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WITCHER [35]
3 years ago
13

How many total moles of ions are released when the following sample dissolves completely in water?

Chemistry
1 answer:
pogonyaev3 years ago
8 0

Answer:

                      8.55 × 10²⁴ Ions

Explanation:

                    Ammonium Chloride is an ionic compound which contains a monatomic anion (Cl⁻ ; Chloride) and a polyatomic cation (NH₄⁺ ; Ammonium).

Hence, when added in water Ammonium Chloride ionizes as;

                                      NH₄Cl   →    NH₄⁺  +  Cl⁻

Hence, we can say that it produces two ions when dissolved in water.

Also,

We know that 1 mole of any substance contains exactly 6.022 × 10²³ particles which is also called as Avogadro's Number. So in order to calculate the number of ions  contained by 7.1 moles of NH₄Cl, we will use following relation to first calculate the number of molecules as;

          Moles  =  Number of Molecules ÷ 6.022 × 10²³ Molecules.mol⁻¹

Solving for Number of Molecules,

          Number of Molecules  =  Moles × 6.022 × 10²³ Molecules.mol⁻¹

Putting values,

          Number of Molecules  =  7.1 mol × 6.022 × 10²³

          Number of Molecules  =  4.27 × 10²⁴ Molecules

So,

As,

                         1 Molecule of NH₄Cl contained  =  2 Ions

So,

              4.27 × 10²⁴ Molecules of NH₄Cl will contain  =  X ions

Solving for X,

                     X =  2 Ions × 4.27 × 10²⁴ Molecules / 1 Molecule

                    X =  8.55 × 10²⁴ Ions

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Ca(OH)2 + 2HNO3 -----> Ca(NO3)2 + H2O

Focus on the NO3. This is an odd problem and you usually do not focus on the complex ion. But this one works easiest if you do.

The problem now is going to be the oxygens. There are 2 with the Calcium and only 1 free one going to the water. (The NO3 has been taken care of in the last step).

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Concentration of sodium propionate = \frac{0.26}{1.20}[/tex]

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                                              = 0.17 mol

Hence, concentration of propionic acid = \frac{0.17}{1.20 L}

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                                                                  = 0.28 mol

Hence, concentration of sodium propionic acid will be calculated as follows.

                        \frac{0.28 mol}{1.20 L}

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                       = 4.87 + log \frac{0.23}{0.14}

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