UNIT 4 BLM Answer Key - Weebly

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UNIT

4 BLM Answer Key

BLM 8.6-1 AMOUNT CONCENTRATIONS OF SOLUTIONS

1. Sample answer: Although a graduated cylinder allows you to measure many different volumes of liquid, the markings are not as precise as the volumetric flask, which is made to measure a certain volume of liquid with high precision. 2. (a) Sample answer: forgetting to tare the mass of the paper; spilling some of the solute when transferring it into a volumetric flask (b) Sample answer: spilling some of the solute when trying to place it into the volumetric flask; adding more solute than is needed 3. n = cV = (0.25 mol/L)(0.465 L) = 0.1162 mol; The molar mass of sodium chloride is 58.44 g/mol. mass = (0.1162 mol)(58.44 g/mol) = 6.791 g mass = 6.8 g 4. The molar mass of potassium nitrate (KNO3) is 101.11 g/mol. n = mass/molar mass = (65.4 g)/ (101.11 g/mol) = 0.64682 mol c = n/V = (0.64682 mol)/ (0.35 L) = 1.848 mol/L c = 1.8 mol/L 5. The molar mass of potassium chloride (KCl) is 74.55 g/mol. n = mass/molar mass = (34.5 g)/(74.55 g/mol) = 0.46278 mol V = n/c = (0.46278 mol)/(0.45 mol/L) = 1.028 L V = 1.0 L

BLM 8.7-1 DILUTIONS OF SOLUTIONS

1. Sample answer: The phrase refers to always adding a small volume of acid solution to a much larger volume of water when diluting the acid solution. Because the dilution of acids is very exothermic, if water is added to acid, the water could heat up and splatter concentrated acid on anyone within range. 2. Amount concentration is inversely proportional to the volume of solution when diluting solutions. 3. A volumetric pipette is used to transfer a precise volume of the concentrated solution into the flask for the dilute solution.

4. (a)

Table 1 Amount Concentrations of Common Stock Acid

Solutions

Volume needed to

Stock acid

Amount concentration

(mol/L)

produce 1.50 L of solution with amount concentration

of 2.0 mol/L (L)

hydrochloric acid, HCl(aq)

12

[0.25 L]

nitric acid, HNO3(aq)

16

[0.19 L]

sulfuric acid, H2SO4(aq)

18

[0.17 L]

(b) cd = ccVc/Vd = (16 mol/L)(0.0100 L)/(0.2500 L) = 0.64 mol/L

BLM 8.Q CHAPTER 8 QUIZ

1. (d); 2. (b) 3. T 4. F. To make a solution, a solute is dissolved in a solvent. 5. F. The dissociation equation for sodium hydroxide is NaOH(s) Na+(aq) + OH?(aq). 6. (a)(ii); (b)(i); (c)(iii) 7. Sample Answer: The solubility of a solid generally increases with increased temperature. The solubility of a gas generally decreases with increased temperature. 8. The molar mass of KCl is 74.55 g/mol. n = mass/molar mass = (25.0 g)/(74.55 g/mol) = 0.33535 mol c = n/V = (0.33535 mol)/(0.2500 L) = 1.3414 mol/L cd = ccVc/Vd = (1.3414 mol/L)(0.0200 L)/(0.2500 L) = 0.10731 mol/L cd = 0.107 mol/L

BLM 9.1-1 WRITING TOTAL AND NET IONIC

EQUATIONS

1. balanced formula equation: Pb(NO3)2(aq) + 2 NaCl(aq) 2 NaNO3(aq) + PbCl2(s) total ionic equation: Pb2+(aq) + 2 NO3?(aq) + 2 Na+(aq) + 2 Cl?(aq) 2 Na+(aq) + 2 NO3?(aq) + PbCl2(s) net ionic equation: Pb2+(aq) + 2 Cl?(aq) PbCl2(s) 2. balanced formula equation: K2SO4(aq) + CaI2(aq)

2 KI(aq) + CaSO4(s)

Unit 4 BLM Answer Key

Copyright ? 2011 by Nelson Education Ltd.

