$A$ catalyst increases the rate of a chemical reaction by

For a certain gaseous reaction,the temperature is increased by $10\,^{\circ}C$ from $25\,^{\circ}C$ to $35\,^{\circ}C$. If the rate of the reaction doubles,what will be the value of the activation energy $(E_a)$?

The rate constant $(K')$ of one reaction is double the rate constant $(K'')$ of another reaction. What is the relationship between the corresponding activation energies of the two reactions (${E_a}'$ and ${E_a}''$)?

Consider the following statements:
$(i)$ Increase in concentration of reactant increases the rate of a zero order reaction
$(ii)$ Rate constant $k$ is equal to $A$ if $E_a = 0$
$(iii)$ Rate constant $k$ is equal to $A$ if $E_a = \infty$
$(iv)$ $\log_e k$ vs $T$ is a straight line
$(v)$ $\log_e k$ vs $1/T$ is a straight line
Which of the following statements are correct?

For a first-order reaction,the time taken to complete $10\%$ at $298\, K$ is equal to the time taken to complete $25\%$ at $308\, K$. The activation energy of the reaction is ................ $kJ/mol$.

Write the Arrhenius equation in the form $\ln \, k = -\frac{E_a}{RT} + \ln \, A$.