Chapter 17 & 18 Worksheets.pptx

C H E M I S T RY I

CHAPTER 17: WS #1

1. Draw three water molecules around the water molecule below. Be sure to draw them in the correct position and label at least two hydrogen bonds.

•   •   •  

2. How much energy (in J) is required to heat 256 g of water from 20.0°C to 99.0°C?

q = mCT

q = 256g(4.18 J/goC)79.0oC

CHAPTER 17: WS #1

3. How many joules are released when 24.0 g of

steam at 100°C condenses to form liquid water at 0°C?

Temperature change and phase change

Heat of vaporization = 2260 J/g C_H2O = 4.18 J/gºC 24.0g 2260 J = 54,200 J

1 g

q = (24.0g)(4.18 J/gºC)(100ºC) = 10,000J 54,200J + 10,000J =

CHAPTER 17: WS #1

4. A 45 g piece of metal was heated to 105 °C and then dropped into a calorimeter containing 177 g of water. The temperature of the water in the calorimeter changed from 11 °C to 24.2 °C.

What is the specific heat of the metal?

• _mCΔT=mCΔT

5. What would be some of the consequences if ice were denser than water (think of environmental effects)?

• _{If ice were denser, it would sink in liquid}

water. This would mean that lakes, streams and rivers would freeze from the bottom up, killing all the fish.

• _{Ice bergs would sink, raising the levels of}

the oceans.

CHAPTER 17: WS #2

1. In the space below, draw molecules of water in ice

crystals and molecules of water in liquid water. Use these to explain why ice has a different density than liquid

water.

•  

•  

•  

2. Which atom in the water molecule is most electronegative? ______________________

•  

3. Complete this statement and then explain it for part b.

• Water has a __________ vapor pressure due to ___________________.

• _____________________________________________________________________ _____________________________________________________________________ _________.

oxygen

low

H-bonding

CHAPTER 17: WS #2

4. Water has a ________________ heat capacity and a __________________ heat of vaporization. As a direct result, coastal areas have ___________________

temperatures.

5. Hydrogen bonding causes…

• _{Water drops to be ___________ in shape.} • _{The vapor pressure of water to _________.} • _{The boiling point of water to be ________.} • _{The freezing point of water to be _______.}

6. A surfactant is a substance that interferes with _________________________ and decreases the

_______________________.

high

high

moderat

e

round

low

high

high

Hydrogen

CHAPTER 17: WS #2

7. Define and give an example of each of the following.

• _{Solution – mixture of solute and solvent; salt}

water

• _{Solvent – the dissolving agent; water}

• _{Solute – the substance being dissolved; salt} • _{Solvation – the process whereby the solute is}

• _{Below is a diagram of a salt formula unit. Draw}

how the salt ionizes and is surrounded by water when it dissolves.

CHAPTER 17: WS#3

1. Write the formulas for these hydrates:

• _{a. sodium sulfate decahydrate: _______________} • _{b. magnesium sulfate heptahydrate: ____________} • _{c. barium hydroxide octahydrate: _____________}

2. Name each hydrate:

• a. SnCl₄∙5H₂O: __________________________

• b. FeSO₄∙7H₂O: ______________________

• c. BaBr₂∙4H₂O: __________________________

• d. FePO₄∙4H₂O: ______________________

Na

SO

10H

2

O MgSO

7H

O

Ba(OH)

8H

2

O

Tin (IV)chloride

pentahydrate

Iron (II)sulfate

heptahydrate

Barium bromide

tetrahydrate

Iron (III)phosphate

CHAPTER 17: WS#3

3. Epsom salt (MgSO₄∙7H₂O) changes to the

monohydrate form at 150°C. Write an equation for the change.

•   MgSO₄∙7H₂O  6 H₂O + MgSO₄∙H₂O

4. Determine the percent by mass of water in Na₂SO₄ ∙ 10H₂O

1. IN THE SPACES PROVIDED, WRITE SATURATED,

UNSATURATED, AND SUPERSATURATED.

• Adrian is making a solution of sodium acetate in water. He adds a

little bit of the sodium acetate, stirs, and it immediately dissolves.

