Predicting the graph of the given reaction 
I have no idea on how to answer this question. Please help.
 A: The concentration must be exponential. Suppose $C(t)$ is the concentration at time $t$ (min). $C(t)$ is of the form
$$
C(t) = A2^{-kt}
$$
where $A$ and $k$ are constants you need to find. Using known data points, you can find that $A = 0.4$ and $k = \frac 1{120}$.
The slope of $C(t)$ at $t = 0$ is $C'(0)$, so compute $C'(t)$ first:
\begin{align}
C'(t) & = -kA 2^{-kt} \log 2 \\
C'(0) & = -\frac 1{120} \cdot 0.4 \cdot \log 2 \approx -0.00231.
\end{align}
Therefore, the tangent line at $0$ would intersect the $x$-axis at
\begin{align}
-\frac{C(0)}{C'(0)} & \approx \frac{0.4}{0.00231} \approx 173.12
\end{align}
A: I'm unsure why this was migrated from chemistry.
The rates of chemical reactions are governed by rate equations, for example (Where $\mathrm{A}$ is the reactant, $k$ is the rate constant, and $n$ is the reaction order which usually must be experimentally determined:
$$rate=-\dfrac{dA}{dt}=k[\mathrm{A}]^n$$
We can solve the differential equation to get a generalized equation for $\mathrm{A}$ as a function of time.
$$-\dfrac{dA}{[\mathrm{A}]^n}=kdt$$
$$-\int_{[\mathrm{A}]_0}^{[\mathrm{A}]_t}\dfrac{dA}{[\mathrm{A}]^n}=\int_0^tkdt=kt$$
If $n=0$ (zeroth order), then the left integral evaluates to:
$$-\int_{[\mathrm{A}]_0}^{[\mathrm{A}]_t}\dfrac{dA}{[\mathrm{A}]^0}=[\mathrm{A}]_0-[\mathrm{A}]_t=kt$$
This equation is linear. Since your graph is not linear, your data is not zerothorder.
If $n=1$ (first order):
$$-\int_{[\mathrm{A}]_0}^{[\mathrm{A}]_t}\dfrac{dA}{[\mathrm{A}]}=\ln{[\mathrm{A}]_t}-\ln{[\mathrm{A}]_0}=\ln{\dfrac{[\mathrm{A}]_t}{[\mathrm{A}]_0}}=-kt$$
Solving for $[\mathrm{A}]_t$ gives you:
$$\dfrac{[\mathrm{A}]_t}{[\mathrm{A}]_0}=e^{-kt}$$
$${[\mathrm{A}]_t}=[\mathrm{A}]_0e^{-kt}$$
Your graph looks like an exponential decay, so you can find the slope of the tangent line at time $t=0$ by differentiating, except you need to know $k$ to evaluate at $t=0$.
$$\dfrac{d}{dt}[\mathrm{A}]_t=\dfrac{d}{dt}[\mathrm{A}]_0e^{-kt}=-k[\mathrm{A}]_0e^{-kt}$$
However, since the half-life of a first order reaction is given by 
$$t_{1/2}=\dfrac{\ln{(2)}}{k}$$
you can solve for $k$ since the half-life is marked on the graph. 
