To calculate the density of an ideal gas, we can use the ideal gas law, which is PV = nRT. However, we need to rearrange it to solve for density (ρ). Density is defined as mass (m) over volume (V), and the number of moles (n) can be expressed as mass (m) over molar mass (M). So, we can rewrite the equation as ρ = m/V = PM/RT.

Given:
- P (pressure) = 1 atm
- R (ideal gas constant) = 0.0821 L atm / (K mol)
- T (temperature) = 30°C = 303.15 K (remember to convert from Celsius to Kelvin)
- M (molar mass of CO) = 28.01 g/mol

Substitute these values into the equation:

ρ = (1 atm * 28.01 g/mol) / (0.0821 L atm / (K mol) * 303.15 K)

Solving this gives ρ ≈ 1.145 g/L. However, we want our answer to three significant figures, so we round to ρ = 1.15 g/L.

Question

To calculate the density of an ideal gas, we can use the ideal gas law, which is PV = nRT. However, we need to rearrange it to solve for density (ρ). Density is defined as mass (m) over volume (V), and the number of moles (n) can be expressed as mass (m) over molar mass (M). So, we can rewrite the equation as ρ = m/V = PM/RT.

Given:
- P (pressure) = 1 atm
- R (ideal gas constant) = 0.0821 L atm / (K mol)
- T (temperature) = 30°C = 303.15 K (remember to convert from Celsius to Kelvin)
- M (molar mass of CO) = 28.01 g/mol

Substitute these values into the equation:

ρ = (1 atm * 28.01 g/mol) / (0.0821 L atm / (K mol) * 303.15 K)

Solving this gives ρ ≈ 1.145 g/L. However, we want our answer to three significant figures, so we round to ρ = 1.15 g/L.

Knowee AI · Accepted Answer

To calculate the density of an ideal gas, we can use the ideal gas law, which is PV = nRT. However, we need to rearrange it to solve for density (ρ). Density is defined as mass (m) over volume (V), and the number of moles (n) can be expressed as mass (m) over molar mass (M). So, we can rewrite the equation as ρ = m/V = PM/RT.

Given:
- P (pressure) = 1 atm
- R (ideal gas constant) = 0.0821 L atm / (K mol)
- T (temperature) = 30°C = 303.15 K (remember to convert from Celsius to Kelvin)
- M (molar mass of CO) = 28.01 g/mol

Substitute these values into the equation:

ρ = (1 atm * 28.01 g/mol) / (0.0821 L atm / (K mol) * 303.15 K)

Solving this gives ρ ≈ 1.145 g/L. However, we want our answer to three significant figures, so we round to ρ = 1.15 g/L.

Calculate to three significant digits the density of carbon monoxide gas at exactly 30°C and exactly 1atm. You can assume carbon monoxide gas behaves as an ideal gas under these conditions.

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