At this point, the solution is unsaturated. Adrian adds more

sodium acetate until it will no longer dissolve and has collected

on the bottom of the beaker. This solution is saturated. Adrian

heats up the beaker and continues stirring. The leftover sodium acetate finally all dissolves. Adrian allows the solution to cool.

The solution is now supersaturated.

• _{What would happen if Adrian added a single crystal of sodium}

acetate to the solution? Why?

• Adrian should see crystallization of the solution. This occurs because a supersaturated solution is not stable.

CHAPTER 18: WS#1

2. WHAT MASS OF AGNO3 CAN BE DISSOLVED IN 250G OF WATER AT 20OC? (222G AGNO

3 /100G H2O)

• 250 g H₂O 222 g AgNO₃ = 560 g AgNO₃ 100 g H₂O

3. YOU ARE GIVEN A CLEAR AQUEOUS SOLUTION CONTAINING KNO₃. HOW WOULD YOU DETERMINE EXPERIMENTALLY IF THE SOLUTION IS UNSATURATED, SATURATED, OR SUPERSATURATED?

• If you add a crystal of KNO₃ and it dissolves, the solution is unsaturated.

• If you add a crystal of KNO₃ and it sinks to the bottom of the container, the solution is saturated.

CHAPTER 18: WS#1

4. Using the graph fill in the paragraph below

• _{On the solubility chart the x-axis represents}

______________ and the y-axis represents

_________________________. The most soluble salt at 0.0 °C is ________, the least soluble is _______. Comparing NaCl to KCl at _____ °C NaCl becomes less soluble then KCl. One salt that is less soluble as temperature increases is ___________. At 90°C, ____ g K₂Cr₂O₇ can be dissolved in 100 g of water. At 30°C a solution of 30g KNO₃ and 100 g H₂O would be________________. At 30°C a solution of 50g KNO₃ and 100 g H₂O would be________________.

temperatu

re

NaNO

solubility

3

KClO

30

Ce

(SO

)

CHAPTER 18: WS#1

5. If the pressure in a closed system increases, the solubility of a gas ____________.

• Fill in the gas particles in the drawing to the right that represents this law.

• What is the equation that represents Henry’s Law? _______________________.

6. What is the effect of pressure on the solubility of gases in liquids?

increase

s

S

/P

= S

/

P

CHAPTER 18: WS#1

7. The solubility of methane in water at 20oC and

1.00 atm pressure is 0.026 g/L. If the

temperature remains constant, what will be the solubility of this gas at the following pressures

• _{0.60 atm b. 1.80 atm}

S

= S

P

P

0.026g/L =

x

1.00 atm

0.60atm

X = 0.016 g/L

S

= S

P

P

0.026g/L =

x

1.00 atm

1.80atm

CHAPTER 18: WS #2

1. FILL IN

THE BLANKS

IN THE

FOLLOWING

TABLE.

SHOW YOUR

WORK IN

THE BLANK.

Moles of

solute

Liters of

solution

Molarity

0.221 mol 1.998 L

0.221/1.998 =

0.111 M

5.48 mol

0.482 L

5.48/0.482 =

11.4 M

CHAPTER 18: WS #2

2. How many mL of a 2.1 M NaOH are needed to make 311.0 mL of 0.20 M NaOH?

•  M₁V₁ = M₂V₂

• (0.20M)(311.0mL) = (2.1M)(V₂) •  V₂ = 30. mL

3. If the percent (m/v) for the solute is 14% and the volume of the solution is 250 mL, what is the mass of solute in the solution?

•  250 mL 14 g solute =

• 100 g solution

CHAPTER 18: WS #2

4. If the solubility of a gas in water is 5.0 g/L when the

pressure of gas above the water is 2.7 atm, what is the pressure of the gas above the water when the solubility of the gas is 2.0 g/L?

