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Alina [70]
3 years ago
15

What is the process by which sand dunes occur?

Chemistry
1 answer:
Arte-miy333 [17]3 years ago
3 0
B............................................................................
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What’s an ionic compound for magnesium and oxide?
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MgO
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Erica is working in the lab. She wants to remove the fine dust particles suspended in a sample of oil. Which method is she most
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Option a is right answer that is reverse osmosis.
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Which of the following statements best provides evidence of the big bang theory?
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Ouhgiuytytufryftrufurtfrthfrthfhtftryfrtfrthffthfrth

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Potassium thiocyanate, KSCN, is often used to detect the presence of Fe3+ ions in solution through the formation of the red Fe(H
Natalija [7]

Answer:

Ferric ions left in the solution at equilibrium is 8.0\times 10^{-6} M.

Explanation:

Moles of ferric nitrate in 0.700 L = n

Volume of the solution = V = 0.700 L

Molarity of ferric nitrate = M = 0.00150 M

n=M\times V=0.00150 M\times 0.700 L=0.00105 mol

Moles of potassium thiocyanate in 0.700 L = n'

Volume of the potassium thiocyanate solution = V' = 0.700 L

Molarity of potassium thiocyanate = M' = 0.200 M

n'=M'\times V'=0.200 M\times 0.700 L=0.140 mol

Molarity of ferric ions after mixing :

1 mol of ferric nitrate gives 1 mol of ferric ions.Then 0.00105 mol ferric nitrate will :

Moles of ferric ions = 0.00105 mol

M_1=\frac{0.00105 mol}{0.700 L+0.700 L}=0.00075 M

Molarity of  thiocyanate ions after mixing :

1 mol of potassium thiocyanate gives 1 mol of thiocyanate ions.Then 0.140 mol potassium thiocyanate will give:

Moles of thiocyanate ions = 0.140 mol

M_2=\frac{0.140 mol}{0.700 L+0.700 L}=0.1 M

Complex equation:

          Fe^{3+}+SCN^-\rightleftharpoons [Fe(SCN)]^{2+}

0.00075 M         0.1 M                        0

At equilibrium:

(0.00075 M -x)    (0.1 M-x)                  x

The formation constant of the given complex =K_f=8.9\times 10^2

K_f=\frac{[[Fe(SCN)]^{2+}]}{[[Fe^{3+}]][SCN^{-}]}

8.9\times 10^2=\frac{x}{(0.00075 M -x)\times (0.1 M-x)}

Solving for x:

x = 0.000742 M

Ferric ions left in the solution at equilibrium :

= (0.00075 M -x) = (0.00075 M - 0.000742 M)= 8.0\times 10^{-6} M

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