The decomposition of dimethylether at 504°C is first order with half-life of 1570 seconds. What fraction of initial amount of dimethylether remain after 1570 seconds?

For a first-order reaction, the half-life ($t_{1/2}$) is related to the rate constant ($k$) by the formula:

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

Rearranging to solve for $k$:

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

Substitute the given half-life:

$$k = \frac{\ln(2)}{1570 \text{ s}}$$ $$k \approx \frac{0.6931}{1570 \text{ s}} \approx 0.0004415 \text{ s}^{-1}$$

The integrated rate law for a first-order reaction is:

$$\ln([A]_t) - \ln([A]_0) = -kt$$

This can be rewritten as:

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

To find the fraction remaining, $\frac{[A]_t}{[A]_0}$, we exponentiate both sides:

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

Substitute the calculated rate constant ($k$) and the given time ($t = 710 \text{ s}$):

$$\frac{[A]_t}{[A]_0} = e^{-(0.0004415 \text{ s}^{-1})(710 \text{ s})}$$ $$\frac{[A]_t}{[A]_0} = e^{-0.313465}$$ $$\frac{[A]_t}{[A]_0} \approx \boxed{\textbf{0.731}}$$