difference as Cl is above Br in the halogens column
Sebaic acid (HOOC-(CH2)8-COOH) is a naturally occuring dicarboxylic acid found in urine. A 200 mg sample of sebaic acid crystals would have the highest solubility in a dilute aqueous solution of:
A) a weak acid
B) a weak base
A) a neutral ionic salt
D) distilled water
B) a weak base
The structure of sebaic acid has polar covalent C=O, C-O, and O-H bonds that readily interact with water via dipole-dipole and hydrogen bonding interactions. Nevertheless, the solubility of sebaic acid in water is still very limited due to the nonpolar C-C and C-H bonds of the central hydrocarbon chain, which is hydrophobic.
Because bases neutralize acids, the solubility of sebaic acid can be increased by adding a base to deprotonate the carboxylic acid groups. The resulting acid-base neutralization reaction converst the carboxylic acid groups into ionic carboxylate salts, which have much higher aqueous solubility because they are charged and highly polar. The added charge increases the attraction to water and overcomes the repulsion from the hydrocarbon.
A concentration cell is constructed using Ni(s) strips as the anode and the cathode with each immersed in a Ni2+ (aq) solution of a different concentration, as shown.
Which statement correctly compares the molar concentrations [Ni2+]anode and [Ni2+]cathode of each half-cell solution and gives the correct direction of electron flow within the cell?
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Ni2+anode < Ni2+ cathode; electrons flow from anode to cathode
Cath = reduction, gaining electrons
Therefore, the anode must be immersed in the less-concentrated soltuion, where Ni2+ can be produced, and the cathode must be immersed in the more-concentrated solution, where Ni2+ can be consumed
Iodine reacts with ClO3- under acidic conditions according to the reaction shown below:
5 CLO3- + 3 I2 + 3 H2O –> H Cl- + 6IO3- + 6 H+
Does I2 act as the oxidizing agent or reducing agent in the reaction?
Reudcing agent because I2 is oxidized
Suppose an experiment is to be performed utilizing a reaction that reduces CLO4- to ClO3-. Which experimental approach can be expected to make the reduction of ClO4- more favorable during the reaction.
A) perform the reaction under acidic conditions
B) perform the reaction under neutral conditions
C) perform the reaction under basic conditions
D) the Ph of the reaction mixture does not make a difference
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A) perform the reaction under acidic conditions
On a frost diagram, the greatest slope is the more favorable reaction.
The slope of acidic conditions (ph 0) is greater than the slope for basic conditions (ph 14)
A sodium atom in an excited state has an electron configuration of 1s2 2s2 2p6 3p1. The excited electron within the configuration was excited from the:
3s orbital
Vacant 3s orbital
The overall reaction for the electrolysis of molten sodium chloride:
2NACl (l) –> 2Na (l) + Cl2 (g)
can be expressed as two separate, simulateanous redox processes
Reaction 1: 2Cl - –> Cl2 (g) 2e-
Reaction 2: 2Na+ + 2e- –> 2Na (l)
Given that the reduction of 2 moles of NaCl (l) requires 2 moles of electrons, a 1:1 ratio, what is the amount of electric charge required to produce a 0.80 mol of Na(L) during the electrolysis?
- 8 faradays
- 80 mol Na (L) x (2 mol NaCl/2 mol Na) = 0.80 mol NaCL
- 80 mol NaCl x (2 mol e- / 2 mol NaCl) = 0.80 mol e-
- 80 mol e- x (1F/1mol e-) = 0.80 faradays
4Nh3 (g) + 5 O2 (g) –> 4 NO (g) + 6H2O (g)
For the balanced reaction, how many grams of nitrogen monoxide are produced when 16.0 g of oxygen react?
12 g
16 g o2 x (1 mol O2 / 32 g O2) x (4 mol NO / 5 mol O2) x (30 g NO / 1 mol NO) = 12 g NO
Which of the following are state functions?
I. Enthalpy
II. Temperature
III. Heat
I and II only
In metal complexes such as copper (II) glycinate, the coordinate number refers to
the number of coordinate bonds formed
coordinate covalent bonds, unlike covalent bonds, are formed between two atoms when both shared electrons are donated by the same atom. Such coordinate bonds are often formed between electron-poor metal ions and molecules called ligands that contain one more more electron-rich atoms with available lone-pair electrons. The coordinately bonded metal and its ligands are called a complex.
The coordination number of a metal complex refers to the number of coordinate bonds formed between the central metal ion and its nearest neighboring atoms.
For example, if bonding interactions from other ligands such as water, both isomers of the copper (II) glycinate complex have a coordination number for 4. the CU 2+ atom has two oxygen atoms and two nitrogen atoms as its nearest neighrbors, forming four coordinate bonds.
A saturated solution of CuF2 dissolved in pure water at 25C has a Cu2+ ion concentration of 7.4 x 10^-3 M. If measured at the same temperature, what is the Cu2+ ion concentration in a saturated solution of CuF2 dissolved in water that also contains 0.20 M NaF?
