ChemistryQ1–2 of 2 questions
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1
ChemistryEasyMCQGUJCET · 2010
The rate of the process doubles with every $10 \ K$ increase in temperature. When the temperature is increased from $303 \ K$ to $353 \ K$,how much will the rate of the process increase?
A
$32$
B
$16$
C
$8$
D
$4$
Solution
(A) The rate of the reaction doubles for every $10 \ K$ rise in temperature.
The total increase in temperature is $\Delta T = 353 \ K - 303 \ K = 50 \ K$.
The number of $10 \ K$ intervals is $n = \frac{50 \ K}{10 \ K} = 5$.
The rate of the process increases by a factor of $2^n$.
Therefore,the increase in the rate is $2^5 = 32$.
Thus,the correct option is $A$.
The total increase in temperature is $\Delta T = 353 \ K - 303 \ K = 50 \ K$.
The number of $10 \ K$ intervals is $n = \frac{50 \ K}{10 \ K} = 5$.
The rate of the process increases by a factor of $2^n$.
Therefore,the increase in the rate is $2^5 = 32$.
Thus,the correct option is $A$.
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2
ChemistryEasyMCQGUJCET · 2010
Which substance is reduced in a dry cell?
A
$NH_4^{+}$
B
graphite
C
$MnO_2$
D
$Zn$
Solution
(C) In a dry cell (Leclanché cell),the cathode reaction involves the reduction of manganese dioxide $(MnO_2)$.
The reaction is: $MnO_2 + NH_4^{+} + e^{-} \rightarrow MnO(OH) + NH_3$.
Here,the oxidation state of $Mn$ changes from $+4$ to $+3$,indicating reduction.
Therefore,$MnO_2$ is the substance that gets reduced.
The reaction is: $MnO_2 + NH_4^{+} + e^{-} \rightarrow MnO(OH) + NH_3$.
Here,the oxidation state of $Mn$ changes from $+4$ to $+3$,indicating reduction.
Therefore,$MnO_2$ is the substance that gets reduced.
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