Answer:
[NH₃] = 14.7 mol/L
Explanation:
28 wt% is a type of concentration that indicates that 28 g of ammonia is contained in 100 g of solution.
Let's determine the amount of ammonia:
28 g . 1 mol / 17.03g = 1.64 moles of NH₃
You need to consider that, when you have density's data it is always referred to solution:
Mass of solution is 100 g, let's find out the volume
0.90 g/mL = 100 g /V
V = 100 g / 0.90mL/g → 111.1 mL
We convert the volume to L → 111.1 mL . 1 L/1000mL = 0.1111 L
mol/L = 1.64 mol/0.1111L → 14.7 M
mol/L = M → molarity a sort of concentration that indicates the moles of solute in 1L of solution
Answer:
14.7 lbs
Explanation:
Air pressure is the weight of the air above us. It is approximately 14.7 pounds or lbs per square inch at sea level. It means that an air column weights 14.7 lbs, 1 square inch in diameter, reaching all the way up to the top of the atmosphere.
Answer:
Kc = [CO2], that is to say it is equal to the concentration of CO2
Explanation:
It is a heterogeneous equilibrium since the substances that participate in the reaction are in different phases
In the heterogeneous limestone decomposition reaction:
CaCO3(s) --> CaO(s) + CO2(g)
The equilibrium constants are:
Kc = [CO2(g)]; Kp = PCO2(g); Kc = Kp (R T)^
−(1−0) = Kp (R T)^
−1
The equilibrium situation is not affected by the amount of solid or liquid, as long as these substances are present.
The equilibrium constant is independent of the amounts of solids and liquids in equilibrium.
Answer:
a) 
b) 
Explanation:
Equation of reaction:
Initial pressure 3 1 0
Pressure change 2P 1P 2P
Total pressure = (3-2P) + (1-P) + (2P)
Total Pressure = 3.75 atm
(3-2P) + (1-P) + (2P) = 3.75
4 - P = 3.75
P = 4 - 3.75
P = 0.25 atm
Let us calculate the pressure of each of the components of the reaction:
Pressure of XO2 = 3 - 2P = 3 - 2(0.25)
Pressure of XO2 =2.5 atm
Pressure of O2 = 1 - P = 1 -0.25
Pressure of O2 = 0.75 atm
Pressure of XO3 = 2P = 2 * 0.25
Pressure of XO3 = 0.5 atm
From the reaction, equilibrium constant can be calculated using the formula:
![K_{p} = \frac{[PXO_{3}] ^{2} }{[PXO_{2}] ^{2}[PO_{2}] }](https://tex.z-dn.net/?f=K_%7Bp%7D%20%3D%20%5Cfrac%7B%5BPXO_%7B3%7D%5D%20%5E%7B2%7D%20%7D%7B%5BPXO_%7B2%7D%5D%20%5E%7B2%7D%5BPO_%7B2%7D%5D%20%7D)
Standard free energy:
b) value of k−1 at 27 °C, i.e. 300K
