CHEM 111 Exam 4

Multiple Choice

100

Give the chemical name of P₄F₈

Tetraphosphorus octafluoride 

200

Predict the outcome of the reaction represented by the following equation by using the activity series, and balance the equations. If not reaction occurs, write "No reaction"

Au (s) + NaBr (aq) --> 

NO reaction! Sodium has higher ease of oxidation than Gold and sodium is already the ion!

300

Draw a picture of an aqueous solution of sodium sulfate.

Should have twice the amount of Na⁺ molecules than SO₄^2- molecules 

400

Two aqueous solutions of AgNO₃ and NaCl are mixed. Which What will be the resulting compounds?

AgCl (s) NaNO₃ (aq)

500

Name all of the strong acids and strong bases. 

Acids: HCl, HBr, HI, HNO₃, HClO₃, HClO₄, and H₂SO₄

Bases: LiOH, NaOH, KOH, RbOH, CsOH, Ca(OH)₂, Sr(OH)₂

Precipitation Reactions

100

Complete the following equations.  Note that precipitates are insoluble and are followed by (s).  Species in solution are followed by (aq).  If no reaction occurs, write “No reaction.”

LiCl + RbClO₃ -->

No reaction (both aqueous) 

200

Complete the following equations.  Note that precipitates are insoluble and are followed by (s).  Species in solution are followed by (aq).  If no reaction occurs, write “No reaction.”

AgNO₃ +  NaOH

AgOH(s) +  Na⁺(aq) + NO3^-(aq)

300

Complete the following equations.  Note that precipitates are insoluble and are followed by (s).  Species in solution are followed by (aq).  If no reaction occurs, write “No reaction.”

Fe(NO3)_3(aq) + NaOH_(aq) --> 

1. Fe(NO₃)_3(aq)  + 3 NaOH_(aq) →  Fe (OH)_{3 (s)}+ 3NaNO_3 (aq)

400

Write the molecular, ionic, and net ionic equations of:

BaCl_{2 (aq)} + K₂SO_{4 (aq)} --> 

Molecular:BaCl_{2 (aq)} + K₂SO_{4 (aq)} --> 2KCl _(aq)  + BaSO_4 (s)

Ionic: Ba²⁺ _(aq) + 2Cl^- _(aq) + 2K⁺_(aq) + SO₄^2-_(aq) --> 2K⁺ _(aq) + 2Cl^-_(aq) + BaSO_{4 (s)}

Net Ionic: Ba²⁺ _(aq) + SO₄^2-_(aq) --> BaSO_{4 (s)}

500

Balance the follow equation and write the corresponding ionic and net ionic equations. Indicate spectator ions. 

Molecular: HNO_{3 (aq)} + Ba(OH)_{2 (aq)}--> 

1. 2HNO_{3 (aq)} + Ba(OH)_{2 (aq)} --> 2 H₂O _(l) + Ba(NO₃)_{2 (aq)}

2. 2H⁺ + 2NO₃^- + Ba²⁺ + 2OH^- --> 2H₂O _(l) + Ba²⁺ + 2NO₃^-

3. 2H⁺ + 2OH^- --> 2H₂O _(l)

4. NO₃^- and Ba²⁺

Redox/ Oxidation #'s

100

Give the oxidation number of P in PO₄^3-

+5

200

Give the oxidation number of P in P₄, PF₂, and Ca₃P₂

0, +2, -3

300

For the reaction 2Fe + 3Br₂ --> 2FeBr₃

What species is being oxidized and reduced?

Write the 2 half reactions. 

Oxidized- Fe

Reduced- Br₂

1. 2Fe --> 2Fe⁺³ + 6e^-

2. 6e^- + 3Br₂ --> 6Br^-

Final: 2Fe + 3Br₂--> 2Fe⁺³ + 6Br^-

400

Fe _(s) + Cu(NO₃)_2 (aq) --> Cu _(s) + Fe(NO₃)₂ (aq)

Write half reaction involving copper, write half reaction involving iron. What species is being oxidized and reduced? Give oxidizing and reducing agents. 

1. 2e^- + Cu⁺² (aq) --> Cu (s) RED

2. Fe (s) --> Fe⁺² (aq) + 2e^- OX

3. Fe is being oxidized, reducing agent 

Cu is being reduced, oxidizing agent 

500

Identify the species being oxidized and reduced in each of the following reactions and write the half reactions.

