Answer:
40% of the ammonia will take 4.97x10^-5 s to react.
Explanation:
The rate is equal to:
R = k*[NH3]*[HOCl] = 5.1x10^6 * [NH3] * 2x10^-3 = 10200 s^-1 * [NH3]
R = k´ * [NH3]
k´ = 10200 s^-1
Because k´ is the psuedo first-order rate constant, we have the following:
b/(b-x) = 100/(100-40) ; 40% ammonia reacts
b/(b-x) = 1.67
log(b/(b-x)) = log(1.67)
log(b/(b-x)) = 0.22
the time will equal to:
t = (2.303/k) * log(b/(b-x)) = (2.303/10200) * (0.22) = 4.97x10^-5 s
Answer: False; it contains particles, but that doesn't mean it's a particle itself.
<em>Hope this help. </em>
1 mole of CH4 results to 2 moles of H2O.
If 8.5 liters are to be produced, then (1CH4/2H20)*(8.5L h2o) = 4.25L CH4
Answer:
1.758 x 10⁻¹¹.
Explanation:
∵ pH = - log[H⁺].
∴ 5.47 = -log[H⁺].
log[H⁺] = -5.47.
∴ [H⁺] = 3.388 x 10⁻⁶ M.
∵ [H⁺] = √(Ka.C)
∴ [H⁺]² = Ka.C
<em>∴ Ka = [H⁺]²/C </em>= (3.388 x 10⁻⁶)²(0.653) = <em>1.758 x 10⁻¹¹.</em>
