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.

the density(ratio between mass and volume as m/v) decreases from top to bottom in periodic table group as the size increases and nuclear charge decreases. but in periods density increases from left to right as atom size decreases and nuclear charge increases.

for step 7 table c on the lab "newtons law of action"   the average time it took the car to travel .25 meters   was 1.92 seconds.

the average time it took the car to travel 0.50 meters was 2.61 seconds

hope this : )