Question

answer plsss .....


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Answer 1

|| ANSWER 2 ||

In CuS ( 100G )

mCu = 66.5g , mS = (100g) − (66.5g) = 33.5g

: Ratio of masses (mCu : mO) = (66.5g) : (33.5g)

= 66.5 / 66.5 , 33.5 / 66.5 = 1 : 0.503

NOW ,

In CuO(100g)

mCu = 79.99g , mO = (100g) − (79.9g) = 20.1g

Ratio of masses (mCu : mS)=79.9g : 20.1g

= 79.9 / 79.9 : 20.1 / 79.1 = 1 = 0.251

Thus, the ratio of masses of S to O combining with a given mass of Cu (1 part) is

0.503 : 0251 = 0.503 / 0.251 : 0.251 / 0.251 = 3 : 1

NOW ,

In SO3(100g)

mS = 40g , mO = (100g) − (40g) = 60g

Ratio of masses (mS : mO) = (40g) : (60g)=2 : 3

|| ANSWER 3 ||

The chemical combinations which is given , illustrate the law of reciprocal proportion :-

In KCl, the percentages of K and Cl are 52% and 48% respectively.

In KI, the percentages of K and I are 23.6 % and 76.7 % respectively.

In ICl, the percentages of I and Cl are 77.8 and 22.2 % respectively.

Thus, the masses of I and Cl, which are combining with a fixed mass of K, are also the masses with which I and Cl combine.

Answer 2

Answer:

percentage composition of I and clothing will be 77.8%,22.2%