Answer is: <span>concentration of NOCl is 3.52 M.
</span>
Balanced chemical reaction: 2NOCl(g) ⇄ 2NO(g) + Cl₂<span>(g).
Kc = 8.0.
</span>[NOCl] = 1.00 M; equilibrium concentration.
[NO] = x.
[Cl₂] = x/2; equilibrium concentration of chlorine.<span>
Kc = </span>[Cl₂] ·[NO]² / [NOCl].
8.00 = x/2 · x² / 1.
x³/2 = 8.
x = ∛16.
x = 2.52 M.
co(NOCl) = [NOCl] + x.
co(NOCl) = 1.00 M + 2.52 M.
co(NOCl) = 3.52 M; the initial concentration of NOCl.
We will use boiling point formula:
ΔT = i Kb m
when ΔT is the temperature change from the pure solvent's boiling point to the boiling point of the solution = 77.85 °C - 76.5 °C = 1.35
and Kb is the boiling point constant =5.03
and m = molality
i = vant's Hoff factor
so by substitution, we can get the molality:
1.35 = 1 * 5.03 * m
∴ m = 0.27
when molality = moles / mass Kg
0.27 = moles / 0.015Kg
∴ moles = 0.00405 moles
∴ The molar mass = mass / moles
= 2 g / 0.00405 moles
= 493.8 g /mol
Answer:
%
Explanation:
The ethanol combustion reaction is:
→
If we had the amount (x moles) of ethanol, we would calculate the oxygen moles required:
Dividing the previous equation by x:
We would need 3.30 oxygen moles per ethanol mole.
Then we apply the composition relation between O2 and N2 in the feed air:
Then calculate the oxygen moles number leaving the reactor, considering that 0.85 ethanol moles react and the stoichiometry of the reaction:
Calculate the number of moles of CO2 and water considering the same:
The total number of moles at the reactor output would be:
So, the oxygen mole fraction would be:
%
Answer: X3Y2
Explanation:
The formula is
X has a valency of 2
Y has a valency of 3
So, we interchange the valencies
Therefore, the formula is
X3Y2
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