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laila [671]
3 years ago
7

What is the maximum amount of kci that can dissolve in 200 g of water?

Chemistry
2 answers:
Anuta_ua [19.1K]3 years ago
7 0

The solubility of KCl in water is 34.2 g per 100 g of water.

That means at STP, 34.2 g of KCl is dissolved in 100 g of water.

So, 100 g of water dissolved= 34.2 g of KCl

1 g of water dissolved=\frac{34.2 g of KCl}{100}

200 g of water dissolved=\frac{34.2 g of KCl}{100}\times 200

=68.4 g of KCl

So, the maximum amount of KCl that can dissolve in 200 g of water is 68.4 g.



Svetradugi [14.3K]3 years ago
5 0
The correct answer is 68 g
hope this helps!!
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The amount of heat released by the sample has been 22.54 kJ. Thus, option C is correct.

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q=mc\Delta T

<h3 /><h3>Computation for the heat absorbed</h3>

The iron and calorimeter are in side the closed system. Thus, the energy released by the sample, has been equivalent to the energy absorbed by the calorimeter.

q_{released}=q_{absorbed}\\&#10;q_{released}=m_{calorimeter}\;c_{calorimeter}\;\Delta T

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Learn more about specific heat, here:

brainly.com/question/2094845

6 0
2 years ago
You have 100 mL of a solution of benzoic acid in water; the amount of benzoic acid in the solution is estimated to be about 0.30
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Answer:

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Explanation:

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To determine the amount of acid remaining using the formula:\dfrac{(final \ mass \ of \ solute)_{water}}{(initial \ mass \ of \ solute )_{water}} = (\dfrac{v_2}{v_1+v_2\times k_d})^n

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v_1 = volume of organic solvent = 20-mL

n = numbers of extractions = 4

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k_d = distribution coefficient = 10

∴

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = (\dfrac{100 \ ml}{100 \ ml +20 \ ml \times 10})^4

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = (\dfrac{100 \ ml}{100 \ ml +200 \ ml})^4

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\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = 0.012345

Thus, the final amount of acid left in the water = 0.012345 * 0.30

= 0.00370 g

3 0
3 years ago
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