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MAVERICK [17]
3 years ago
13

In a Compton scattering experiment, an x-ray photon scatters through an angle of 17.4 from a free electron that is initially at

rest. The electron recoils with a speed of 2180 km/s. Calculate (a) the wavelength of the incident photon and (b) the angle through which the electron scatters.
Chemistry
1 answer:
skelet666 [1.2K]3 years ago
5 0

Answer:

D that is

Explanation:

You might be interested in
Oxidation state of Nitrogen in N2O5​
balandron [24]

Explanation:

Oxidation state of Nitrogen in N2O5 is +5

7 0
1 year ago
Consider the following reaction: A(g)⇌2B(g). Find the equilibrium partial pressures of A and B for each of the following differe
Illusion [34]

Answer:

a. Kp=1.4

P_{A}=0.2215 atm

P_{B}=0.556 atm

b.Kp=2.0 * 10^-4

P_{A}=0.495atm

P_{B}=0.00995 atm

c.Kp=2.0 * 10^5

P_{A}=5*10^{-6}atm

P_{B}=0.9999 atm

Explanation:

For the reaction  

A(g)⇌2B(g)

Kp is defined as:

Kp=\frac{(P_{B})^{2}}{P_{A}}

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:

Kp=\frac{(1-2x)^{2}}{x}

Working with this equation:

x*Kp=(1-2x)^{2} - -> x*Kp=4x^{2}-4x+1- - >4x^{2}-(4+Kp)*x+1=0

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:

x=\frac{-b(+-)*\sqrt{(b^{2}-4ac)}}{2a} (equation 1)

We just take the positive values from the solution since negative partial pressures don´t make physical sense.

Kp = 1.4

x_{1}=\frac{(1.4+4)+\sqrt{(-(1.4+4)^{2}-4*4*1)}}{2*4}=1.128

x_{1}=\frac{(1.4+4)-\sqrt{(-(1.4+4)^{2}-4*4*1)}}{2*4}=0.2215

With x1 we get a partial pressure of:

P_{A}=1.128 atm

P_{B}=1-2*1.128 = -1.256 atm

Since negative partial pressure don´t make physical sense x1 is not the solution for the system.

With x2 we get:

P_{A}=0.2215 atm

P_{B}=1-2*0.2215 = 0.556 atm

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

P_{A}=0.505atm

P_{B}=1-2*0.505 = -0.01005 atm

Not sense.

With x2

P_{A}=0.495atm

P_{B}=1-2*0.495 = 0.00995 atm

X2 is the solution for this equation.  

Kp=2*10^5

X1=50001

X2=5*10^{-6}

With x1

P_{A}=50001atm

P_{B}=1-2*50001=-100001atm

Not sense.

With x2

P_{A}= 5*10^{-6}atm

P_{B}=1-2*5*10^{-6}= 0.9999 atm

X2 is the solution for this equation.  

8 0
3 years ago
Can I get help with this, please!!
kondor19780726 [428]

Pressure in the submarine when the temperature is changed to 293 K is 108.9 K Pa

Explanation:

Pressure in the submarine = 108.9 kPa

Volume, V = 2.4 * 10^5 L

Pressure, P = 116k Pa

Temperature, T = 312 K

Ideal gas law: PV = nRT  or  n = PV / RT

So, moles of gas, n =116 KPa * 2.4 * 10 ^5L / 8.314 LK Pa K^-1 *312 K

= 1.073 *10^4 mol

when temperature is changed to 293K,

PV = nRT  or P = nRT / V

=1.073 *10^4 mol *8.314 LK Pa mol^-1 K^-1 *293 K / 2.4*10^5L

=108.9 K Pa

Pressure in the submarine when the temperature is changed to 293 K is 108.9 K Pa

8 0
3 years ago
Which of the following best describes the neutron?
Setler [38]
The particles that combined in the middle of the structure best describes neutron as neutron is always present in the middle of atomic structure
7 0
3 years ago
Which substance can be decomposed by chemical means?
sweet-ann [11.9K]
Are u sure this is the right option? Well, antimony can be decomposed. Including octane.
5 0
3 years ago
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