11047Ag → 0.je + 110 48Cd For the nuclear reaction Starting with 200. g of Ag-110 in a container, how long will it take to decay to only 6.25 g ? The half-life of Ag-110 is 252 days. times. For each half-life we divide by 2. So, in order to reach 6.25 g we have to divide 200. g by 2 The number of half-lives is

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Chapter1: Chemical Foundations
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For the nuclear reaction: 

\[ ^{110}_{47} \text{Ag} \rightarrow ^0_{-1}\text{e} + ^{110}_{48}\text{Cd} \]

Starting with 200 g of Ag-110 in a container, how long will it take to decay to only 6.25 g? The half-life of Ag-110 is 252 days.

For each half-life, we divide by 2. So, to reach 6.25 g, we have to divide 200 g by 2 [____] times.

The number of half-lives is [____].

Each half-life is 252 days, so we multiply this by the number of half-lives. The time required is thus [____] days.

The total mass in the container is still about [____] g. Most of the mass is now [____].

Options:

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  
s. 1512  

t. 1260  
u. 1008  
v. 756  
w. 504
Transcribed Image Text:For the nuclear reaction: \[ ^{110}_{47} \text{Ag} \rightarrow ^0_{-1}\text{e} + ^{110}_{48}\text{Cd} \] Starting with 200 g of Ag-110 in a container, how long will it take to decay to only 6.25 g? The half-life of Ag-110 is 252 days. For each half-life, we divide by 2. So, to reach 6.25 g, we have to divide 200 g by 2 [____] times. The number of half-lives is [____]. Each half-life is 252 days, so we multiply this by the number of half-lives. The time required is thus [____] days. The total mass in the container is still about [____] g. Most of the mass is now [____]. Options: 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 s. 1512 t. 1260 u. 1008 v. 756 w. 504
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