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3 years ago

Calculate the molar solubility of AgCl in 0.05M NaCl. (Ksp = 1.77 x 10-10)

Chemistry

1 answer:

Aneli [31]3 years ago

6 0

Answer: 3.54 x 10^{-9}

Explanation:

For this Case we have keep in mind the effect of the Common Ion, which is Cl.

You can find the amount of silver ions at equilibrium from Ksp since You already know the chloride ion concentration. Keep in mind that only a few amount of silver chloride will be dissolved. This means the chloride concentration will stay in 0.05 M, Then I can use Ksp to solve for the silver ion concentration. So you can use the following equation.

AgCl Ksp = Ksp/[Cl^-]= (1.77 x 10^{-10})/(0.05) = 3.54 x 10^{-9}

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3 years ago

The heat of vaporization of water is 40.66 kJ/mol. How much heat is absorbed when 3.11 g of water boils at atmospheric pressure?

Roman55 [17]

Answer:

The amount of heat that is absorbed when 3.11 g of water boils at atmospheric pressure is 7.026 kJ.

Explanation:

A molar heat of vaporization of 40.66 kJ / mol means that 40.66 kJ of heat needs to be supplied to boil 1 mol of water at its normal boiling point.

To know the amount of heat that is absorbed when 3.11 g of water boils at atmospheric pressure, the number of moles represented by 3.11 g of water is necessary. Being:

H: 1 g/mole
O: 16 g/mole

the molar mass of water is:

H₂O= 2* 1 g/mole + 16 g/mole= 18 g/mole

So: if 18 grams of water are contained in 1 mole, 3.11 grams of water in how many moles are present?

$moles of water=\frac{3.11 grams*1 mole}{18 gramos}$

moles of water= 0.1728

Finally, the following rule of three can be applied: if to boil 1 mole of water at its boiling point it is necessary to supply 40.66 kJ of heat, to boil 0.1728 moles of water, how much heat is necessary to supply?

$heat=\frac{0.1728 moles*40.66 kJ}{1 mole}$

heat= 7.026 kJ

<u><em>The amount of heat that is absorbed when 3.11 g of water boils at atmospheric pressure is 7.026 kJ.</em></u>

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