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erastovalidia [21]
3 years ago
7

Calculate the mass, in grams, of cucl2 (mw = 134.452 g/mol) required to prepare 250.0 ml of a 6.11 % w/v cu2 (mw = 63.546 g/mol)

solution.
Chemistry
1 answer:
Sever21 [200]3 years ago
6 0

6.11% w/v of Cu2+ implies that 6.11 g of Cu2+ is present in 100 ml of the solution

therefore,  250 ml of the solution would have: 250 ml * 6.11 g/100 ml = 15.275 g

# moles of Cu2+ = 15.275 g/63.546 g mole-1 = 0.2404 moles

1 mole of CuCl2 contain 1 mole of Cu2+ ion

Hence, 0.2404 moles of Cu2+ would correspond to 0.2404 moles of CuCl2

Molar mass of CuCl2 = 134.452 g/mole

The mass of CuCl2 required = 0.2404 moles * 134.452 g/mole = 32.32 grams

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You have 4 moles of oxygen gas in a flask. 4 moles of helium gas is added. What happens to the total pressure of the gases in th
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Answer: The correct option is (c). The total pressure doubles.

Solution:

Initially,  only 4 moles of oxygen gas were present in the flask.

p_{O_2}=Tp_1\times X_{O_2}  (X_{O_2}=\frac{4}{4}) ( according to Dalton's law of partial pressure)

p_{O_2}=Tp_1\times 1=Tp_1....(1)

Tp_1= Total pressure when only oxygen gas was present.

Final total pressure when 4 moles of helium gas were added:

X'_{O_2}=\frac{4}{8}=\farc{1}{2},X_{He}=\frac{4}{8}=\frac{1}{2}

partial pressure of oxygen in the mixture :

Since, the number of moles of oxygen remains the same, the partial pressure of oxygen will also remain the same in the mixture.

p_{O_2}=Tp_2\times X'_{O_2}=Tp_2\times \frac{1}{2}

Tp_2= Total pressure of the mixture.

from (1)

Tp_1=Tp_2\times X'_{O_2}=Tp_2\times \frac{1}{2}

On rearranging, we get:

Tp_2=2\times Tp_1

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1. Draw a wedge/dash structure for trans-1,2-dimethylcyclohexane.
Angelina_Jolie [31]

Answer:

The structure with the ring flipped is the most stable

Explanation:

We have the  trans 1,2 - dimethylcyclohexane. With the wedge/dash structure we could not figure is this form is stable (If we do a comparison with the cis structure). But when we do a chair structure and ring flipped structure, this is easier to look.

The picture attached shows the structures, they are labeled as 1, 2 and 3, according to this problem.

In the chair structure, according to the picture below, you can see that both methyls are heading in the axial positions of the ring (One facing upward and the other downward). This is pretty stable, however, when the methyls are in those positions, the methyl position 1, can undergoes an 1,3 diaxial interactions with the hydrogens atoms (They are not drawn, but still are there), so this interaction makes this structure a little less stable that it can be.

On the other side, the ring flipped structure, we can see that both methyls are in the equatorials positions of the ring, and in these positions, it can avoid the 1,4 diaxial interactions with the hydrogens atoms, making this structure the most stable structure.

Hope this helps

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