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3 years ago

The addition of which substance will decrease the solubility of AgNO3 in water?

Chemistry

2 answers:

eduard3 years ago

8 0

Answer: The correct answer is Option 3.

Explanation:

Common ion effect is defined as the effect which occurs on equilibrium when a common ion (an ion which is already present in the solution) is added to a solution. This effect decreases the solubility of a solute.

In silver nitrate, the ions present in the solution are: $Ag^+\text{ and }NO_3^-$

For the given options:

Option 1: 0.1 M potassium chloride

The ions formed by the ionization of this solution are $K^+\text{ and }Cl^-$ . As, no common ions are present. So, addition of this substance will not decrease the solubility of silver nitrate solution.

Option 2: 0.1 M barium chloride

The ions formed by the ionization of this solution are $Ba^{2+}\text{ and }Cl^-$ . As, no common ions are present. So, addition of this substance will not decrease the solubility of silver nitrate solution.

Option 3: 0.1 M sodium nitrate

The ions formed by the ionization of this solution are $Na^+\text{ and }NO_3^-$ . As, nitrate ions are present which is a common ion. So, addition of this substance will decrease the solubility of silver nitrate solution.

Option 4: 0.1 M sodium carbonate

The ions formed by the ionization of this solution are $Na^+\text{ and }CO_3^{2-}$ . As, no common ions are present. So, addition of this substance will not decrease the solubility of silver nitrate solution.

Hence, the correct answer is Option 3.

sertanlavr [38]3 years ago

5 0

3 is correct because the .1 moles of the NO3 will be better option to decrease the solubility of the Silver Nitrate.

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Answer:

$\Delta S^{0}$ for the given reaction is -99.4 J/K

Explanation:

Balanced reaction: $\frac{1}{2}N_{2}(g)+\frac{3}{2}H_{2}(g)\rightarrow NH_{3}(g)$

$\Delta S^{0}=[1mol\times S^{0}(NH_{3})_{g}]-[\frac{1}{2}mol\times S^{0}(N_{2})_{g}]-[\frac{3}{2}mol\times S^{0}(H_{2})_{g}]$

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Plug in all the standard entropy values from available literature in the above equation:

$\Delta S^{0}=[1mol\times 192.45\frac{J}{mol.K}]-[\frac{1}{2}mol\times 191.61\frac{J}{mol.K}]-[\frac{3}{2}mol\times 130.684\frac{J}{mol.K}]=-99.4J/K$

So, $\Delta S^{0}$ for the given reaction is -99.4 J/K

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3 years ago

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3 years ago

If 842 grams of sodium hydroxide reacts with 750.0 grams of aluminum, how many grams of aluminum hydroxide should theoretically

Phantasy [73]

548.55 grams of aluminum hydroxide should theoretically form.

Explanation:

Balanced equation for the reaction:

3 NaOH + Al ⇒ Al(OH)3 +3 Na

DATA GIVEN:

mass of NaOH = 842 grams, atomic mass =39.9 grams/mole

mass of Al = 750 grams, atomic mass = 26.9 grams/mole

aluminum hydroxide theoretical yield = ?

Moles of NaOH reacted

number of moles = $\frac{mass}{atomic mass of 1 mole}$

putting the values in the equation

NaOH = $\frac{842}{39.9}$

= 21.1 MOLES OF NaOH

Al = $\frac{750}{26.9}$

= 27.8 moles

from the equation

from 3 moles of NaOH 1 mole of Al(OH)3 is produced

21.1 moles of NaOH will react to give x moles of Al(OH)3

$\frac{1}{3}$ = $\frac{x}{21.1}$

7.03 moles of Al(OH)3 is formed.

and

1 mole of Al(OH)3 is formed from 1 mole of Al in the reaction

so, 27.8 Moles will react to give give 27.8 moles of Al(OH)3 limiting reagent of the given reaction is NaOH

mass of Al(OH)3 =7.03 x 78 (atomic mass of Al(OH)3)

= 548.55 grams

theoretical yield from the given data is 548.55 grams

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