Answer:
86.3 g of N₂ are in the room
Explanation:
First of all we need the pressure from the N₂ in order to apply the Ideal Gases Law and determine, the moles of gas that are contained in the room.
We apply the mole fraction:
Mole fraction N₂ = N₂ pressure / Total pressure
0.78 . 1 atm = 0.78 atm → N₂ pressure
Room temperature → 20°C → 20°C + 273 = 293K
Let's replace data: 0.78 atm . 95L = n . 0.082 . 293K
(0.78 atm . 95L) /0.082 . 293K = n
3.08 moles = n
Let's convert the moles to mass → 3.08 mol . 28g /1mol = 86.3 g
Answer: When an acidic solution is diluted with water the concentration of H + ions decreases and the pH of the solution increases towards 7.
Explanation:
Explanation:
Starting moles of ethanol acid = 0.020 mol
At the equilibrium 50 % of the ethanol acid molecules reacted
∴ Moles of ethanol acid reacted = 0.020 mol * 50 %/100 %
= 0.010 mol
Moles of ethanol acid remain = 0.020 mol + 0.010 mol = 0.010 mol
Moles of the product
gas formed are calculated as
0.010 mol CH3COOH * 1 mol
/ 2 mol CH3COOH
= 0.005 mol 
Therefore at the equilibrium total moles of gas present in the vessel are 0.010 mol CH3COOH and 0.005 mol 
That is total gas moles at equilibrium = 0.010 mol + 0.005 mol = 0.015 mol
Now Calculate the pressure :
0.020 mol gas has pressure of 0.74 atm therefore at the same condition what will be the pressure exerted by 0.015 mol gas
P1/n1 = P2/n2
P2 = P1*n2 / n1
= 0.74 atm * 0.015 mol / 0.020 mol
= 0.555 atm
There are 1000 meters in 1 Kilometer.
Hope this helps you. :)
Answer is: sulfuric acid is the limiting reactant.
Chemical reaction: 3H₂SO₄ + 2Al(OH)₃ → Al₂(SO₄)₃ + 6H₂O.
m(H₂SO₄) = 34 g.
n(H₂SO₄) = m(H₂SO₄) ÷ M(H₂SO₄).
n(H₂SO₄) = 34 g ÷ 98 g/mol.
n(H₂SO₄) = 0,346 mol.
m(Al(OH)₃) = 33 g.
n(Al(OH)₃) = 33 g ÷ 78 g/mol.
n(Al(OH)₃) = 0,423 mol.
From chemical reaction: n(H₂SO₄) : n(Al(OH)₃) = 3 : 2.