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Reika [66]
2 years ago
9

13. An aerosol spray can of deodorant with a volume of 0.410 L contains 3.0 g of propane gas (CH3) as propellant. What is the pr

essure in the can at 20°C? VRT 3. og festa​
Chemistry
1 answer:
qaws [65]2 years ago
8 0

Answer: The pressure in the can is 4.0 atm

Explanation:

According to ideal gas equation:

PV=nRT

P = pressure of gas = ?

V = Volume of gas = 0.410 L

n = number of moles = \frac{\text {given mass}}{\text {Molar mass}}=\frac{3.0g}{44.1g/mol}=0.068mol

R = gas constant =0.0821Latm/Kmol

T =temperature =20^0C=(20+273)K=293K

P=\frac{nRT}{V}

P=\frac0.068mol\times 0.0820 L atm/K mol\times 293K}{0.410L}=4.0atm

Thus the pressure in the can is 4.0 atm

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First convert celcius to Kelvin.

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2 years ago
How many atoms are in 13.97 liters of water vapor at STP
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<span>Let's </span>assume that water vapor has ideal gas behavior. <span>
Then we can use ideal gas formula,
PV = nRT<span>

</span><span>Where, P is the pressure of the gas (Pa), V is the volume of the gas (m³), n is the number of moles of gas (mol), R is the universal gas constant ( 8.314 J mol</span></span>⁻¹ K⁻¹) and T is temperature in Kelvin.<span>
<span>
</span>P = 1 atm = 101325 Pa (standard pressure)
V = 13.97 L = 13.97 x 10</span>⁻³ m³<span>
n = ?
R = 8.314 J mol</span>⁻¹ K⁻¹<span>
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<span>
By substitution,
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Hence, the moles of water vapor at STP is 0.624 mol.

According to the </span></span>Avogadro's constant, 1 mole of substance has 6.022 × 10²³ particles.
<span>
Hence, number of atoms in water vapor = 0.624 mol x </span>6.022 × 10²³ mol⁻¹
<span>                                                                = 3.758 x 10</span>²³<span>

</span>
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