Solve this question in the way that I can explain to my lecturer

For a first-order reaction, the half-life ($t_{1/2}$) is the time required for the concentration of a reactant to decrease to half of its initial value. This is a constant value for a given first-order reaction and is related to the rate constant ($k$) by the following equation:

$$t_{1/2} = \frac{\ln(2)}{k}$$

Here, $\ln(2)$ is the natural logarithm of 2, which is approximately 0.6931.
We are given the half-life $t_{1/2} = 1670 \text{ s}$. To find the rate constant $k$, we rearrange the formula:

$$k = \frac{\ln(2)}{t_{1/2}}$$

Substitute the given value:

$$k = \frac{0.6931}{1670 \text{ s}}$$ $$k \approx 0.0004150 \text{ s}^{-1}$$

This value of $k$ represents how quickly the reaction proceeds.

To find the amount of reactant remaining after a certain time, we use the integrated rate law for a first-order reaction. This equation relates the concentration of the reactant at time $t$ ($[A]_t$) to its initial concentration ($[A]_0$), the rate constant ($k$), and the time ($t$):

$$\ln\left(\frac{[A]_t}{[A]_0}\right) = -kt$$

The term $\frac{[A]_t}{[A]_0}$ represents the fraction of the reactant remaining at time $t$. To solve for this fraction, we need to remove the natural logarithm ($\ln$). We do this by taking the exponential (base $e$) of both sides of the equation:

$$e^{\ln\left(\frac{[A]_t}{[A]_0}\right)} = e^{-kt}$$

This simplifies to:

$$\frac{[A]_t}{[A]_0} = e^{-kt}$$

Now, we substitute the calculated rate constant $k = 0.0004150 \text{ s}^{-1}$ and the given time $t = 4710 \text{ s}$:

$$\frac{[A]_t}{[A]_0} = e^{-(0.0004150 \text{ s}^{-1} \times 4710 \text{ s})}$$

First, calculate the product in the exponent:

$$0.0004150 \times 4710 = 1.9545$$

So, the equation becomes:

$$\frac{[A]_t}{[A]_0} = e^{-1.9545}$$

Finally, calculate the value:

$$\frac{[A]_t}{[A]_0} \approx 0.1416$$

This means that after $4710 \text{ s}$, approximately 0.1416 (or $14.16\%$) of the initial diethyl ether remains.

Note: The value 0.750 was not used in these calculations. The constant used in the half-life formula is $\ln(2) \approx 0.6931$.

The fraction of diethyl ether remaining after $4710 \text{ s}$ is approximately $\boxed{\text{0.1416}}$.