• S₁ = S₂ • P₁ P₂

• 5.0 g/L = 2.0 g/L

• 2.7 atm x atm

• X = 1.1 atm

5. Calculate the molarity of a solution that has 21.4 g of NaCl dissolved in 1200 mL of solution.

•  21.4 g NaCl 1 mol NaCl = 0.366 mol NaCl

• 58.5 g NaCl

6. What is the number of moles of solute in 460 mL of a 5.0 M solution?

• _{460 mL 1 L 5.0 mol =} • _{1000mL 1 L}

• _{2.3 mol}

BELLWORK

• _{List the values.}

• Heat of vaporization of water

• Heat of fusion of water

• _Define:

• Solute

• Efflorecscent

CHAPTER 18: WS#3

1. Solve for the molarity of the following substances.

• a. 400 g CuSO₄ in a 4.00 L solution

• 400g CuSO4 1 mol CuSO₄ = 2.50 mol or 3 mol

•   159.6 g CuSO₄

•  3 mol CuSO₄/4.00L = 0.8 M

• b. 0.060 mol NaHCO₃ in 1500 mL solution

•   _{1500 ml Solution 1 liter = 1.500 L} •   _{1000 ml}

CHAPTER 18: WS#3

2. You have the following stock solutions available: 2.00M

NaCl, 4.0M KNO₃, and 0.50M MgSO₄. Calculate and

describe how you would make the following solutions.

• a. 500.0 mL of 0.500M NaCl

• (500.0ml)(0.500M)=(x ml)(2.00M) 

•  x = 125 ml so add 125ml of 2.00M NaCl to enough water to make 500ml of solution.

•   b. 2.0 L of 0.20M MgSO₄

•  (2.0L)(0.20M)=(x L)(0.50M) 

•  x = 0.80L so add 0.80L of 0.50M MgSO₄ to enough water to make 2 L of solution.

• c. 50.0 mL of 0.20M KNO₃

•  (50.0ml)(0.20m)=(x ml)(4.0M) 

•  x = 2.5 ml so add 2.5ml of 4.0M KNO₃ to enough water to make 50ml of solution.

• d. 1.4 L of 0.500M NaCl

• (1.4L)(0.500M)=(x L)(2.00M) 

CHAPTER 18: WS#3

3. What is the concentration, in percent

(m/v), of a solution with 75 g K

SO

in

1500 ml solution?

•

75g K

SO

x 100 =

5.0%

CHAPTER 18: WS#3

4. CALCULATE THE MOLES AND GRAMS OF SOLUTE IN EACH SOLUTION.

a. 1.0 L of 0.50M NaCl 1.0 L NaCl 0.50 mol NaCl = 0.50 mol NaCl 1 L NaCl

0.50 mol NaCl 58.6 g NaCl = 29 g NaCl 1 mol NaCl

b. 5.0 X 102 mL of 2.0M KNO

3 500. ml KNO3 1L KNO3 2.0 mol = 1.0 mol

KNO₃

1000 ml 1L

1.0 mol KNO₃ 101.1 g KNO₃ = 1.0 x 102 _KNO 3

1 mol KNO₃

c. 250 mL of 0.10M CaCl₂ 250 ml CaCl₂ 1 L CaCl₂ 0.10 mol CaCl₂ = 0.025 mol

1000 ml 1 L CaCl₂ CaCl₂ 0.025 mol CaCl₂ 111.1 g CaCl₂ = 2.8 g CaCl₂

1 mol CaCl₂

d. 2.0 L of 0.30M Na₂SO₄ 2.0L Na₂SO₄ 0.30 mol Na₂SO₄ = 0.60 mol Na₂SO₄

1 L Na₂CO₄

5. Magnesium metal reacts with a 65.0 ml of 1.50M zinc nitrate solution to produce the products shown below.

Mg (s) + Zn(NO₃)₂ (aq)  Mg(NO₃)₂ (aq) + Zn (s)

Assuming the reaction goes to completion, how many grams of magnesium are consumed by this reaction?

65.0 ml Zn(NO3)2 1 L Zn(NO3)2 1.50 molZn(NO3)2 1 mol Mg 24.3g Mg

1000L Zn(NO3)2 1 L Zn(NO3)2 1 mol Zn(NO3)2

1 mol Mg

= 2.37 g Mg

ANSWERS TO WORK OUT PROBLEMS IN

REVIEW.



• _{#5. 13,680 J}

• _{#7. a. 22, d. 43.45%} • _{#9. a. 56.5 g}

• _{#11. 19.1M, 12M, 0.60M}

• _{#12. a. 466g, b. 2.36M, 466g} • _{#13. 35L Ag}