Question

for the nuclear reaction $^{110}_{47}ag \to ^{0}_{-1}e+^{110}_{48}cd$. starting with 200. g of ag - 110 in a container, how many grams of ag - 110 remains after 756 days? the half - life of ag - 110 is 252 days. the number of half - lives in 756 days is 1. for each half - life we divide by 2. so, we divide 200. g by 2 2 times and there are 3 g of ag - 110 remaining. the total mass in the container is still 4 g. most of the mass is now 5. a. 200. b. 110 c. 100. d. 50.0 e. 25.0 f. 12.5 g. 6.25 h. ag - 110 i. cd - 110 j. electron k. 756 l. 1 m. 2 n. 3 o. 4 p. 5 q. 6 r. 7

Explanation:

Step1: Calculate number of half - lives

Number of half - lives $n=\frac{756}{252}=3$.

Step2: Calculate remaining mass of Ag - 110

Initial mass $m_0 = 200$ g. After $n$ half - lives, remaining mass $m=m_0\times(\frac{1}{2})^n$. Substituting $m_0 = 200$ g and $n = 3$, we get $m=200\times(\frac{1}{2})^3=200\times\frac{1}{8}=25.0$ g.

Step3: Total mass in container

According to the law of conservation of mass, the total mass in the container remains the same as the initial mass of Ag - 110, which is 200 g. Since Ag - 110 decays to Cd - 110 and an electron, and the mass of the electron is negligible compared to the mass of the nuclei, most of the mass is now Cd - 110.

Answer:

1. N. 3
2. N. 3
3. E. 25.0
4. A. 200.
5. I. Cd - 110