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.
HEY THERE!
THE ANSWER IS: the properties of an ideal gas are: An ideal gas consists of a large number of identical molecules. The volume occupied by the molecules themselves is negligible compared to the volume occupied by the gas. The molecules obey Newton's laws of motion, and they move in random motion.
CREDITS:<span>physics.bu.edu/~duffy/py105/Idealgas.htm</span>
Tabulations of chemical elements differing in their organization from the traditional seen periodic system
Answer:
A = 349 g.
Explanation:
Hello there!
In this case, since the radioactive decay kinetic model is based on the first-order kinetics whose integrated rate law is:

We can firstly calculate the rate constant given the half-life as shown below:

Therefore, we can next plug in the rate constant, elapsed time and initial mass of the radioactive to obtain:

Regards!