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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

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H2SO4 is referred to as a strong acid and is denoted as option A.

<h3>What is an Acid?</h3>

This refers to any substance which tastes sour when in water and changes the color of blue litmus paper to red. It is usually very corrosive and are used in industries for different functions.

H2SO4 is referred to as a strong acid because it dissociates completely in its aqueous solution or water.

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Answer:

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Explanation:

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Answer:

A chemical reaction in which an uncombined element replaces an element that is part of a compound is called a simple substitution reaction or simple displacement reaction.

Explanation:

A simple substitution reaction or simple displacement reaction, called single-displacement reaction, is a reaction in which an element of a compound is substituted by another element involved in the reaction. The starting materials are always pure elements and an aqueous compound. And a new pure aqueous compound and a different pure element are generated as products. The general form of a simple substitution reaction is:

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So, in a Single replacement reaction an uncombined element replaces an element.

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