Three samples of a solid substance composed of elements A and Z were prepared. The first contained 4.31 g A and 7.70 g Z. The second sample was 35.9% A and 64.1% Z. It was observed that 0.718 g A reacted with Z to form 2.00 g of the third sample. Show that these data illustrate the law of definite composition.

The percentage composition of the elements of the compound in the three samples is the same.

Explanation:

The law of definite proportions states that all pure samples of a particular chemical compound contain the same elements in the same proportion by mass.

Sample A:

Mass of A = 4.31 g; mass of Z = 7.70 g

Total mass of sample = 12.01

Percentage mass of A in the sample = (4.31 * 100)/12.01 = 35.9 %

Percentage mass of Z in the sample = (7.70 * 100)/12.01 = 64.1 %

Sample B:

Percentage mass of A in the sample = 35.9 %

Percentage mass of Z in the sample = 64.1 %

Sample C:

Mass of A = 0.718 g; Total mass of sample = 2.00 g

mass of Z = mass of sample - mass of A = 2.00 g - 0.718 g = 1.282 g

Percentage mass of A in the sample = (0.718 * 100)/2.00 = 35.9 %

Percentage mass of Z in the sample = (1.282 * 100)/2.00 = 64.1 %

From the calculations, it can be seen that the percentage composition of the elements in the compound is the same for the three samples.

a). in this case, dh> 0 and sd> 0. if t becomes larger, the term "-tds" will become more negative. at some point, it is possible that -dts will be more negative than the dh is positive, and the reaction will become spontaneous. this is probably the correct answer, but you should look at all of the other possibilities and make sure of that.