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Assignment 10 Problems with Solution - Chemistry I |, Assignments of Chemistry

Material Type: Assignment; Class: Chemistry 1 - Introduction; Subject: Chemistry; University: Albright College; Term: Forever 1989;

Typology: Assignments

Pre 2010

Uploaded on 12/04/2009

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Chemistry 105A - Chapter 10 Problems - DUE IN CLASS 12/2/2009
Show your work and complete the following problems. Work independently but take advantage
of your notes and textbook as necessary just no other human help.
1. A sample of gas (24.2 g) initially at 4.00 atm was compressed from 8.00 L to 2.00 L at
constant temperature. Determine the pressure after the compression.
1 1 2 2
1 1
2
2
P V = P V
4.00 atm 8.00 L
P V
P = = 16 atm
V 2.00 L
2. A sample of H2 gas (12.28 g) occupies 100.0 L at 400.0 K and 2.00 atm. A sample
weighing 9.49 g occupies what volume at 353 K and 2.00 atm.
1 1 2 2
1 2
2 2
2 1
1 1
P nT
PV = nRT =
R V
n T n T
=
V V
1 mole
9.49 g × 353.0 K
2 g
n T
V = V = × 100.0 L
n T 1 mole
12.28 g × 400.0 K
2 g
= 68.2 L
You would not have to convert to mo
L atm
K mole
les since this cancels in the numerator and denominator.
OR with the use of the ideal gas constant:
1 mole
9.49 g × 0.08206 353.0 K
2.016 g
nRT
V = = 68.2 L
P 2 atm
3. Automobile air bags use the decomposition of sodium azide as their source of gas for
rapid inflation: 2NaN3 (s) 2Na(s) + 3N2(g)
Answers Chapter 10 problems 1
pf2

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Chemistry 105A - Chapter 10 Problems - DUE IN CLASS 12/2/

Show your work and complete the following problems. Work independently but take advantage

of your notes and textbook as necessary just no other human help.

  1. A sample of gas (24.2 g) initially at 4.00 atm was compressed from 8.00 L to 2.00 L at

constant temperature. Determine the pressure after the compression.

1 1 2 2

1 1

2

2

P V = P V

P V^ 4.00 atm 8.00 L

P = = 16 atm

V 2.00 L

  1. A sample of H 2 gas (12.28 g) occupies 100.0 L at 400.0 K and 2.00 atm. A sample

weighing 9.49 g occupies what volume at 353 K and 2.00 atm.

1 1 2 2

1 2

2 2

2 1

1 1

P nT

PV = nRT =

R V

n T n T

=

V V

1 mole

9.49 g × 353.0 K

n T^ 2 g

V = V = × 100.0 L

n T (^) 1 mole

12.28 g × 400.0 K

2 g

= 68.2 L

You would not have to convert to mo

L atm

K mole

les since this cancels in the numerator and denominator.

OR with the use of the ideal gas constant:

1 mole

9.49 g × 0.08206 353.0 K

nRT 2.016 g

V = = 68.2 L

P 2 atm

  1. Automobile air bags use the decomposition of sodium azide as their source of gas for

rapid inflation: 2NaN 3 (s)  2Na(s) + 3N 2 (g)

Answers Chapter 10 problems 1

What mass (g) of NaN 3 is required to provide 40.0 L of N 2 at 25C and 763 torr?

l atm^2

K mole

3

2 3

2

3 3

3

PV

n =

RT

1 atm

763 torr 40.0 L)

760 torr = = 1.64 mole N

0.0821 298.15 K

2 mole NaN

1.64 mole N = 1.09 mole NaN

3 mole N

65 g

1.09 mole NaN 71.1 g NaN

1 mole NaN

  1. A gas mixture of Ne and Ar has a total pressure of 4.00 atm and contains 16 mol of gas.

If the partial pressure of Ne is 2.75 atm, how many moles of Ar are in the mixture?

t

Ne

Ne

Ne

t

Ar Ar

t

t Ne Ar Ar

Ar

Ar

Ar

t

P = 4.00 atm

P = 2.75 atm

P

P

P

P

0.3125 16 mole = 5 mole Ar

OR

P = 4.00 atm = P + P 2.75 + P

P = 1.25 atm

P 1.25 atm

Χ = = 0.

P 4 atm

0.3125 16 mole =

5 mole Ar

Answers Chapter 10 problems 2