How many hours will it take for the concentration of methyl isonitrile to drop to 14.0 %% of its initial value?
1 answer:
This is an incomplete question, here is a complete question.
The rearrangement of methyl isonitrile (CH₃NC) to acetonitrile (CH₃NC) is a first-order reaction and has a rate constant of 5.11 × 10⁻⁵ s⁻¹ at 472 K. If the initial concentration of CH₃NC is 3.00 × 10⁻² M :
How many hours will it take for the concentration of methyl isonitrile to drop to 14.0 % of its initial value?
Answer : The time taken will be, 10.7 hours
Explanation :
Expression for rate law for first order kinetics is given by:
where,
k = rate constant =
t = time passed by the sample = ?
a = let initial amount of the reactant = 100
a - x = amount left after decay process = 14 % of 100 = 14
Now put all the given values in above equation, we get
Therefore, the time taken will be, 10.7 hours
You might be interested in
Answer:
B. The student chose the correct tile, but needs to flip the tile to make the units cancel
Explanation:
Based on the reaction:
2AgNO₃(aq) + Cu(s) → 2Ag(s) + Cu(NO₃)₂ (aq)
<em>2 moles of AgNO₃ react per mole of Cu producing 2 moles of Ag and 1 mole of Cu(NO₃)₂</em>
Thus, if you want to produce 6.75moles of Cu(NO₃)₂ you need:
= 13.50 moles of AgNO₃ are needed
Thus, if you analize the tile shown by the student:
<em>B. The student chose the correct tile, but needs to flip the tile to make the units cancel</em>
The distance covered will be:
The correct answer is D).
The correct answer is D).