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Preparation of Carboxylic Acids

When two mole...

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When two moles of [Co(NH3)5Cl]Cl2 is treated with excess silver nitrate solution, the number of moles of silver chloride formed is

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Q. 0.02 mole of

[C o (N H_{3})_{5} B r] C l_{2}

and 0.02 mole of

[C o (N H_{3})_{5} C l] S O_{4}

are present in 200 cc of a solution X. The number of moles of the preceipitates Y and Z that are formed when the solution X is treated with excess silver nitrate and excess barium chloride are respectively :

Q. When 0.01 mole of a cobalt complex is treated with excess silver nitrate solution, 4.305 g silver chloride is precipitated. The formula of the complex is:

Q. The number of chloride ions which would be precipitated, when

C r C l_{3} .4 N H_{3}

is treated with silver nitrate solution:

Q. 2.66 g chloride of a metal when treated with silver nitrate solution give 2.87 g of silver chloride. 3.37 g of another chloride of the same metal give 5.74 g of silver chloride when treated with silver nitrate solution. Show that the results are in agreement with a law of chemical combination.

Q. What do you observes when:
Hydrogen chloride gas is passed through Silver nitrate solution and the product thus form treated with excess of Ammonium hydroxide?

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