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Elena-2011 [213]
3 years ago
6

An Olympic-size pool is 50.0 m long and 25.0 m wide.

Chemistry
1 answer:
Lostsunrise [7]3 years ago
8 0
1. To calculate the volume of the pool you need to convert the length, width and height into one same unit. The volume would be:
volume= length * width * height
volume= 50m *25m *(48ft *0.3048m)= 18288m^3
volume in gallons= 18288m^3 * (264.172 gallons/m^3)= 483.118 gallons

2. The mass would be found by multiplying density with volume. You will need to some unit conversion to get the mass in kilogram. Remember that 1g= 1000kg and 1m^3= 1,000,000ml
mass= volume * density
mass= 18288m^3 * (10^6 ml^3/m^3) * 1.0 g/mL / (1000g/kg)= 18,288,000 kg
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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
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Answer : The equilibrium concentration of SCN^- in the trial solution is 4.58\times 10^{-8}M

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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}

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The limiting reagent is, SCN^-

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

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Using Beer-Lambert's law :

A=\epsilon \times C\times l

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

For trial solution:

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[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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