Homework Answers
Molar mass of O2 = 32 g/mol
mass(O2)= 41.7 g
use:
number of mol of O2,
n = mass of O2/molar mass of O2
=(41.7 g)/(32 g/mol)
= 1.303 mol
Molar mass of CO2,
MM = 1*MM(C) + 2*MM(O)
= 1*12.01 + 2*16.0
= 44.01 g/mol
mass(CO2)= 4.5 g
use:
number of mol of CO2,
n = mass of CO2/molar mass of CO2
=(4.5 g)/(44.01 g/mol)
= 0.1022 mol
Total mol gas = 1.303 mol + 0.1022 mol
= 1.405 mol
Given:
V = 8.1 L
n = 1.405 mol
T = 29.0 oC
= (29.0+273) K
= 302 K
use:
P * V = n*R*T
P * 8.1 L = 1.405 mol* 0.08206 atm.L/mol.K * 302 K
P = 4.30 atm
Answer: 4.30 atm
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We were unable to transcribe this imageWe were unable to transcribe this imageWe were unable to transcribe this image
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A mixture of CS2(g) and excess O2(g) is placed in a 10 L
reaction vessel at 100.0 ∘C and a pressure of 3.10 atm . A spark
causes the CS2 to ignite, burning it completely, according to the
equation:
CS2(g)+3O2(g)→CO2(g)+2SO2(g)
After reaction, the temperature returns to 100.0 ∘C, and the
mixture of product gases (CO2, SO2, and unreacted O2) is found to
have a pressure of 2.45 atm .
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and G for O2
(g) <--> 2 O (g) at 4000 K. Assume
ideal gases.
We were unable to transcribe this imageWe were unable to transcribe this imageWe were unable to transcribe this image
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