Answer:
the answer you have marked is correct.
C is correct.
Answer:
Explanation:
<u>1) Equilibrium equation (given):</u>
- 2CH₂Cl₂ (g) ⇄ CH₄ (g) + CCl₄ (g)
<u>2) Write the concentration changes when some concentration, A, of CH₂Cl₂ (g) sample is introduced into an evacuated (empty) vessel:</u>
- 2CH₂Cl₂ (g) ⇄ CH₄ (g) + CCl₄ (g)
A - x x x
<u>3) Replace x with the known (found) equilibrium concentraion of CCl₄ (g) of 0.348 M</u>
- 2CH₂Cl₂ (g) ⇄ CH₄ (g) + CCl₄ (g)
A - 0.3485 0.348 0.348
<u>4) Write the equilibrium constant equation, replace the known values and solve for the unknown (A):</u>
- Kc = [ CH₄ (g) ] [ CCl₄ (g) ] / [ CH₂Cl₂ (g) ]²
- A² = 56.0 / 0.348² = 462.
Answer:
K, the rate constant = 9.73 × 10^(-1)/s
Explanation:
r = K × [A]^x × [B]^y
r = Rate = 1.07 × 10^(-1)/s
K = Rate constant
A and B = Concentration in mol/dm^-3
A = 0.44M
B = 0.11M
x = Order of reaction with respect to A = 0
y = Order of reaction with respect to B = 1
Solving, we get
r/([A]^x × [B]^y) = K
K = 1.07 × 10^(-1)/s/(0.44^0 × 0.11^1)= 0.9727
K = 0.9727
From the equation above the reacting ratio of KClO3 to O2 is 2:3 therefore the number of moles of oxygen produced is ( 4 x3)/2 = 6 moles since four moles of KClO3 was consumed
mass=relative formula mass x number of moles
That is 32g/mol x 6 moles =192grams