<span>The rate of effusion of a gas is inversely proportional to the square root of the molecular weight of the species. Now there will be differences among isotopomers but neglecting these and taking the avg mol wt of N2 = 28 and Xe = 132;
Rate(N2)/Rate(Xe) = sqrt (132/28) = 2.17</span>
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Answer:
3.9mole/liter
Explanation:
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Answer:
We will use the combined gas equation.
The final volume V₂=931.74 mL
Explanation:
The combined gas equation shows the interrelationship between the volume, the pressure and the absolute temperature of a fixed mass of an ideal gas,
P₁V₁/T₁=P₂V₂/T₂
Because we are looking for the volume after the initial conditions havre been varied, we will call this volume, volume 2 ( V₂)
Let us make it the subject of the equation.
V₂=(P₁V₁T₂)T₁P₂
P₁=1.50 ATM
P₂=1.20 ATM
V₁=800 mL
V₂=?
T₁=(20+273)K=293 K
T₂=0+273 K=273 K
V₂=(1.50 ×800×273)/(293×1.2)
V₂=931.74 mL
Balanced chemical reaction happening here is:
3Mg(s) + N₂(g) → Mg₃N₂(s)
<u>moles of product formed from each reactant:</u>
2.0 mol of N2 (g) x <u> 1 mol Mg₃N₂ </u> = <u>2 mol Mg₃N₂</u>
1 mol N2
and
8.0 mol of Mg(s) x <u> 1 mol Mg₃N₂ </u> = 2.67 mol Mg₃N₂
3 mol Mg
Since N2 is giving the least amount of product(Mg₃N₂) ie. 2 mol Mg₃N₂
N2 is the limiting reactant here and Mg is excess reactant.
Hence mole of product formed here is 2 mol Mg₃N₂
molar mass of Mg₃N₂
= 3 Mg + 2 N
= 101g/mol
mass of product(Mg₃N₂) formed
= moles x Molar mass
= 2 x 101
= 202g Mg₃N₂
<u>202g of product are formed from 2.0 mol of N2(g) and 8.0 mol of Mg(s).</u>
<u> </u> The following are indicators of chemical changes:
Change in Temperature
Change in Color
Formation of a Precipitate
