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ludmilkaskok [199]
3 years ago
7

Calculate the concentration of ammonium nitrate in a solution prepared by dissolving 3.20 g of the salt in enough water to make

100. mL of solution, then diluting 2.00 mL of this solution to a volume of 25.00 mL.
Chemistry
1 answer:
Tpy6a [65]3 years ago
3 0

Answer:

.032 M .

Explanation:

Molecular weight of ammonium nitrate is 80  .

3.2 g = 3.2 / 80 moles

= .04 moles

volume = 100 mL = 0.1 L

Molarity of 100 mL solution = .04 moles / 0.1 L

= 0.4 M solution.

Now 2 mL solution of 0.4 M is diluted to a volume of 25 mL .

Using the formula S₁ V₁ = S₂V₂

0.4 M x 2 mL = S₂ x 25 mL

S₂ = .4 x 2 / 25

= .032 M

Hence required concentration is .032 M .

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Answer:

30.17 × 10²³ atoms

Explanation:

Given data:

Number of moles of lead = 5.01 mol

Number of atoms = ?

Solution:

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The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

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