UNIT 1 REVIEW ANSWERS

1. Gas because it doesn’t have a definite size, shape nor density. Thus, the temperature and pressure ahs the greatest effect on it.

2. Solid has definite shape, very close distance between molecules and very slow molecules and definite volume. Liquids, on the other hand take the shape of the container, close distance between molecules, slow molecules. Whereas gases takes the shape of the container and fills it up, has no definite volume, very far apart and very fast molecules.

3.

a. qualitative because it can’t be measured

b. quantitative because it can be measured

c. qualitative because based on 5 senses

d. quantitative because it can be measured

e. qualitative because based on 5 senses has no numbers associated with it.

4.

a. chemical because it describes how gold reacts with air

b. physical because determined without altering chemical composition

c. physical because determined without altering chemical composition

d. chemical because it altered the chemical composition of gas

e. physical because only change stateà liquid

5.

a. physical because no new substance is formed

b. chemical because change in energy/heat

c. chemical because change in energy

d. chemical because change in smell and precipitate formed

e. physical because only excite electrons

6.

a. chemical because chemicals are mixed in the firefly which produce “cold light’

b. chemical because new substances are formed by photosynthesis to make the tree larger.

c. chemical because burning of the wax (combining it with oxygen) produces light

d. physical because ice turns into water, it’s still water

e. chemical because change in light and heat

f. physical because only excite electrons

7. Pure substance has a definite composition, mixtures have a variable composition.

8. Pure substances can’t be separated by physical means like distillation, fractional distillation.

9. Solutionà can be separated by distillation and physical means. As well, it contains at least 2 compounds.

10. d, A mixture of ice and water is heterogeneous because it has 2 phases (solid and liquid). All the other are homogeneous solutions (1 phase).

11.

a. a metal alloy; i.e. copper and zinc

b. apple juice; water and apple juice

c. air; nitrogen and oxygen

d. Kool Aid; Kool Aid power and water

e. Soda pop; Pepsi

12. A proton and a neutron is found in the nucleus of an atom. Proton has a positive change and a neutron has neutral. Electrons orbit the nucleus in the atom with a negative charge.

13.

a. neutrons: 33 p: 27 e: 27

b. n : 52 p: 38 e: 38

c. n : 61 p: 47 e: 47

d. n :125 p: 82 e: 82

e. n :125 p: 85 e: 85

f. n :146 p: 92 e: 92

14. they are called metalloids

15.

a. 5

b. 5

c. 5

16. •

a. Ga •

•

•

b. • Ge •

•

•

c. • As ••

•

••

d. • Se •

••

17. Sodium has 11 electrons and only 1 valence electron. When it bonds with another element, it will lose 1 electron or gain 7 electrons. Losing 1 electron is much easier than gaining g 7 electrons, so it will always form a +1 charge. Calcium ahs 20 electrons and only 2 valence electrons, it’ll prefer to lose 2 electrons than gain 6, therefore it’ll form a +2 charge.

18.

a. loses 2e

b. gains 1e

c. gains 3e

d. loses 3e

e. gains 2e

f. n/a

19.

a. nitrogen

b. fluorine/chlorine/bromine/iodine/astatine

c. carbon

20.

a. Au

b. S

c. Al

d. Rn

e. Zn

f. Sr

21.

a. O

b. Se

c. Mg

d. Xe

e. Ne

f. P

22. Stable octet is an element whose outer shell is full; noble gases all have 8 valence electrons.

23. Noble gases are very unreactive/ stable because they have a full valence electron. They do not want to give/lose any of its electrons. They have a stable octet.

24. Metal and a non-metal element are involved in ionic bonding.

25. For oxygen, it has 6 valence electrons, meaning it wants to gain 2 electrons. A potassium ion has one valence electron to lose. To balance up, 2 potassium ion will satisfy the oxygen ion to make a full valence.

For question 26 and 27 the website won’t let me down load the diagrams my apologies.

