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ankoles [38]
2 years ago
11

What is the volume of a gas at 756.14 Torr. If the volume was 588.56 mL at 588.56 torr.

Chemistry
2 answers:
dimaraw [331]2 years ago
8 0

Answer: 458.12 mL

Explanation:

Solving it with Boyle's law:

P1V1 = P2V2

Putting values

588.56 mL × 588.56 torr = 756.14 torr × V2

Rearranging the equation

V2 = 588.56 mL × 588.56 torr / 756.14 torr

torr get cancelled

Simplifying further

V2 = 458.12 mL

creativ13 [48]2 years ago
5 0

Answer:

Explanation:

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A standard solution of FeSCN2+ is prepared by combining 9.0 mL of 0.20 M Fe(NO3)3 with 1.0 mL of 0.0020 M KSCN . The standard so
Xelga [282]

Answer : The equilibrium concentration of SCN^- in the trial solution is 4.58\times 10^{-8}M

Explanation :

First we have to calculate the initial moles of Fe^{3+} and SCN^-.

\text{Moles of }Fe^{3+}=\text{Concentration of }Fe^{3+}\times \text{Volume of solution}

\text{Moles of }Fe^{3+}=0.20M\times 9.0mL=1.8mmol

and,

\text{Moles of }SCN^-=\text{Concentration of }SCN^-\times \text{Volume of solution}

\text{Moles of }SCN^-=0.0020M\times 1.0mL=0.0020mmol

The given balanced chemical reaction is,

Fe^{3+}(aq)+SCN^-(aq)\rightleftharpoons FeSCN^{2+}(aq)

Since 1 mole of Fe^{3+} reacts with 1 mole of SCN^- to give 1 mole of FeSCN^{2+}

The limiting reagent is, SCN^-

So, the number of moles of FeSCN^{2+} = 0.0020 mmole

Now we have to calculate the concentration of FeSCN^{2+}.

\text{Concentration of }FeSCN^{2+}=\frac{0.0020mmol}{9.0mL+1.0mL}=0.00020M

Using Beer-Lambert's law :

A=\epsilon \times C\times l

where,

A = absorbance of solution

C = concentration of solution

l = path length

\epsilon = molar absorptivity coefficient

\epsilon and l are same for stock solution and dilute solution. So,

\epsilon l=\frac{A}{C}=\frac{0.480}{0.00020M}=2400M^{-1}

For trial solution:

The equilibrium concentration of SCN^- is,

[SCN^-]_{eqm}=[SCN^-]_{initial}-[FeSCN^{2+}]

[SCN^-]_{initial} = 0.00050 M

Now calculate the [FeSCN^{2+}].

C=\frac{A}{\epsilon l}=\frac{0.220}{2400M^{-1}}=9.17\times 10^{-5}M

Now calculate the concentration of SCN^-.

[SCN^-]_{eqm}=[SCN^-]_{initial}-[FeSCN^{2+}]

[SCN^-]_{eqm}=(0.00050M)-(9.17\times 10^{-5}M)

[SCN^-]_{eqm}=4.58\times 10^{-8}M

Therefore, the equilibrium concentration of SCN^- in the trial solution is 4.58\times 10^{-8}M

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