Answer:
B
Explanation:
Not sure, but that one makes the most since
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:
0.256 L
Explanation:
We should use the following formula:
concentration (1) × volume (1) = concentration (2) × volume (2)
concentration (1) = 0.82 M NaOCl
volume (1) = ?
concentration (2) = 0.21 M NaOCl
volume (2) = 1 L
volume (1) = [concentration (2) × volume (2)] / concentration (1)
volume (1) = [0.21 / 1] / 0.82 = 0.256 L
6.2 grams of CO2 = 1.408786739226764 moles
1. Find its coordination figure/coordination number of central atom (CF)
Ev = Vallence electron of central atom
Σe = electrons donated from substituents
Terminal O gives 0 electrons, hence Σe = 3 x 0
charge = charge of the compound
2. Find EP (electron pairs) and LP (lone pairs)
LP = CF - EP
3. Draw the skeleton with octet substituents (top right figure)
4. Find formal charge for each atoms (Qf)
5. Write formal charge near atom in skeleton
6. Enjoy