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Stels [109]
2 years ago
11

if B is completely insoluble in water. Your description should include the volume of solvent required.6b) Assuming that 2 mg of

the impurity B are present along with 100 mg of A, describe how you can purify A if B has the same solubility behavior as A. Will one crystallization produce pure A
Chemistry
1 answer:
Misha Larkins [42]2 years ago
8 0

Answer:

1st step : mix the mixture with water that way A will dissolve while B will remain insoluble.

2nd step :To get B from the solution, filter the mixture and get B

3rd step : To get A from the solution evaporate the new solution

Explanation:

Assuming 2mg of impurity B to be present

100 mg of A is present as well

<u>Method of purifying A given that B is of same solubility </u>

Dissolve the 100 mg of A with 30 mL

1st step : mix the mixture with water that way A will dissolve while B will remain insoluble.

2nd step :To get B from the solution, filter the mixture and get B

3rd step : To get A from the solution evaporate the new solution

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Select the correct answer. When an atom in a reactant loses electrons, what happens to its oxidation number? A. Its oxidation nu
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C. Its oxidation number increases.

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g Consider an ideal atomic gas in a cylinder. The upper part of the cylinder is a moveable piston of negligible weight. The heig
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A cylindrical weight with a mass of 3 kg is dropped onto the piston from a height of 10 m. The entropy of the gas is 1.18 J/K and the change in the entropy of the environment is -1.18 J/K.

A cylindrical weight with a mass (m) of 3 kg is dropped, that is, its initial velocity (u) is 0 m/s and travels 10 m (s). Assuming the acceleration (a) is that of gravity (9.8 m/s²). We can calculate the velocity (v) of the weight in the instant prior to the collision with the piston using the following kinematic equation.

v^{2} = u^{2} + 2as = 2 (9.8 m/s^{2} ) (10m) \\\\v = 14 m/s

The object with a mass of 3 kg collides with the piston at 14 m/s, The kinetic energy (K) of the object at that moment is:

K = \frac{1}{2} m v^{2} = \frac{1}{2} (3kg) (14m/s)^{2} = 294 J

The kinetic energy of the weight is completely converted into heat transferred into the gas cylinder. Thus, Q = 294 J.

Given all the process is at 250 K (T), we can calculate the change of entropy of the gas using the following expression.

\Delta S_{gas} = \frac{Q}{T} = \frac{294 J}{250K} = 1.18 J/K

The change in the entropy of the environment, has the same value but opposite sign than the change in the entropy of the gas. Thus, \Delta S_{env} = -1.18 J/K

A cylindrical weight with a mass of 3 kg is dropped onto the piston from a height of 10 m. The entropy of the gas is 1.18 J/K and the change in the entropy of the environment is -1.18 J/K.

Learn more: brainly.com/question/22655760

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