A5.0 g sample of copper(ll) oxide containing an inert contaminant is analyzed and found to contain 3.0 g copper. using molar masses of 64 g/mol for copper and 16 g/mol for oxygen, which of the following is the percent purity of the copper(ii) oxide sample? elimination tool a 80% 75% c 60% 50%

The percent purity is 75%

Explanation:

To obtain the purity of the copper (II) oxide sample there is necessary to calculate the amount of copper in sample. And then, by stoichiometry, the real amount of CuO.

The amount of copper in sample is:

3,0g × (1 mol Cu / 64 g) = 0,047 mol of Copper ≡ 0,047 mol Copper oxide

The copper moles are the same for CuO.

0,047 mol CuO × ( 80g / 1 mol CuO) = 3,75 g CuO

The real amount of copper (ll) oxide in the sample is 3,75 g.

Thus, the percent purity is:

3,75g of CuO / 5,0g of Sample × 100 = 75%

I hope it helps!

n+h→nh3

we must remember that nitrogen and hydrogen are both diatomic molecules in their standard gas form. this adjusts the equation to

n2+h2→nh3

now we need to adjust coefficients in order to balance the atoms on each side of the equation. currently we have 2 atoms of nitrogen and 2 atoms of hydrogen on the reactant side and 1 atom of nitrogen and 3 atoms of hydrogen on the product side.

we can balance the hydrogens by placing a coefficient of 2 in front of ammonia and a coefficient of 3 in front of the hydrogen.

n2+3h2→2nh3

this gives us 6 hydrogen on each side and coincidentally the nitrogens now equal 2 on each side.

43.452

step-by-step explanation:

284 times .153 = 43.452