CHEM125 - CHAPTER 04 - REGULAR

CHAPTER 4 REGULAR HOMEWORK

Resources:

Have you found videos, websites, or explanations that helped you understand this chapter? Let us know and we’ll add them to “Resources” part of this page for other students to use.

The order may vary but all questions are available. Use CTRL + F to find all the questions.

Question 1

When a solution of ammonium carbonate and magnesium chloride are mixed, what precipitate if any is formed?

Select one:

a. Magnsium carbonate, MgCO₃ (s) Correct

b. Ammonium chloride, NH₄Cl (s)

c. Magnsium carbonate, MgCO₃ (s) and Ammonium chloride, NH₄Cl (s)

d. No precipitate

Explanation

(NH4)2(CO3) + MgCl2 —> MgCO3 + 2 NH4Cl One of these things is not like the other… Hint its a solid Another hint its insoluble Yet another hint it’s to the left

Question 2

When a solution of nitric acid and sodium hydroxide are mixed, what precipitate if any is formed?

Select one:

a. No precipitate. This is a neutralization reaction. Water is formed and nitrate and sodium ions move throughout the solution. Correct

b. Sodium nitrate, NaNO₃ (s) and water are formed

c. Sodium nitrate, NaNO₃ (s)

d. No reaction

Explanation

Acid + base == salt and water formed

Question 3

When a solution of beryllium sulfate and calcium hydroxide are mixed, what precipitate if any is formed?

Select one:

a. Beryllium hydroxide, Be(OH)₂ (s) and calcium sulfate, CaSO₄ (s) Correct

b. Beryllium hydroxide, Be(OH)₂ (s)

c. calcium sulfate, CaSO₄ (s)

d. No precipitate

Explanation

Be(SO4) + Ca(OH)2 —> CaSO4 + Be(OH)2 According to the solubility rules neither product is soluble making them both solids aka precipitates because the two regents are liquids/aqueous.

Question 4

When a solution of sodium sulfide and iron (III) chloride are mixed, what precipitate if any is formed?

Select one:

a. Iron (III) sulfide, Fe₂S₃ (s) Correct

b. Sodium chloride, NaCl (s)

c. Iron (III) sulfide, Fe₂S₃ (s) and Sodium chloride, NaCl (s)

d. No precipitate

Explanation

3Na2S + 2FeCl3 –> Fe2S3 + 3NaCl Fe2S3 is the only insoluble product

Question 5

When a solution of ammonium carbonate and lead (II) acetate are mixed, what precipitate if any is formed?

Select one:

a. lead (II) carbonate, PbCO₃(s) Correct

b. Ammonium acetate, NH₄CH₃COO (s)

c. lead (II) carbonate, PbCO₃(s) and Ammonium acetate, NH₄CH₃COO (s)

d. No precipitate

Explanation

(NH4)2CO3 + Pb(CH3COO)2 –> 2NH4(CH3COO) + Pb(CO3) only one precipitate.. you guess it its PbCO3

Question 6

When a solution of potassium hydroxide and zinc chloride are mixed, what precipitate if any is formed?

Select one:

a. KCl (s) and Zn(OH)₂ (s)

b. Zn(OH)₂ (s) Correct

c. KCl (s)

d. No precipitate

Explanation

Writing out the formula always helps and is good practice. 2KOH + ZnCl2 —-> 2 KCl + Zn(OH)2 KCl is aq meaning aqueous and soluble in this solution whilst Zn(OH)2 is s meaning solid and insoluble in this solution

Question 7

Aqueous solutions of cobalt (II) bromide and sodium hydroxide are mixed. Correctly match the following:

CoBr₂(aq) + 2NaOH(aq) –>2NaBr(aq) + Co(OH)₂(s)	Answer 1
Co²⁺(aq) + 2Br^–(aq) + Na⁺(aq) + OH^–(aq) –> 2Na⁺(aq) + 2Br^–(aq) +Co(OH)₂(s)	Answer 2
Co²⁺(aq) + OH^–(aq) –>Co(OH)₂(s)	Answer 3
Na⁺(aq) and Br^–(aq)	Answer 4

Question 8

What mass (in g) of sodium sulfate, Na₂SO₄, is needed to prepare 380. mL of a 1.20 M solution?

Answer:

Explanation

3 Sig. Fig.

120 moles Na2SO4/ 1L * 0.380 L = 0.456 mol Na2SO4 0.456 mol Na2SO4 = Xg/ (142.04 g/mol Na2SO4) X= 64.77 g Na2SO4

Question 9

What is the molarity of the sodium ion in sodium sulfate, Na₂SO₄, prepared by dissolving 6.28 g of sodium sulfate in enough water to form 1.50 L of solution?

