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Xelga [282]
2 years ago
10

What is the concentration of a solution that is made by diluting 50.0 mL of a 0.40 M NaCl solution to a final volume of 1000.0 m

L?​
Chemistry
1 answer:
Novosadov [1.4K]2 years ago
3 0

Answer:

The correct answer is 0.020 M.

Explanation:

To solve this problem, we can use the equation M1V1 = M2V2, where M represents the molarity of the solution and V represents the volume of solution.  Since we are given that the original solution is 50 mL and 0.40 M, these values are V1 and M1, respectively.  The solution after dilution has a volume of 1000 mL, so this value is V2.  We are solving for the molarity after dilution, which represents M2.

If we plug in the values specified above, we get the following:

M1V1 = M2V2

(0.40 M)(50.0 mL) = (M2)(1000.0 mL)

Solving for M2, we get:

M2 = 0.020 M

Notice that our answer has 2 significant figures because 0.40 has 2 significant figures, the least of any values given in the problem.

Therefore, the answer is 0.020 M.

Hope this helps!

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2.1mol/L

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Number of moles = 0.21 moles

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Typical "hard" water contains about 2.0 x 10–3 mol of Ca2+ per liter. Calculate the maximum concentration of fluoride ion that c
malfutka [58]

Answer:

[F^-]_{max}=4x10{-3}\frac{molF^-}{L}

Explanation:

Hello,

In this case, for the described situation, we infer that calcium reacts with fluoride ions to yield insoluble calcium fluoride as shown below:

Ca^{+2}(aq)+2F^-(aq)\rightleftharpoons CaF_2(s)

Which is typically an equilibrium reaction, since calcium fluoride is able to come back to the ions. In such a way, since the maximum amount is computed via stoichiometry, we can see a 1:2 mole ratio between the ions, therefore, the required maximum amount of fluoride ions in the "hard" water (assuming no other ions) turns out:

[F^-]_{max}=2.0x10^{-3}\frac{molCa^{2+}}{L}*\frac{2molF^-}{1molCa^{2+}}  \\

[F^-]_{max}=4x10{-3}\frac{molF^-}{L}

Best regards.

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