It would be more beneficial than acid rain because it is non-corrosive. Its pH is neutral. It would help to dilute acid and alkaline materials and would wash away impurities. Animals, including humans, can drink pure water, but acid rain is harmful. Plants and the soil can accept pure water which aid in supplying plants with water they can absorb directly, and water can dissolve certain nutrients that can then be more readily absorbed by plants, vegetation and microbes living in the soil. Pure water can be stored (reservoir) without the need for filtering or purifying, so reducing costs in providing drinking water. Pure water is part of the natural cycle. It evaporates leaving no deposit.
Answer:
0.01
Explanation:
volume of 22.4 L (22,400 ml) at s.t.p. 250 ml out of 22,400 ml is 0.01 rounded, therefore I assume that is the answer. About 0.01 mole occupies 250 cm3 at s.t.p.
Answer:
NaOH is the limiting reactant.
204.9 g of sodium phosphate are formed.
51.94 g of excess reactant will remain.
Explanation:
The reaction that takes place is:
- H₃PO₄ + 3NaOH → Na₃PO₄ + 3H₂O
First we <u>convert the mass of both reactants to moles</u>, using their <em>respective molar masses</em>:
- H₃PO₄ ⇒ 175 g ÷ 98 g/mol = 1.78 mol
- NaOH ⇒ 150 g ÷ 40 g/mol = 3.75 mol
1.78 moles of H₃PO₄ would react completely with (1.78 * 3) 5.34 moles of NaOH. There are not as many NaOH moles so NaOH is the limiting reactant.
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We <u>calculate the produced moles of Na₃PO₄</u> using the <em>limiting reactant</em>:
- 3.75 mol NaOH *
= 1.25 mol Na₃PO₄
Then we <u>convert moles into grams</u>:
- 1.25 mol Na₃PO₄ * 163.94 g/mol = 204.9 g
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We calculate how many H₃PO₄ moles would react with 3.75 NaOH moles:
- 3.75 mol NaOH *
= 1.25 mol H₃PO₄
We substract that amount from the original amount:
- 1.78 - 1.25 = 0.53 mol H₃PO₄
Finally we <u>convert those remaining moles to grams</u>:
- 0.53 mol H₃PO₄ * 98 g/mol = 51.94 g
Answer : The vapor pressure of solution is 23.67 mmHg.
Solution:
As the relative lowering of vapor pressure is directly proportional to the amount of dissolved solute.
The formula for relative lowering of vapor pressure will be,

where,
= vapor pressure of pure solvent (water) = 23.76 mmHg
= vapor pressure of solution= ?
= mass of solute (sucrose) = 12.25 g
= mass of solvent (water) = 176.3 g
= molar mass of solvent (water) = 18.02 g/mole
= molar mass of solute (sucrose) = 342.3 g/mole
Now put all the given values in this formula ,we get the vapor pressure of the solution.


Therefore, the vapor pressure of solution is 23.67 mmHg.