Answer:

Explanation

62.8 g Na2SO4/ (142.04 g/mol) = 0.04421 mol Na2SO4 0.04421 mol Na2SO4/ 1.50 L = 0.029475 M Na2SO4 2(0.029475) = 0.05895 M Na+

Question 10

What is the molarity of the carbonate ion in sodium carbonate, Na₂CO₃, prepared by dissolving 6.6 g of sodium carbonate in enough water to form 0.200 L of solution?

Answer:

Explanation

6.6 g Na2CO3/ (105.99 g/mol) = 0.06227 mol Na2CO3 0.06227 mol NaCO3/ 0.200 L = 0.31135 M Na2CO3 0.31134 M CO3(2-) [1:1 ratio]

Question 11

Potassium permanganate (KMnO₄) solutions are used for the determination of iron in samples of unknown concentration. As a laboratory assistant, you are supposed to prepare 1200. mL of a 0.300 M KMnO₄ solution. What mass of KMnO₄, in grams, do you need?

Answer:

Explanation

3 Sig. Fig.

0.300 mol KMnO4/ 1L * 1.2 L = 0.36 mol KMnO4

0.36 mol KMnO4 (158.034 g/mol) = 56.89g

Question 12

What is the molar concentration of a solution of sulfuric acid, H₂SO₄, if 3.52 mL react completely with 0.973 g of NaOH which was dissolved in 200 mL of water? Hint given in feedback

Answer:

Explanation

Hint–to get the right answer you first need to write the balanced neutralization equation.

0.973 g NaOH/ 39.997 g/mol NaOh= 0.224343 mol /2 L = 0.0122M

0.0122 M / 0.00352 L = 3.46 mol

Question 13

Pre question hint: Sodium hydroxide is combined with magnesium chloride. What kind of reaction is this and what solid forms? (Check those statements which are true.)

Select one or more:

a. This is a combustion reaction.

b. This is a neutralization reaction.

c. This is a precipitation reaction. Correct

d. Ths solid is Mg(OH)₂. Correct

e. Ths solid is MgOH.

f. Ths solid is NaCl.

Explanation

If it was a combustion reaction carbon dioxide and oxygen would be involved and they are not and it is not a neutralization reaction as an acid and a base haven’t reacted to create a salt and water.

Question 14

56.0 mL of 4.00 M sodium hydroxide is combined with 41.0 mL of 1.50 M magnesium chloride. What mass in grams of solid forms? (Hint: the total or overall reaction is most useful in doing calculations.)

Answer:

Explanation

Hint, this is a limiting reactant problem. To get the correct answer the overall balanced equation is needed. 2 NaOH + MgCl2 –> NaCl + Mg(OH)2 0.056 L *4M = 0.224 mol NaOH 0.041 L * 1.50 M = 0.00615 mol MgCl2 0.0615 mol MgCl2 * 2 mol NaOH/ 1 mol MgCl2 = 0.123 mol NaOh 0.224 mol > 0.123 mol MgCl2 = limiting reagent 0.0615 mol Mg(OH)2 is made 1:1 ration with MgCl2 0.0615 mol Mg(OH)2 * 58.32 g/mol Mg(OH)2 = 3.586 g Mg(OH)2

Question 15

What is the molarity of a solution of calcium hydroxide, Ca(OH)₂, if 138 mL is neutralized with 44.86 mL of a 0.403 M HCl? Hint given in feedback

Answer:

Explanation

Remember to first balance the equation.

Ca(OH)2 + 2HCl –> CaCl2 + 2H2O

0.04486 L * 0.403 M = 0.01807 mol HCl

1:2 ration with Ca(OH)2

0.00903929 mol = .138L * X M

0.0655= M

Question 16

Potassium acid phthalate, KHC₈H₄O₄, is a crystalline solid that is available in a state of high purity, making it an excellent choice as a standard acid. What is the concentration of a NaOH solution, if it takes 26.00 mL to neutralize a 0.0206 g sample of the acid? For the neutralization reaction, KHC₈H₄O₄(aq) –> K⁺(aq) + H⁺(aq) + C₈H₄O₄^2-(aq) Hint regarding acid is given in feedback

Answer:

Explanation

Note, there is 1 mole H⁺ for each mole of acid. KHC8H4O4 —> K+ + H+ + C8H4O4 2- 0.206 g acid / ( 204.22 g/mol acid) = 0.00010087 mol 1:1 ratio 0.00010087 mol / 0.026 L = 0.003879 M

Question 17

Warm-up for the next question. What reaction takes place in question 18?

