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Lubov Fominskaja [6]
3 years ago
11

A compound has a percent composition of 15.24% sodium (molar mass = 22.99

Chemistry
1 answer:
RUDIKE [14]3 years ago
8 0

The empirical formula of the compound obtained from the question given is NaBrO₃

<h3>Data obtained from the question </h3>
  • Sodium (Na) = 15.24%
  • Bromine (Br) = 52.95%
  • Oxygen (O) = 31.81%
  • Empirical formula =?

<h3>How to determine the empirical formula </h3>

The empirical formula of the compound can be obtained as illustrated below:

Divide by their molar mass

Na = 15.24 / 22.99 = 0.663

Br = 52.95 / 79.90 = 0.663

O = 31.81 / 16 = 1.988

Divide by the smallest

Na = 0.663 / 0.663 = 1

Br = 0.663 / 0.663 = 1

O = 1.988 / 0.663 = 3

Thus, the empirical formula of the compound is NaBrO₃

Learn more about empirical formula:

brainly.com/question/24297883

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Answer:

The specific heat capacity of the unknown metal is C = 0.6991 J/g°C = 0.1671 cal/g°C

Explanation:

Heat lost by the unknown metal is equal to the heat gained by the water and aluminium cup.

Given,

Mass of unknown metal = 77 g

Initial Temperature of unknown metal = 99°C

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Heat gained by Aluminium cup

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Heat lost by unknown metal = (Heat gained by water) + (Heat gained by Aluminium cup)

5621C = 3,767.4 + 162.00448 = 3,929.40448

5621C = 3,929.40448

C = (3,929.40448 ÷ 5621) = 0.6991 J/g°C

C = 0.6991 J/g°C = 0.1671 cal/g°C

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