4.0 x 10^-5 M
Ksp = [Cu2+] [F-]^2
= x(0.20)^2 = 0.040x
In passage, Ksp of CuF2 = 1.6 x 10^-6
1.6 x 10^-6 = 0.040x
x = 4.0 x 10^-5 M
Which of the following will increase the amount of CuF2 (s) that dissolves in water and raise the Cu 2+ concentration in the solution? Note: CuOH2 has a Ksp = 2.2 x 10^-20
A) adding dilute nitric acid
B) adding dilute sodium hydroxide
C) diluting the solution with aqueous potassium flouride
D) Evaporating some solution to decrease the solution volume
A) adding dilute nitric acid
Adding acid = adding H+, bond with F- , equilibrium then shifts towards products to form more F-
If excess methylamine CH3Nh2 is added to a saturated aqueous CuF2 solution, the amount of CuF2 that dissolves will:
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increase
excessed added, decreases concentration of Cu2-+, equilibrium shifts towards products to form more Cu2+
Which of the following element groups is not included in the representative elements?
A) group 1
B) group 3
C) group 13
D) group 18
B) group 3
representative elements = groups 1-2 and groups 13-18
A galileo thermometer consists of a cylinder filled with a variable solution mixture and several sealed glass bulbs partially filled with water. The density of the solution in the cylinder varies significantly with temperature, but the change in the density of the water inside each bulb is negligible. If one of the bulbs has the same density as the cylinger solution at 25C, how will the bulb’s position and the density of the cylinder solution change as the temperature decreases from 25C to 20C?
The cylinder solution density will increase, and the glass bulb will float
The exact density (m/v) can vary due to minute changes in volume because materials tend to expand when heated and contract when cooled. Materials can also physically arrange according to density. For example, when a solid object is placed into a liquid, the object will float if it has a density less than that of the liquid, but it will sink if its density is greater than that of the liquid.
Initially at 25C, the mass-to-volume ratio (ie, density) of the sealed glass bulb exactly mataches that of the solution in the cylinder. As the temperature of the solution in the cylinder decreases, the liquid contracts slightly and the density of the cylinder solution increases (same mass occupying a smaller, contracted volume)> because the glass bulb acts like an isolated system, its density changes very little compared with that of the cynlinder solution. Accordingly, the density of the cooling solution in the cylinder becomes greater than that of the bulb, causing the bulb to float.
Boyle’s law
Pressure proportional to 1/V
Inverse relationship between V and P for a fixed number of molecules of an ideal gas at a constant temperature
Charles Law
directly proportional relationship between Volume and Temp
Avogadro’s Law
directly proportional relationship between the number of moles and the volume of an ideal gas
Gay-Lussac’s law
directly proportional relationship between the pressure and absolute temperature of a fixed number of molecules of an ideal gas
Suppose that a multiplace chamber has three times the volume of a monoplace chamber, and both are pressurized to 1.5ATA with 100% O2 at 29C. Compared to that of the monoplace chamber, the number of moles of oxygen in the multiplace chamber would be:
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3 times higher
When pressure and temperature are constant, the volume of an ideal gas is directly proportional to the number of moles of the gas
Assume that the pressure in a 1500L monopalce chamber is increased from 1 ATA to 2 ATA. Which expression gives the number of moles of oxygen gas added to the chamber if the temperature remains constant at 27C?
A) (4.5x10^5)R
B) 10/R
C) 5/R
D) 0.2R
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C) 5/R
PV=nRT
1atm(1500L) = n1R(300K)
n1=5/2
PV=nRT
2atm(1500L) = n2R(300K)
n2=10/R
n2-n1 = 10/R - 5/R = 5/R moles added to the chamber
In a protein showed in Figure 2, the folding of region I is the result of:
A) salt bridge interactions between amino acid side chains
B) covalent bonding between peptide segments
C) hydrophobic interactions between amino acid side chains
D) hydrogen bonding between amino acids in the peptide chain
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D) hydrogen bonding between amino acids in the peptide chain
In region I of the protein shown, the partial positive charge from the dipole of the N-H bond is attracted to the partial negative charge from the dipole of the C=O bond adjacent to the peptide bonds within the protein structure. This forms strong noncovalent hydrogen bonding interactions, which cause the b-sheet formation in the protein
If MgCO3 undergoes a double replacement reaction in 6M Hcl, and one of the resulting products then undergoes a decomposition reaction, the gas that evolves is
A) hydrogen
B) carbon dioxide
C) oxygen
D) magnesium oxide
B) carbon dioxide
Double replacement : MgCO3 + 2HCl –> MgCl2 + H2CO3
Decomposition: H2Co3 –> H2O + CO2
transition states are seen graphically as
higher energy maxima (peaks)
When sugar is burned completeley, the following unbalanced combustion reaction takes place:
C12H22O11 (s) + O2 (G) –> Co2 (g) + H2O (g) + heat
If the reaction is balanced, the ratio of Co2 to H2O is
12 to 11
Compound 1 = CH3CH2NH2
Compound 2 = Ch3NHCH3
Given that the boiling point of Compound 1 is 18C, the boiling point of compound 2 would most likely be:
A) 8C
B) 18C
C) 89C
D) 100C
A) 8C , less NH bonds, less hydrogen bonding capability, weaker IMF
Isolated system
no exchange of heat or matter
open system
both heat and matter are exchanged
closed sytem
only heat is exchanged, not matter
invalid system
only matter is exchanged, not heat
