A radioactive substance decays according to the formula A(t) = A₀e⁻ᵏᵗ. If 90% of the substance remains after 2 hours, find the decay constant k.

The formula for radioactive decay is $A(t) = A_0e^{-kt}$.
We are given that after two hours ($t=2$), $0.9 A_0$ remains, so $A(2) = 0.9 A_0$.
Substitute these values into the formula:

$$0.9 A_0 = A_0e^{-k(2)}$$

$$\ln(0.9) = \ln(e^{-2k})$$

Using the logarithm property $\ln(e^x) = x$:

$$\ln(0.9) = -2k$$

$$k = \frac{\ln(0.9)}{-2}$$ $$k = -\frac{\ln(0.9)}{2}$$

$$k \approx -\frac{-0.1053605}{2}$$ $$k \approx 0.05268025$$

Rounding to two decimal places:

$$k \approx 0.05$$

The value of $k$ is:

$$k = 0.05$$