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Elodia [21]
2 years ago
13

Two samples of sodium chloride with different masses were decomposed into their constituent elements. One sample produced 2.55 g

of sodium and 3.93 g of chlorine. Being consistent with the law of constant composition, also called the law of definite proportions or law of definite composition, which set of masses could be the result of the decomposition of the other sample
Chemistry
1 answer:
Charra [1.4K]2 years ago
3 0

Answer:

4.71 g of sodium and 7.25 g of chlorine

Explanation:

<em>Note: The question is incomplete. the complete question is given below:</em>

<em>Two samples of sodium chloride with different masses were decomposed into their constituent elements. One sample produced 2.55 g of sodium and 3.93 g of chlorine. Being consistent with the law of constant composition, also called the law of definite proportions or law of definite composition, which set of masses could be the result of the decomposition of the other sample?</em>

<em>4.71 g of sodium and 3.30 g of chlorine</em>

<em>4.71 g of sodium and 7.25 g of chlorine</em>

<em>4.71 g of sodium and 1.31 g of chlorine</em>

<em>4.71 g of sodium and 13.7 g of chlorine</em>

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

This means that irrespective of the source of any sample of a pure chemical compound, the constituents elements are always combined in the same mass ratio.

In the first sample, the mass ratio of Sodium to chlorine is given below:

mass of sodium = 2.55 g

mass of chlorine = 3.93 g

mass ratio; sodium to chlorine = 2.55 / 3.93 = 0.65

From the set of masses give above, we can determine the result of the decomposition of the second sample.

4.71 g of sodium and 3.30 g of chlorine

mass ratio; sodium to chlorine = 4.71 / 3.30 = 1.43

4.71 g of sodium and 7.25 g of chlorine

mass ratio; sodium to chlorine = 4.71 / 7.25 = 0.65

4.71 g of sodium and 1.31 g of chlorine

mass ratio; sodium to chlorine = 4.71 / 1.31 = 3.59

4.71 g of sodium and 13.7 g of chlorine

mass ratio; sodium to chlorine = 4.71 / 13.7 = 0.34

From the results above, the correct set of masses for the second sample is 4.71 g of sodium and 7.25 g of chlorine

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The force of attraction between a divalent cation and a divalent anion is 1.64 x 10-8 N. If the ionic radius of the cation is 0.
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The radius of the anion is 7.413 nm

<h3>How to calculate the force of attraction between charges</h3>

The force of attraction (F) is given by the formula:

  • F = (1/4π∈r²)(Zc*e)(Za*e)

where:

∈ = permittivity of free space = 8.85*10⁻¹⁵ F/m

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Therefore based on the equation of force of attraction:

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