Solved: "Three compounds containing potassium and oxygen are compared...."

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elvishwitchxyp

Answered question

2022-01-10

Three compounds containing potassium and oxygen are compared. Analysis shows that for each 1.00 g of O, the compounds have 1.22 g, 2.44 g, and 4.89 g of K, respectively. Show how these data support the law of multiple proportions.

Answer & Explanation

servidopolisxv

Beginner2022-01-11Added 27 answers

Starting with the compound with the least mass of potassium, create a ratio for the rest of the compounds:

1.00 g O = 1.22 g K

Then, divide the next greater value of mass of K by the least value for the mass of K

2.44 g K \div 1.22 g K = 2

Repeat with the next greater value of mass of K

4.89 g K \div 1.22 g K = 4.008 \approx 4

The established ratio was that for each 1.22 gK there is 100 g O, thus the quotients acquired are the ratio of oxygen per each mass of potassium presented
a.) For every 1.22 g K there is 1.00 g O;

\frac{1}{1}

b.) For every 2.44 g K there is 2 of 1.00 g O;

\frac{1}{2}

c.) For every 4.89 g K there is approximately 4 of 1.00 g of O;

\frac{1}{4}

sukljama2

Beginner2022-01-12Added 32 answers

The law of proportion states that elements combine in whole number ratios.
The gram readings for K are multiples of each other, both in grams and moles.
Let us compare the ratios:

\frac{2.44 g r a m s}{1.22 g r a m s} = 2

\frac{4.89 g r a m s}{2.44 g r a m s} = 2

Therefore, Potassium always combines with Oxygen ina ratio of 2 is to 1.

alenahelenash

Expert2023-05-26Added 556 answers

Result:
The given data, with the ratios of potassium to oxygen being

1.22

2.44

, and

4.89

, supports the law of multiple proportions.
Solution:
Let's consider the three compounds containing potassium (

K

) and oxygen (

O

). The given data states that for every

1.00

g of oxygen, the compounds contain

1.22

2.44

g, and

4.89

g of potassium, respectively.
Let's calculate the ratios of potassium to oxygen in each compound:
Compound 1:

Ratio of K to O = \frac{1.22 g}{1.00 g} = 1.22

Compound 2:

Ratio of K to O = \frac{2.44 g}{1.00 g} = 2.44

Compound 3:

Ratio of K to O = \frac{4.89 g}{1.00 g} = 4.89

Now, let's analyze the ratios we obtained:
The ratio of potassium to oxygen in Compound 2 (

2.44

) is exactly double the ratio in Compound 1 (

1.22

). Similarly, the ratio in Compound 3 (

4.89

) is exactly double the ratio in Compound 2 (

2.44

).
These ratios are all whole numbers, and the ratios between them are also whole numbers. This observation supports the law of multiple proportions, which states that elements combine in ratios of small whole numbers to form compounds.

star233

Skilled2023-05-26Added 403 answers

To show how these data support the law of multiple proportions, let's consider the ratios of potassium to oxygen in each compound.
Compound 1:

\frac{mass of K}{mass of O} = \frac{1.22 g}{1.00 g} = 1.22

Compound 2:

\frac{mass of K}{mass of O} = \frac{2.44 g}{1.00 g} = 2.44

Compound 3:

\frac{mass of K}{mass of O} = \frac{4.89 g}{1.00 g} = 4.89

We can observe that the ratios are in simple whole number multiples of each other. The ratio between the mass of potassium and oxygen in Compound 2 is twice the ratio in Compound 1, and the ratio in Compound 3 is four times the ratio in Compound 1. This is consistent with the law of multiple proportions, which states that when two elements combine to form multiple compounds, the ratios of the masses of one element that combine with a fixed mass of the other element will be in small whole number ratios.
Therefore, these data support the law of multiple proportions.

karton

Expert2023-05-26Added 613 answers

Let's assume the three compounds are represented by the formulas

K_{x} O_{y}

K_{2 x} O_{2 y}

, and

K_{4 x} O_{4 y}

, respectively.
According to the given data, the mass of oxygen (O) in each compound is 1.00 g.
For the first compound,

K_{x} O_{y}

, the mass of potassium (K) is 1.22 g. Therefore, the ratio of K to O can be calculated as:

\frac{mass of K in K_{x} O_{y}}{mass of O in K_{x} O_{y}} = \frac{1.22 g}{1.00 g} = 1.22

For the second compound,

K_{2 x} O_{2 y}

, the mass of potassium (K) is 2.44 g. Hence, the ratio of K to O is:

\frac{mass of K in K_{2 x} O_{2 y}}{mass of O in K_{2 x} O_{2 y}} = \frac{2.44 g}{1.00 g} = 2.44

Finally, for the third compound,

K_{4 x} O_{4 y}

, the mass of potassium (K) is 4.89 g. The ratio of K to O is:

\frac{mass of K in K_{4 x} O_{4 y}}{mass of O in K_{4 x} O_{4 y}} = \frac{4.89 g}{1.00 g} = 4.89

By examining the ratios, we can observe that they are in simple integer multiples of each other: 1.22, 2.44, and 4.89. This finding aligns with the law of multiple proportions, which states that when two elements combine to form different compounds, the ratio of masses of one element that combine with a fixed mass of the other can be expressed as a ratio of small whole numbers.
Therefore, the given data supports the law of multiple proportions.

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