Quiz 5 Chapters 4 & 7

Question 1

Solution ?
A) Solute?
B) Solvent

Answer 1

Solution is a homogeneous (uniform) [liquids, gases, solids] mixture made up of 2 or more substances

A) Solute - present in smaller amounts
B) Solvent- present in larger amounts

Question 2

What is the solvent in an aqueous solution?

Answer 2

Water

Question 3

Strong Electrolyte

Answer 3

A type of aqueous solution
-completely ionize in solution (all molecules break up into ions)
3 types
1. Ionic compound (metal +nonmetal)
2. Strong Acid (HCl,HBr,HI,HNO3, HClO4, H2SO4) only these
3. Strong Base- group 1 Hydroxide & (Ba(OH)2

Question 4

Weak Electrolyte

Answer 4

Partly ionize in solution (some ions, some molecules)
A) Weak Acids (all other than the 7 strong)
B) Weak Bases - all bases other than Group 1 Hydroxides and Barium Hydroxide

Question 5

Non-Electrolyte

Answer 5

None of the solute dissociated into ions (only molecules remain)
Ex. Ar, Ne, Vitamin C (C6H8O6)

Question 7

What is it called when molecules break up into ions

Answer 7

Ionize, dissociate, or break apart

Question 8

3 types of aqueous solutions

Answer 8

1. strong electrolyte
2. Weak electrolyte
3. Non-electrolyte

Question 9

Nonelectrolyte

Answer 9

Nothing breaks apart, only molecules left

Anything that is not a strong acid or base

Ar, Ne, C6H8O6-vitamin c

Question 10

Creates a bright light bc it’s a good conductor of electricity

Answer 10

Strong electrolyte like NaCl(aq)

Question 11

Causes light bulb to glow dimly

Answer 11

Weak electrolyte
Ex. HF is a weak acid
—->
<—- double headed arrow represents weak electrolytes

HF(aq) —>
<——- H+(aq) + F-(aq)

Question 12

No light conducted

Answer 12

Non-electrolyte
Like sugar.. doesn’t break apart at all in water, it just dissolves

Question 13

Precipitation Reaction

Answer 13

Insoluable Product (the precipitate) that separates from the solution. - a solid that builds up at the bottom of a test tube

Question 14

Soluble Compounds

Answer 14

Contain alkali metal ions
Li, Na, K, Rb, Cs
Ammonia ion (NH4+)
Nitrates (NO3)-
Bicarbonates (HCO3-)
Chlorates (ClO3-)
Halides (Cl-) (Br-) (I-)
Sulfates (SO2)4-

Question 15

Insoluble Exceptions that form a white precipitate

Answer 15

Halides of Ag+, Hg2/2+, Pb2+
Sulfates of Ag+, Ca2+, Sr2+, Ba2+, Hg2/2+, Pb2+

Question 16

Insoluble Compounds - incapable of being dissolved

Answer 16

Carbonates CO2/3-, phosphates PO3/4-, chromates CrO2/4-, Sulfides S2-, Hydroxides OH-

Question 17

Soluble Exceptions - dissolves into ions

Answer 17

Compounds containing alkali metal ions and the ammonium ion
Compounds containing alkali metal ions and the Ba2+ ion

Question 18

Soluble compounds - able to dissolve

Answer 18

Nitrates NO3-, bicarbonates HCO3-, Chlorates ClO3-
Halides Cl-,Br-, I-
Sulfates SO3/4-

Question 19

Insoluble Compounds - not able to dissolve

Answer 19

Carbonates CO2/3-, phosphates PO3/4-, chromates CrO2/4-, Sulfides S2-, Hydroxides OH-

Question 20

Pb(NO3)2 (aq) + NaI (aq) —->

Lead II Nitrate & Sodium Iodide

Answer 20

A joins with D, C joins with D

PbI2(s) + 2Na(NO3)(aq)

Question 21

Ionic equations includes spectator ions and keeps solids together

Answer 21

Pb+2 + 2NO3- + 2Na+ + 2I- ——>

—-> PbI2(s) + 2Na+ + 2NO3-

Question 22

Net ionic equation (no spectators)

Answer 22

Pb+2 + 2I- —> PbI2(s)

Question 23

Oxidation Reduction Reactions

Answer 23

Electron transfer reactions

Zn(s) + CuSO4(aq) —> ZnSO4(aq) +Cu(s)

Question 24

Oxidation Reduction Reaction

Answer 24

Elemental Free Form :

