Reaction of Cu(NO₃)₂ with each salt is as follow,
1) with KNO₃;
Cu(NO₃)₂ + KNO₃ → Cu(NO₃)₂ + KNO₃
Both salt products are water soluble.
2) With CuSO₄;
Cu(NO₃)₂ + CuSO₄ → CuSO₄ + Cu(NO₃)₂
Again both Salt products are water soluble.
3) With K₂SO₄;
Cu(NO₃)₂ + K₂SO₄ → CuSO₄ + 2 KNO₃
Again both salt products are water soluble.
4) With K₂S;
Cu(NO₃)₂ + K₂S → CuS + 2 KNO₃
In this case CuS is water insoluble, hence precipitates out.
Result:
Option-4 is the correct answer.
Hydrogen gas, oxygen gas and water
4H2O (l) ---> 2H2O (l) + O2 (g) + 2H2 (g)
5 moles because molarity signifies number of moles dissolved in One litre of water
Answer:
20 molecules of oxygen gas remains after the reaction.
Explanation:

Molecules of ethyne = 52
Molecules of oxygen gas = 150
According to reaction, 2 molecules of ethyne reacts with 5 molecules of oxygen gas.
Then 52 molecules of ethyne will react with:
of oxygen gas.
As we can see that we have 150 molecules of oxygen gas, but 52 molecules of ethyne will react with 130 molecules of oxygen gas. So, this means that ethyne is a limiting reagent and oxygen gas is an excessive reagent.
Remaining molecules of recessive reagent = 150 - 130 = 20
20 molecules of oxygen gas remains after the reaction.
Answer: Mass of
produced in this reaction was 6.56 grams
Explanation:
According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

Mass or reactants = Mass of
+ mass of
= 16.00 + 64.80 = 80.80 g
Mass of products = mass of aqueous solution + mass of
+ = 74.24 + x g
Mass or reactants = Mass of products
80.80 g = 74.24 + x g
x = 6.56 g
Thus mass of
produced in this reaction was 6.56 grams