D, since it can hold up to 10 electrons
Hello I a m s o m o n e I n t h e w o r l d
Answer:
P = 0.0009417 atm
Or,
P = 9.417 × 10⁻⁴ atm
Or,
P = 0.0954157 kPa
Or,
P = 0.715677 mmHg (Torr)
Explanation:
Data Given:
Moles = n = 3.2 mol
Temperature = T = 312 K
Pressure = P = ?
Volume = V = 87 m³ = 87000 L
Formula Used:
Let's assume that the gas is acting as an Ideal gas, the according to Ideal Gas Equation,
P V = n R T
where; R = Universal Gas Constant = 0.082057 atm.L.mol⁻¹.K⁻¹
Solving Equation for P,
P = n R T / V
Putting Values,
P = (3.2 mol × 0.082057 atm.L.mol⁻¹.K⁻¹ × 312 K) ÷ 87000 L
P = 0.0009417 atm
Or,
P = 9.417 × 10⁻⁴ atm
Or,
P = 0.0954157 kPa
Or,
P = 0.715677 mmHg (Torr)
molar mass of methane CH4
= C + 4 H
= 12.0 + 4 x 1.008
= 12.0 + 4.032
= 16.042g/mol
7.31 x 10^25 molecules x <u> 1 mole CH4 </u> = 121.43 moles
6.02 x 10^23 CH4 molecules
121.43 moles CH4 are present.
