Answer:
A catalyst is a chemical substance that alters the rate of chemical reaction not consumed by the reaction. Hence, a catalyst can be recovered chen unchanged at the ends of chemical reaction. Catalyst can be divided into two typ the basis whether it speeds up or slowdowns the rate of chemical reaction. The positive catalyst and negative catalyst.
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:
K = [ HOCl ] . [HgO. HgCl2] / [Cl2]^2 [H2O] [HgO]^2
Explanation:
The law of Mass Action states that, at constant temperature, the rate of reaction is proportional to the active masses of each of the reactants.
The reaction above is a reversible reaction and the law of mass action also applies to it.
The rate of reaction from left-to-right reaction = r1 = k. [Cl2]^2 [H2O] [HgO]^2
Rate of reaction from right - to - left r2 = k. [hocl]^2 [HgO . hgcl2]
Then at equilibrium,
r1 = r2
k1/k2 = [HOCl ]^2 [HgO. HgCl2] / [Cl2]^2 [H2O] [HgO]^2 = K
where K is the equilibrium constant for the reaction.
Answer:
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