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soldi70 [24.7K]
3 years ago
7

I have a buddy who recycles electronics, and isolates metals from the connector pins electrical boards. He isolates gold, for ex

ample, and purifies it the best he can, then sells it along with his other scrap metal. This last go around he was able to isolate 3.00 g of Au with a process he claims results in a 80.0% yield. If he is correct, what was his theoretical yield
Chemistry
1 answer:
Kisachek [45]3 years ago
3 0

Answer:

Theoretical yield = 3.75g

Explanation:

Percent yield is defined as one hundred times the ratio between actual yield and theoretical yield. The expression is:

Percent yield = Actual Yield / Theoretical Yield * 100

In the problem, your actual yield was 3-00g.

Percent yield is 80.0%.

Solving for theoretical yield:

80% = 3.00g / Theoretical yield * 100

Theoretical yield = 3.00g / 80.0% * 100

<h3>Theoretical yield = 3.75g</h3>
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What is the answers and pls show work if possible!!
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D = m / V


It even gives you the density of gold in the problem. Major hint. Once you know the volume (using V = m / D) then you can calculate the height (thickness) from the equation...


V = L x W x H

Volume = Length x Width x Height


start by converting 200.0 mg into grams

1000 mg = 1 g

200. mg x (1 g / 10^3 mg) = 0.200 g


V = m / D

V = 0.200 g / (19.32 g/cm^3)

V = 0.01035 cm^3


Convert 2.4 ft and 1 ft to cm

2.4 ft x (12 in / 1 ft) x (2.54 cm / 1 in) = 73.15 cm

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Compute the height (thickness)

V = LxWxH

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H = 4.64 x 10^-6 cm


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Knowing the atomic radius of gold, I might have asked my students for the minimum number of gold atoms in this thickness of gold. This would assume that the gold atoms are all in a row. This would give the minimum number of gold atoms.


Atomic radius gold = 174 pm

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46.4 nm x (1 m / 10^9 nm) x (10^12 pm / 1 m) x (1 Au atom / 248 pm) = 133 atoms of gold


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