26. OH^-1 is linear

27. Boron is an exception BF₃ is trigonal planar and nonpolar

GeH₄ is tetrahedral and nonpolar

SCl₂ is bent/ v shape and polar

ClO^-1 is linear (it is an ion therefore polarity is not done because it is a charged particle)

H₂Se is bent/v shape and slightly polar

28.

a. FeCl3	b. FeO
c. CoBr2	d. SnCl4
e. Li2S	f. MgBr2
g. Al2S3	h. Mg(OH)2
i. Fe2(CO3)3	j. CBr4
k. P2S5	l. SnCl2
m. Fe2(SO4)3	n. VBr2
o. PbO2	p. Au(CN)3
q. Lithium Sulphide	r. Phosphorous Trihydride
s. Gallium Iodide	t. Aluminum Sulphide
u. Boron nitride	v. Ammonium Dichromate
w. Cupric Nitrate	x. Barium Hydrogen Carbonate
y. Plumous Acetate	z. Copper (I) Chlorite
aa. Mercury Dichromate	bb. Thallium (II) Phosphate
cc. Copper Hydrogen Carbonate	dd. Iron (II) Chlorate
ee. Mercury Phosphate	ff. Gold (II) Perchlorate
hh. Cobalt Dichromate

29.

a. Zn + 2 HCl à ZnCl₂ + H₂

b. 2 Mg + CO₂ à 2 MgO + C

c. Al₂O₃ + 3 H₂ à 2 Al + 3 H₂O

d. Cu(NO₃)₂ + H₂S à CuS + 2 HNO₃

e. CaO + 2 HNO₃ à Ca(NO₃)₂ + H₂O

30.

a. Ca(AlO₂)₂ + 8 HCl à 2 AlCl₃ + CaCl₂ + 4 H₂O

b. 9 O₂ + 2 Sb₂S₃ à 2 Sb₂O₃ + 6 SO₂

c. 4 Cr + 3 O₂ à 2 Cr₂O₃

d. 2 C₂H₆ + 7 O₂ à 4 CO₂ + 6 H₂O

e. 2 F₂ + 4 NaOH à O₂ + 4 NaF + 2 H₂O

f. 4 NH₃ + 5 O₂ à 6 H₂O + 4 NO

g. 3 Cl₂ + 2 CrBr₃ à 3 Br₂ + 2 CrCl₃

h. Sr(IO₃)₂ à SrI₂ + 3 O₂

i. Al₂(SO₄)₃ + 6 NH₃ + 6 H₂O à 2 Al(OH)₃ + 3 (NH₄)₂SO₄

j. 3 SiO₂ + 4 Al à 3 Si + 2 Al₂O₃

k. 4 C₇H₅N₃O_6(s) + 21 O_{2 (g)} à 28 CO_{2 (g)} + 6 N_{2 (g)} + 10 H₂O _(g)

31.

a. Yes, single displacement

b. No, because Mg is not reactive enough to displace H from H₂O

c. No, because Mg is more reactive than N on the activity series

d. Yes, Zn will displace Ni

e. No, because Zn is not reactive enough to displace H from H₂O

f. Yes, Sr is reactive enough to displace H₂O

g. Yes, Al is more reactive than Zn

h. No, because Al is more reactive than Zn

i. Yes, Fe is above H

j. No, because Ag is not reactive enough to displace H from HCl

32.

a. 2 K(s) + 2 HCl(aq) à H₂ ↑ + 2 KCl (aq)

b. 2 Na (s) + 2 H₂O (l) à 2 Na(OH) + H₂ ↑

c. 2 Mg (s) + O₂ (g) à 2 MgO

d. Mg (s) + CuSO₄ (aq) à Cu(s) + MgSO₄ (aq)

e. Ni (s) + Pb(NO₃)₂ (aq) à Pb(s) + Ni(NO₃)₂(aq)

f. Pb (s) + H₂SO₄(aq) à H₂↑ + PbSO₄(aq)

g. Fe(s) + SnCl₂ (aq) à Sn(s) + FeCl₂(aq)

h. Ca(s) + H₂SO₄(aq) à H₂↑ + CaSO₄(aq)

i. 3 Na(l) + AlCl₃(s) à Al(s) + 3 NaCl(aq)

