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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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How many moles of tungsten (W,183.85 g/lol are in 415 grams of tungsten?
vladimir1956 [14]

Given mass of tungsten, W = 415 g

Molar mass of tungsten, W = 183.85 g/mol

Calculating moles of tungsten from mass and molar mass:

415 g * \frac{1 mol}{183.85 g} = 2.26 mol W

7 0
1 year ago
A solution is made by dissolving 58.125 g of sample of an unknown, nonelectrolyte compound in water. The mass of the solution is
e-lub [12.9K]

Answer:

molecular weight (Mb) = 0.42 g/mol

Explanation:

mass sample (solute) (wb) = 58.125 g

mass sln = 750.0 g = mass solute + mass solvent

∴ solute (b) unknown nonelectrolyte compound

∴ solvent (a): water

⇒ mb = mol solute/Kg solvent (nb/wa)

boiling point:

  • ΔT = K*mb = 100.220°C ≅ 373.22 K

∴ K water = 1.86 K.Kg/mol

⇒ Mb = ? (molecular weight) (wb/nb)

⇒ mb = ΔT / K

⇒ mb = (373.22 K) / (1.86 K.Kg/mol)

⇒ mb = 200.656 mol/Kg

∴ mass solvent = 750.0 g - 58.125 g = 691.875 g = 0.692 Kg

moles solute:

⇒ nb = (200.656 mol/Kg)*(0.692 Kg) = 138.83 mol solute

molecular weight:

⇒ Mb = (58.125 g)/(138.83 mol) = 0.42 g/mol

8 0
2 years ago
A 75 lb (34 kg) boy falls out of a tree from a height of 10 ft (3 m). i. What is the kinetic energy of the boy when he hits the
Jobisdone [24]

Answer:

Kinetic energy of boy just before hitting the ground is \approx1000 J.

Speed of boy just before hitting the ground is 7.67 m/s

or 17.16 mi/hr.

Explanation:

Given that:

Mass of boy = 75lb = 34 kg

Height, h = 10ft = 3m

To find:

Kinetic energy of boy when he hits the ground.

<em>As per law of conservation of energy The potential energy gets converted to kinetic energy.</em>

<em></em>

<em></em>\therefore<em> </em>Kinetic energy at the time boy hits the ground = Initial potential energy of the boy when he was at the Height 'h'

The formula for potential energy is given as:

PE = mgh

Where m is the mass

g is the acceleration due to gravity, g = 9.8 m/s^2

h is the height of object

Putting all the values:

PE = 34 \times 9.8 \times 3 \approx 1000\ J

Hence, Kinetic energy is \approx1000 J.

Formula for Kinetic energy is:

KE = \dfrac{1}{2}mv^2

where m is the mass and

v is the speed

Putting the values and finding v:

1000 = \dfrac{1}{2}\times 34 \times v^2\\\Rightarrow v^2 = 58.82\\\Rightarrow v = 7.67\ m/s

Given that:

1 m = 1.094 yd and 1 mi = 1760 yd

\Rightarrow 1609\ m = 1\ mi

Converting 7.67 m/s to miles/hour:

\dfrac{7.67 \times 3600}{1609}=17.16\ mi/h

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Ostrovityanka [42]

Answer:D

Explanation:

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A farmer had an accident and spilled a chemical into his pond. Several days later, he notices many dead frogs around the pond. T
andrezito [222]
B. the frogs are a limiting factor for the gnats
the frogs limit the reproduction of the gnats, and therefore with less frogs the gnat population can increase
3 0
2 years ago
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