Preview

test 1

Satisfactory Essays
Open Document
Open Document
2294 Words
Grammar
Grammar
Plagiarism
Plagiarism
Writing
Writing
Score
Score
test 1
Version 017 – Exam 1 – Laude – (52455)
This print-out should have 30 questions.
Multiple-choice questions may continue on the next column or page – find all choices before answering.
001

1

003 6.0 points
What would be pH of a 1 M solution of a very expensive weak base, unobtainamine, whose pKb is 3.4?

6.0 points
1. 12.3 correct

Consider a reaction with ∆Hrxn =
997 kJ · mol−1 . Which of the following pairs of K values and temperatures is possible for this reaction?
1. K1 = 15.7, T1 = 138 K;
K2 = 1.57 × 1038 , T2 = 153 K correct
2. K1 = 11.5, T1 = 128 K;
K2 = 3.83, T2 = 145 K
3. K1 = 3.1, T1 = 103 K;
K2 = 3.1, T2 = 340 K

2. 14
3. 1.7
4. 7
5. 0
Explanation:
The equation
[OH− ] = (Kb · Cb )0.5 can be algebraically rearranged to pOH = 0.5(pKb + pCb )

4. K1 = 9.76, T1 = 102 K;
K2 = 3.74 × 10−61 , T2 = 116 K
Explanation:
For and endothermic reaction, K is directly related to T. For an exothermic reaction, K is inversely related to T. K is constant with temperature only for ∆Hrxn = 0 kJ.
002 6.0 points
Which combination of ∆G and K is possible?
1. ∆G = −29.4 J, K = 1.01 correct
2. ∆G = −77.6 kJ, K = 0.981
3. ∆G = 47.3 J, K = 3.02 × 1013
4. ∆G = 54.7 kJ, K = 1.03
5. ∆G = −82.5 J, K = 3.15 × 10−10
Explanation:
If ∆G is positive, K must be less than 1; if ∆G is negative, K must be greater than 1.
Also, if ∆G is small (compared to RT), then
K will be close to 1, and if ∆G is large, K will be much smaller or greater than 1.

= 0.5(3.4 + 0) = 1.7 pH = 14 − pOH = 12.3
004 6.0 points
Consider the chemical equation below:
N2 O4 (g) + 4H2 (g) ←→ 4H2 O(g) + N2 (g)
What would Keq be for this reaction?
1. Keq =
2. Keq

[H2 O]4 [N2 ] correct [N2 O4 ][H2 ]4

[N2 O4 ][H2 ]4
=
[H2 O]4

3. Keq =

[H2 O]4
[N2 O4 ][H2 ]4

4. Keq =

[N2 O4 ][H2 ]4
[H2 O]4 [N2 ]

Explanation:
For a reaction of the form: aA(s) + bB(aq) ↔ cC(ℓ) + dD(g)
The equilibrium constant will take the form: [D]d
Keq =
[B]b

You May Also Find These Documents Helpful

  • Satisfactory Essays

    4.) The individual with genotype AaBbCCDdEE can make many kinds of gametes. Which of the following is the major reason?…

    • 495 Words
    • 1 Page
    Satisfactory Essays
  • Satisfactory Essays

    The 3 types of photographs taken at crime scenes are overview, intermediate and close-up photographs. Overview photographs show crime scene in wide angles taken at different angles to show entries and exits to the crime scene. Intermediate photographs are taken closer to the evidence, but still show area surrounding the evidence. Close-up photographs focus on an injury, a weapon, or any piece of…

    • 786 Words
    • 2 Pages
    Satisfactory Essays
  • Powerful Essays

    Calorimetry Lab

    • 1355 Words
    • 6 Pages

    We were given KOH and KNO3 as our salts. We weighed the calorimeter and water then subtracted accordingly o get the mass of the calorimeter and that of the water. The lab instructed on the amounts of water and salt to be used. The initial temperature of the water in the KOH reaction was 22.8 degrees Celsius and the final temperature of the mixture determined from the graph was 37.0 degrees Celsius. The graph of the KOh showed it to be an exothermic salt and the KNO3 to be and endothermic salt because of the change in the temperature indicating either absorbing or giving off of heat by the salts. This the change in temperature of the solution of the KOH and water is + 28880 J and the heat change of the salt is 82.6 J. Thus the total heat change, in the reaction is -2960 J. To determine the DeltaHs (J/g salt), divide the totalt heat of change by the measure mass of the salt resulting in -590 J/g. The same process was followed for KNO3 but this solution had a -5.9 degree Celsius temperature change this producing a negative total heat change of water and the slat in the solution giving a +13000 J in the reaction. DeltaHs is then determines to be 250 J/g. The expected value of DeltaHs of KNO3 is +354 J/g and for KOH, -1026 J/g. The error for the KNO3 is -27.5% and for the KOH it is 42.5%. These errors are high but not uncommon for the experiment. They are caused because the salts are in a solid form and thus after the peak of temperature of solution is found, the remainder of the graph is flawed because salt is still in the process of dissolving. This dissolving still of the salts is giving off or absorbing heat thus making the extrapolated change in the temperature of solution to be small. The temperature of solution being too small results in the total heat change being too small, giving the DeltaHs value as in the case of KOH too high (not negative enough) and in the KNO3…