UNIT 2 REVIEW ANSWERS

1. [(39.0U)( 93.1)] + [ (41.0u)( 6.9)] = 39u

100 100

2. [(203.0U)( 29.50)] + [ (205.0u)( 70.50)] = 204.4u

100 100

1. H₂ S

2.0g/mol + 32.1g/mol = 34.1 g/mol

2.0g/mol X 100% 32.1g/mol X 100%

34.1 g/mol 34.1 g/mol

H₂ = 5.9% S = 94.1%

2. Na₂ S O₄

46.0g/mol + 32.1g/mol + 64.0 g/mol = 142.1 g/mol

46.0g/mol 32.1g/mol 64.0 g/mol

142.1 g/mol 142.1 g/mol 35.6 g/mol

Na₂ = 32.4% S = 22.6% O_{4 =} 45.0%

3. Cu H₁₀ S

48.0g/mol + 10.0g/mol + 32.1 g/mol = 90.1 g/mol

48.0g/mol 10.0g/mol 32.1 g/mol

90.1 g/mol 90.1 g/mol 90.1 g/mol

Cu = 53.3% H₁₀ = 11.1% S ₌ 35.6%

4. .
1. C₂H₅OH
   6H = 6 (1.0g/mol) = 6.0g/mol
   2C = 2(12.0g/mol) = 24.0g/mol
   1O = 1(16.0g/mol) = 16.0 g/mol +
   Mm = 46.0 g/mol
2. Al₂(SO₄)₃
   2Al = 2(27.0g/mol) = 54.0g/mol
   3S = 3(32.1g/mol) = 96.3g/mol
   12O = 12(16.0g/mol) = 192.0g/mol +
   Mm = 342.3g/mol
3. NH₄NO₃
   N = 14.0g/mol = 14.0g/mol
   4H = 4(1.0g/mol) = 4.0 g/mol
   N = 14.0g/mol = 14.0g/mol
   3O = 3(16.0g/mol) = 48.0g/mol +
   Mm = 80.0g/mol

4. Cu = Mm = 63.5g/mol

1. (7.10mol)(6.02x10²³molecules/mol) = 4.27x10²⁴ molecules
2. (7.10mol)(6.02x10²³molecules/mol)(4atoms/molecule) = 1.71x10²⁵ atoms of hydrogen
6. .
1. n = 7.45x10²² molecules
   6.02x10²³ molecules/mol
   n = 0.1238 mol
   Mm = [6(12.0g/mol) + 12(1.0g/mol) + 6(16.0g/mol)]
   Mm = 180.0g/mol
   
   n = m/Mm
   m = n(Mm)
   m = (0.1238 mol)(180.0g/mol)
   m = 22.3g
   
   Therefore, the mass of C₆H₁₂O₆ (glucose) is 22.3g.

2. n = 3.11x1024 molecules
   6.02x1023 molecules/mol
   n = 5.166 mol
   Mm = [4(12.0g/mol) + 9(1.0g/mol) + 32.1g/mol + 1.0g/mol]
   Mm = 90.1g/mol
   n = m/Mm
   m = n(Mm)
   m = (5.166mol)(90.1g/mol)
   m = 465g
   
   Therefore, the mass of C₄H₉SH is 465 g.
1. n = m = 33.0g
   Mm [40.1g/mol + 2(1.0g/mol) + 2(12.0g/mol) + 6(16.0g/mol)]
   n = 33.0g = 0.2036mol
   162.1g/mol
   (0.2036mol)(6.02x1023 molecules/mol) = 1.23x10²³ molecules = 1.35x10²⁴ atoms
2. n = m = 256g
   Mm [87.6g/mol + 2(14.0g/mol) + 6(16.0g/mol)]
   n = 256g = 1.210 mol
   211.6g/mol
   (1.210mol)(6.02x1023 molecules/mol) = 7.28x10²³ molecules = 6.56x10²⁴ atoms

a. In 100 g of compound, there are:

63.53g of Fe 36.47g of S

n = m = 63.53g = 1.136mol n = m = 36.47g = 1.136mol
Mm 55.9g/mol Mm 32.1g/mol

Therefore the empirical formula is FeS.

b. In 100g of compound, there are:

21.6g of Na 33.3 g of Cl 45.1g of O

n = m = 21.6g = 0.939mol n = m = 45.1g = 0.944mol n = m = 45.1g =2.82 mol
Mm 23.0g/mol Mm 35.5g/mol Mm 16.0g/mol

n _Na : n_Cl : n_O

0.939: 0.944: 2.82

1 : 1: 3

Therefore the empirical formula is NaClO₃.

c. In 100g of compound there are

26.52g Cr 24.52g S 48.96g O

n = m = 26.52g = 0.51mol n = 24.52g = 0.764mol n = 48.96g = 3.06mol

Mm 52.0g/mol 32.1g/mol 16.0g/mol

n_Cr : n_S : n_O

0.51 : 0.764 : 3.06

1 : 1.5 : 6 × 2 to get rid of decimal

2 : 3 : 12

Therefore the empirical formula is Cr₂S₃O₁₂ [Cr₂(SO₄)₃]

d. In 100 g of compound there are

63.1 g C 11.92g H 24.97 g F

n = m = 63.1 g = 5.26 mol n = 11.92g = 11.92 mol n = 24.97 g = 1.31 mol

Mm 12.0g/mol 1.0g/mol 19.0g/mol

n_C : n_H : n_F

5.26 : 11.92: 1.31

4 : 9 : 1

Therefore the empirical formula is C₄H₉F

9. In 100 g of compound, there are:

48.28g of C 5.05g of H 14.14g of H 32.32g of O

n = m n = m n = m n = m
Mm Mm Mm Mm

= 48.48g = 5.05g = 14.14g = 32.32g

12.0g/mol 1.0g/mol 14.0g/mol 16.0g/mol

= 4.04 mol = 5.05 mol = 1.01 mol = 2.02mol

C : H : N : O

4.04 mol 5.05 mol 1.01 mol 2.02mol

1.01 mol 1.01 mol 1.01 mol 1.01 mol

4 5 1 2

Therefore, the empirical formula is C₄H₅NO₂

Mm(compound) = 198.0g/mol = 2 Therefore 2 is the multiplier.

Mm(empirical) 99.0g/mol

Therefore the molecular formula is C₈H₁₀N₂O₄

10. In 100 g of compound, there are:

23.8g of C 5.99g of H 70.2g of Cl

n = m n = m n = m
Mm Mm Mm

= 23.8g = 5.99g = 70.2g

12.0g/mol 1.0g/mol 35.5g/mol

= 1.983 mol = 5.99 mol = 1.977 mol

C : H : Cl

1.983 mol 5.99 mol 1.977 mol

1.977 mol 1.977 mol 1.977 mol

1 3 1

Therefore, the empirical formula is CH₃Cl

Mm(compound) = 50.5g/mol = 1 Therefore 1 empirical formula in molecule

Mm(empirical) 50.5g/mol

Therefore the molecular formula is CH₃Cl

11. -

a) 2 NaOH_(aq) + H₂SO_4(aq) à Na₂SO_4(aq) + 2 H₂O_(l)

2 mol 1mol 1 mol 2 mol

N_NaOH = N_Na2SO4

2mol 1mol

N_NaOH = 2mol X 8.61

1mol

N_NaOH = 17.2 mol

Therefore, 17.2 moles of NaOH are required.

b) 2 NaOH_(aq) + H₂SO_4(aq) à Na₂SO_4(aq) and 2 H₂O_(l)

2 mol 1mol 1 mol 2 mol

N_H2SO4 = N_Na2SO4

1mol 1mol

m = N_Na2SO4 X Mm_H2SO4

m = 4.77 mol X 98.1g/mol

m = 468g

Therefore, 468g of H₂SO₄ are required.

12. Li₃N_(s) + 3 H₂O_(l) à 3 LiOH_(s) + NH_3(g).

1mol 1mol

N_Li3N = N_NH3

1 mol 1 mol

M_NH3 = M_Li3N X Mm_NH3

Mm_Li3N

M_NH3 = 4.84g X 17.0g/mol

34.7 g/mol

M_NH3 = 2.37g

Therefore 2.37g of NH₃ are formed.

13. CH_4(g) + 2 O_2(g) à CO_2(g) + 2 H₂O_(g)

6.4g Xg

a) N_CH4 = N_O2

1 mol 2 mol

M_O2 = 2mol x M_CH4 X Mm_O2

Mm_CH4 1mol

M_O2 = 2mol x 6.4g X 32.0g/mol

16.0g/mol

M_O2 = 25.6g

Therefore 25.6g of O₂ are required.

b) N_CO2 = N_CH4

1 mol 1 mol

M_CO2 = M_CH4 X Mm_CO2

Mm_CH4

M_CO2 = 6.4g X 44.0g/mol

16.0g/mol

M_CO2 = 17.6g

Therefore 17.6g of CO₂ will be produced

14. CaO_(s) + 2 HCl_(aq) à CaCl_2(aq) + H₂O_(l)

60.4g 69.0g Xg

If all CaO is used

N_CaO = N_CaCl2

1 mol 1 mol

M_CaCl2 = M_CaO X Mm_CaCl2

Mm_CaO

M_CaCl2 = 60.4g X 111.1g/mol

56.1g/mol

M_CaCl2 = 120g [excess]

If all HCl is used

N_HCl = N_CaCl2

2 mol 1 mol

M_CaCl2 = 1mol X M_HCl X Mm_CaCl2

Mm_HCl 2mol

M_CaCl2 = 1mol X 69.0g X 111.1g/mol

36.5 g/mol 2mol

M_CaCl2 = 105g [limiting]

Therefore HCl is limiting and 105g of CaCl₂ will be produced.

