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Gasoline use contributes to air pollution
Gasoline is a toxic and highly flammable liquid. The vapors given off when gasoline evaporates and the substances produced when gasoline is burned (carbon monoxide, nitrogen oxides, particulate matter, and unburned hydrocarbons) contribute to air pollution. Burning gasoline also produces carbon dioxide, a greenhouse gas.
Answer:
<u>136.67 g of Na3PO4 i</u>s required to create 100 gram of NaOH.
Explanation:
The balanced equation:

1 mole Na3PO4 = 164 g/mole (Molar mass)
1 mole NaOH = 40 g/mole (Molar mass)
Now,
1 mole of Na3PO4 produce = 3 mole of NaOH
164 g/mol of Na3PO4 produce = 3(40) g/mol of NaOH
or
120 g/mol of NaOH is produced from = 164 g/mol of Na3PO4
1 g/mol of NaOH is produced from =

100 grams of NaOH is produced from =
gram of Na3PO4
calculate,
= 136.67 g
A) in pure water :
by using ICE table:
According to the reaction equation:
BaCrO4(s) → Ba^2+(aq) + CrO4^2-(aq)
initial 0 0
change +X +X
Equ X X
when Ksp = [Ba^2+][CrO4^2-]
by substitution:
2.1 x 10^-10 = X* X
∴X = √2.1 x 10*-10
∴X = 1.4 x 10^-5
∴ the solubility = X = 1.4 X 10^-5
B) In 1.6 x 10^-3 m Na2CrO4
by using ICE table:
According to the reaction equation:
BaCrO4(s) → Ba^2+(aq) + CrO4^2-(aq)
initial 0 0.0016
Change +X +X
Equ X X+0.0016
when Ksp = [Ba^2+][CrO4^2-]
by substitution:
2.1 x 10^-10 = X*(X+0.0016) by solving for X
∴ X = 1.3 x 10^-7
∴ solubility =X = 1.3 x 10^-7
Explanation:
The given precipitation reaction will be as follows.

Here, AgCl is the precipitate which is formed.
It is known that molarity is the number of moles present in a liter of solution.
Mathematically, Molarity = 
It is given that volume is 1.14 L and molarity is 0.269 M. Therefore, calculate number of moles as follows.
Molarity = 
0.269 M = 
no. of moles = 0.306 mol
As molar mass of AgCl is 143.32 g/mol. Also, relation between number of moles and mass is as follows.
No. of moles = 
0.307 mol = 
mass = 43.99 g
Thus, we can conclude that mass of precipitate produced is 43.99 g.