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zysi [14]
3 years ago
7

If an aqueous solution of ___________ is added to a mixture of barium nitrate and silver nitrate, the barium ion will precipitat

e, but the silver ion would remain in solution.
Chemistry
1 answer:
kkurt [141]3 years ago
4 0

Answer:

sodium fluoride

Explanation:

When we add sodium fluoride to a solution of a given mixture, we get barium fluoride from Ba, which would be an insoluble salt, and silver fluoride from Ag, which would be a soluble salt. 

The solubility rule will be used to determine the barium salt that forms as a precipitate and leaves Ag+ salt in the solution.

From the following equations, we will see that the precipitate is formed in Ba but Ag remains dissolved in the solution.

Ba(NO_3)_2{(aq)} + 2NaF_{(aq)} \to BaF_{2(s)} +2NaNo_{3(aq)}

AgNo_{3(aq)} +NaF_{(aq)} \to AgF_{(aq)} + NaNO_{(aq)}

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The neutralization of a 0.455 g sample of recrystallized aspirin required 38.6 mL of 0.100 M NaOH solution. An additional 25.0 m
Schach [20]

Answer:

E) Two of the above statements are true.

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The options are:

A) Before the solution is titrated with HCl it is pink and when the color changes from pink to colorless, the moles of H*(aq) equals the moles of OH"(aq) used in the hydrolysis of the neutralized aspirin. <em>TRUE. </em>Before the solution is titrated, there is an excess of NaOH (Basic solution, phenolphtalein is pink). Then, at equivalence point, after the addition of HCl, the pH is acidic and phenolphtalein is colorless.

B) Before the solution is titrated with HCl it is colorless and when the color changes from colorless to pink, the moles of H*(aq) equals the excess moles of OH(aq) added. <em>FALSE. </em>As was explained, before the titration, the solution is pink.

C) 25.0 mL of 0.100 M NaOH was added to the sample to hydrolyze the neutralized aspirin in the solution. The titration with HCl allows us to determine the moles of excess OH(aq) added. Once we determine the moles of excess OH(aq), we can determine moles of OH"(aq) used in the hydrolysis of the neutralized aspirin, which is equal to the moles of aspirin in the recrystallized aspirin. <em>TRUE. </em>Aspirin requires an excess of base (NaOH) for a complete dissolution (Hydrolysis). Then, we add H+ as HCl to know the excess moles of OH-. As we know the added moles of OH-, we can find the moles of OH that reacted = Moles of aspirin.  

D) We can determine the moles of aspirin in the recrystallized aspirin by titrating with the 0.100 M NaOH to the neutralization point. The purpose of the hydrolysis of the neutralized aspirin and the back-titration with the 0.100 M HCl is to confirm the moles of aspirin in the recrystallized aspirin. <em>FALSE. </em>NaOH can be added directly unyil neutralization point because, initially, aspirin can't be dissolved completely

E) Two of the above statements are true. <em>TRUE</em>

<em></em>

Right option is:

<h3>E) Two of the above statements are true.</h3>
3 0
3 years ago
In the laboratory, a student combines 28.3 mL of a 0.489 M manganese(II) acetate solution with 12.5 mL of a 0.339 M manganese(II
Eddi Din [679]

Answer:

M = 0.441 M

Explanation:

In this case, we have two solutions that involves the Manganese II cation;

We have Mn(CH₃COOH)₂ and MnSO₄

In both cases, the moles of Mn are the same in reaction as we can see here:

Mn(CH₃COO)₂ <-------> Mn²⁺ + 2CH₃COO⁻

MnSO₄ <------> Mn²⁺ + SO₄²⁻

Therefore, all we have to do is calculate the moles of Mn in both solutions, do the sum and then, calculate the concentration with the new volume:

moles of MnAce = 0.489 * 0.0283 = 0.0138 moles

moles MnSulf = 0.339 * 0.0125 = 0.0042 moles

the total moles are:

moles of Mn²⁺ = 0.0138 + 0.0042 = 0.018 moles

Finally the concentration: 12.5 + 28.3 = 40.8 mL or 0.0408 L

M = 0.018 / 0.0408

M = 0.441 M

This would be the final concentration of the manganese after the mixing of the two solutions

7 0
4 years ago
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