15. 2 Al_(s) + 3 H₂SO_4(aq) à Al₂(SO₄)_3(aq) + 3 H_2(g).

6.71g 12.95g Xg

If all Al is used

N_Al = N_Al2(SO4)3

2 mol 1 mol

M_Al2(SO4)3 = 1mol X M_Al X Mm_Al2(SO4)3

Mm_Al 2mol

M_CaCl2 = 1mol X 67.1g X 342.3g/mol

27.0 g/mol 2mol

M_CaCl2 = 42.5g [excess]

If all H₂SO₄ is used

N_H2SO4 = N_Al2(SO4)3

3 mol 1 mol

M_Al2(SO4)3 = 1mol X M _H2SO4 X Mm_Al2(SO4)3

Mm_H2SO4 3mol

M_CaCl2 = 1mol X 12.95g X 342.3g/mol

98.1g/mol 3 mol

M_CaCl2 = 15.1g [limiting]

H₂SO₄ is limiting. Therefore 15.1g of Al₂(SO₄)₃ are produced

UNIT 3 REVIEW ANSWERS

1. A non-electrolyte will not conduct electricity when it is dissolved in water because it does not dissociate into ions, they remain totally in molecular form.

2. Water is a polar molecule and when an ionic compound (KBr) is placed in water, the slightly negative end of water will be attracted to the positive ion of the KBr. The slightly positive end of water will behave similarly with the negative ion of KBr. These attractions will pull the ions or polar molecules into solution and away from crystal and then completely surround the ions with water molecules (hydration).

3. CaCl₂ à Ca⁺²_(aq) + 2 Cl^-_(aq) shows that calcium chloride is decomposed/dissociated into ions calcium and chloride when dissolved in water.

4. Polar solvents dissolve polar or ionic solutes because of the intermolecular attractions between dipoles of the solute and the dipoles of the solvent. Nonpolar solutes don’t dissolve in polar solvents because there is no intermolecular attraction between the particles to promote solution formation. They want to London force bond. There is nothing for the dipole to be attracted to.

5. Ionic substances don’t dissolve in a nonpolar substance because there is no intermolecular attraction between the particles to promote solution formation. Ionic substances require a dipole and the nonpolar use London forces.
6. .
1. MgCl_2(s) à Mg⁺²_(aq) + 2 Cl^-1_(aq)
2. Ba(OH)₂ à Ba⁺²_(aq) + 2 OH^-1_(aq)
3. Al₂(SO₄)₃ à 2 Al⁺³_(aq) + 3 SO₄^-2_(aq)

4. AlBr₃ and LiOH
   equation: AlBr₃ + 3 LiOH à Al(OH)₃↓ + 3 LiBr
   ionic: Al⁺³ + 3 Br^-1 + 3 Li⁺¹ + 3 OH^-1 à Al(OH)₃↓ + 3 Li⁺¹ + 3 Br^-1
   net ionic: Al⁺³ + 3 OH^-1 à Al(OH)₃↓
   spectator ions: 3 Br^-1 and 3 Li⁺¹

5. Fe(NO₃)₂ and (NH₄)₂CO₃
   equation: Fe(NO₃)₂ + (NH₄)₂CO₃ à FeCO₃↓ + 2 NH₄NO₃
   ionic: Fe⁺² + 2 NO₃^-1 + 2 NH₄⁺¹ + CO₃^-2 à FeCO₃↓ + 2 NH₄⁺¹ + 2 NO₃^-1
   net ionic: Fe⁺² + CO₃^-2à FeCO₃↓
   spectator ions: 2 NO₃^-1 and 2 NH₄⁺¹

6. KBr and MgCl₂
   equation: 2 KBr + MgCl₂ à 2 KCl + MgBr₂
   ionic: 2 K⁺¹ + 2 Br^-1 + Mg⁺² + 2 Cl^-1 à 2 K⁺¹ + 2 Cl^-1 + Mg⁺² + 2 Br^-1
   net ionic: no reaction
   spectator ions: 2 K⁺¹ and 2 Br^-1 and Mg⁺² and 2 Cl^-1

7. Na₃PO₄ and CaCl₂
   equation: 2 Na₃PO₄ + 3 CaCl₂ à NaCl + Ca₃(PO₄)₂↓
   ionic: 6 Na⁺¹ + 2 PO₄^-3 + 3 Ca⁺² + 6 Cl^- à 3 Na⁺¹ + 6 Cl^- + Ca₃(PO₄)₂↓
   net ionic: 3 Ca⁺² + 2 PO₄^-3 à Ca₃(PO₄)₂↓↓
   spectator ions: 6 Na⁺¹ + 6 Cl^-1

Given:

m = 100g

V = 10.0L

Mm = [2(23.0) + 32.1 + 4(16.0)] g/mol

Mm = 142.1 g/mol

7. n_solid = n_solution

m/Mm = CV

C = M × 1

Mm V

C = 100g × 1

142.1 10.0L

Therefore the concentration is about 7.04 × 10 ־²M.

