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.
Answer:
The energy of the particles increase and the molecules move more quickly.
Explanation:
The molecules are moving from a solid (barely moves, molecules close together) to a liquid (molecules slide past each other and take any shape), so molecules are moving more and have more energy
the mass percent of sugar in this solution is 46%.
Answer:
Solution given:
mass of solute=34.5g
mass of solvent=75g
mass percent=
=
Answer:
506.912 L
Explanation:
From the question given above, the following data were obtained:
Number of mole of O₂ = 22.63 moles
Volume of O₂ =?
Recall:
1 mole of a gas occupy 22.4 L at STP.
With the above information, we obtained the volume occupied by 22.63 moles of O₂ as follow:
1 mole of O₂ occupied 22.4 L at STP.
Therefore, 22.63 moles of O₂ will occupy = 22.63 × 22.4 = 506.912 L at STP.
Thus, 22.63 moles of O₂ is equivalent to 506.912 L.
I think your anwser ahould be b!
