once the activity of americium in a smoke det...
once the activity of americium in a smoke detector falls below 5 percent of its original activity, its no longer able to effectively detect smoke and trigger the alarm. if the half - life of americium - 241 is approximately 432.2 years, how many years will it take before the americium - 241 is unable to detect smoke?\na) 1,868 years\nb) 2 years\nc) 20 years\nd) 10 years\nquestion 9 (5 points)\nthe primary role of coenzyme a in aerobic respiration is to\na) speed up the citric acid cycle.\nb) transport electrons to fad.\nc) activate acetyl coa and other molecules in the citric acid cycle.\nd) kick off the transport electron cascade.\nquestion 10 (5 points)\nthe internal combustion engine takes advantage of the ability of the combustion reaction of gasoline to do ______ and move a piston up and down.\na) enthalpy\nb) work\nc) entropy\nd) free energy
Answer
### Question 8
# Explanation:
## Step1: Recall radioactive - decay formula
The radioactive - decay formula is $N = N_0(\frac{1}{2})^{\frac{t}{T_{1/2}}}$, where $N$ is the final amount, $N_0$ is the initial amount, $t$ is the time elapsed, and $T_{1/2}$ is the half - life. We want to find $t$ when $N = 0.05N_0$ and $T_{1/2}=432.2$ years.
## Step2: Substitute values into the formula
Substitute $N = 0.05N_0$ into $N = N_0(\frac{1}{2})^{\frac{t}{T_{1/2}}}$, we get $0.05N_0=N_0(\frac{1}{2})^{\frac{t}{432.2}}$. Since $N_0\neq0$, we can cancel out $N_0$ on both sides, resulting in $0.05 = (\frac{1}{2})^{\frac{t}{432.2}}$.
## Step3: Take the natural logarithm of both sides
$\ln(0.05)=\ln((\frac{1}{2})^{\frac{t}{432.2}})$. Using the property of logarithms $\ln(a^b)=b\ln(a)$, we have $\ln(0.05)=\frac{t}{432.2}\ln(\frac{1}{2})$.
## Step4: Solve for $t$
$t = 432.2\times\frac{\ln(0.05)}{\ln(\frac{1}{2})}$. We know that $\ln(0.05)\approx - 2.9957$ and $\ln(\frac{1}{2})\approx - 0.6931$. Then $t = 432.2\times\frac{-2.9957}{-0.6931}\approx432.2\times4.322\approx1868$ years.
# Answer:
A) 1,868 years
### Question 9
# Brief Explanations:
Coenzyme A (CoA) plays a crucial role in aerobic respiration by activating acetyl - CoA and other molecules in the citric acid cycle. It helps in the transfer of acetyl groups.
# Answer:
C) activate acetyl CoA and other molecules in the citric acid cycle
### Question 10
# Brief Explanations:
In an internal combustion engine, the combustion of gasoline is used to do work. The expanding gases from the combustion push the piston up and down, which is a form of mechanical work.
# Answer:
B) work