Solution with a pH of 3 has 10⁻³ moles of H⁺, solution with a pH of 4 has 10⁻⁴ moles of H⁺ and solution with a pH of 5 has 10⁻⁵ moles of H⁺ (in dm³) so the solution with a pH of 3 has 10 times more H⁺ ions than the solution with a pH of 4 and 100 times more H⁺ ions than the solution with a pH of 5.
The balanced reaction for combustion is as follows ;
2C₂H₅OH + 6O₂ ---> 4CO₂ + 6H₂O
the stoichiometry of C₂H₅OH to O₂ is 2:6
that means 2 mol of C₂H₅OH reacts with 6 mol of O₂.
when 1 mol of C₂H₅OH reacts with 6/2 mol of O₂,
then 0.3020 mol of C₂H₅OH reacts with - 6/2 x 0.3020
therefore number of O₂ moles reacted = 0.91 mol
Answer:
Time = 0.929s = 0.93s (2 s.f)
Explanation:
Rate constant, k = 34.1 M^-1s^-1
Initial Concentration, [A]o = 0.100M
Time = ?
Final Concentration [A] = 0.0240M
The parameters are represented in the following equation as;
1/[A] = kt + 1/[A]o
kt = 1/[A] - 1/[A]o
kt = 1/0.0240 - 1/0.1
kt = 31.67
t = 31.67 / 34.1
t = 0.929s = 0.93s (2 s.f)
Answer:
a) First-order.
b) 0.013 min⁻¹
c) 53.3 min.
d) 0.0142M
Explanation:
Hello,
In this case, on the attached document, we can notice the corresponding plot for each possible order of reaction. Thus, we should remember that in zeroth-order we plot the concentration of the reactant (SO2Cl2 ) versus the time, in first-order the natural logarithm of the concentration of the reactant (SO2Cl2 ) versus the time and in second-order reactions the inverse of the concentration of the reactant (SO2Cl2 ) versus the time.
a) In such a way, we realize the best fit is exhibited by the first-order model which shows a straight line (R=1) which has a slope of -0.0013 and an intercept of -2.3025 (natural logarithm of 0.1 which corresponds to the initial concentration). Therefore, the reaction has a first-order kinetics.
b) Since the slope is -0.0013 (take two random values), the rate constant is 0.013 min⁻¹:

c) Half life for first-order kinetics is computed by:

d) Here, we compute the concentration via the integrated rate law once 1500 minutes have passed:

Best regards.