Mm = [2(23.0) + 32.1 + 4(16.0)] g/mol

Mm = 142.1 g/mol

8. n_solid = n_solution

m/Mm = CV

m = CVMm

m = (0.050mol/L) (0.250L) (180.0g/mol)

m = 2.25g

Therefore there is 2.25g of C₆H₁₂O₆

Mm = [63.5 +32.1 + 9(16.0) + 10 (1.0)]

Mm = 249.6 g/mol

9. n_solid = n_solution

m/Mm = CV

m = CVMm

m = (3.0mol/L)(2.0L)(249.6g/mol)

m = 1497.6g

Therefore the 1498 g of CuSO₄ ۰ 5H₂0 is needed.

10. n_solid = n_solution

m/Mm = CV

m = CVMm

m = (0.320 mol/L) ( 0.250L)[3(23) + 31.0 + 4(16.0) g/mol]

m = 13.12g

Weigh out 13.12g of Na₃PO₄. Place it in a volumetric flask, rinse paper with water to remove all solute. Dissolve in less than 200mL of H₂O. Once all solute is dissolved fill to the 200mL mark and mix thoroughly then use.

11. n_c = n_d

C_cV_c = C_dV_d

V_d = C_cV_c

C_d

V_d = (0.15 mol/L) ( 0.200L)

0.95mol/L

V_d = 0.0316L

Therefore 31.6ml is required.

12. n_c = n_d

C_cV_c = C_dV_d

C_d = C_cV_c

V_d

C_d = (0.55 mol/L) ( 0.055L)

0.250mol/L

C_d = 0.12mol/L

Therefore the new concentration would be 0.12mol/L.

13. n_c = n_d

C_cV_c = C_dV_d

V_d = C_cV_c

C_d

V_d = (1.50 mol/L) ( 0.0500L)

0.45mol/L

V_d = 0.17L

Therefore the final volume will be 0.17L.

14. 23.50ml 16.75ml

2 NaOH + H₂SO₄ à Na₂SO₄ + 2 H₂O

2mol 1mol

n_NaOH = n_H2SO4

2mol 1mol

C_cV_c = C_dV_d

2mol

C_c = (0.180mol/L) (0.01675L) X 2mol

1mol 0.02350L

C_c = 0.257mol/L

Therefore the concentration of sulfuric acid is 0.257mol/L

15. 17.35ml 35.65ml

Ba(OH)₂ +2 HNO₃ à 2 H₂O + Ba(NO₃)₂

1mol 2mol

n_b = n_a

1mol 2mol

C_bV_b = C_aV_a

1mol 2mol

C_b = (0.150mol/L) (0.03565L) X 1mol

2mol 0.01735L

C_c = 0.154mol/L

Therefore the concentration of barium hydroxide is 0.154mol/L

a. Ag⁺¹, Sr⁺², Ca⁺²

__Cl^-1___│___(HCl)_

║ │

AgCl Sr⁺², Ca⁺²

__OH^-1___│___(NaOH)__

║ │

Ca(OH)2 Sr⁺²___________

SO₄^-2___│ (H₂SO₄)__

║ ┴

SrSO₄

b. Cu⁺² , Ca^+2, Ag⁺¹

__Cl^-1___│___(HCl)_

║ │

AgCl Cu⁺², Ca⁺²

__ SO₄^-2___│___(H₂SO₄)_

║ │

CaSO₄ Cu⁺²

OH^-1___│ (NaOH)__

║ ┴

Cu(OH)₂

a. pH = -log[H⁺¹]

pH = -log 8.2X10^-3M

pH = 2.08

b. pH = -log[H⁺¹]

pH = -log 7.3X10^-6M

pH = 5.14

a. [H⁺¹] = 10^-pH

[H⁺¹] = 10^-2.50

[H⁺¹] = 3.16 X 10^-3

b. [H⁺¹] = 10^-pH

[H⁺¹] = 10^-6.25

[H⁺¹] = 5.62 X 10^-7

UNIT 4 REVIEW ANSWERS

1)

a. ⁶³Cu₂₉ + ²D₁ à ⁶¹Ni₂₈ + ⁴He₂

b. ¹n₀ + ¹⁰B₅ à ⁷Li₃ + ⁴He₂

c. ²³⁸U₉₂ à ^234Th₉₀ + ⁴He₂

d. ⁷Li₃ + ¹H₁ à 2 ⁴He₂

e. ²³⁵U₉₂ + ¹n₀ à ¹⁴³Ba₅₆ + ⁸⁹Kr₃₆ + 3 ¹n₀

f. ¹⁷O₈ + ⁰β_-1 à ¹⁴C₆ + ³He₁

g. Tritium

2) Fission occurs when a heavier atom breaks apart into 2 lighter atoms.

