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tatiyna
3 years ago
6

In the electrolysis of molten fei3, which product forms at the cathode?

Chemistry
1 answer:
harina [27]3 years ago
6 0
  Then  product  forms   at  the  cathode is  Fe  metal

Fei3   
undergo   dissociate  to  form  fe^  3+    +   3I^-

Fe^3+  move  to  the  negative   electrode(cathode)  where   they  are   reduced   to  metallic  iron  by  gaining 3e-

that  is  Fe^3+(aq)   +  3e-  = Fe(s)
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What is the molarity of a solution that contains 87.75g of NaCI in 500.ml of solution
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Answer:

M = 3.0 mol/L.

Explanation:

  • We can calculate the molarity of a solution using the relation:

<em>M = (mass x 1000) / (molar mass x V)</em>

  • M is the molarity "number of moles of solute per 1.0 L of the solution.
  • mass is the mass of the solute (g) (m = 87.75 g of NaCl).
  • molar mass of NaCl = 58.44 g/mol.
  • V is the volume of the solution (ml) (V = 500.0 ml).

∴ M = (mass x 1000) / (molar mass x V) = (87.75 g x 1000) / (58.44 g/mol x 500.0 ml) = 3.0 mol/L.

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Answer:

                    Percent by mass of water is 56%

Explanation:

                    First of all calculate the mass of hydrated compound as,

Mass of Sodium = Na × 2 = 22.99 × 1 = 45.98 g

Mass of Sulfur = S × 1 = 32.06 × 1 = 32.06 g

Mass of Oxygen = O × 14 = 16 × 14 = 224 g

Mass of Hydrogen = H × 20 = 1.01 × 20 = 20.2 g

Mass of Na₂S0₄.10H₂O = 322.24 g

Secondly, calculate mass of water present in hydrated compound. For this one should look for the coefficient present before H₂O in molecular formula of hydrated compound. In this case the coefficient is 10, so the mass of water is...

Mass of water = 10 × 18.02

Mass of water = 180.2 g

Now, we will apply following formula to find percent of water in hydrated compound,

           %H₂O  =  Mass of H₂O / Mass of Hydrated Compound × 100

Putting values,

                                      %H₂O  = 180.2 g / 322.24 g × 100

                                           %H₂O =  55.92 % ≈ 56%

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