The solution 550 ml total and first we will find the amount of alcohol. 3% = 0.03 550 ml x .03 = 16.5 ml alcohol
Then to find the amount of water used, we just have to subtract the amount of alcohol from the total volume
550 ml total - 16.5 ml alcohol = 533.5 ml water
Solving this chemistry is a little bit hard because the question didn't give some important detailed.
So first, there are a couple problems with your question.
We will just need to know which direction will it proceed to reach equilibrium.
Your expression for Kc (and Qc ) for the reaction should be:
Kc = [C] / [A] [B]^2
You have not provided a value for Kc, so a value of Qc tells you absolutely nothing. Qc is only valuable in relation to a numerical value for Kc. If Qc = Kc, then the reaction is at equilibrium. If Q < K, the reaction will form more products to reach equilibrium, and if Q > Kc, the reaction will form more reactants.
Answer:
9.2
Explanation:
Let's do an equilibrium chart of this reaction:
2NO(g) + O₂(g) ⇄ 2NO₂(g)
4.9 atm 5.1 atm 0 Initial
-2x -x +2x Reacts (stoichiometry is 2:1:2)
4.9-2x 5.1-x 2x Equilibrium
The mole fraction of NO₂ (y) can be calculated by the Raoult's law, that states that the mole fraction is the partial pressure divided by the total pressure:
y = 2x/(4.9 - 2x + 5.1 -x + 2x)
0.52 = 2x/(10 - x)
2x = 5.2 -0.52x
2.52x = 5.2
x = 2.06 atm
Thus, the partial pressure at equilibrium are:
pNO = 4.9 -2*2.06 = 0.78 atm
pO₂ = 5.1 - 2.06 = 3.04 atm
pNO₂ = 2*2.06 = 4.12 atm
Thus, the pressure equilibrium constant Kp is:
Kp = [(pNO₂)²]/[(pNO)²*(pO₂)]
Kp = [(4.12)²]/[(0.78)²*3.04]
Kp = [16.9744]/[1.849536]
Kp = 9.2
Answer:
41.17g
Explanation:
We are given the following parameters for Flourine gas(F2).
Volume = 5.00L
Pressure = 4.00× 10³mmHG
Temperature =23°c
The formula we would be applying is Ideal gas law
PV = nRT
Step 1
We find the number of moles of Flourine gas present.
T = 23°C
Converting to Kelvin
= °C + 273k
= 23°C + 273k
= 296k
V = Volume = 5.00L
R = 0.08206L.atm/mol.K
P = Pressure (in atm)
In the question, the pressure is given as 4.00 × 10³mmHg
Converting to atm(atmosphere)
1 mmHg = 0.00131579atm
4.00 × 10³ =
Cross Multiply
4.00 × 10³ × 0.00131579atm
= 5.263159 atm
The formula for number of moles =
n = PV/RT
n = 5.263159 atm × 5.00L/0.08206L.atm/mol.K × 296K
n = 1.0834112811moles
Step 2
We calculate the mass of Flourine gas
The molar mass of Flourine gas =
F2 = 19 × 2
= 38 g/mol
Mass of Flourine gas = Molar mass of Flourine gas × No of moles
Mass = 38g/mol × 1.0834112811moles
41.169628682grams
Approximately = 41.17 grams.
