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3 years ago

Please Help!!

Chemistry

1 answer:

Oksi-84 [34.3K]3 years ago

4 0

Answer:

2K + Br2 --> 2KBr is the answer

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What is the volume of 33.25g of butane gas at 293 C and 10.934 kPa?

PSYCHO15rus [73]

Answer:

V = 240.79 L

Explanation:

Given data:

Volume of butane = ?

Temperature = 293°C

Pressure = 10.934 Kpa

Mass of butane = 33.25 g

Solution:

Number of moles of butane:

Number of moles = mass/ molar mass

Number of moles = 33.25 g/ 58.12 g/mol

Number of mole s= 0.57 mol

Now we will convert the temperature and pressure units.

293 +273 = 566 K

Pressure = 10.934/101 = 0.11 atm

Volume of butane:

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K

T = temperature in kelvin

V = nRT/P

V = 0.57 mol × 0.0821 atm.L/ mol.K ×566 K / 0.11 atm

V = 26.49 L/0.11

V = 240.79 L

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