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.
Your answer is right.
Important elements to consider:
- to use the balanced equation (which you did)
- divide the masses of each compound by the correspondant molar masses (which you did)
- compare the theoretical proportions with the current proportions
Theoretical: 2 mol of Na OH : 1 mol of CuSO4
Then 4 mol of NaOH need 2 mol of CUSO4.
Given that you have more than 2 mol of of CUSO4 you have plenty of it and the NaOH will consume first, being this the limiting reagent.
Calcium will loose one electron. Fluorine will gain one electron. Lithium will loose one electron. Argon will not loose any because it already has a full valence level. Aluminium will loose 3 electrons.
Answer:
1. When you see the moon, think of the whereabouts of the sun
2. The moon rises in the east and sets in the west, each and every day
3. The moon takes about a month (one month) to orbit the Earth
4. The moon’s orbital motion is toward the east
Explanation:
