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OverLord2011 [107]
2 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]2 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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For the reaction A (g) → 2 B (g), K = 14.7 at 298 K. What is the value of Q for this reaction at 298 K when ∆G = -20.5 kJ/mol?
harina [27]

Answer:

Q= 245 =2.5 * 10^2

Explanation:

ΔG = ΔGº + RTLnQ, so also ΔGº= - RTLnK

R= 8,314 J/molK, T=298K

ΔGº= - RTLnK = - 6659.3 J/mol = - 6.7 KJ/mol

ΔG = ΔGº + RTLnQ → -20.5KJ/mol = - 6.7 KJ/mol + 2.5KJ/mol* LnQ

→ 5.5 = LnQ → Q= 245 =2.5 * 10^2

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Burka [1]

Better than i am and very precice


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2 years ago
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A laboratory utilizes a mixture of 10% dimethyl sulfoxide (DMSO) in the freezing and long-term storage of embryonic stem cells.
Mars2501 [29]

Answer:

The correct answer is "1.0100".

Explanation:

Let the volume of mixture be 100 ml.

then,

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DMSO will be:

= 10\times 1.1004

= 11.004 \ g

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3 0
2 years ago
What volume of a 0.160 m solution of koh must be added to 450.0 ml of the acidic solution to completely neutralize all of the ac
inna [77]
An  acidic  solution  is 0.1M  in  HCl  and  0.2  H2so4. volume  is  equal  to  no  of  moles  divided  by  molarity.  
number  of  moles  of  HCl is  450ml x 0.1  divided  by 1000  which is equal  to 0.045 moles
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Number  of  moles  of  H2so4  is  450ml x 0.2  divided  by  1000  which  is  equal  to  0.09  moles
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4 0
2 years ago
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If a large marshmallow has a volume of 2.75 in3 and density of 0.242 g/cm3, how much would it weigh in grams? 1 in3=16.39 cm3.
Anestetic [448]
<span>
•   </span>Volume of the marshmallow:

V = 2.75 in^3          (but, 1 in^3 = 16.39 cm^3)

V = 2.75 × 16.39 cm^3

V = 2.75 × 16.39 cm^3

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•   Density:

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<span>•  </span>Mass:

m = d × V

m = (0.242 g/cm^3) × (45.0725 cm^3)

m = (0.242 g/cm^3) × (45.0725 cm^3)

m = 10.907545 g

m ≈ 10.9 g   <——<span>—  this is the answer.


I hope this helps. =)
</span>
4 0
2 years ago
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