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References Use the References to access important values if needed for this question. A 4.88 gram sample of carbon dioxide ga

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Answer #1


Molar mass of CO2,
MM = 1*MM(C) + 2*MM(O)
= 1*12.01 + 2*16.0
= 44.01 g/mol


mass(CO2)= 4.88 g

use:
number of mol of CO2,
n = mass of CO2/molar mass of CO2
=(4.88 g)/(44.01 g/mol)
= 0.1109 mol

Given:
P = 2.78 atm
V = 852.0 mL
= (852.0/1000) L
= 0.852 L
n = 0.1109 mol

use:
P * V = n*R*T
2.78 atm * 0.852 L = 0.1109 mol* 0.08206 atm.L/mol.K * T
T = 260 K
= (260 - 273) oC
= -13 oC
Answer: -13 oC

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