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OverLord2011 [107]
3 years ago
14

How many atoms of gold are in a pure gold coin weighing 12.0g?

Chemistry
1 answer:
grigory [225]3 years ago
3 0
12.0g x 1 mol / 63.546g = 0.188839581mol 

<span>So, for every 1 mole, we have 6.022 x 10^23 of whatever we're measuring. This gives us a conversion factor of (1 mole / 6.022 x 10^23 atoms) or (6.022 x 10^23 atoms / 1 mole).
</span>
0.188839581 mol x (6.022 x 10^23 atoms) / 1 mol = 1.137191955 x 10^23 

<span>Remember from before that we are limited to 3 significant figures. Since our calculations are complete, we can now round down to: 1.14 x 10^23 </span>

<span>That should be your answer!

Hope it helps!
xo</span>
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Sonbull [250]

Answer:

2Ag⁺ (aq)  + CrO₄⁻² (aq) ⇄  Ag₂CrO₄ (s) ↓

Ksp = [2s]²  . [s] → 4s³

Explanation:

Ag₂CrO₄ → 2Ag⁺  + CrO₄⁻²

Chromate silver is a ionic salt that can be dissociated. When we have a mixture of both ions, we can produce the salt which is a precipitated.

2Ag⁺ (aq)  + CrO₄⁻² (aq) ⇄  Ag₂CrO₄ (s) ↓ Ksp

That's the expression for the precipitation equilibrium.

To determine the solubility product expression, we work with the Ksp

Ag₂CrO₄ (s)  ⇄ 2Ag⁺ (aq)  + CrO₄⁻² (aq)   Ksp

                          2 s                 s

Look the stoichiometry is 1:2, between the salt and the silver.

Ksp = [2s]²  . [s] → 4s³

 

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