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Harrizon [31]
1 year ago
9

3. ml of 1.00 M LiOH to 500.00 ml of 0.075 M of LiOH (Dilution Problem)​

Chemistry
1 answer:
castortr0y [4]1 year ago
6 0

The volume (in mL) of the 1.00 M LiOH needed to prepare 500.00 mL of 0.075 M of LiOH is 37.5 mL

<h3>How do I determine the volume (in mL) of LiOH needed?</h3>

Dilution formula is given as follow:

M₁V₁ = M₂V₂

Where

  • M₁ is the molarity of stock solution
  • V₁ is the volume of stock solution
  • M₂ is the molarity of diluted solution
  • V₂ is the volume of diluted solution

Applying the above formula, we obtain the volume (in mL) of LiOH needed to prepare 500.00 mL of 0.075 M of LiOH. This is illustrated below:

  • Molarity of stock solution of LiOH (M₁) = 1.00 M
  • Volume of diluted solution of LiOH (V₂) = 500.00 mL
  • Molarity of diluted solution of LiOH (M₂) = 0.075 M
  • Volume of stock solution of LiOH needed (V₁) =?

M₁V₁ = M₂V₂

1 × V₁ = 0.075 × 500

V₁ = 37.5 mL

Thus, the volume (in mL) of LiOH needed is 37.5 mL

Learn more about volume:

brainly.com/question/24159217

#SPJ1

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Answer:

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Explanation:

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Thus, the above equation can written as:

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This implies that rate is inversely proportional to the molar mass of a gas i.e the lighter the gas, the faster the rate and the heavier the gas, the slower the rate of diffusion.

Now, to obtain the answer to the question, let us determine the molar mass of each gas.

This is illustrated below:

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Molar mass of Ar = 40 g/mol

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Summary

Gas >>>>>>> Molar mass

Kr >>>>>>>>> 84 g/mol

NO >>>>>>>> 30 g/mol

Ar >>>>>>>>> 40 g/mol

N2O >>>>>>> 44 g/mol

From the above table, we can see the lightest gas is NO.

Therefore, NO will escape through the hole in the balloon at the highest rate.

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