Answer:
0.44 moles
Explanation:
Given that :
A mixture of water and graphite is heated to 600 K in a 1 L container. When the system comes to equilibrium it contains 0.17 mol of H2, 0.17 mol of CO, 0.74 mol of H2O, and some graphite.
The equilibrium constant ![K_c= \dfrac{[CO][H_2]}{[H_2O]}](https://tex.z-dn.net/?f=K_c%3D%20%20%5Cdfrac%7B%5BCO%5D%5BH_2%5D%7D%7B%5BH_2O%5D%7D)
The equilibrium constant 
The equilibrium constant 
Some O2 is added to the system and a spark is applied so that the H2 reacts completely with the O2.
The equation for the reaction is :
Total mole of water now = 0.74+0.17
Total mole of water now = 0.91 moles
Again:
![0.03905 = \dfrac{[0.17+x][x]}{[0.91 -x]}](https://tex.z-dn.net/?f=0.03905%20%3D%20%20%5Cdfrac%7B%5B0.17%2Bx%5D%5Bx%5D%7D%7B%5B0.91%20-x%5D%7D)
0.03905(0.91 -x) = (0.17 +x)(x)
0.0355355 - 0.03905x = 0.17x + x²
0.0355355 +0.13095
x -x²
x² - 0.13095
x - 0.0355355 = 0
By using quadratic formula
x = 0.265 or x = -0.134
Going by the value with the positive integer; x = 0.265 moles
Total moles of CO in the flask when the system returns to equilibrium is :
= 0.17 + x
= 0.17 + 0.265
= 0.435 moles
=0.44 moles (to two significant figures)
The Boyle-Mariotte's law or Boyle's law is one of the laws of gases that <u>relates the volume (V) and pressure (P) of a certain amount of gas maintained at constant temperature</u>, as follows:
PV = k
where k is a constant.
We can relate the state of a gas at a specific pressure and volume to another state in which the same gas is at different P and V since the product of both variables is equal to a constant, according to the Boyle's law, which will be the same regardless of the state of the gas. In this way,
P₁V₁ = P₂V₂
Where P₁ and V₁ is the pressure and volume of the gas to a state 1 and P₂ and V₂ is the pressure and volume of the same gas in a state 2.
In this case, in the state 1 the gas occupies a volume V₁ = 100 mL at a pressure of P₁ = 150 kPa. Then, in the state 2 the gas occupies a volume V₂ (that we must calculate through the boyle's law) at a pressure of P₂ = 200 kPa. Substituting these values in the previous equation and clearing V₂, we have,
P₁V₁ = P₂V₂ → V₂ = 
→ V₂ = 
→ V₂ = 75 mL
Then, the volume occupied by the gas at 200 kPa is V₂ = 75 mL
Water has h bonding
H-H
Sodium fluoride
I think
Answer:
pH = 13.09
Explanation:
Zn(OH)2 --> Zn+2 + 2OH- Ksp = 3X10^-15
Zn+2 + 4OH- --> Zn(OH)4-2 Kf = 2X10^15
K = Ksp X Kf
= 3*2*10^-15 * 10^15
= 6
Concentration of OH⁻ = 2[Ba(OH)₂] = 2 * 0.15 = 3 M
Zn(OH)₂ + 2OH⁻(aq) --> Zn(OH)₄²⁻(aq)
Initial: 0 0.3 0
Change: -2x +x
Equilibrium: 0.3 - 2x x
K = Zn(OH)₄²⁻/[OH⁻]²
6 = x/(0.3 - 2x)²
6 = x/(0.3 -2x)(0.3 -2x)
6(0.09 -1.2x + 4x²) = x
0.54 - 7.2x + 24x² = x
24x² - 8.2x + 0.54 = 0
Upon solving as quadratic equation, we obtain;
x = 0.089
Therefore,
Concentration of (OH⁻) = 0.3 - 2x
= 0.3 -(2*0.089)
= 0.122
pOH = -log[OH⁻]
= -log 0.122
= 0.91
pH = 14-0.91
= 13.09
Answer:
Due to gravitational Force the water exerts more pressure at "ground floor" than at "2nd floor".
Explanation:
