Aluminium Sulfide
According to rules the positive specie is named first and the negative specie is named last.
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:
Chemical reactions often involve changes in energy due to the breaking and formation of bonds. Reactions in which energy is released are exothermic reactions, while those that take in heat energy are endothermic.
Explanation:
Answer:
The molar solubility of YF₃ is 4.23 × 10⁻⁶ M.
Explanation:
In order to calculate the molar solubility of YF₃ we will use an ICE chart. We identify 3 stages: Initial, Change and Equilibrium and we complete each row with the concentration of change of concentration. Let's consider the solubilization of YF₃.
YF₃(s) ⇄ Y³⁺(aq) + 3 F⁻(aq)
I 0 0
C +S +3S
E S 3S
The solubility product (Ksp) is:
Ksp = [Y³⁺].[F⁻]³= S . (3S)³ = 27 S⁴
![S=\sqrt[4]{Ksp/27} =\sqrt[4]{8.62 \times 10^{-21} /27}=4.23 \times 10^{-6}M](https://tex.z-dn.net/?f=S%3D%5Csqrt%5B4%5D%7BKsp%2F27%7D%20%3D%5Csqrt%5B4%5D%7B8.62%20%5Ctimes%2010%5E%7B-21%7D%20%20%2F27%7D%3D4.23%20%5Ctimes%2010%5E%7B-6%7DM)
A) Deer
D) Grass
Hope that helps you out (: