Answer:
Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.
Explanation:
Volume of NaOH = 1.7 ml = 0.0017 L
Molarity of NaOH = 0.0811 M
Moles of NaOH = n
n = 0.0001378 mol
According to reaction, 2 mol of NaOH neutralize 1 mol of sulfuric acid.
Then 0.0001378 mol of NaOH will neutralize:
of sulfuric acid.
Concentration of sulfuric acid in the acid rain sample: x
Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.
There are 6.022*10^23 molecules in 1 mole of carbon
So how many will moles will be 7.87*20^7?
Let the required number of moles be ‘x’.
1 mole ———6.022*10^23
x moles———7.87*10^7
(Cross multiplication)
x=7.87*10^7/6.022*10^23
Therefore x=1.3*10^-16
Answer:It is a mixture
Explanation:
If it is pure sugar it’s neither but if it has water it is a homogeneous mixture
Answer:
59.2 grams
Explanation:
We are given that 70.4% of the weight of the total 200 g of the concentration is made up of nitric acid, the remaining information is not required to solve the problem. Since water and nitric acid are the only components of the solution, the total weight of water is given by:
There are 59.2 grams of water in this solution.
