Jay’s CHE 132 Archived Resources

Hi everyone! My name is Jay and I will be your SI leader for Chemistry 132! Make sure to keep looking over exam material!

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Below are some resources that I think are helpful for CHE 132 students . If you have any questions come to an SI session or leave a comment below. You can also upload any of your helpful resources in the comment section!

My Session Schedule, Winter 2016

The document below should help you get more refreshed with significant figures!

Download (PDF, 11KB)


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Below is an Exam I Review Worksheet!

Download (DOCX, 19KB)

Exam I Worksheet Key
CHE 132 1

CHE 132 2

CHE 132 3

CHE 132 4

CHE 132 5

CHE 132 6

CHE 132 7

Exam II Review Worksheet

Download (DOCX, 31KB)

Exam II Review Worksheet Key

Download (DOCX, 32KB)


Review #1
review #2
review #3
Review #4

Exam III Worksheet

Download (DOCX, 14KB)

Exam III Key

Download (DOCX, 14KB)

Wednesday 8:30-9:30am, Richardson Library 105

Friday 8:30-9:30am, Richardson Library 105

Office Hour – Monday 8:30-9:30am, Richardson Library 105 [/box]

Can’t attend any of my sessions? Check out other CHE 132 SI leaders’ schedules here.

About Me

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Videos and Handouts

16 thoughts

  1. Please Help!

    #1
    The data shown below were collected for the following second-order reaction:
    Cl(g)+H2(g)→HCl(g)+H(g
    Temperature (K) Rate Constant (L/mol⋅s)
    90 0.00357
    100 0.0773
    110 0.956
    120 7.781
    Use an Arrhenius plot to determine the activation barrier for the reaction. AND
    Use an Arrhenius plot to determine the frequency factor for the reaction.

    #2
    At 473 K, for the elementary reaction
    2NOCl(g)k1⇌k−12NO(g)+Cl2(g)
    k1=7.8×10−2L/mols and k−1=4.7×102L2/mol2s
    A sample of NOCl is placed in a container and heated to 473 K. When the system comes to equilibrium, [NOCl] is found to be 0.29 mol/L .

    Calculate the concentration of NO and Cl2

    #3
    Suppose that a catalyst lowers the activation barrier of a reaction from 121 kJ/mol to 59 kJ/mol .
    By what factor would you expect the reaction rate to increase at 25 ∘C? (Assume that the frequency factors for the catalyzed and uncatalyzed reactions are identical.)

  2. Having trouble with this problem on Assignment 12
    Exercise 15.119

    At 473 K, for the elementary reaction
    2NOCl(g)k1⇌k−12NO(g)+Cl2(g)
    k1=7.8×10−2L/mols and k−1=4.7×102L2/mol2s
    A sample of NOCl is placed in a container and heated to 473 K. When the system comes to equilibrium, [NOCl] is found to be 0.45 mol/L .

  3. Hello guys wondering if anyone can help me with questions 8-9 and 19 in the mastering. Honestly have no clue

  4. Couldn’t figure this one out last night. Could someone help me?
    What is the boiling point of an aqueous solution that has a vapor pressure of 19.0 torr at 25 degrees Celsius?
    (P^o h2O =23.78 torr; Kb =0.512 degrees C/m)

  5. Question 9 on the practice exam is confusing me. Even when I add up all of the given numbers you put in the answer key and divide by 4 to get Dc=o, my answer doesn’t add up to your answer of 837kJ/mol. Can someone help me?

    Dc=o = 6(410 kJ/mol) + 2 (350 kJ/mol) – 6(460 kJ/mol) + 3(498 kJ/mol) – 1453 kJ/mol

    Dc=o = ?

    Also, ΔH= -1453 kJ/mol is given in the problem. So when you manipulate the equation to isolate Dc=o, aren’t you supposed to end up with a +1453kJ/mol in the equation?

  6. Question 9 on the practice exam is confusing me. Even when I add up all of the given numbers you put in the answer key and divide by 4 to get Dc=o, my answer doesn’t add up to your answer of 837kJ/mol. Can someone help me?

    Dc=o = 6(410 kJ/mol) + 2 (350 kJ/mol) – 6(460 kJ/mol) + 3(498 kJ/mol) – 1453 kJ/mol

    Dc=o = ?

    Also, Δ = -1453 kJ/mol is given in the problem. So when you manipulate the equation to isolate Dc=o, aren’t you supposed to end up with a +1453kJ/mol in the equation?

  7. For question 3 on the practice exam, how do we know that ΔHreactants = 0 ?

    I guess I just don’t understand what I’m supposed to do.

  8. Anyone know how to do this problem? This is in the Exam review.

    1.) Calcium metal reacts with water to produce hydrogen gas:
    Ca + 2H2O→Ca(OH)2 + H2
    In a given experiment, 825 mL of the gas is collected over water at 25oC at a pressure of 0.967 atm. If the partial pressure of water at 25oC is 0.0313 atm, calculate the mass of H2 produced.

  9. Answer to 10.95 Assignment 3

    Na(s) –> Na(g) —> Na+(g)
    109kJ + 496kJ = 605kJ for this process

    1/2 Cl2(g) –> Cl(g) –> Cl-(g)
    1/2(243)kJ + (-349kJ) = -227.5kJ for this process

    Finally, both ionic gases combine together to form NaCl(s) and this is known as the lattice energy, which is what we’re solving for. The overall equation will look like this:

    -411kJ = 605kJ – 227.5 + E(lattice)

    where the term on the left side corresponds to the enthalpy of formation and it is the overall process (summation of all the steps):

    Na(s) + 1/2Cl2(g) –> NaCl(s) H = -411kJ

    So, the lattice energy will be:

    E(lattice) = -411kJ – 605kJ + 227.5kJ = -788.5kJ

  10. I’m having trouble figuring out what I’m doing wrong for Exercise 11.44 with feedback from mastering Chemistry:
    A weather balloon is inflated to a volume of 26.3 L at a pressure of 734 mmHg and a temperature of 26.5 ∘C. The balloon rises in the atmosphere to an altitude where the pressure is 365 mmHg and the temperature is -16.1 ∘C.
    Assuming the balloon can freely expand, calculate the volume of the balloon at this altitude.
    I used V=(nRT)/P but it keeps saying incorrect.

  11. I’m having some trouble on Assignment 3 Problem 10.95. The problem states,

    Use the Born-Haber cycle and data from Appendix IVB and Chapters 4 and 10 in the textbook to calculate the lattice energy of NaF.(ΔHsub for sodium is 107.5 kJ/mol.)
    Express your answer as an unteger.

    I found the values below
    – Hf = -576.6 kJ
    – Hsub = 107.5 kJ/mol
    – Hdiss = 1/2(79.38) kJ
    – Hionization = 496 kJ
    – H4eaffinity = -328 kJ

    The answer i got was -892 kJ/mol but the computer told me “Not quite. Check through your calculations; you may have made a rounding error or used the wrong number of significant figures.” Please advice.

  12. Has anybody done Mastering Chemistry’s Exercise 10.88 Part B? I have done the following approach:

    (3H2O (l) + 2CO2 (g) ) – (C2H6O + 3O2 (g))

    (3(285.8) + 2(-393.5)) – (-235.2 +3(0))

    -1512.4 – (-235.2) = -1277.2

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