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2 years ago

If 625 ml of pure water is added to 344 ml of 1.675 m na2so4, what is the new concentration of na2so4?

1 answer:

8 0

Molar concentration is the number of Na₂SO₄ moles that are dissolved in 1 L of solution.
when 625 mL is added to 344 mL of Na₂SO₄, then the number of moles that are dissolved in 344 mL remains constant but volume increases. Therefore concentration now decreases.
Molar concentration of Na₂SO₄ is 1.675 M
the number of Na₂SO₄ moles in 1000 mL - 1.675 mol
then number of Na₂SO₄ moles in 344 mL - 1.675 mol / 1000 x 344 = 0.5762 mol
the new volume - 344 mL + 625 mL = 969 mL
concentration = number of moles / volume
therefore new concentration = 0.5762 mol / 0.969 L = 0.595 mol/L
new concentration of Na₂SO₄ is 0.595 M

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3 years ago

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Use this equation for the following problems: 2NaN3 --> 2Na+3N2

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It is clear from the balanced equation that 2.0 moles of NaN₃ are decomposed to 2.0 moles of Na and 3.0 moles of N₂.

Q1: How many grams of NaN₃ are needed to make 23.6L of N₂?

Density of N₂ = 0.92 g/L which means that every 1.0 L of N₂ contains 0.92 g of N₂.

Now, we can get the mass of N₂ in 23.6 L N₂ using cross multiplication:

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Answer:

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