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2 years ago

The maximum amount of nickel(II) cyanide that will dissolve in a 0.220 M nickel(II) nitrate solution is...?

Chemistry

1 answer:

sweet [91]2 years ago

7 0

Answer : The maximum amount of nickel(II) cyanide is $5.84\times 10^{-12}M$

Explanation :

The solubility equilibrium reaction will be:

$Ni(CN)_2\rightleftharpoons Ni^{2+}+2CN^-$

Initial conc. 0.220 0

At eqm. (0.220+s) 2s

The expression for solubility constant for this reaction will be,

$K_{sp}=[Ni^{2+}][CN^-]^2$

Now put all the given values in this expression, we get:

$3.0\times 10^{-23}=(0.220+s)\times (2s)^2$

$s=5.84\times 10^{-12}M$

Therefore, the maximum amount of nickel(II) cyanide is $5.84\times 10^{-12}M$

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An unknown amount of acid can often be determined by adding an excess of base and then back-titrating the excess. A 0.3471−g sam

MAXImum [283]

Explanation:

The given data is as follows.

Mass of mixture = 0.3471 g

As the mixture contains oxalic acid and benzoic acid. So, oxalic acid will have two protons and benzoic acid has one proton.

This means oxalic acid will react with 2 moles of NaOH and benzoic acid will react with 1 mole of NaOH.

Hence, moles of NaOH in 97 ml = $\frac{0.1090 \times 97}{1000}$

= $10.573 \times 10^{-3} mol$

Moles of HCl in 21.00 ml = $\frac{0.2060 \times 21}{1000}$

= $4.326 \times 10^{-3}$ mol

Therefore, total moles of NaOH that reacted are as follows.

$10.573 \times 10^{-3} mol$ - $4.326 \times 10^{-3} mol$

= $6.247 \times 10^{-3}$ mol

So, total 3 mole of NaOH will react with 1 mole of mixture. Therefore, number of moles of NaOH reacted with benzoic acid is as follows.

$\frac{6.247 \times 10^{-3}}{3}$

= $2.082 \times 10^{-3}$ mol

Since, molar mass of NaOH is 40 g/mol. Therefore, calculate the mass of NaOH as follows.

$2.082 \times 10^{-3} mol \times 40 g/mol$

= $83.293 \times 10^{-3}$ g

= 0.0832 g

Whereas molar mass of benzoic acid is 122 g/mol.

Therefore, 40 g NaOH = 122 g benzoic acid

So, 0.0832 g NaOH = $\frac{122 g}{40 g} \times 0.0832 g$

= 0.253 g

Hence, calculate the % mass of benzoic acid as follows.

$\frac{0.253 g}{0.3471 g} \times 100$

= 73.10%

Thus, we can conclude that mass % of benzoic acid is 73.10%.

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3 years ago

DUPLICATE. The half-equivalence point of a titration occurs half way to the end point, where half of the analyte has reacted to

Law Incorporation [45]

At the half equivalence point [HA] = [A-] and pH = pKa

if Ka is 5.2e-5 then pKa = pH = 4.28

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3 years ago

How many grams of sodium sulfide can be produced when 45.3 g Na react with 105 g S?

RoseWind [281]

The chemical reaction would be as follows:

2Na + S → Na2S

We are given the amount of the reactants to be used in the reaction. We use these to calculate the amount of product. We do as follows:

45.3 g Na ( 1 mol / 22.99 g ) = 1.97 mol Na
105 g S ( 1 mol / 32.06 g ) = 3.28 mol S

The limiting reactant would be Na. We calculate as follows:

1.97 mol Na ( 1 mol Na2S / 2 mol Na ) (78.04 g / mol ) = 76.87 g Na2S produced

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2 years ago

How many moles are in 525 g of ammonia, NH3? 8940.75 M 0.03 M 30.83 M

victus00 [196]

Answer:

30.83 M

Explanation:

17.03052 re in one mole. So, if you multiply it by 30.83, you will get 535 g of ammonia.

In fact, the detailed answer is 30.827009392549122.

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slamgirl [31]

Explanation:

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