Answer :
At 5 PM, approximately 395 g of substance X remained.
To find out how much substance was left at 5 PM, we can use the concept of exponential decay, which is typically represented by the equation:
[tex]\[N(t) = N_0 \times e^{-\lambda t}\][/tex]
Where:
- [tex]\(N(t)\)[/tex] is the amount of substance at time [tex]\(t\)[/tex].
- [tex]\(N_0\)[/tex] is the initial amount of substance.
- [tex]\(e\)[/tex] is the base of the natural logarithm (approximately 2.71828).
- [tex]\(\lambda\)[/tex] is the decay constant (which can be determined from the half-life of the substance).
- [tex]\(t\)[/tex] is the time elapsed.
Given that 100 g remained after 6 hours, we can set up the equation:
[tex]\[100 = 1000 \times e^{-6\lambda}\][/tex]
We need to find the value of \(\lambda\) to solve this equation. We can use the fact that the half-life of the substance is the time it takes for half of it to decay. Let's say the half-life is [tex]\(t_{1/2}\)[/tex], then we know:
[tex]\[500 = 1000 \times e^{-\lambda t_{1/2}}\][/tex]
Since [tex]\(t_{1/2}\)[/tex] is the half-life, at that time, the amount remaining is half of the initial amount. Solving this equation for [tex]\(\lambda\)[/tex], we find:
[tex]\[e^{-\lambda t_{1/2}} = \frac{1}{2}\][/tex]
[tex]\[-\lambda t_{1/2} = \ln{\left(\frac{1}{2}\right)}\][/tex]
[tex]\[-\lambda t_{1/2} = -\ln{2}\][/tex]
[tex]\[\lambda t_{1/2} = \ln{2}\][/tex]
[tex]\[\lambda = \frac{\ln{2}}{t_{1/2}}\][/tex]
Given that [tex]\(t_{1/2}\)[/tex] is the half-life, it is the time taken for the substance to reduce to half of its initial amount. Therefore, [tex]\(t_{1/2} = 6\)[/tex] hours.
[tex]\[\lambda = \frac{\ln{2}}{6}\][/tex]
Now, we can use this value of \(\lambda\) to find out how much substance was left at 5 PM, which is [tex]\(5 - 12 = 5\)[/tex] hours after 12 PM.
[tex]\[N(5) = 1000 \times e^{-\frac{\ln{2}}{6} \times 5}\][/tex]
Now, let's calculate this.
[tex]\[N(5) = 1000 \times e^{-\frac{\ln{2}}{6} \times 5}\][/tex]
[tex]\[N(5) = 1000 \times e^{-\frac{\ln{2}}{6} \times 5}\][/tex]
[tex]\[N(5) \approx 1000 \times e^{-\frac{\ln{2}}{6} \times 5}\][/tex]
[tex]\[N(5) \approx 1000 \times e^{-\ln{\left(2^{\frac{5}{6}}\right)}}\][/tex]
[tex]\[N(5) \approx 1000 \times e^{-\ln{\left(\sqrt[6]{32}\right)}}\][/tex]
[tex]\[N(5) \approx 1000 \times e^{-\ln{2.52}}\][/tex]
[tex]\[N(5) \approx 1000 \times e^{-0.927}\][/tex]
[tex]\[N(5) \approx 1000 \times 0.395\][/tex]
[tex]\[N(5) \approx 395\][/tex]
