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garik1379 [7]
3 years ago
5

It has been estimated that 8.0 x 10^4 tons of gold (Au) have been mined. Assume gold costs $1350 per troy ounce. What is the tot

al worth of this quantity of gold? (1 troy ounce =31.103 g)
Chemistry
1 answer:
vodomira [7]3 years ago
4 0

Answer:

\$ 3.14\times 10^{12} is the total worth of this quantity of gold.

Explanation:

Estimated mass of gold =M = 8.0\times 10^4 tons

1 troy ounce = 31.103 g

1 ton = 907185 g

1 g= \frac{1}{907185} ton

31.103 g= 31.103\times \frac{1}{907185} ton=3.4285\times 10^{-5} ton

1 troy ounce =3.4285\times 10^{-5}ton

1 ton = \frac{1}{3.4285\times 10^{-5}}\text{troy ounce}

=2.9167\times 10^4 \text{troy ounce}

M = 8.0\times 10^4 tons=8.0\times 10^4\times 2.9167\times 10^4 \text{troy ounce}

M = 2.333\times 10^{9} troy ounces

Cost of 1 troy ounce gold = $1350

Cost of  2.333\times 10^{9} troy ounces of gold =

\$1350\times 2.333\times 10^{9} =\$ 3.14\times 10^{12}

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Given the following reaction:

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Reaction rate is also equal to the rate of formation of products divided by their coefficients:

r = \frac{\Delta [NO_2]}{4 \Delta t} = \frac{\Delta [O_2]}{\Delta t}

Let's find the rate of disappearance of the reactant firstly. This would be found dividing the change in molarity by the change in time:

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r_{NO_2} = \frac{\Delta [NO_2]}{\Delta t}

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-\frac{\Delta [N_2O_5]}{2 \Delta t} = \frac{\Delta [NO_2]}{4 \Delta t}

If we multiply both sides by 4, that is:

-\frac{4 \Delta [N_2O_5]}{2 \Delta t} = \frac{\Delta [NO_2]}{\Delta t}

This yields:

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