    • 1355 Words
    • 6 Pages
    Powerful Essays
  • Satisfactory Essays

    test 4

    • 1357 Words
    • 14 Pages

    If capital is fixed at two units, what is the marginal product of the fourth unit of labor?…

    • 1357 Words
    • 14 Pages
    Satisfactory Essays
  • Good Essays

    ACT Test 1

    • 2751 Words
    • 28 Pages

    Directions: Each question has five answer choices. Choose the best answer for each question, and…

    • 2751 Words
    • 28 Pages
    Good Essays
  • Powerful Essays

    Exam 1

    • 1953 Words
    • 8 Pages

    is the process by which we learn about our culture by watching and imitating the behaviors…

    • 1953 Words
    • 8 Pages
    Powerful Essays
  • Satisfactory Essays

    Exam 1

    • 5433 Words
    • 22 Pages

    | In Business 1.0 approach, messages are scripted by designated communication, approved by someone in authority, distributed through selected channels, and delivered without modification to a passive audience. In the 2.0 approach, customers and other stakeholders participate in, influence, and often take control of conversations in the marketplace. The audience must be active listeners in the 2.0 approach.…

    • 5433 Words
    • 22 Pages
    Satisfactory Essays
  • Satisfactory Essays

    Exam 1

    • 410 Words
    • 2 Pages

    The first midterm exam, which counts for 20% of your course grade, will be on Thursday, October 2, in class. Blue books are not necessary; there is no need to bring a blue book to class on exam day. You will be asked to complete the exam on the bubble sheet (provided) and on the exam.…

    • 410 Words
    • 2 Pages
    Satisfactory Essays
  • Powerful Essays

    Exam 1

    • 968 Words
    • 5 Pages

    2) The index of openness for a nation that had $300 million in exports, $200 million in imports, and GDP of $1,000 million would be…

    • 968 Words
    • 5 Pages
    Powerful Essays
  • Good Essays

    Test Dilemma

    • 587 Words
    • 3 Pages

    Q1) How would you characterize Timberland’s exercise of its corporate power in society? Is Timberland engaging multiple stakeholders in its business operations? If so, hw?…

    • 587 Words
    • 3 Pages
    Good Essays
  • Satisfactory Essays

    Exam 1

    • 3667 Words
    • 8 Pages

    The following is an example of a constrained optimization model, Use this model to answer the…

    • 3667 Words
    • 8 Pages
    Satisfactory Essays
  • Satisfactory Essays

    Test 1 Assignment

    • 296 Words
    • 2 Pages

    2. What kind of feedback is the honest and ethical response a receiver provides to the messages of others?…

    • 296 Words
    • 2 Pages
    Satisfactory Essays
  • Powerful Essays

    5 Conclusion The concept of energy storage is discussed its types, advantage of latent heat storage over the sensible storage, the material used in latent heat storage i.e., phase changing materials (PCM). Some case studies are considered for more knowledge about the PCM concept. Its application in solar thermal storage, use in heat recovery system and numerical analysis of such systems are considered. The important drawback of PCM, its thermal conductivity seems to be a huge hurdle to be applied in common use, besides cost consideration, space availability and the…

    • 3007 Words
    • 13 Pages
    Powerful Essays
  • Good Essays

    Buckling of columns

    • 499 Words
    • 2 Pages

    These are the hand written notes for Bucking of columns/Euler's theory for Statics @ Design Analysis for Engineering, 3rd Year.…

    • 499 Words
    • 2 Pages
    Good Essays
  • Better Essays

    ADVANCED LABORATORY I FALL, 2000 ADIABATIC CALORIMETRY Reference: S,G,&N Exp. 8 in Chp. VI. Objectives: (1) Determine the integral heat of solution for finite amounts of KNO3 dissolved in water. Extrapolate the measurement to infinite dilution to obtain the differential heat of dilution. (2) Incorporate the concepts learned in your Electronics laboratory to amplify and accurately measure temperature changes associated with solution formation. (3) Use the A/D capabilities of a computer to record voltage as a function of time (e.g. a strip chart recorder). Pre-lab assignment and questions: (1) Thoroughly read the "Principles of Calorimetry" section of Chp. VI (pgs. 145 - 151). (2) After thoroughly analyzing the circuits described in Chapter XVII of your textbook, describe in detail how the operational amplifier circuit used in this experiment works. Be sure to consider both parts of the circuit, and how they work in combination. If dRT /dT is approximately 100 (ohms per °C) what do you expect the voltage change (Vo ) to be for a 10o C temperature change? Assume the following values of voltages and resistances: V1 = -50 mV; RH = 2 kΩ; RF = 100 kΩ; R2 = 1 kΩ. By analysis of the circuit, you know the relationship between RH and R1 before an addition of KNO3. (3) Prepare your pre-lab abstract as described in your syllabus. Background: (I) Heats of Solution. In this experiment the adiabatic calorimeter is used to measure the heat of solution of KNO3 in water. The generalized reaction for this solvation is: A + x S → A xS (solution at m concentration) Where: A is pure solute S is pure solvent which is mixed to form a solution m is the molal concentration.…

    • 1882 Words
    • 8 Pages
    Better Essays