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Andreas93 [3]
2 years ago
11

How many grams and moles of gold are present in 8.09 x 10^28 atoms of gold?

Chemistry
1 answer:
Blizzard [7]2 years ago
4 0

Answer:

Explanation:

how many grams and moles of gold are present in 8.09 x 10^28 atoms of gold?

there are 6.02 x 10^23 molecules or atoms in 1 mole

of a compound or element

8.09 x 10^28 /(6.02 x 10^23) = 1.34 x 10^5 moles of Au

1 mole of gold (see periodic table for #79) 197 g

so 1.34 x 10^5 moles of Au weigh

197 x 1.34 x 10^5 = 2.64 x 10^7 g

your little gold treasure is worth

Gold Price Per Gram $59.87 x 2.64 x 10^7=

$$$$ 1.58 x 10^9

=$ 1,580,000,000

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