I need help on these chemistry problems

15. 2.5 M

16. 2.5 moles

17. 0.128 M

18. 56 g

Explanation:

15. Determination of the molarity of NaOH.

Mole of NaOH = 2.5 moles

Volume = 1 L

Molarity =?

Molarity = mole /Volume

Molarity of NaOH = 2.5 / 1

Molarity of NaOH = 2.5 M

16. Determination of the number of mole of solute.

Volume = 1 L

Molarity = 2.5 M

Mole of HCl =?

Molarity = mole /Volume

2.5 = mole of HCl /1

Mole of HCl = 2.5 moles.

17. Determination of the molarity of Na₂S

We'll begin by calculating the number of mole in 20 g of Na₂S. This can be obtained as follow:

Mass of Na₂S = 20 g

Molar mass of Na₂S = (23×2) + 32

= 46 + 32

= 78 g/mol

Mole of Na₂S =?

Molar = mass / molar mass

Mole of Na₂S = 20 / 78

Mole of Na₂S = 0.256 mole

Finally, we shall determine the molarity of Na₂S. This can be obtained as follow:

Mole of Na₂S = 0.256 mole

Volume = 2 L

Molarity of Na₂S =?

Molarity = mole /Volume

Molarity of Na₂S = 0.256 / 2

Molarity of Na₂S = 0.128 M

18. Determination of the mass of solute.

We'll begin by calculating the number of mole solute (CH₄) in the solution. This can be obtained as follow:

Volume = 1 L

Molarity of CH₄ = 3.5 M

Mole of CH₄ =?

Molarity = mole / Volume

3.5 = mole of CH₄ / 1

Mole of CH₄ = 3.5 moles

Finally, we shall determine the mass of the solute (CH₄). This can be obtained as follow:

Mole of CH₄ = 3.5 moles

Molar mass of CH₄ = 12 + (4×1)

= 12 + 4

= 16 g/mol

Mass of CH₄ =?

Mole = mass /Molar mass

3.5 = Mass of CH₄ / 16

Cross multiply

Mass of CH₄ = 3.5 × 16

Mass of CH₄ = 56 g

b.final

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