The volume at STP : 9.856 L
<h3>Further explanation</h3>
Given
Mass of ethane : 13.21 g
Required
The volume at STP
Solution
Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). At STP, Vm is 22.4 liters / mol.
mol ethane(C2H6) :
= mass : molar mass
= 13.21 g : 30 g/mol
= 0.44
Volume at STP :
= 0.44 x 22.4 L
= 9.856 L
We use the gas law named Charle's law for the calculation of the second temperature. The law states that,
V₁T₂ = V₂T₁
Substituting the known values,
(0.456 L)(65 + 273.15) = (3.4 L)(T₁)
T₁ = 45.33 K
According to Grahams law the rate of effusion of a gas is inversely proportional to the square root of it's molecular weight. The rate of diffusion is the measure of rate at which two gases mix. From this law we can say that for the two gases carbon monoxide and carbon dioxide, the rate of effusion of carbon monoxide is greater than that of carbon dioxide, this is because carbon monoxide is lighter (28 g) compared to carbon dioxide (44 g).
How does the law of conservation of mass apply to this reaction: C2H4 + O2 → H2O + CO2?
B ia the correct one .
Hydrogen only uses the first energy shell, which holds 2 electrons, not 8.