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${H_2}O$,${H_2}S$, and $S{O_2}$ follow the law of reciprocal properties.
Answer
561.6k+ views
Hint: We know that the law of reciprocal proportions was planned by Jeremias Ritcher It states that, when two dissimilar elements unite disjointedly with the same weight of a third element, the ratio of the masses in which they do so are also the identical or a simple multiple of the mass ratio in which they unite.
Complete step by step answer:
If part A combines with element B associated additionally with C, then, if B and C combine together, the proportion by weight during which they are doing so are going to be merely involving the weights of B and C which severally combine with a relentless weight of A.
In ${H_2}O$, the magnitude relation of plenty of Hydrogen and Oxygen is \[2:16\].
In $S{O_2}$, the ratio of masses of Sulfur and Oxygen is $32:32$.
Therefore, the ratio of masses of Hydrogen and Sulfur which mix with a hard and fast mass of gas are going to be \[4:32\].i.e., \[1:8\] .When Hydrogen and Sulfur unite together, they type ${H_2}S$within which the ratio of masses of hydrogen and sulfur is \[2:32\] i.e.,\[1:16\].
The two ratios are concerning one another as \[2:1\] , i.e., they're integer multiple of every other. Thus, the ratio of plenty of Hydrogen and Sulfur with a hard and fast mass of gas could be an integer multiple of the magnitude relation within which Hydrogen and Sulfur combine together.
Therefore the given statement is true..
Note: As an example, one gram of metal is discovered to mix with either \[1.54grams\] of chlorine gas or \[5.52grams\] of iodine. (These quantitative relations correspond to the fashionable formulas NaCl and NaI). The ratio of those 2 weights is \[\dfrac{{5.52}}{{1.54}} = 3.58\].
It’s also observed that one gram of gas reacts with \[1.19g\] of iodine. This quantitative relation of \[1.19\] obeys the law as a result of it's an easy fraction of \[3.58\] (This is because it corresponds to the formula $ICl_3$, that is one celebrated compound of iodine and chlorine.). Likewise, hydrogen, carbon, and chemical elements follow the law of reciprocal proportions.
Complete step by step answer:
If part A combines with element B associated additionally with C, then, if B and C combine together, the proportion by weight during which they are doing so are going to be merely involving the weights of B and C which severally combine with a relentless weight of A.
In ${H_2}O$, the magnitude relation of plenty of Hydrogen and Oxygen is \[2:16\].
In $S{O_2}$, the ratio of masses of Sulfur and Oxygen is $32:32$.
Therefore, the ratio of masses of Hydrogen and Sulfur which mix with a hard and fast mass of gas are going to be \[4:32\].i.e., \[1:8\] .When Hydrogen and Sulfur unite together, they type ${H_2}S$within which the ratio of masses of hydrogen and sulfur is \[2:32\] i.e.,\[1:16\].
The two ratios are concerning one another as \[2:1\] , i.e., they're integer multiple of every other. Thus, the ratio of plenty of Hydrogen and Sulfur with a hard and fast mass of gas could be an integer multiple of the magnitude relation within which Hydrogen and Sulfur combine together.
Therefore the given statement is true..
Note: As an example, one gram of metal is discovered to mix with either \[1.54grams\] of chlorine gas or \[5.52grams\] of iodine. (These quantitative relations correspond to the fashionable formulas NaCl and NaI). The ratio of those 2 weights is \[\dfrac{{5.52}}{{1.54}} = 3.58\].
It’s also observed that one gram of gas reacts with \[1.19g\] of iodine. This quantitative relation of \[1.19\] obeys the law as a result of it's an easy fraction of \[3.58\] (This is because it corresponds to the formula $ICl_3$, that is one celebrated compound of iodine and chlorine.). Likewise, hydrogen, carbon, and chemical elements follow the law of reciprocal proportions.
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