Answer : 0.8663 Kg of chalcopyrite must be mined to obtained 300 g of pure Cu.
Solution : Given,
Mass of Cu = 300 g
Molar mass of Cu = 63.546 g/mole
Molar mass of 
= 183.511 g/mole
- First we have to calculate the moles of Cu.
The moles of Cu = 4.7209 moles
From the given chemical formula, we conclude that the each mole of compound contain one mole of Cu.
So, The moles of Cu = Moles of = 4.4209 moles
- Now we have to calculate the mass of .
Mass of = Moles of × Molar mass of = 4.4209 moles × 183.511 g/mole = 866.337 g
Mass of = 866.337 g = 0.8663 Kg (1 Kg = 1000 g)
Therefore, 0.8663 Kg of chalcopyrite must be mined to obtained 300 g of pure Cu.
Scientists should control most possible variables in experiments to get the most valid and correct data. If many variables are included in experiments it is more difficult to interpret what is causing a different outcome.
Answer:
The mole ratio of the cation and the anion in a precipitate is a simple fraction. ( im sorry if this dosent help a lot.)
Explanation:
Answer:
[OH-] = 3.0 x 10^-19 M
Explanation:
[H3O+][OH-] = Kw
Kw = 1.0 x 10^-14
[H3O+][OH-] = 1.0 x 10^-14
[OH-] = 1.0 x 10^-14 / 3.3 x 10^4 = 3.0 x 10^-19