Fusion occurs when two light atoms combine to form a heavier atom

3) .

a. ²²⁶₈₈Ra à ⁴₂He + ²²²₈₆Rn

b. ²¹⁰₈₁Tl à ⁰_-1e + ²¹⁰₈₂Pb

c. ¹³₇N à ⁰_-1e + ¹³₆C

d. ¹⁰₄Be à ⁰_-1e + ¹⁰₃Li

e. ¹⁸₉F à ⁰₁e + ¹⁸₁₀Ne

f. ²³⁸₉₂U à ⁴₂He + ²³⁴₉₀^Th

g. ²⁴⁴₉₅Am + ⁰_-1e à ²⁴⁴₉₄Pu + x-rays

h. ²⁵⁸₁₀₀Fm à ¹₀n + ²⁵⁷₁₀₀Fm

Gases

1) P₁V₁ = P₂V₂
V₂ = (P₁V₁)/P₂
V₂ = (110kPa)(220mL)/55.0kPa
V₂ = 440 mL
Therefore, the volume would be 440 mL

2) P₁V₁ = P₂V₂
P₂ = (P₁V₁)/V₂
P₂ = (300kPa)(100mL)/225mL
P₂ = 133 kPa
Therefore, the pressure is approximately 133 kPa

3) P₁V₁ = P₂V₂
P₁ = (P₂V₂)/ V₁
P₁ = (0.400kPa)(0.360mL)/300mL Note: 1000 P = 1 kPa
P₁ = 4.8 x 10^-4 kPa
The original pressure of nitrogen is 4.8 x 10^-4 kPa

4) P₁V₁ = P₂V₂
V₂ = (P₁V₁)/P₂
V₂ = (200kPa)(6.00L)/400kPa
V₂ = 3.00 L
Therefore, the volume would be 3.00L

5) Let there be 10.0 L of gas to start with and 5.00L of gas compressed
P₁V₁ = P₂V₂
P₂ = (P₁V₁)/V₂
V₁ = 2V₂
P₂ = (150kPa)(2V₂)/ V₂
P₂ = 300kPa
Therefore, the final pressure would be 300 kPa.

6) .

a. 27°C + 273 = 300K

b. 273°C + 273 = 546K

c. -162° + 273 = 111K

d. 727°C + 273 = 1000K

e. -150°C + 273 = 123K

7) .

a. 0K – 273 = -273°C

b. 273K – 273 = 0°C

c. 1000K – 273 = 727°C

d. 225K – 273 = -48°C

e. 400K – 273 = 127°C

f. 298K – 273 = 25°C

8) V₁/T₁ = V₂/T₂
V₂ = V₁T₂/T₁
V₂ = (2.00L)(323K)/298K
V₂ = 2.17L
Therefore, the volume is 2.17 L

9) V₁/T₁ = V₂/T₂
T₂ = T₁V₂/V₁
T₂ = (300K)(317mL)/127mL
T₂ = 749K
749K – 273 = 476°C
Therefore, the final temperature is 476°C

P = 33.0 KPa

n = ½ mol

T = -35°C + 273 K = 238K

R = 8.31 KPa۰L / mol۰K

10) PV = nRT

V = nRT

P

= (0.500mol) (8.31 KPa۰L / mol۰K) (232K)

33.0KPa

= 30.0L

Therefore the volume occupied is 30. 0L.

P = 103 KPa

T = 78°C + 273 K = 351 K

V = 0.570 L

R = 8.31 KPa۰L / mol۰K

11) PV = nRT

n = PV

RT

= (103KPa) (0.570L)

(8.31 KPa۰L/mol۰K) (351K)

= 2.01 × 10־² mol

Therefore there was 2.01 × 10־² moles of gas

Mm = 12.0 + 4.0

= 16.0 g/mol

12) PV = m × RT

Mm

P = m × RT

Mm V

P = 6.40g . 8.31KPa۰K . 300K

16.0g/mol mol۰K 0.570L

P = 186KPa

Therefore the pressure exerted is 186KPa.

