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.
Ammonia undergoes combustion with oxygen to produce nitric oxide and water. The volume of the oxygen required to react with 720 ml of ammonia is 900 ml.
<h3>What is volume?</h3>
Volume is the area occupied by the substance and is the ratio of the mass to the density.
At STP, 1 mole of gas occupies 22.4 L of volume
Given,
Volume of ammonia reacted = 0.720 L
The combustion reaction is shown as,

From the stoichiometry of the reaction, it can be said that,
L of ammonia reacts with
L of oxygen gas.
So, 0.720 L of ammonia will react with:

Therefore, the volume of oxygen required is 900 mL.
Learn more about volume here:
brainly.com/question/14090111
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Answer : The energy of the photon emitted is, -12.1 eV
Explanation :
First we have to calculate the
orbit of hydrogen atom.
Formula used :

where,
= energy of
orbit
n = number of orbit
Z = atomic number of hydrogen atom = 1
Energy of n = 1 in an hydrogen atom:

Energy of n = 2 in an hydrogen atom:

Energy change transition from n = 1 to n = 3 occurs.
Let energy change be E.

The negative sign indicates that energy of the photon emitted.
Thus, the energy of the photon emitted is, -12.1 eV