Answer:-
95 grams
Explanation:-
Let the mass of water to be added be M
So total mass = 5 + M
So 5% of this solution has 5g of NaCl by mass.
∴ (M+5) x (5/100) = 5
M+5 = 5 x 100/5
M+5=100
M= 100-5
95
So amount of water to be added is 95 gram
It is a reaction between an element and a compound that results in a different element and compound
Answer:
GOOD MORNING DIDI be happy
Answer:

Explanation:
Hello,
In this case, the undergoing chemical reaction is:

Next, we identify the limiting reactant by computing the available moles of ethane and the moles of ethane consumed by 60.0 grams of oxygen:

Thus, we notice there are less available moles, for that reason, the ethane is the limiting reactant. Finally, we can compute the produced moles of water by:

Best regards.