It allows you to determine the relation between the reactants and the products.
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.
The amount
per 100 g is:
38.7 %
calcium = 38.7g Ca / 100g compound = 38.7g
19.9 %
phosphorus = 19.9g P / 100g compound = 19.9g
41.2 %
oxygen = 41.2g O / 100g compound = 41.2g
The molar amounts of calcium,
phosphorus and oxygen in 100g sample are calculated by dividing each element’s
mass by its molar mass:
Ca = 38.7/40.078
= 0.96
P = 19.9/30.97
= 0.64
O = 41.2/15.99
= 2.57
C0efficients
for the tentative empirical formula are derived by dividing each molar amount
by the lesser value that is 0.64 and in this case, after that multiply wih 2.
Ca = 0.96 /
0.64 = 1.5=1.5 x 2 = 3
P = 0.64 /
0.64 = 1 = 1x2= 2
O = 2.57 /
0.64 = 4= 4x2= 8
Since, the
resulting ratio is calcium 3, phosphorus 2 and oxygen 8
<span>So, the
empirical formula of the compound is Ca</span>₃(PO₄)₂
Answer:
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