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OverLord2011 [107]
4 years ago
6

Gold(ill) hydroxide is used in medicine, porcelain making, and gold plating. It is quite insoluble in aqueous solution. Which of

the following substances, if any, can increase the solubility of this compound in water?a) Ca(OH)2(aq) b) NH3(aq) c) NaCl(aq) d) HBr(aq) e) Gold(III) hydroxide is an ionic solid compound; its solubility is not affected by any of the above reagents.
Chemistry
1 answer:
erik [133]4 years ago
6 0

Answer:

NH3(aq)

Explanation:

Gold III hydroxide is an inorganic compound also known as auric acid. It can be dehydrated at about 140°C to yield gold III oxide. Gold III hydroxide is found to form precipitates in alkaline solutions hence it is not soluble in calcium hydroxide.

However, gold III hydroxide forms an inorganic complex with ammonia which makes the insoluble gold III hydroxide to dissolve in ammonia solution. The equation of this complex formation is shown below;

Au(OH)3(s) + 4 NH3(aq) -------> [Au(NH3)4]^3+(aq) + 3OH^-(aq)

Hence the formation of a tetra amine complex of gold III will lead to the dissolution of gold III hydroxide solid in aqueous ammonia.

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Explanation:

You would have to add up the atomic masses of all the compounds in the compound, making sure you include how many molecules of each are in the compound

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Cu = 63.55 u

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(Atomic mass of Cu) + (Atomic mass of S) + 4(Atomic Mass of O)

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Which is a conjugate pair in the following equilibrium? H2C2O4(aq) + HPO4(aq) 2- HC2O4(aq)- + H2PO4(aq)-
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