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Mkey [24]
4 years ago
10

WORTH 99 POINTS PLEASE ANSWER!!! The table shows the solubility of two substances in water at 20 °C. Solubility of Substances Ma

ss (kg) Solubility of Sodium Chloride (g per 100 g of water) Solubility of Lead Nitrate (g per 100 g of water) 100 L K 200 36 54 Part 1: What will be the values of L and K for 100 kg of each substance? Part 2: Explain your answer for Part 1.
Chemistry
2 answers:
DaniilM [7]4 years ago
8 0

Answer:

L is equal to 18 and K is equal to 27

If the solubility for L in 200kg of water is 36g then to find the solubility in 100kg of water, you would divided 36 by 2 to find half. You would do the same to find K, if in 200kg of water K's solubility is 54, you would divided 54 by 2 to get 27.

Explanation:

also you only set this question to 5 points not 99, if this helps at all could you mark me brainliest? :) have a great day!

Lunna [17]4 years ago
3 0

Answer:

the difference between k and l is that (k) is more soluble than (L) because the mass of 200 has lower grams than 100.

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<u>Answer:</u> The energy released for the decay of 3 grams of 230-Thorium is 2.728\times 10^{-15}J

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First we have to calculate the mass defect (\Delta m).

The equation for the alpha decay of thorium nucleus follows:

_{90}^{230}\textrm{Th}\rightarrow _{88}^{226}\textrm{Ra}+_2^{4}\textrm{He}

To calculate the mass defect, we use the equation:

Mass defect = Sum of mass of product - Sum of mass of reactant

\Delta m=(m_{Ra}+m_{He})-(m_{Th})

\Delta m=(225.9771+4.008)-(229.9837)=1.4\times 10^{-3}amu=2.324\times 10^{-30}kg

(Conversion factor: 1amu=1.66\times 10^{-27}kg )

To calculate the energy released, we use Einstein equation, which is:

E=\Delta mc^2

E=(2.324\times 10^{-30}kg)\times (3\times 10^8m/s)^2

E=2.0916\times 10^{-13}J

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We need to calculate the energy released for the decay of 3 grams of thorium. By applying unitary method, we get:

As, 230 grams of Th release energy of = 2.0916\times 10^{-13}J

Then, 3 grams of Th will release energy of = \frac{2.0916\times 10^{-13}}{230}\times 3=2.728\times 10^{-15}J

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3 years ago
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