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skelet666 [1.2K]
4 years ago
10

Which substance is most soluble in water? 1) (NH4)3PO4 2)Cu(OH)2 3) AG2SO4 4) CACO3

Chemistry
2 answers:
ivann1987 [24]4 years ago
8 0

Answer: Option (1) is the correct answer.

Explanation:

When dissolved in water, ammonium phosphate (NH_{4})_{3}PO_{4} completely dissociates into ions and it gets attached to the opposite ends of water.

Therefore, it is soluble in water.

Whereas silver sulfate (Ag_{2}SO_{4}) slightly dissolves into water as it gives very less Ag^{+} ions. Therefore, silver sulfate has low solubility.

On the other hand, both Ca(OH)_{2} and CaCO_{3} are insoluble in water.

Thus, we can conclude that out of the given options, (NH_{4})_{3}PO_{4} is most soluble in water.

valina [46]4 years ago
4 0
The answer is Cu(OH)2
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What is solubility? If a substance completely dissolves in water, does it have high or low solubility?
svet-max [94.6K]

Answer:solubility is  the ability to dissolve or break down something, especially in water

Explanation:

if something dissolves in water, it will have a high dissolvability

8 0
4 years ago
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Calculate [H3O+] and [OH−] for each of the following solutions at 25 ∘C given the pH. pH= 8.74, pH= 11.38, pH= 2.81
Gnom [1K]

Answer:

Explanation:

Given parameters;

pH  = 8.74

pH = 11.38

pH = 2.81

Unknown:

concentration of hydrogen ion and hydroxyl ion for each solution = ?

Solution

The pH of any solution is a convenient scale for measuring the hydrogen ion concentration of any solution.

It is graduated from 1 to 14

      pH = -log[H₃O⁺]

      pOH = -log[OH⁻]

 pH + pOH = 14

Now let us solve;

   pH = 8.74

             since  pH = -log[H₃O⁺]

                           8.74 =  -log[H₃O⁺]

                           [H₃O⁺] = 10⁻^{8.74}

                             [H₃O⁺]  = 1.82 x 10⁻⁹mol dm³

       pH + pOH = 14

                 pOH = 14 - 8.74

                  pOH = 5.26

                  pOH = -log[OH⁻]

                     5.26  = -log[OH⁻]

                     [OH⁻] = 10^{-5.26}

                      [OH⁻] = 5.5 x 10⁻⁶mol dm³

2.  pH = 11.38

             since  pH = -log[H₃O⁺]

                           11.38 =  -log[H₃O⁺]

                           [H₃O⁺] = 10⁻^{11.38}

                             [H₃O⁺]  = 4.17 x 10⁻¹² mol dm³

           pH + pOH = 14

                 pOH = 14 - 11.38

                  pOH = 2.62

                  pOH = -log[OH⁻]

                     2.62  = -log[OH⁻]

                     [OH⁻] = 10^{-2.62}

                      [OH⁻] =2.4 x 10⁻³mol dm³

3. pH = 2.81

             since  pH = -log[H₃O⁺]

                           2.81 =  -log[H₃O⁺]

                           [H₃O⁺] = 10⁻^{2.81}

                             [H₃O⁺]  = 1.55 x 10⁻³ mol dm³

           pH + pOH = 14

                 pOH = 14 - 2.81

                  pOH = 11.19

                  pOH = -log[OH⁻]

                     11.19  = -log[OH⁻]

                     [OH⁻] = 10^{-11.19}

                      [OH⁻] =6.46 x 10⁻¹²mol dm³

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Explanation:

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