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Q.
In a zero order reaction for every $10^{\circ}$C rise of temperature, the rate is doubled. If the temperature is increased from $10^{\circ}$C to $100^{\circ}$C, the rate of the reaction will become .
For 10$^{\circ}$ rise in temperature, n = 1
so rate = $2^n =2^1 =2$
When temperature is increased from 10$^{\circ}$C to 100$^{\circ}$C, change in temperature
$ =100-10 =$90$^{\circ}$C
i.e. $ n=9$
So, rate = $2^9$ = 512 times
Alternate method with every 10$^{\circ}$ rise in temperature, rate becomes double,
so $\frac{r'}{r}=2^{\big(\frac{100-10}{10}\big)} =2^9=$ 512 times.