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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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A compressed gas cylinder is filled with 5270 g of argon gas. The pressure inside the cylinder is 2050 psi at a temperature of 1
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Answer:

A compressed gas cylinder is filled with 5270 g of argon gas.

The pressure inside the cylinder is 2050 psi at a temperature of 18C.

The valve to the cylinder is opened and gas escapes until the pressure inside the cylinder is 650. psi and the temperature are 26 C.

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Explanation:

First, calculate the volume of argon gas that is present in the gas cylinder by using the ideal gas equation:

Mass of Ar gas is --- 5270g.

The number of moles of Ar gas:

Number of moles of Ar gas=\frac{given mass of Ar}{its atomic mass} \\                                             =\frac{5270g}{39.948g/mol} \\                                             =131.9mol

Temperature T=(18+273)K=291K

Pressure P=2050psi

2050* 0.0680atm\\\\=139.4atm\\

Volume V=?

PV=nRT\\139.4atm * V=131.9mol *0.0821L.atm.mol-1.K-1 * 291K\\=>V=\frac{131.9mol *0.0821L.atm.mol-1.K-1 * 291K}{139.4atm} \\=>V=22.6L

Using this volume V=22.6L

Pressure=650psi=44.2atm

Temperature T= (26+273)K=299K

calculate number of moles "n" value:

PV=nRT\\=>n=\frac{PV}{RT} \\=>n=\frac{44.2atm*22.6L}{0.0821L.atm.mol^-1K^-1* 299K} \\=>n=40.7mol

Mass of 40.7mol of Ar gas:

mass of Ar gas=number of moles * its atomic mass\\\\                        =40.7mol* 39.948g/mol\\\\                        =1625.8g

Answer:

The mass of Ar gas becomes 1625.8g.

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