Poisonous carbon monoxide gas is a product of incomplete burning of a carbon containing fuel. What is the volume of 1.63 moles of CO gas at 11 °C and 600 mm Hg?

Please show work!! I am having trouble with this topic ngl

48.11L

Explanation:

The equation below shows how limited supply of air results in incomplete combustion of carbon which then leads to the formation of carbon monoxide gas

2C(s) + O2(g) ---> 2CO(g)

Given that:

Volume of CO (V) = ?

Temperature of CO gas (T) = 11°C

[Convert 11°C to Kelvin by adding 273

11°C + 273 = 284K]

Pressure of CO (P) = 600 mmHg

[convert 600 mmHg to atmosphere

If 760 mmHg = 1 atm

600 mmHg = 600/760 = 0.789 atm]

Number of moles of CO gas (n) = 1.63 moles

Molar gas constant (R) is a constant with a value of 0.082 atm L K-1 mol-1

Then, apply the formula for ideal gas equation

pV = nRT

0.789 atm x V = 1.63 moles x 0.082 atm L K-1 mol-1 x 284K

0.789 atm•V = 37.96 atm•L

Divide both sides by 0.789 atm to get V

0.789 atm•V/0.789 atm = 37.96 atm•L/0.789 atm

V = 48.11L

Thus, the volume of CO gas is 48.11 litres

hey there!

boiling point is the temperature at which the vapor pressure of the liquid equals atmospheric pressure. the vapor pressure of solvent molecules is decreased when a solute is added, so a higher temperature is required to increase the number of solvent molecules in the gas phase above the liquid.

at the freezing point, the vapor pressures of the solid and liquid are equal; a lower temperature is needed to reduce the number of solvent particles above the liquid.

hope this you

tobey