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alina1380 [7]
3 years ago
6

Read the descriptions below of two substances and an experiment on each. Decide whether the result of the experiment tells you t

he substance is a pure substance or a mixture, if you can.
Sample A is of a coarse grey powder with a faint unpleasant smell. of the powder is put into a funnel lined with a sheet of thick paper. Distilled water is poured slowly over the powder. Most of the powder disappears, but of a gritty black sand-like material is left on the surface of the paper. Pouring more water over the black material doesn't change how much of it there is.

Sample B is a solid yellow cube with a total mass of 50.0 g. The cube is put into a beaker filled with 250. mL of water. The cube collapses into a small pile of orange powder at the bottom of the beaker. When this powder is filtered out, dried and weighed, it has a total mass of 29.9 g. If the experiment is repeated with 500. mL of water, the powder that's left over has a mass of 10.0 g.
Chemistry
1 answer:
Lelu [443]3 years ago
8 0

Answer:

Sample A - mixture

Sample B- Mixture

Explanation:

Looking at sample A, we can see that as water was poured over sample A, the sample was separated into its components as the powder disappeared leaving behind a gritty black sand-like material on the surface of the paper. A separation of the mixture has taken place.

In sample B, we can clearly see that it is a mixture because the amount of solid recovered is much less than the total mass of the solid put into the beaker. The sample must have been separated into its components.

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First step is to calculate the mass of Ag in each compound separately:
From the periodic table: 
molar mass of Ag is 107.87 gm
molar mass of Cl is 35.45 gm
molar mass of Br is 79.9 gm
For AgCl, mass % of Ag = [107.87/143.32] x 100 = 75.26%
For AgBr, mass % of Ag = [107.87/187.77] x 100 = 57.45 %

Second step is to calculate the mass % of each compound in the mixture:
Assume mass % of AgCl is y and that of AgBr is (1-y) as the total percentage is 100% or 1
0.6094 = 0.7526 y + 0.5745 (1-y)
y = 0.8716
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mass % of chlorine = (1-0.6094)(0.8716) x 100 = 34.044%
mass % of bromine = 39.04 - 34.044 = 5.056%
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