Zn(s) + Cu2+(aq) + (SO4)2-(aq) —>
—> Zn+2(aq) + (SO4)-2(aq) + Cu(s)

Question 25

Net ionic equation

Answer 25

No spectators Zn(s) + Cu+2(aq)—> Zn+2(aq) + Cu(s)

Question 26

2 half reactions of oxidation & reduction

Answer 26

To balance these equations we have to add 2 electrons to the positive sides Zn(s)—> Zn+2 +2e- oxidation ( 0 to +2) 2e- + Cu+2—-> Cu reduction (+2 to 0) The first 1 starts with Zinc 0 Element form The 2nd ends with Copper 0 element form

Question 27

Oxidation Numbers

Answer 27

-Anything in the Elemental Free Form =0 Sum of oxidation numbers adds up to the element charge (if neutral it’s 0 oxidation) O is -2 except in H2O2 & (O2)-2 - H is +1 except with Li and Na it is -1 - oxidation #s can be fractional

Question 29

Types of Redox Reactions 1. Combination 2. Decomposition

Answer 29

1. Combination - 2 elements/simple compounds form 1 compound A+B=C Ex. 2Mg (0-element form) + O2 (0) =2MgO Mg goes from 0 (element) to +2oxidation O from 0 (element form) to -2 REDUCTION 2. Decomposition - compound breaks apart C—> A+B 2KClO3–> 2KCL+3O2 3. Disproportionation -1 element oxidized & reduced Cl2+2OH->ClO+Cl-+H2O 4. Displacement (metal)—> next

Question 30

Displacement reaction

Answer 30

K+ NaCl—> KCl+Na Na+ KCl—> no reaction AuCl2 + Ag—> AgCl2 +Au AgCl2 + Au—> in reaction

Question 31

Molarity = moles/Liter solution What is the Molality of .6 moles of NaCl in 200 milliliter solution

Answer 31

[NaCl] brackets means concentration 200 milliliters = .200 liters .6 moles /.200 liters = 3.00 moles per liter

Question 32

What is the Molality of 600grams of NaCl in 200 milliliter solution

Answer 32

1st convert grams to moles .600divided by 58.5g/mol NaCl=.0103 mol of NaCl 200 mL-> .200 liters .0103mol/.2 L= .0513 mol/L

Question 33

Standardizing a solution - preparing a solution of known Molarity Create 500mL of 1.78 moles K2Cr2O7

Answer 33

1.78 moles K2Cr2O7* 294.2g/mol *.500L= 262g K2Cr2O7 (3sig figs) Steps 1. weight the solute and transfer to 500 mL volumetric flask with funnel 2. Add water to dissolve solid 3. Fill exactly to line with water

Question 34

How would you prepare 100mL of 0.17094 M NaCl solution ?

Answer 34

100mL—>.100L*.17094 Moles/L * 58.4g = .998g NaCl Per mol NaCl 1. Add about 1gram NaCl to 100mL volumetric flask 2. Add a little water to dissolve the NaCl 3. Fill water to the fill line

Question 35

Dilution Prepare 100mL of 2.00 M H2SO4 starting with 7.00 M stock solution

Answer 35

M1V1=M2V2 2.00(100)=7.00(?) 200/7=28.571 = 28.6 mL Measure 28.6 mL stock solution Pour into 100 mL volumetric flask Fill exactly to line with water

Question 36

Quantitative Analysis Gravimetric Original sample = 0.5662g Cl compound Treated with excess AgNO3 Forms 1.0882 g. AgCl precipitate What’s the % mass of Cl in original compound?

Answer 36

Mass of precipitate formed/total mass of a whole precipitate x mass of what ur finding (Cl) take that # and divide by the original sample # to get the percent Start with the mass of the precipitate 1.0882g AgCl x 1 mole AgCl/ x 35.45gCl 143.4g AgCl =.2690g Cl. Divide this by what the whole original sample was .2690g/.5662gx100 = 47.51% Cl

Question 37

Acid Base Titration

Answer 37

Buret with an acid or base that we know Unknown in a flask below it Open valve to allow solution to enter flask and react Put solution on a magnetic plate with a stir bar that stirs the solution as the drops are added (Add until acid has completely reacted with base) Use phenolphthalein paper (colorless in acid/pink in base) Take the given grams of KHP divide by the grams in 1 mole of KHP Take your answer and divide it by the # of Liters of the neutralizer (mL->L move decimal to places left. Liters should be a smaller # bc more Milliliters in a liter

