Q. 8 - An Introduction to General, Organic, and Biological Chemistry

Question

A sterling silver bracelet, which is $92.5\%$ silver by mass, has a volume of $25.6 c{m}^3$ and a density of $10.2 g/c{m}^3$. ($2.10, 4.5, 5.2, 5.4$)

Sterling silver is $92.5\%$ silver by mass.

a. What is the mass, in kilograms, of the bracelet?

b. Determine the number of protons and neutrons in each of the two stable isotopes of silver:

${}_{47}{}^{107}\mathrm{Ag}$   ${}_{47}{}^{109}\mathrm{Ag}$

c. $\mathrm{Ag}-112$ decays by beta emission. Write the balanced nuclear equation for the decay of $\mathrm{Ag}-112$.

d. A $64.0-\mu Ci$ sample of $\mathrm{Ag}-112$ decays to $8.00 \mu Ci$ in $9.3 h$. What is the half-life of $\mathrm{Ag}-112$?

Step-by-Step Solution

Answer

(a) The mass is $0.2612 kg$

(b) The number of protons and neutrons in each of the two stable isotopes of silver i.e. ${}_{47}{}^{107}\mathrm{Ag}$ and ${}_{47}{}^{109}\mathrm{Ag}$ are $47, 60$ and $47, 62$ respectively.

(c) The balanced nuclear equation for the decay is $107.868 amu$

(d) The half-life is $53.934 amu$

Part (a) Step 1: Given information

We need to find the mass of the bracelet i.e. 

Part (a) Step 2: Explanation

We know that

Mass is equal to product of density and volume

$10.2\times 25.6=261.12 g=0.2612 kg$

Part (b) Step 1: Given information

We need to find no. of protons and neutrons in each of the two stable isotopes of silver

Part (b) Step 2: Explanation

In ${}_{47}{}^{107}Ag$

No. of protons is equal to atomic no. i.e. $47$

And no. of neutrons are $107-47=60$

In ${}_{47}{}^{109}\mathrm{Ag}$

No. of protons is equal to atomic no. i.e. $47$

And no. of neutrons are $109-47=62$

Part (c) Step 1: Given information

We need to find the balanced nuclear equation for the decay 

Part (c) Step 2: Explanation

We know that

The balanced nuclear equation for the decay 

$0.5184\times 106.905+0.4816\times 108.905=55.42+52.449=107.868 amu$

Part (d) Step 1: Given information

We need to find the half-life 

Part (d) Step 2: Explanation

We know that

Half-life  is, 

$\frac{107.868}{2}=53.934 amu$