Answer:
The percent by mass of copper in the mixture was 32%
Explanation:
The ammount of HNO₃ used is:
mol HNO₃ = volume * concentration
mol HNO₃ = 0.015 l * 15.8 mol/l
mol HNO₃ = 0.237 mol
According to the reaction, 4 mol HNO₃ will react with 1 mol Cu and produce 1 mol Cu²⁺. Since we have 0.237 mol HNO₃, the amount of Cu that could react would be (0.237 mol HNO₃ * 1 mol Cu / 4 mol HNO₃) 0.06 mol. This reaction would produce 0.060 mol Cu²⁺, however, only 0.010 mol Cu²⁺ were obtained, indicating that only 0.010 mol Cu were present in the mixture. This means that the acid was in excess, so we can assume that all copper present in the mixture has reacted.
Since 0.010 mol of Cu²⁺ were produced, the amount of Cu was 0.01 mol.
1 mol of Cu has a mass of 63.5 g, then 0.01 mol has a mass of:
0.01 mol Cu * 63.5 g / 1 mol = 0.635 g.
Since this amount was present in 2.00 g mixture, the amount of copper in 100 g of the mixture will be:
100 g(mixture) * 0.635 g Cu / 2 g(mixture) = 32 g
Then, the percent by mass of Cu (which is the mass of Cu in 100 g mixture) is 32%
2 tablets every 6 hours. The patient will receive four doses of 2 tablets. The physician will need 8 tablets for one days treatment.
We are given the molecular mass of tetraoxosulphate(VI) acid as 98 g/mol. The molecular formula of this is H₂SO₄ which is also known as sulfuric aid. The vapour density of a substance is the density of a vapour in relation to hydrogen gas. Therefore, it the mass of a volume of gas, divided by the mass of the same volume of hydrogen gas. We can use the following formula:
vapour density = molar mass of gas/ molar mass of H₂
vapour density = 98 g/mol H₂SO₄ / 2 g/mol H₂
vapour density = 49
The vapour density of tetraoxosulphate(IV) acid is found to be 49. It should also be noted that the vapour density is a unitless quantity.
Answer:
236.8 g
Explanation:
From the reaction equation;
CaC2 + 2H2O(I) --------> C2H2(g) + Ca(OH)2(aq)
Since;
1 mole of CaC2 yields 1 mole of Ca(OH)2
It follows that 3.20 moles of CaC2 also yields 3.2 moles of Ca(OH)2
Mass of Ca(OH)2 = number of moles * molar mass
molar mass of Ca(OH)2 = 74 g/mol
Mass of Ca(OH)2 = 3.20 moles * 74 g/mol =
Mass of Ca(OH)2 = 236.8 g