Answer:
By giving electricity to copper compound solution.
Explanation:
Electrolysis is one of the major way of refined copper. The copper containing solution has two electrodes.i) positive electrodes called anode. ii) negative electrodes called cathode. When electricity is pass into the copper containing solution electrolysis process is starts and impure copper is formed in anode and pure copper is formed in cathode.
So, We can get pure copper in cathode through electrolysis.
I Hope this will be helpful for you.
If this is helpful for you .Then choose this answer as brainliest answer.
Hence the resulting concentration of Li+ is 1M. Hope it helps.
Answer:
13.8 moles of water produced.
Explanation:
Given data:
Moles of KMnO₄ = 3.45 mol
Moles of water = ?
Solution:
Chemical equation:
16HCl + 2KMnO₄ → 2KCl + 2MnCl₂ + 5Cl₂ + 8H₂O
Moler ratio of water and KMnO₄:
KMnO₄ : H₂O
2 : 8
3.45 : 8/2×3.45 = 13.8 mol
Hence, 3.45 moles of KMnO₄ will produced 13.8 mol of water.
Answer:
a)there would be no reaction
Explanation:
The activity series of metals has many functions. The one applicable to this problem is that it can be used to determine whether a reaction will occur or not. Also, based on the positions of metals in the series, we can know how reactive a metal is compared to another.
In a single displacement reaction, a metal replaces another metal based on their position on the activity series. Metals that are higher in the series are generally more reactive than others below them and so will displace them.
Would aluminum replace magnesium to form a new compound or would there be no reaction?
Magnesium is higher than aluminum in the activity series. Therefore it is more reactive than aluminum. No reaction will occur.
Answer:
Percentage by mass of oxygen = 76.20% (Approx)
Explanation:
Given:
HNO3
H=1, N=14, O=16]
Find:
Percentage by mass of oxygen
Computation:
HNO3
Total mass = 1 + 14 + 3(16)
Total mass = 63
Mass of oxygen = (3)(16) = 48
Percentage by mass of oxygen = [48/63]100
Percentage by mass of oxygen = 76.20% (Approx)
