The answer u are looking for is b
Answer:
It will take 5492 seconds to electroplate 0.5 mm of gold on an object .
Explanation:
Mass of gold = m
Volume of gold = v
Surface area on which gold is plated = 
Thickness of the gold plating = h = 0.5 mm = 0.05 cm
1 mm = 0.1 cm
Density of the gold = 
Moles of gold = 
According to reaction, 1 mole of gold required 3 moles of electrons,then 0.152 moles of gold will require :
of electrons
Number of electrons = N =
Charge on single electron = 
Total charge required = Q
Amount of current passes = I = 8 Ampere
Duration of time = T
It will take 5492 seconds to electroplate 0.5 mm of gold on an object .
Unit ppm stands for parts per million. in terms of mass, ppm is equivalent to mg/kg.
since 1 kg is 10⁻⁶ mg, 1 kg is equivalent to million mg.
therefore mg/kg is also ppm.
there are 1.3 g of Ca ions in 3100 kg
if 3100 kg contains - 1.3 g of Ca
then 1 kg contains - 1.3 g / 3100 kg
then Ca ions - 0.42 x 10⁻³ g/kg
Ca ion concentration - 0.42 mg/kg
therefore Ca ion concentration is 0.42 ppm
Answer:
The percent by mass of copper in the mixture was 32%
Explanation:
The ammount of HNO₃ used is:
mol HNO₃ = volume * concentration
mol HNO₃ = 0.015 l * 15.8 mol/l
mol HNO₃ = 0.237 mol
According to the reaction, 4 mol HNO₃ will react with 1 mol Cu and produce 1 mol Cu²⁺. Since we have 0.237 mol HNO₃, the amount of Cu that could react would be (0.237 mol HNO₃ * 1 mol Cu / 4 mol HNO₃) 0.06 mol. This reaction would produce 0.060 mol Cu²⁺, however, only 0.010 mol Cu²⁺ were obtained, indicating that only 0.010 mol Cu were present in the mixture. This means that the acid was in excess, so we can assume that all copper present in the mixture has reacted.
Since 0.010 mol of Cu²⁺ were produced, the amount of Cu was 0.01 mol.
1 mol of Cu has a mass of 63.5 g, then 0.01 mol has a mass of:
0.01 mol Cu * 63.5 g / 1 mol = 0.635 g.
Since this amount was present in 2.00 g mixture, the amount of copper in 100 g of the mixture will be:
100 g(mixture) * 0.635 g Cu / 2 g(mixture) = 32 g
Then, the percent by mass of Cu (which is the mass of Cu in 100 g mixture) is 32%