Question 38

Light spectrum

Answer 38

GXUVIMT Gamma xray uv visible infrared microwave tv waves Visible 400-700nm

Question 39

Wavelength

Answer 39

Lamda - distance between 2 adjacent peaks in a wave

Question 40

Frequency

Answer 40

# of peaks that pass a point per second Frequency is the birds beak called nu Measured in Hz per seconds or s^-1

Question 41

Amplitude

Answer 41

2a vertical distance from midline to peak or trough

Question 42

Frequency (u)

Answer 42

u= lamda X Nu

Question 43

Speed =

Answer 43

Distance/time= distance/wavelength X wave/time (frequency = nu)

Question 45

What is the maximum # of electrons the s orbital can hold?

Answer 45

2

Question 46

Where are the p orbitals found?

Answer 46

P orbitals start at the second principal energy level (n=2), not all levels.

Question 47

Orbital Filling- P and d orbitals in the same principal energy level

Answer 47

P orbitals ARE FILLED BEFORE D orbitals within the same principal energy level due to the order of orbital filling.

Question 48

The Aufbau Principle states that...?

Answer 48

Electrons enter the lowest available energy level. The Aufbau Principle dictates how electrons fill atomic orbitals, starting with the lowest energy level and progressing to higher levels.

Question 49

The Pauli Exclusion Principal States that?

Answer 49

Electrons in the same orbital must have opposite spins according to the Pauli exclusion principle.

Question 50

Principal energy levels get closer together as they get further from the nucleus. Explain:

Answer 50

- The 4s sublevel is filled before the 3d sublevel. - The 3rd principal energy level can hold up to 18 electrons. - Orbitals are not always filled in numerical order (e.g., 4s before 3d).

Question 51

What is the electronic configuration of an atom of an element with atomic number 8?

Answer 51

1s2,2s2,2p4 This configuration accounts for all 8 electrons of an oxygen atom, which has an atomic number of 8.

Question 52

Which one of the following is not the electronic configuration of atom of a noble gas? 1s2 1s22s2 1s22s22p6 1s22s22p63s23p6

Answer 52

1s2 is Helium, a noble gas. 1s22s2 ends with 2s2 and is not a full p-orbital, so it does not represent a noble gas. 1s22s22p6 is Neon, a noble gas. 1s22s22p63s23p6 is Argon, a noble gas. Therefore, 1s22s2 is not the electronic configuration of a noble gas.

Question 53

What is the order of filling orbitals according to the Aufbau principle?

Answer 53

1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p

Question 54

What does Hund's Rule state?

Answer 54

Electrons will fill the orbitals with 1 electron in each box before adding a second electron to each box. It adds 1 to each box until all boxes are full, then goes back around to add the second to each box. (Maximizes spin, minimize repulse) Hund's Rule states that electrons in similar energy orbitals remain unpaired as far as possible. This means that electrons will fill degenerate orbitals (orbitals with the same energy) singly before pairing up. This minimizes electron-electron repulsion and maximizes the total spin.

Question 55

Which of the following atoms has three unpaired electrons: B, C, N, or O?

Answer 55

- **Nitrogen (N):** Has 3 unpaired electrons. -Boron (B): Has 1 unpaired electron. - Carbon (C): Has 2 unpaired electrons. - Oxygen (O): Has 2 unpaired electrons.

Question 56

Why do electrons enter the 4s sub-level before the 3d sub-level?

Answer 56

The 4s orbital has a lower energy. Electrons are filled into orbitals in order of increasing energy. The 4s orbital has lower energy than the 3d orbitals, so it is filled first.

Question 57

Which of the atom pairs both have only three unpaired electrons in their d orbitals? Ti and V Ti and Co V and Cr V and Co

Answer 57

V and Co**: Co has 3 unpaired electrons, and V does as well. This is the correct answer.

Question 58

Which of the following atoms has the greatest number of unpaired electrons ? Ti V Cr Mn

Answer 58

Chromium (Cr) has the greatest number of unpaired electrons, totaling 6 unpaired electrons. Looking at the rest: Titanium (Ti): Atomic number 22. Electron configuration: [Ar] 4s² 3d². It has 2 unpaired electrons in the 3d subshell. Vanadium (V): Atomic number 23. Electron configuration: [Ar] 4s² 3d³. It has 3 unpaired electrons in the 3d subshell. Manganese (Mn): Atomic number 25. Electron configuration: [Ar] 4s² 3d⁵. Manganese has 5 unpaired electrons in the 3d subshell.