Select one:

a. It is a neutralization reaction.

b. It is a precipitation reaction, forming KNO₃ (s).

c. It is a precipitation reaction, forming AgCl (s). Correct

d. It is a precipitation reaction, forming K₂NO₃ (s).

e. It is a precipitation reaction, forming AgCl₂ (s).

Explanation

A solid/ precipitate is formed from two liquids making the reaction a precipitate reaction. And the solid that is formed is AgCl and is not shown in the net ionic equation as it is soluble.

Question 18

A solution is prepared by placing 10.9 g of KCl in a 0.800 L volumetric flask and adding water to dissolve the solid, then filling the flask to the mark. What is the molarity of an AgNO₃ solution if 29.1 mL of the KCl solution react exactly with 34.0 mL of the AgNO₃ solution? The hint is given in feedback

Answer:

Explanation

Hint, first find the molarity of the KCl, then use the balanced equation to find the molarity of the silver nitrate.

10.9 g KCl / (74.55 g/mol KCl) = 0.1462 mol KCl / 0.8 L = 0.1822763 M

0.182763 M * 0.0291 L += 0.005318 moles = 0.034 * X M

0.156 M

1:1 ratio

KCl + AgNO3 –> KNO3 + AgCl

Question 19

Sodium thiosulfate, Na₂S₂O₃, is used as a fixer in photographic film developing. The amount of Na₂S₂O₃ in a solution can be determined by a titration with iodine, I₂, according to the equation: 2Na₂S₂O₃(aq) + I₂(aq) –> Na₂S₄O₆ +2NaI(aq). Calculate the concentration of the Na₂S₂O₃ solution if 28.70 mL of a 0.2210 M I₂ solution react exactly with a 100.0 mL sample of the Na₂S₂O₃solution. Use 4 sig. fig. Aside, the end of the titration is determined by the color. NaI is a pale yellow and I₂ is a deep purple. Just when the purple color persists, all the iodine that can react has reacted.

Answer:

Explanation

4 Sig. Fig.

0.2210 M I2 * 0.287 L = 0.0063427 moles I2

2:1 ratio

0.0063427(2)= 0.0126854 = .100 L * X M

M= 0.12685

Question 20

A 0.3120 g sample of a compound made up of aluminum and chlorine is dissolved in 150 mL H₂O. When excess AgNO₃ is added, 1.006 g of AgCl is precipitated. What is the number of moles Cl in the aluminum chloride sample?

Answer:

Explanation

AlCl3 + 3 AgNO3 –> 3 AgCl + Al(NO3)3 1.006 g AgCl/ (143.32 g/mol ) = 0.00701926 mol 3:1 AlCl3 0.0071926 mol Cl 0.002339753 mol Al

Question 21

What is the number of moles of aluminum in the sample?

Answer:

Question 23

AlCl3 + 3 AgNO3 –> 3 AgCl + Al(NO3)3 1.006 g AgCl/ (143.32 g/mol ) = 0.00701926 mol 3:1 AlCl3 0.002339753 mol Al

Question 22

How many chlorine atoms are in the empirical Formula of the aluminum chloride compound? For example, MgCl₂ has two atoms of Cl for each atom of Mg. You could also say the same thing with moles.

Answer:

Explanation

Question 23

What is the oxidation state of P in PO₄ ^3-

Answer:

Explanation

O: -2 Overall charge: 3- P + 4(-2)= -3 P : +5

Question 24

What is the oxidation state of P in PF5?

Answer:

Explanation

Florine is in Group 7A oxidation charge: -1 So P + 5(-1)= 0 P = +5

Question 25

What is the oxidation state of Cl in HClO4?

Answer:

Explanation

H: +1 O: -2 +1 + Cl + 4(-2) = 0 Cl= +7

Question 26

What is the oxidation state of P in PO₄ ^3-

Answer:

Explanation

O: -2 Overall charge: 3- P + 4(-2)= -3 P : +5

Question 27

What is the oxidation state of O in OF₂?

Answer:

Explanation

F: -1 O + 2*(-1)= 0 O = +2

Question 28

What is the oxidation state of oxygen in KO2?

Select one:

a. 0

b. 2

c. -2

d. -1.5

e. -0.5

f. 1

Explanation

K : +1 +1 + 2(Oxygen)= 0 Oxygen: -.5

Question 29

What is the oxidation state of xenon in XeO2F2?

Select one:

a. 6

b. 2

c. 3

d. 4

e. 1

f. 5

Explanation

O : -2 F: -1 Overall charge of molecule: 0 Xe + (2 x -2)+(2 x -1) = 0 Xe = 6