13)

a. n = m

Mm

= 1.00g

2(1.0g/mol)

= 0.500mol

b. 0.500mol × (6.02X10²³)molecules/mol = 3.01 X 10²³ molecules

c. 0.500mol(6.02X10²³)molecules/mol(2 atoms/molecules) = 6.02 X 10²³ atoms

d. 22.4L/mol × 0.500mol = 11.2L

14)

PV = nRT

PV = m × RT

P = 40.5 KPa

V = 1.09L

T = 127°C + 273 K = 400K

R = 8.31 KPa۰L / mol۰K

Mm

m = PVMm

RT m =(40.5KPa) (1.09L) × 44.0g/mol

(8.31 KPa۰L/mol۰K) (400K)

m = 0.584g

Therefore there is 0.584g of CO₂

15) 2C₈H_18(g) + 25 O_2(g) à 16 CO_2(g) + 18 H₂O_(g)

xL P = 100KPa

V = 0.500mL

T = 323K

Since TP are constant Avogadros hypothesis

V_C8H18 = V_O2

2mol 25mol

V_C8H18 = V_O2 × 2mol

25mol

V_C8H18 = 0.500mol × 2mol

25mol

V_C8H18 = 0.0400L

Therefore 40.0mL of Octane will react

16) 2 KClO_3(s) à 2 KCl_(s) + 3 O_2(g)

Xg P = 101.3KPa

V = 5.00L

T = 273K

n_KClO3 = n_O2

2mol 3mol

3mol* m _KClO3 = 2mol × PV

Mm_KClO3 RT

m _KClO3 = 2mol × 101.3KPa× 5.00L × 122.6g/mol

(8.31 KPa۰L/mol۰K) (273K) 3mol

m _KClO3 = 18.2g

Therefore 18.2g of KCl will be required

17) 2 C₆H_6(l) + 15 O_2(g) à 12 CO_2(g) + 6 H₂O_(l).

Xg P = 101.3KPa

V = 5.00L

T = 273K

N_C6H6 = n_O2

2mol 15mol

15mol× m_C6H6 = 2mol × PV

Mm_C6H6 RT

m_C6H6 = 2mol × 101.3KPa × 25.0L × 78.0g/mol

(8.31 KPa۰L/mol۰K) (273K) 15mol

m_C6H6= 11.6g

Therefore 11.6g of benzene will be required

18) #particles - in container is constant

- outside container is constant

Volume – constant because the balloon is fully inflated

Temperature – increases due t the sun

Pressure – inside the balloon the pressure increases due to the increasing temperature and the

volume being unable to change. The increasing pressure causes the balloon to pop.

19) #particles - in container is constant, even though the can is empty there is some residual gas in the

can

- outside the can s constant (atmospheric pressure)

Volume – constant because the rigid can has a fixed volume

Temperature – increases due to heat from the fire

Pressure – increasing inside the can due to the heat energy causing the molecules to move faster

and heat the sides of the can more frequently. Eventually the pressure will cause the

can to explode throwing hot metal everywhere.

UNIT 5 REVIEW ANSWERS

6-ethyl - 3 - methyldecane 1- butyl –3 -chlorocyclobutane

2, 5, 9-tribromo-1,6–dicloro-3– nonyne 1,7–dibromo–2-heptene

1 - bromo-2, 3-dicloro-5-propylcyclohexane 1-bromo-4-chloro-2,6-dimethyl-3, 5-

dipropylcyclohexane

#2 a) b)

c) d)

e)

f)

#3 a)

1-hexene cyclohexane

trans-2-hexene cis-2-hexene

b) Structural isomers are compounds with the same chemical formula but completely different chemical structures, while geometric isomers are compounds with the same chemical formula, but different orientation of atoms across a double bond. Atoms cannot rotate about a double bond, so hydrogen or other attachments can have two different arrangements on carbons that are double bonded – on the same side (cis-) or opposite sides (trans-) of the double bond.

c) The cis- isomer has a higher boiling point then the trans- isomer, since the dipoles on the trans- isomer cancel out, causing there to be no net dipole, while the dipoles on the cis- isomer do not cancel out, causing the molecule to be more polar, and experience stronger IMF’s then the trans- isomer. Because of the higher IMF’s on the cis- isomer, it will be harder to pull these molecules apart, and will require more energy, thus meaning that the cis- isomer will have a higher boiling point.