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dezoksy [38]
2 years ago
11

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

Chemistry
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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Explanation:

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3 years ago
(a) Show that the pressure exerted by a fluid P (in pascals) is given by P = hdg, where h is the column of the fluid in metres,
Komok [63]

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B) THE DEPTH OF THE LAKE IS 0.060 m

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P1 = Unknown

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2. Calculate the depth of the lake:

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Pressure = 1.633 atm = 1.633 * 101, 325 Nm^2 = 165, 463.725 Nm^2

Density =  1.02 g/cm3 = 1.02 * 10^3 kg/m^3

Acceleration = 9.8 m/s^2

So therefore, the length in metres is:

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6 0
3 years ago
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malfutka [58]

Answer:

d-  334 kJ/g.

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You can detect it from the units of the different choices.

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<em>So, the right choice is: d-  334 kJ/g.</em>

5 0
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