Answer:
a. Kp=1.4


b.Kp=2.0 * 10^-4


c.Kp=2.0 * 10^5


Explanation:
For the reaction
A(g)⇌2B(g)
Kp is defined as:

The conditions in the system are:
A B
initial 0 1 atm
equilibrium x 1atm-2x
At the beginning, we don’t have any A in the system, so B starts to react to produce A until the system reaches the equilibrium producing x amount of A. From the stoichiometric relationship in the reaction we get that to produce x amount of A we need to 2x amount of B so in the equilibrium we will have 1 atm – 2x of B, as it is showed in the table.
Replacing these values in the expression for Kp we get:

Working with this equation:

This last expression is quadratic expression with a=4, b=-(4+Kp) and c=1
The general expression to solve these kinds of equations is:
(equation 1)
We just take the positive values from the solution since negative partial pressures don´t make physical sense.
Kp = 1.4


With x1 we get a partial pressure of:


Since negative partial pressure don´t make physical sense x1 is not the solution for the system.
With x2 we get:


These partial pressures make sense so x2 is the solution for the equation.
We follow the same analysis for the other values of Kp.
Kp=2*10^-4
X1=0.505
X2=0.495
With x1


Not sense.
With x2


X2 is the solution for this equation.
Kp=2*10^5
X1=50001

With x1


Not sense.
With x2


X2 is the solution for this equation.
Answer:
= 331.81 g
Explanation:
Molarity is calculated by the formula;
Molarity = Moles/volume in liters
Therefore;
Moles = Molarity ×Volume in liters
= 0.35 M × 1.575 L
= 0.55125 Moles
But; Molar mass of Ba3(PO4)2 is 601.93 g/mol
Thus;
Mass = 0.55125 moles × 601.93 g/mol
<u>= 331.81 g</u>
<span>The two techniques for separating an insoluble solid from a liquid are filtration and centrifuging. Filtration relies on the fact that the solid particles are smaller than the filter paper pores which allow the tiny molecules of liquid to pass through. With solids which dissolve in a liquid solvent, evaporation is commonly used to recover the solid and distillation is used to recover the solvent.</span>