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oee [108]
3 years ago
8

Leon uses a pressure gauge to measure the air pressure in one of his car tires. The gauge shows that the pressure is 220 kilopas

cals. The temperature is 297 K, and the outdoor air is at standard pressure. If the tire contains 4.8 moles of air, what is the volume of the tire? The volume of the car tire is __ liters.
Chemistry
2 answers:
ollegr [7]3 years ago
4 0

Answer:

The volume of the car tire is 53.93 Liters.

Explanation:

Pressure of the gas in car tire =P=220kPa=2.17 atm

(1 kPa = 0.009869 atm)

Temperature of the gas in tire =T = 297 K

Moles of air in the tire = n = 4.8 moles

Volume of the gas in the tire= V

Using an ideal gas equation:

PV=nRT

V=\frac{nRT}{P}=\frac{4.8 mol\times 0.0821 atm L/mol K\times 297 K}{2.17 atm}

V = 53.93 L

The volume of the car tire is 53.93 Liters.

mamaluj [8]3 years ago
3 0

Answer:

53.8 L

Explanation:

Ideal gas law

PV=nRT

must be for volume so we arrange to V=nRT/P

V= (4.8)(8.31)(297)/220

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Depending on the reaction, we could monitor the progress towards equilibrium by observing __________.
Alex Ar [27]

Depending on the reaction, we could monitor the progress towards equilibrium by observing the concentration of the reactant and the product are equal with time.

<h3>What is equilibrium?</h3>

Equilibrium is a stage of reaction in which the rate of forwarding reaction is equal to the rate of backward reaction and equilibrium is stable at the reversible state of mode.

The concentration of reactant and product must also be equal or the same as the time then only it can be an equilibrium reaction.

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4 0
2 years ago
How many moles of argon are contained in 58 L of At at STP?
Harman [31]

Answer:

n = 2.58 mol

Explanation:

Given data:

Number of moles of argon = ?

Volume occupy = 58 L

Temperature = 273.15 K

Pressure = 1 atm

Solution:

The given problem will be solve by using general gas equation,

PV = nRT

P= Pressure

V = volume

n = number of moles

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

T = temperature in kelvin

1 atm × 58 L = n × 0.0821 atm.L/ mol.K × 273.15 K

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n = 58 atm.L / 22.43 atm.L/ mol

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8 0
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