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REY [17]
3 years ago
9

A 11.97g sample of NaBr contains 22.34 % Na by mass. Considering the law of constant composition (definite proportions), how man

y grams of sodium does a 9.51g sample of sodium bromide contain?
Chemistry
1 answer:
kap26 [50]3 years ago
3 0
<h3>Answer:</h3>

2.125 g

<h3>Explanation:</h3>

We have;

  • Mass of NaBr sample is 11.97 g
  • % composition by mass of Na in the sample is 22.34%

We are required to determine the mass of 9.51 g of a NaBr sample.

  • Based on the law of of constant composition, a given sample of a compound will always contain the sample percentage composition of a given element.

In this case,

  • A sample of 11.97 g of NaBr contains 22.34% of Na by mass
  • Therefore;

A sample of 9.51 g of NaBr will also contain 22.345 of Na by mass

  • But;

% composition of an element by mass = (Mass of element ÷ mass of the compound) × 100

  • Therefore;

Mass of the element = (% composition of an element × mass of the compound) ÷ 100

Therefore;

Mass of sodium = (22.34% × 9.51 g) ÷ 100

                         = 2.125 g

Thus, the mass of sodium in 9.51 g of NaBr is 2.125 g

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<h3>Avogadro's hypothesis </h3>

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<h3>How to determine the atoms in 0.58 mole of Mg </h3>

1 mole of Mg = 6.02×10²³ atoms

Therefore,

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brainly.com/question/26141731

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