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Margarita [4]
3 years ago
13

How many milligrams of sodium sulfide are needed to completely react with 25.00 ml of a 0.0100 m aqueous solution of cadmium nit

rate, to form a precipitate of cds(s)?
Chemistry
2 answers:
german3 years ago
8 0

<u>Answer:</u> The mass of sodium sulfide needed is 195000 mg.

<u>Explanation:</u>

To calculate the moles of cadmium nitrate, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Molarity of cadmium nitrate = 0.0100 M

Volume of cadmium nitrate = 25.0 mL = 0.025 L   (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

0.0100mol/L=\frac{\text{Moles of cadmium nitrate}}{0.025L}\\\\\text{Moles of cadmium nitrate}=2.5\times 10^{-4}mol

The chemical reaction of cadmium nitrate with sodium sulfide, the equation follows:

Cd(NO_3)_2+Na_2S\rightarrow CdS+2NaNO_3

By Stoichiometry of the reaction:

1 mole of cadmium nitrate reacts with 1 mole of sodium sulfide.

So, 2.5\times 10^{-4}mol of cadmium nitrate will react with = \frac{1}{1}\times 2.5\times 10^{-4}=2.5\times 10^{-4}mol of sodium sulfide.

  • To calculate the mass of sodium sulfide, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Moles of sodium sulfide = 2.5\times 10^{-4}mol

Molar mass of sodium sulfide = 78 g/mol

Putting values in above equation, we get:

2.5\times 10^{-4}mol=\frac{\text{Mass of sodium sulfide}}{78g/mol}\\\\\text{Mass of sodium sulfide}=195g

Now, converting this mass into milligrams, we use the conversion factor:

1 g = 1000 mg

So, 195g=195\times 1000=195000mg

Hence, the mass of sodium sulfide needed is 195000 mg.

NARA [144]3 years ago
5 0
Na₂S(aq) + Cd(NO₃)₂(aq) = CdS(s) + 2NaNO₃(aq)

v=25.00 mL
c=0.0100 mmol/mL
M(Na₂S)=78.046 mg/mmol

n(Na₂S)=n{Cd(NO₃)₂}=cv

m(Na₂S)=M(Na₂S)n(Na₂S)=M(Na₂S)cv

m(Na₂S)=78.046*0.0100*25.00≈19.5 mg
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In a titration of 47.41 mL of 0.3764 M ammonia with 0.3838 M aqueous nitric acid, what is the pH of the solution when 47.41 mL +
Volgvan

<u>Answer:</u> The pH of the solution is 1.136

<u>Explanation:</u>

To calculate the moles from molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

  • <u>For ammonia:</u>

Molarity of ammonia = 0.3764 M

Volume of ammonia = 47.41 mL = 0.04741 L   (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

0.3764mol/L=\frac{\text{Moles of ammonia}}{0.04741L}\\\\\text{Moles of ammonia}=0.01784mol

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Volume of ammonia = (47.41 + 10.00) mL = 57.41 mL= 0.05741 L

Putting values in above equation, we get:

0.3838mol/L=\frac{\text{Moles of nitric acid}}{0.05741L}\\\\\text{Moles of nitric acid}=0.02203mol

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For the reaction of ammonia with nitric acid, the equation follows:

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Completion        0     0.0042        0.0178

As, the solution of the reaction is made from strong acid which is nitric acid and the conjugate acid of weak base which is ammonia. So, the pH of the reaction will be based totally on the concentration of nitric acid.

To calculate the pH of the reaction, we use the equation:

pH=-\log[H^+]

where,

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Putting values in above equation, we get:

pH=-\log(0.0731)\\\\pH=1.136

Hence, the pH of the solution is 1.136

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