From the information provided in the question, the mass of gold that can be obtained from this sample is 31.5 g of Au.
Given that the reduction of gold III occurs thus;
Au^3+(aq) + 3e ------> Au(s)
Number of moles of gold(III) chloride dihydrate = mass/molar mass = 54.9 g /339.35 g/mol = 0.16 moles
Number of moles of Au^3+ in AuCl3.2H20 = 0.16 moles
Now;
1 mole of Au has a mass of 197 g/mol
0.16 moles of Au has a mass of 0.16 moles × 197 g/mol/1 mole
= 31.5 g of Au.
The mass of gold that can be obtained from this sample is 31.5 g of Au.
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Answer:
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Answer:
the answer is correct part another is a part
Explanation:
c part
Answer:
Kinetic energy
Explanation:
Objects or particles under motion posses energy in the form of kinetic energy which is directly proportional to the square of the velocity of the given particle.
E= 1/2mv² where m is the mass of the particle and v the velocity of the particle.