Unit 4 Blackline Master Answer Key

total ionic equation: 2 K+(aq) + SO42?(aq) + Ca2+(aq) + 2 I?(aq) 2 K+(aq) + 2 I?(aq) + CaSO4(s) net ionic equation: Ca2+(aq) + SO42?(aq) CaSO4(s) 3. balanced formula equation: CuSO4(aq) + Zn(s) Cu(s) + ZnSO4(aq) total ionic equation: Cu2+(aq) + SO42?(aq) + Zn(s) Cu(s) + Zn2+(aq) + SO42?(aq) net ionic equation: Cu2+(aq) + Zn(s) Cu(s) + Zn2+(aq) 4. balanced formula equation: 2 Li3PO4(aq) + 3 MgCl2(aq) 6 LiCl(aq) + Mg3(PO4)2(s) total ionic equation: 6 Li+(aq) + 2 PO43?(aq) + 3 Mg2+(aq) + 6 Cl?(aq) 6 Li+(aq) + 6 Cl?(aq) + Mg3(PO4)2(s) net ionic equation: 2 PO43?(aq) + 3 Mg2+(aq) Mg3(PO4)2(s)

BLM 9.5-1 SOLVING SOLUTION STOICHIOMETRY PROBLEMS

1. V = 0.16 L ? (0.60 mol/L) ? (1 mol Na2SO4/1 mol CaI2) ? (1 L/0.45 mol) = 0.2133 L = 210 mL 2. V = 0.32 L ? (0.50 mol/L) ? (2 mol K3PO4/3 mol ZnCl2) ? (1 L/0.30 mol) = 0.3556 L = 360 mL 3. (a) Pb(NO3)2(aq) + 2 KCl(aq) 2 KNO3(aq) + PbCl2(s) (b) V = 5.40 L ? (0.050 mol/L) ? (2 mol KCl/1 mol Pb(NO3)2) ? (1 L/0.40 mol) = 1.35 L 4. m = 1.25 L ? (0.70 mol/L) ? (1 mol Ag2SO4/1 mol CuSO4) ? (311.80 g/1 mol) = 273 g OR m = 1.10 L ? (0.80 mol/L) ? (1 mol Ag2SO4/2 mol AgNO3) ? (311.80 g/1 mol) = 137 g; The mass of precipitate expected is 137 g.

BLM 9.Q CHAPTER 9 QUIZ

1. (a); 2. (d) 3. T 4. F. Ions responsible for water hardness are calcium, magnesium, and iron. 5. F. Water treatment produces safe, drinkable tap water. 6. (a)(iii); (b)(i); (c)(ii) 7. Water contaminants can be physical, biological, or chemical. 8. Improved technology allows us to detect lower concentrations of chemicals in water, so more tests detect the presence of pharmaceuticals.

10.1-1 DESCRIBING ACIDS AND BASES

Acid

Anion

formula formula

Anion name

Acid name

1 HCl(aq)

Cl?

chloride

hydrochloric acid

2 H2SO3(aq)

SO32?

sulfite

sulfurous acid

3 HNO3(aq)

NO3?

nitrate nitric acid

4 H3PO3(aq)

PO33?

phosphite

phosphorous acid

5 H2CO3(aq)

CO32?

carbonate

carbonic acid

6. Ca(OH)2(aq) Ca2+(aq) + 2 OH?(aq) 7. KOH(aq) K+(aq) + OH?(aq) 8. Ba(OH)2(aq) Ba2+(aq) + 2 OH?(aq)

9. Sample answer: The acidic or basic properties of the

substances are only seen when the substances dissolve in

water and become ions.

10. Acids that do not contain oxygen have names that begin

with hydro-.

10.3-1 SOLVING TITRATION PROBLEMS

1. Sample answer: The initial volume reading is subtracted

from the final volume reading to show how much titrant

was released from the burette.

2. Sample answer: The pH range at which the indicator

changes colour must be close to the equivalence point and

occur during the steepest part of the pH curve.

3. (a) KHP(aq) + NaOH(aq) H2O(l) + NaKP(aq) (b) nNaOH = 0.415 g KHP ? (1 mol/204.23 g) ? (1 mol NaOH/1 mol KHP) = 2.0320 ? 10?3 mol NaOH; cNaOH = nNaOH/VNaOH = 2.0320 ? 10?3 mol NaOH/1.654 ? 10?2 L =

0.123 mol/L

(c) H2SO4(aq) + 2 NaOH(aq) 2 H2O(l) + Na2SO4(aq); nacid = 3.664 ? 10?2 L NaOH ? (0.123 mol/1 L) ? (1 mol H2SO4/2 mol NaOH) = 2.2534 ? 10?3 mol H2SO4; cacid = nacid/Vacid = 2.2534 ? 10?3 mol H2SO4/2.760 ? 10?2 L = 8.16 ? 10?2 mol/L

BLM 10.Q CHAPTER 10 QUIZ

1. (d); 2. (c) 3. F. A weak base only partly dissociates in water. 4. T 5. T 6. (a)(ii); (b)(iii); (c)(i) 7. high molecular mass, pure, stable 8. The solution that has a pH of 4 is 10 000 times more acidic than a solution that has a pH of 8.

Unit 4 BLM Answer Key

Copyright ? 2011 by Nelson Education Ltd.

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