Answer:
k = 1.3 x 10⁻³ s⁻¹
Explanation:
For a first order reaction the integrated rate law is
Ln [A]t/[A]₀ = - kt
where [A] are the concentrations of acetaldehyde in this case, t is the time and k is the rate constant.
We are given the half life for the concentration of acetaldehyde to fall to one half its original value, thus
Ln [A]t/[A]₀ = Ln 1/2[A]₀/[A]₀= Ln 1/2 = - kt
- 0.693 = - k(530s) ⇒ k = 1.3 x 10⁻³ s⁻¹
Answer:
-608KJ/mol
Explanation:
3 C2H2(g) -> C6H6(g)
ΔHrxn = ΔHproduct - ΔHreactant
ΔHrxn= ΔHC6H6 - 3ΔHC2H2
ΔHrxn = 83 - 3(230)
ΔHrxn = -608
In order to answer this question, the units of volume must be consistent. In this problem, we decide the unit m3 to be uniform. Option A is equal to 12 m3, option b is equal to 1.2x10^8/100^3 or 120 m3. Option C is 2.0 x10^4/ 10^3 or 20 m3. Option D is 1.2x10^8/ 1000^3 or 0.12 m3. The greatest volume is option b. 120 m3.
The answer is: the mass of carbon is 420.6 grams.
m(C₈H₁₈) = 500 g; mass of octane.
M(C₈H₁₈) = 114.22 g/mol; molar mass of octane.
n(C₈H₁₈) = m(C₈H₁₈) ÷ M(C₈H₁₈).
n(C₈H₁₈) = 500 g ÷ 114.22 g/mol.
n(C₈H₁₈) = 4.38 mol; amount of octane.
In one molecule of octane, there are eight carbon atoms:
n(C) = 8 · n(C₈H₁₈).
n(C) = 8 · 4.38 mol.
n(C) = 35.02 mol; amount of carbon.
m(C) = 35.02 mol · 12.01 g/mol.
m(C) = 420.6 g; mass of carbone.