Answer:
After 190 s the concentration of X will be 0.0396 M
Explanation:
[tex]ln\frac{A_t}{A_0} = -kt[/tex]
Where;
Xt is the concentration of X at a time t
X₀ is the initial concentration of X
k is rate constant = 1.7×10⁻² s⁻¹
t is time = 190s
ln(Xt/X₀) = -( 1.7×10⁻²)t
ln(Xt/1.0) = -( 1.7×10⁻²)190
ln(Xt/1.0) = -3.23
[tex]\frac{X_t}{1} = e^{-3.23}[/tex]
Xt = 0.0396 M
Therefore, after 190 s the concentration of X will be 0.0396 M
