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2 years ago

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There are 31.1 grams (g) in 1 troy ounce (ozt). How many ozt of gold are extracted from 1 short ton of average Nevada ore

1 answer:

devlian [24]2 years ago

4 0

This problem is providing us with the mass equivalent to one troy ounce. Thus, the troy ounces of gold in one short ton of average Nevada ore is required and found to be the 0.103 otz according to the following dimensional analysis.

<h3>Dimensional analysis</h3>

In chemistry, a raft of problems do not always provide an equation in order to be solved yet dimensional analysis can be applied, so as to obtain the desired amount in the required units.

Thus, since this problem asks for try ounces in an average Nevada ore, which has 3.2 grams of gold per short ton of ore, one can solve the following setup in order to obtain the required answer in otz:

$\frac{3.2gAu}{1short-ton}*1short-ton*\frac{1otz}{31.1g} \\$

Where the short tons are cancelled out as well as the grams, in order to obtain:

$0.103 otz$

Learn more about dimensional analysis: brainly.com/question/10874167

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