Answer:
[tex]Q (t) = \frac{90}{7t +2 }[/tex]
[tex]Q(2) = 5.63 \ g[/tex]
Step-by-step explanation:
Let Q(t) be the first substance at time t. We know that the rate of change is directly proportional to [tex]Q^2[/tex].
Hence we can write it as:
[tex]\frac{dQ}{dt} \propto Q^2\\\\\frac{dQ}{dt} = kQ^2[/tex]
Step 1: Integrating the function to form an equation:
[tex]\frac{dQ}{dt} = kQ^2\\\\\frac{dQ}{Q^2} = k \ dt\\\\\int \frac{dQ}{Q^2} = \int k \ dt\\\\- \frac{1}{Q} = kt + C[/tex] ----------------------------------------------- equation (1)
Step 2: Substitute the value Q(0) = 45 in equation 1 to find value of C.
[tex]-\frac{1}{Q} = kt + C\\\\-\frac{1}{45} = k(0) + C\\\\-\frac{1}{45} = C\\\\\\-\frac{1}{Q} = kt - \frac{1}{45}\\[/tex]---------------------------------------------- equation (2)
Step 3: Substitute the value Q(1) = 10 in equation 2 to find the value of k.
[tex]-\frac{1}{Q} = kt - \frac{1}{45}\\\\-\frac{1}{10} = k(1) - \frac{1}{45}\\\\-\frac{1}{10} - \frac{1}{45} = k \\\\\frac{2-9}{90} = k\\\\-\frac{7}{90} = k[/tex]
Step 4: Form the equation and simplify it.
[tex]-\frac{1}{Q} = -\frac{7}{90} t - \frac{1}{45}\\ \\\frac{1}{Q} = \frac{7}{90}t + \frac{2}{90}\\\\\frac{1}{Q} = \frac{7t + 2}{90}\\\\\\\bullet \ cross - multiply \ the \ equation \\\\Q = \frac{90}{7t + 2}[/tex]
Step 5: To find the amount of substance after 2 hours, we substitute the value of t with 2.
[tex]Q = \frac{90}{7t + 2}\\\\Q = \frac{90}{7(2) + 2}\\\\Q = \frac{90}{14+ 2}\\\\Q = \frac{90}{16}\\\\Q = 5.625 \approx 5.63[/tex]
The amount of substance after 2 hours is 5.63 grams.
