What is the composition, in atom percent, of an alloy that contains a) 44.5 lbm of silver, b) 84.7 lbm of gold, and c) 7.3 lbm of Cu? The atomic weights for silver, gold, and copper are, respectively, 107.87, 196.97, and 63.55 g/mol.

The atom percent of silver, gold and copper in the alloy is 43.08 %, 44.93 % and 11.99 % respectively

Explanation:

To convert the given masses into grams, we use the conversion factor:

1 lb = 453.6 grams

To calculate the number of moles, we use the equation:

   ......(1)

To calculate the atom percent of substance in sample, we use the equation:

    ......(2)

where,

= Avogadro's number

Moles of Silver:

Mass of silver = 44.5 lb = 20185.2 grams

We know that:

Molar mass of silver = 107.87 g/mol

Putting values in equation 1, we get:

Moles of Gold:

Mass of gold = 84.7 lb = 38419.9 grams

We know that:

Molar mass of gold = 196.97 g/mol

Putting values in equation 1, we get:

Moles of Copper:

Mass of copper = 7.3 lb = 3311.3 grams

We know that:

Molar mass of copper = 63.55 g/mol

Putting values in equation 1, we get:

Total moles of the sample =

For Silver:

Moles of silver = 187.12 moles

Total moles = [187.12 + 195.05 + 52.10] = 434.27 moles

Putting values in equation 2, we get:

For Gold:

Moles of gold = 195.05 moles

Total moles = [187.12 + 195.05 + 52.10] = 434.27 moles

Putting values in equation 2, we get:

For Copper:

Moles of copper = 52.10 moles

Total moles = [187.12 + 195.05 + 52.10] = 434.27 moles

Putting values in equation 2, we get:

Hence, the atom percent of silver, gold and copper in the alloy is 43.08 %, 44.93 % and 11.99 % respectively

First, make all of your units the same.
420 mmHg x 1 atm/760 mmHg = 0.553 atm

Then use one of the gas laws to solve.
P1V1=P2V2

Plug in the known values and solve for the one remaining.
(2.15atm)(2.00L)=(0.553atm)(V2)
4.3=0.5533V2
V2=4.3/0.5533
V2=7.77 Liters