a) This part requires plotting a graph on graph paper, which cannot be done here. You would plot Temperature on the x-axis and Rate of reaction on the y-axis, then draw a smooth curve through the points. b) The optimum temperature is the temperature at which the enzyme exhibits the highest rate of reaction. From the table, the highest rate of reaction is $4.8$, which occurs at both $35^{\circ}\text{C}$ and $38^{\circ}\text{C}$. The optimum temperature for this enzyme is between $35^{\circ}\text{C}$ and $38^{\circ}\text{C}$. $$\boxed{\text{Between } 35^{\circ}\text{C} \text{ and } 38^{\circ}\text{C}}$$ c) To find the temperature at which the rate of reaction was $1.4$, we look at the table: At $20^{\circ}\text{C}$, the rate is $1.5$. At $10^{\circ}\text{C}$, the rate is $0.5$. The rate $1.4$ is between $0.5$ and $1.5$. We can estimate by interpolation. The rate of reaction $1.4$ is slightly less than $1.5$, which occurs at $20^{\circ}\text{C}$. Assuming a linear change between $10^{\circ}\text{C}$ and $20^{\circ}\text{C}$: Rate change from $0.5$ to $1.5$ is $1.0$ over a temperature change of $10^{\circ}\text{C}$. This means an increase of $0.1$ in rate per $1^{\circ}\text{C}$ increase in temperature. To reach a rate of $1.4$ from $0.5$, the rate needs to increase by $1.4 - 0.5 = 0.9$. The corresponding temperature increase would be $\frac{0.9}{0.1} = 9^{\circ}\text{C}$. So, the temperature would be $10^{\circ}\text{C} + 9^{\circ}\text{C} = 19^{\circ}\text{C}$. $$\boxed{19^{\circ}\text{C}}$$ d) i. Between $5^{\circ}\text{C}$ and $35^{\circ}\text{C}$, the rate of reaction increases with temperature. This is because as temperature rises, the kinetic energy of both enzyme and substrate molecules increases. This leads to more frequent and more energetic collisions between the enzyme's active site and the substrate molecules, increasing the likelihood of successful enzyme-substrate complex formation and thus accelerating the reaction rate. That's 2 down. 3 left today — send the next one.