Phosphorus forms several compounds with oxygen, including diphosphorus trioxide and diphosphorus pentoxide. a decomposition of a sample of diphosphorus trioxide forms 1.29 g phosphorus to every 1.00 g oxygen. the decomposition of a sample of diphosphorus pentoxide forms 0.775 g phosphorus to every 1.00 g oxygen. show that these results are consistent with the law of multiple proportions.

Results are consistent with law of multiple proportions.

Explanation:

Dalton's law of multiple proportions states that the elements can be combined to form more than one compound and do so in such a way that if we keep the quantity of one element as fixed, the quantity of the other element is added in such a way that they are obtained integer proportions.

Molar mass of oxygen is 16g per mole, molar mass of phosphorus is 31g per mole.

In the ratio between P and O is , that is consistent with the experimental result.

and for the ratio between P and O is that is also consistent with experimental results. In conclusion we can say that the results obtained are consistent with the law of multiple proportions.

answer; ///i do believe that the correct answer is (b); ///particle composed of two or more atoms bonded together;

part 1: the half-life of the element is 22 years.

part 2: it will take 132.02 year for the sample to decay from 308.0 g to 4.8125 g.

explanation:

part 1: there are two ways to determine the half life time.

first way: the half life time can be determined directly from its definition as it is the time required for the sample to decay to its half concentration.

from the data the initial concentration was 45.0 g and decayed to its half value (22.5 g) in 22.0 years.

second way: the radio active decay is considered first order reaction so we can use the laws of first order reactions.

k = (1/t) ln (a/a-x), where k is the rate constant of the reaction,

t is the time of the reaction

a is the initial concentration of the reactant

a-x is the remaining concentration at time (t)

a = 45.0 g the initial concentration at t = 0 year.

at t = 22 year, a-x = 22.5 g

then k = (1/22 year) ln (45.0 g/ 22.5 g) = 0.0315 year-1.

the half-life time (t1/2) = ln 2/ k = 0.693/ k =0.693/ 0.0315 = 22 years.

part 2:

using the same law k = (1/t) ln (a/a-x),

using the given data; a = 308.0 g and a-x = 4.8125 g and the calculated k from the part 1 (k = 0.0315 year-1)

t = (1/k) ln (a/a-x) = (1/0.0315) ln (308.0/ 4.8125) = 132.02 years