We need to prepare 0.0021 M solution of C2SO4. But there are only 80.7 grams of the chemical available. What is the maximum volume that can be prepared?

320 L

Explanation:

We'll begin by calculating the number of mole in 80.7 g of C₂SO₄. This can be obtained as follow:

Mass of C₂SO₄ = 80.7 g

Molar mass of C₂SO₄ = (12×2) + 32 + (16×4)

= 24 + 32 + 64

= 120 g/mol

Mole of C₂SO₄ =?

Mole = mass /molar mass

Mole of C₂SO₄ = 80.7 / 120

Mole of C₂SO₄ = 0.6725 mole

Finally, we shall determine the volume. This can be obtained as follow:

Mole of C₂SO₄ = 0.6725 mole

Molarity = 0.0021 M

Volume =?

Molarity = mole / Volume

0.0021 = 0.6725 / Volume

Cross multiply

0.0021 × Volume = 0.6725

Divide both side by 0.0021

Volume = 0.6725 / 0.0021

Volume ≈ 320 L

Thus, the volume that can be prepared is approximately 320 L.

out of these options:

ability to burn paper

pure sodium reacting violently with water

the boiling point of water

helium does not tend to react with anything

i would say this is a physical property:

helium does not tend to react with anything

why? :

i said helium does not tend to react with anything because well, it reacts to ! but ability to burn paper, pure sodium reacting violently with water, & the boiling point of water are chemical or

physical properties: the properties of a substance that can be observed without changing the chemical makeup of the substance

chemical properties: properties of a substance relating to the chemical nature and reactivity of the substance

some sources quote the lethal dose, for an average 170lbs opiate-naive individual is between 75 and 375 mg