Answer:
k = 0.0306 min-1
Explanation:
The table is given as;
Time, Concentration
0 1.48
5 1.27
10 0.98
15 0.84
The integrated rate law for a first order reaction is given as;
ln [A] = -kt + ln [Ao]
where;
[A] = Final Concentration
[Ao] = Initial Concentration
k = rate constant
t = time
In the table, taking the first two sets of values;
t = 5
k = ?
[Ao] = 1.48
[A] = 1.27
Inserting into the equation;
ln(1.27) = - k (5) + ln(1.48)
ln(1.27) - ln(1.48) = -5k
-0.1530 = -5k
k = -0.1530 / -5
k = 0.0306 min-1
168.96 g of carbon dioxide (CO₂)
Explanation:
The chemical reaction representing the combustion of acetylene:
2 C₂H₂ (g) + 5 O₂ (g)→ 4 CO₂ (g) + 2 H₂O (g)
number of moles = mass / molecular weight
number of moles of acetylene (C₂H₂) = 50 / 26 = 1.92 moles
Taking in account the stoichiometry of the chemical reaction, we devise the following reasoning:
if 2 moles of acetylene (C₂H₂) produces 4 moles of carbon dioxide (CO₂)
then 1.92 moles of acetylene (C₂H₂) produces X moles of carbon dioxide (CO₂)
X = (1.92 × 4) / 2 = 3.84 moles of carbon dioxide (CO₂)
mass = number of moles × molecular weight
mass of carbon dioxide (CO₂) = 3.84 × 44 = 168.96 g
Learn more about:
combustion of hydrocarbons
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