2 As (s) + 3 Cl₂ (g) → 2 AsCl₃

Oxidized: 2 As (s)--> 2 As⁺³ + 6e^-

Reduced: 6e^- + 3Cl_{2 (g)}--> 6 Cl^-

Concentrations of Solutions

100

1. Find the pH of 14.0 M HCl

2. Calculate the hydronium ion concentration of solution with a pH of 6.91

1. -1.146

2. 1.2 x 10^-7

200

If a 20.00 mL of a 1.296 M stock solution of Na₂CO₃ are diluted to 300. mL, what is the molarity of the new solution?

M1V1=M2V2

(0.020 L)(1.296M) = (0.300L) (M2)

M2=0.0864

300

A solution is made by dissolving 76.87 g of Mg(NO₃)_{2 (}formula mass 148.31 g/mol) in a total volume of solution being 500.0 mL. What is the molarity of solution?

76.87 g x (1mol/ 148.31g)= 0.5183 mol / 0.500L =1.037 M

400

What is the molar concentration of nitrate in Ba(NO₃)₂ if M= 0.8501 

2 x 0.8501= 1.700M= [NO₃^-]

500

A volume of 35.2 mL of a 1.56 M KMnO₄ solution is mixed with 43.8 mL of 0.756 M KMnO_4. Calculate the concentration of the final solution.

35.2 mL x (1L/1000mL) x (1.56 mol/L)= 0.054912 mol

43.8 mL x (1L/1000 mL) x (0.756 mol/L) = 0.03311 mol

0.054912 + 0.03311= 0.0880248 mol /0.079 L= 1.11M (final answer) 

35.2 mL + 43.8 mL= 79 mL= 0.079L

Ch. 10 Calculations

100

What is the final temperature after 5.87 x 10² Joules is absorbed by 15.0 g of water at 25 degrees C?

q= mc delta T

t_f= (q/ mc) + t_i = (5.87 x 10² J/ 15.0 g x 4.184 J     g^-1 C^-1 ) + 25.0 degrees C= 34.4 degrees C

200

Given the thermochemical equation 

 2C₆H_{14 (l)} + 19 O_2 (g) --> 12 CO_2 (g) + 14H₂O _(l) 

How much heat is given off when 15.67g of hexane (86.18 g/mol) undergoes complete combustion? 

Delta H= -4194 kJ/mol

15.67 g C₆H₁₄ x (1mol/86.18 g) x (-4194 kJ/ 2 mol)= -381.3 kJ

300

Calculate the enthalpy of the reaction using the following standard enthalpies of formation:

2NO_(g)+O_2(g)⇋2NO_2(g)

• NO(g): 90.25 kJ/mol
• NO2(g): 33.18 kJ/mol

Delta H reaction: summation delta H formation of products- summation delta H formation of reactants 

ΔH^oreaction=(2mol)(33.18kJ/mol)−[(2mol)(90.25 kJ/mol)+(1mol)(0kJ/mol)]

ΔH^oreaction=−114.1kJ

400

Given the following equations and Delta H values, determine the enthalpy of reaction (kJ) at 298 K for the reaction:

B₂H_{6 (g)} + 6Cl₂ (g) --> 2BCl_{3 (g)} + 6HCl _(g)

1. BCl_{3 (g)} + 3 H₂O _(l) --> H₃BO₃ (g) + 3 HCl (g) [delta H/ kJ = -112.5]

2. B₂H_{6 (g)} + 6 H₂O _(l) --> 2 H₃BO_{3 (g)} + 6H_{2 (g)}    [delta H/kJ = -493.4]

3.  1/2 H_{2 (g)}  +  1/2 Cl₂ (g) --> HCl (g)                [delta H/kJ = -92.3]

1. -2*

2.

3. 12*

-1376 kJ

500

Given the following equations and Delta H values, determine the enthalpy of reaction (kJ) at 298 K for the reaction:

N₂H_{4 (l) +} 2H₂O_{2 (g)} --> N_{2 (g)} + 4H₂O (l) 

1. N₂H_{4 (l)} + O_{2 (g)} --> N_{2 (g)} + 2H₂O _(l)                    [delta H/kJ = -622.3]

2. H_2(g) + 1/2 O_{2 (g)} --> H₂O _(l)                                             [delta H/kJ = -285.8]

3. H₂O_{2 (g)} --> H_2(g) + O_{2 (g)}

 [delta H/kJ = -187.8]

1. no change 

2. 2*

3. -2*

Answer: -818.3 kJ