Answer:
The mole fraction of NaOH in an aqueous solution that contain 22.9% NaOH by mass=0.882
Explanation:
We are given that
Aqueous solution that contains 22.9% NaOH by mass means
22.9 g NaOH in 100 g solution.
Mass of NaOH(WB)=22.9 g
Mass of water =100-22.9=77.1
Na=23
O=16
H=1.01
Molar mass of NaOH(MB)=23+16+1.01=40.01
Number of moles =
Using the formula
Number of moles of NaOH

Molar mass of water=16+2(1.01)=18.02g
Number of moles of water

Now, mole fraction of NaOH
=

=0.882
Hence, the mole fraction of NaOH in an aqueous solution that contain 22.9% NaOH by mass=0.882
Answer:
1) Maximun ammount of nitrogen gas: 
2) Limiting reagent: 
3) Ammount of excess reagent: 
Explanation:
<u>The reaction </u>

Moles of nitrogen monoxide
Molecular weight: 


Moles of hydrogen
Molecular weight: 


Mol rate of H2 and NO is 1:1 => hydrogen gas is in excess
1) <u>Maximun ammount of nitrogen gas</u> => when all NO reacted


2) <u>Limiting reagent</u>:
3) <u>Ammount of excess reagent</u>:


Answer:
Longer hydrocarbon molecules have a stronger intermolecular force. More energy is needed to move them apart so they have higher boiling points . This makes them less volatile and therefore less flammable
The balanced equation for the above reaction is as follows;
2S + 3O₂ --> 2SO₃
Stoichiometry of O₂ to SO₃ is 3:2
O₂ is the limiting reactant and S is provided in excess. since O₂ is the limiting reactant, the whole amount is consumed in the reaction and amount of product formed depends on amount of limiting reactant present.
Number of O₂ moles reacted- 4 g / 32 g/mol = 0.125 mol
3 mol of O₂ forms 2 mol of SO₃
therefore when 0.125 mol of O₂ reacts number of SO₃ moles - 2/3 x 0.125 mol
Number of SO₃ moles formed - 0.0833 mol
Answer is 4) 0.08 mol
Answer:
7.28 mol
Explanation:
2 NaOH + H₂SO₄ = 2 H₂O + Na₂SO₄ -------------------(1)
mole fraction for the reaction is;
2 : 1 = 2 : 1
Number of moles of H₂SO₄ = 7.28 mol
1 mol of H₂SO₄ shall form 1 mole of Na₂SO₄
therefore,
7.28 mol of H₂SO₄ shall form 7.28 mole of Na₂SO₄