Docsity
Docsity
Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity

Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan

Guidelines and tips
Guidelines and tips
Sell on Docsity
Sell on Docsity
Docsity AI
Docsity AI
Log inSign up

Prepare for your exams

Study with the several resources on Docsity

Find documents

Prepare for your exams with the study notes shared by other students like you on Docsity

Search for your university

Find the specific documents for your university's exams

Docsity AINEW

Summarize your documents, ask them questions, convert them into quizzes and concept maps

Explore questions

Clear up your doubts by reading the answers to questions asked by your fellow students

Earn points to download

Earn points by helping other students or get them with a premium plan

Share documents
download point icon20 Points

For each uploaded document

Answer questions
download point icon5 Points

For each given answer (max 1 per day)

All the ways to get free points
Get points immediately

Choose a premium plan with all the points you need

Study Opportunities

Choose your next study program

Get in touch with the best universities in the world. Search through thousands of universities and official partners

Community

Ask the community

Ask the community for help and clear up your study doubts

Free resources

Our save-the-student-ebooks!

Download our free guides on studying techniques, anxiety management strategies, and thesis advice from Docsity tutors

Solutions for Exam One - Principles of Chemistry I | CH 301, Exams of Chemistry

The University of Texas at AustinChemistry
Prof. Cynthia A. Labrake

Material Type: Exam; Professor: Labrake; Class: PRINCIPLES OF CHEMISTRY I; Subject: Chemistry; University: University of Texas - Austin; Term: Fall 2017;

Typology: Exams

2016/2017

Uploaded on 12/20/2017

koofers-user-x1k
koofers-user-x1k 🇺🇸

4

(1)

10 documents

1 / 4

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
271
version last name first name signature
LaBrake CH301 Exam 1 Fall 2017
50090
R= 0.08206 L atm/mol K
R= 62.36 L torr/mol K
R= 0.08314 L bar/mol K
R= 8.314 J/mol K
NA= 6.022 ×1023 mol−1
1 atm = 1.01325 ×105Pa
1 atm = 760 torr
1 atm = 14.7 psi
1 bar = 105Pa
1 in = 2.54 cm
1 lb = 453.6 g
1 gal = 3.785 L
ρwater = 1.00 g/mL
ρmercury = 13.6 g/mL
NOTE: Please keep your Exam copy intact (all pages still stapled). You must turn in your exam
copy plus your bubble sheet, and scratch paper.
pf3
pf4

Partial preview of the text

Download Solutions for Exam One - Principles of Chemistry I | CH 301 and more Exams Chemistry in PDF only on Docsity!

version last name first name signature

LaBrake CH301 Exam 1 Fall 2017

50090

R = 0.08206 L atm/mol K

R = 62.36 L torr/mol K

R = 0.08314 L bar/mol K

R = 8.314 J/mol K

NA = 6. 022 × 1023 mol−^1

1 atm = 1.01325 × 105 Pa

1 atm = 760 torr

1 atm = 14.7 psi

1 bar = 10^5 Pa

1 in = 2.54 cm

1 lb = 453.6 g

1 gal = 3.785 L

ρwater = 1.00 g/mL

ρmercury = 13.6 g/mL

NOTE: Please keep your Exam copy intact (all pages still stapled). You must turn in your exam

copy plus your bubble sheet, and scratch paper.

This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering.

001 10.0 points You fill a balloon with 2.25 L of air at 33◦C and 1.00 atm pressure. You then take this balloon to a higher altitude where the tem- perature is 20◦C and the pressure is 0. atm. What is the new volume of the balloon?

  1. 2.34 L correct
  2. 2.55 L
  3. 2.16 L
  4. 1.98 L

Explanation: Using the combined gas law:

P 1 V 1 T 1 =^

P 2 V 2 T 2

T 1 = 33 + 273 = 306 K

T 2 = 20 + 273 = 293 K

(1.00 atm)(2.25 L) 306 K =^

(0.920 atm)(V 2 ) 293 K

V 2 = 2.34 L

002 10.0 points You have a sample of Ne gas at a certain temperature, pressure, and volume. If you double the volume and decrease the absolute temperature (in Kelvin) by half, how will the pressure change?

  1. The new pressure will be double the orig- inal pressure.
  2. The new pressure will be one fourth the original pressure. correct
  3. The new pressure will be the same as the original pressure.
  4. The new pressure will be half the original pressure.
  5. The new pressure will be four times the original pressure. Explanation: Doubling the volume of a sample of gas will halve the pressure. Halving the absolute tem- perature of a gas will also halve the pressure. Doing both will make the pressure one fourth the original pressure.

003 10.0 points A sample of methane gas (CH 4 ) occupies a volume of 11.5 L at 1 atm and 273 K. What is the mass of the gas?

  1. 7.56 g
  2. 0.513 g
  3. 8.23 g correct
  4. 16.0 g Explanation: The sample of gas is at STP. 1 mole of a gas at STP occupies 22.4 L. x mol 11 .5 L =^

1 mol 22 .4 L x = 0.513 mol CH 4 0 .513 mol CH 4 × (^16) 1 mol CH.04 g CH 44 = 8.23 g CH 4

004 10.0 points Consider the following UNBALANCED reac- tion:

?Al 2 S 3 + ?NH 4 NO 3 → ?Al(NO 3 ) 3 + ?(NH 4 ) 2 S What is the sum of the coefficients in the balanced reaction?

  1. 15
  2. 12 correct
  3. 4
  4. 8
  5. 11 Explanation:

009 10.0 points How many grams of NO can be produced from the following reaction if 60.2 grams of NO 2 is consumed?

3NO 2 (g) + H 2 O(l) → 2HNO 3 (aq) + NO(g)

  1. 0.131 g
  2. 13.1 g correct
  3. 118 g
  4. 277 g

Explanation:

60 .2g NO 2 ×

1 mol NO 2 46 .01g NO 2

×

1 mol NO 3 mol NO 2

×

30 .01g NO 1 mol NO

= 13.1g NO

010 10.0 points Express 642 kg in terms of grams.

  1. 0.642 g
    1. 42 × 108 g
    1. 42 × 105 g correct
  4. 642 g
    1. 42 × 1011 g

Explanation:

642 kg ×

1 g 0 .001 kg

= 6. 42 × 105 g