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Anestetic [448]
3 years ago
11

When you need to produce a variety of diluted solutions of a solute, you can dilute a series of stock solutions. A stock solutio

n has a significantly higher concentration of the given solute (typically 101 to 104 times higher than those of the diluted solutions). The high concentration allows many diluted solutions to be prepared using minimal amounts of the stock solution. What volume of a 6.57 M stock solution do you need to prepare 500. mL of a 0.0490 M solution of HCl
Chemistry
1 answer:
nignag [31]3 years ago
4 0

Answer:

3.73 mL

Explanation:

To solve this problem we can use the equation C₁V₁=C₂V₂, where:

  • C₁ = 6.57 M
  • V₁ = ?
  • C₂ = 0.0490 M
  • V₂ = 500 mL

We<u> input the data</u>:

  • 6.57 M * V₁ = 0.0490 M * 500 mL

And <u><em>solve for V₁</em></u>:

  • V₁ = 3.73 mL

So 3.73 mL of a 6.57 M stock solution would be required to prepare 500 mL of a 0.0490 M solution.

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A 36.165 mg36.165 mg sample of a chemical known to contain only carbon, hydrogen, sulfur, and oxygen is put into a combustion an
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Answer:

The empirical formula for the compound is C6H12SO2.

Explanation:

We'll begin by writing out what was given from the question. This is shown:

Let us consider the First experiment:

Mass of the compound = 36.165 mg

Mass of CO2 = 64.425 mg

Mass of H2O = 26.373 mg

Data obtained from the Second experiment:

Mass of compound = 47.029 mg

Mass of SO2 = 20.32 mg

Next, we'll determine the mass of C, H and S. This is illustrated below:

Molar Mass of CO2 = 12 + (2x16) = 44g/mol

Mass of C in CO2 = 12/44 x 64.425 Mass of C = 17.57 mg

Molar Mass of H2O = (2x1) + 16 = 18g/mol

Mass of H in H2O = 2/18 x 26.373

Mass of H = 2.93 mg

Molar Mass of SO2 = 32 + (16x2) = 64g/mol

Mass of S in SO2 = 32/64 x 20.32

Mass of S = 10.16 mg

At this stage, it is important we determine the percentage composition of C, H, S and O. This is illustrated below:

% of C = 17.57/36.165 x 100 = 48.58%

% of H = 2.93/36.165 x 100 = 8.10%

% of S = 10.16/47.029 x 100 = 21.60%

% of O = 100 - (48.58 + 8.1 + 21.6)

% of O = 21.72%

Now we can easily obtain the empirical formula for the compound by doing the following.

Step 1:

Divide by their molar mass

C = 48.58/12 = 4.0483

H = 8.10/1 = 8.1

S = 21.60/32 = 0.675

O = 21.72/16 = 1.3575

Step 2:

Divide by the smallest:

C = 4.0483/0.675 = 6

H = 8.1/0.675 = 12

S = 0.675/0.675 = 1

O = 1.3575/0.675 = 2

From the calculations made above, empirical formula for the compound is C6H12SO2

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