New version page

UT CH 301 - Exam 1 KEY

Documents in this Course
Exam 3

Exam 3

8 pages

Exam 2

Exam 2

5 pages

Exam 1

Exam 1

5 pages

Kinetics

Kinetics

12 pages

Exam

Exam

7 pages

Quiz 1

Quiz 1

3 pages

Syllabus

Syllabus

13 pages

CH 301

CH 301

2 pages

Load more

This preview shows page 1-2 out of 6 pages.

View Full Document
View Full Document

End of preview. Want to read all 6 pages?

Upload your study docs or become a GradeBuddy member to access this document.

View Full Document
Unformatted text preview:

001version last name first name signatureBIBERDORF CH301unique: 48815, 48830Exam 1Fall 2019Sep 18, 2019Wed 5:30pm - 7pmRemember to refer to the Periodic Table handout that is separate from this exam copy.Version 001 – Exam 1 - F19 – biberdorf – (48815) 2This print-out should have 16 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.001 5.0 pointsA sample of methane (CH4) has a compress-ibility factor of 1 at a pressure of 400 atm.Estimate the compressibility factor at a pres-sure of 600 atm.1. 0.62. 1.4 correct3. 04. 1Explanation:The compressibility factor i s the ratio ofP V/nRT . When the value is greater thanone, repulsions dominate. When the value isless than one, attractions dominate. Repul-sions are more likely to dominate at pressureshigher t han the ideal pressure. Therefore, theval ue should be the only answer choice that isgreater than one.002 5.0 points2 moles of A and 2 moles of B are added to aflexible-walled (elastic) container and allowedto react at STP according to the balancedchemical equation:A(g) + 2B(g) → C(g)If the reaction continues until one of the reac-tants runs out, and the initial volume (beforethe reaction sta r ted) was 89.6 L, what is thefinal volume when the reaction is over?1. 112 L2. 22.4 L3. 67.2 L4. 89.6 L5. 44.8 L correctExplanation:Before the reaction there are 4 moles of gas inthe container (2 mol A + 2 mol B). Then 2moles of B react with 1 m o le of A, leaving 0moles of B, 1 mole of A in excess, and 1 moleof C produced by the reaction. Therefore,when the reaction is over, there are 2 molesof gas in t he container (1 mo l A + 1 mol C).Since there’s half the number of moles afterthe reaction as there were before it started,the volume after the reaction is half of whatit was before the reaction. 89.6 L / 2 = 44.8L. Alternatively, you can simpl y consider thefact t hat there are two moles of gas in thecontainer after the reaction at STP. 2 moles x22.4 L/mol at STP = 44.8 L003 5.0 points17.3 mol es of compound A and 25.8 molesof compound B are mixed at STP a nd reactaccording to the equation:2A(g) + 3B(g) → 3C(g) + 2D(g)Product C is then isolated at STP from theother components of the mix ture. What isthe volume of C?1. 9 65 L2. 2 5.95 L3. 2 5.8 L4. 6 7.2 L5. 5 78 L correctExplanation:This is a limiting reactant problem.17.3 mol A ×3 mol C2 mol A= 25.95 mol C25.8 mol B ×3 mol C3 mol B= 25.8 mol CYou have enough compound A to make 25.95moles of compound C, but only enough ofcompound B to make 25.8 moles of compo undC. Therefore, compound B is the limiting re-actant and 25.8 moles of compound C will beVersion 001 – Exam 1 - F19 – biberdorf – (48815) 3formed. At STP, 1 mole of gas has a volumeof 22.4 L.25.8 mol × 22.4Lmol= 578 L004 5.0 pointsWhat volume of HF gas is needed to produce1550 L of CF4gas according to the reaction:CCl2F2(g) + HF(g) → CF4(g) + HCl(g)Assume all the ga ses are present at the sametemperature and pressure and CCl2F2is avail-able in excess.1. 775 L2. 22.4 L3. 1550 L4. 3100 L correct5. 44.8 LExplanation:Balancing the reaction, we get:CCl2F2(g) + 2HF(g) → CF4(g) + 2HCl(g)2 moles of HF are needed for every 1 mole ofCF4produced. For ideal gases at the sametemperature and pressure, the mole ratio ofthe ga ses is equal to the volume ratio, so 2 L ofHF are needed for every 1 L of CF4produced.1550 L CH4×2 L HF1 L CH4= 3100 L HF needed005 5.0 pointsSelect the following incorrect statement aboutideal gases.1. Gases at t he same temperature have thesame average kinetic energy.2. Gases do not attract or repel each other.3. All gas particles have negligible volumewith respect to the size of the container.4. A ll collisions are inelastic; energy is trans-ferred during collisions. correctExplanation:Each statement here is an assumption wemake to treat a gas as ideal. The one that isfalse is that all colli sions are inelastic. In or-der to treat a gas as ideal, we must assumethat energy is not lost during coll isions, whichwould be a perfectly elastic collision.006 5.0 pointsWhich of the explanations below best de-scribes the relationship between pressure andvolume for an ideal gas at constant tempera-ture?1. Pressure and volume are directly pro-portional. An increase in the volume of thecontainer results in an increase in the kineticenergy of the molecules which creates higherpressure.2. Pressure and volume are inversely pro-portional. When you expand the volume ofa container the walls pull the gas moleculesoutward, resulting in a pressure drop in thecenter.3. Pressure and volume are independent ofone another since pressure is measured in at-mospheres and volume is measured in liters.4. Pressure and volume are inversely prop or-tional. If the same gas is in a larger container,the molecules will hit the wall s less frequently,resulting in lower pressure. correct5. Pressure and volume are directly propor-tional. When you compress a gas, the gasbecomes louder.Explanation:007 5.0 pointsWhich of the followi ng gases has a mass den-sity of 0.819 g/L at 780 torr a nd 308 K?1. S2Version 001 – Exam 1 - F19 – biberdorf – (48815) 42. H23. Kr4. Ne correctExplanation:Use the density to sol ve for molar mass to giveyou the identity of the gas:MM =ρRTP(0.819 g/L)(62.36 L torr/mol K)(308 K)780 torr= 20.17 g/molThis matches up most closely to neon gas.Note: this equation was derived from the idealgas law by dividing both sides by volume,mult iplying by molar mass, and then isolatingmolar mass.008 5.0 pointsIdentify the unknown gas that has a rate ofeffusion that is 1.07 times faster than oxygengas.1. Ar2. P3. Li24. Cl25. F26. CO correctExplanation:Solve for the molar mass of t he unknown gasto identify it:1.07 =pMO2/Mi1.07 =p32/MiMi= 28 g/molThis molar mass can only correspond to COof the choices.009 5.0 pointsAir has a mass density of about 1.18 grams perliter at room temperature (25◦C) and 1 atmpressure. What is the mass density of the gasin a helium balloon at the same temperatureand pressure?1. 0 .047 g/L2. 1 .18 g/L3. 1 .95 g/L4. 0 .59 g/L5. 0 .16 g/L correctExplanation:moles = n =massmolar massPV =massMMRTmass density =massV=P (MM)RTMMHe= 4 g/molmass density =P (MM)RT=(1 atm)(4 g/mol)(0.08206)(298 K)mass density = 0 .16 g/L010 5.0 pointsAn experiment shows that the vel ocity of ni-trogen gas ranges from 5 to 1 250 m/s at 300 K.In a separate experiment, 1 mole of an un-know n gas is stored in a 3 L container at300 K. After taking a few measurements, youfind


View Full Document
Loading Unlocking...
Login

Join to view Exam 1 KEY and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view Exam 1 KEY and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?