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Licemer1 [7]
3 years ago
5

A student takes a sample of KOH solution and dilutes it with 100.00 mL of water. The student determines that the diluted solutio

n is 0.046 M KOH, but has forgotten to record the volume of the original sample. The concentration of the original solution is 2.09 M. What was the volume of the original sample?
Chemistry
1 answer:
Firlakuza [10]3 years ago
8 0

Answer:

2.2mL

Explanation:

First, let us analyse what was given from the question:

C1 = 2.09M

V1 =?

C2 = 0.046M

V2 =100mL

Using dilution formula (C1V1 = C2V2), we can calculate the volume of the original solution as follows:

C1V1 = C2V2

2.09 x V1 = 0.046 x 100

Divide both side by the coefficient of V1 ie 2.09, we have:

V1 = (0.046 x 100) / 2.09

V1 = 2.2mL

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Answer:

Empirical formula (which matches the molecular formula) is = PbC₈H₂₀

Explanation:

Our sample: 60 g of tetraethyl lead

In order to determine the compound empirical formula we need the centesimal composition:

(Mass of element / Total mass) . 100 =

(38.43 g lead / 60g ) . 100 = 64.05%

(17.83 g C / 60g) . 100 = 29.72%

(3.74 g H / 60g) . 100 = 6.23 %

These % are the mass of the elements in 100 g of compound. Let's find out the moles of them:

64.05 g / 207.2 g/mol = 0.309 moles

29.72 g / 12 g/mol = 2.48 moles

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Next, we divide the moles, by the lowest value of them (0.309)

0.309 / 0.309 = 1 mol Pb

2.48 / 0.309 = 8 mol C

6.23 / 0.309 = 20 mol H

There, we have our formula PbC₈H₂₀

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How can you use a fossil to indemnify the environment where the organism lived
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