What pressure in kPa is exerted by 94.60 g CO2 at 57.00 ℃ in 0.7500 L?

7859.98 KPa

Explanation:

Step 1:

Data obtained from the question. This includes the following:

Mass of CO2 = 94.60g

Temperature (T) = 57℃

Volume (V) = 0.75 L

Pressure (P) =?

Step 2:

Determination of the number of mole of CO2.

Mass of CO2 = 94.60g

Molar Mass of CO2 = 12 + (2x16) = 12 + 32 = 44g/mol

Number of mole of CO2 =?

Number of mole = Mass/Molar Mass

Number of mole of CO2 = 94.60/44

Number of mole of CO2 = 2.15 moles

Step 3:

Conversion of celsius temperature to Kelvin temperature.

Temperature (Kelvin) = temperature (celsius) + 273

Temperature = 57℃ + 273 = 330K

Step 4:

Determination of the pressure. This is illustrated below:

The pressure can be obtained by using the ideal gas equation as shown below:

PV = nRT

Temperature (T) = 330K

Volume (V) = 0.75 L

Number of mole (n) = 2.15 moles

Gas constant (R) = 0.082atm.L/Kmol

Pressure (P) =?

PV = nRT

0.75 x P = 2.15 x 0.082 x 330

Divide both side by 0.75

P = (2.15 x 0.082 x 330)/0.75

P = 77.572 atm

Step 5:

Conversion of the pressure in atm to KPa. This is illustrated below:

1 atm = 101.325 KPa

Therefore, 77.572 atm = 77.572x101.325 = 7859.98 KPa

i think it is true since you can test the answer and can use observations to complete it.

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

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