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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:

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Which of the following processes does our Sun and other stars in our galaxy use to create energy?
Usimov [2.4K]

Answer:

fusion reaction

The simple answer is that the sun, like all stars, is able to create energy because it is essentially a massive fusion reaction. Scientists believe that this began when a huge cloud of gas and particles (i.e. a nebula) collapsed under the force of its own gravity – which is known as Nebula Theory.

Explanation:

hope this helps

5 0
2 years ago
A gas system contains 2.00 moles of O2 and CO2 gas, has an initial temperature of 25.0 oC and is under 1.00 atm of pressure. If
Ainat [17]

Answer:

V_2=52.2L

Explanation:

Hello there!

In this case, by bearing to to mind the given conditions, it is firstly possible to determine the initial volume of the closed system via the ideal gas equation:

PV=nRT\\\\V=\frac{2.00mol*0.08206\frac{atm*L}{mol*K}*298.15K}{1.00 atm} \\\\V=48.9L

Which is V1 in the Charles' law:

\frac{T_2}{V_2} =\frac{T_1}{V_1}

And of course, T1 is 298.15 (25+273.15). Therefore, by solving for V2 as the final volume, we obtain:

V_2=\frac{V_1T_2}{T_1}\\\\V_2=\frac{48.9L*(45+273.15)K}{(25+273.15)K}  \\\\V_2=52.2L

Best regards!

8 0
3 years ago
Calculate the initial rate for the formation of C at 25 ∘C, if [A]=0.50M and [B]=0.075M.Express your answer to two significant f
N76 [4]

The question is incomplete, here is the complete question:

Calculate the initial rate for the formation of C at 25°C, if [A]=0.50 M and [B]=0.075 M. Express your answer to two significant figures and include the appropriate units.Consider the reaction

A + 2B ⇔ C

whose rate at 25°C was measured using three different sets of initial concentrations as listed in the following table:

The table is attached below as an image.

<u>Answer:</u> The initial rate for the formation of C at 25°C is 2.25\times 10^{-2}Ms^{-1}

<u>Explanation:</u>

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

A+2B\rightleftharpoons C

Rate law expression for the reaction:

\text{Rate}=k[A]^a[B]^b

where,

a = order with respect to A

b = order with respect to B

  • Expression for rate law for first trial:

5.4\times 10^{-3}=k(0.30)^a(0.050)^b ....(1)

  • Expression for rate law for second trial:

1.1\times 10^{-2}=k(0.30)^a(0.100)^b ....(2)

  • Expression for rate law for third trial:

2.2\times 10^{-2}=k(0.50)^a(0.050)^b ....(3)

Dividing 2 by 1, we get:

\frac{1.1\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.30)^a(1.00)^b}{(0.30)^a(0.050)^b}\\\\2=2^b\\b=1

Dividing 3 by 1, we get:

\frac{2.2\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.50)^a(0.050)^b}{(0.30)^a(0.050)^b}\\\\4.07=2^a\\a=2

Thus, the rate law becomes:

\text{Rate}=k[A]^2[B]^1       ......(4)

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

5.4\times 10^{-3}=k[0.30]^2[0.050]^1\\\\k=1.2M^{-2}s^{-1}

Calculating the initial rate of formation of C by using equation 4, we get:

k=1.2M^{-2}s^{-1}

[A] = 0.50 M

[B] = 0.075 M

Putting values in equation 4, we get:

\text{Rate}=1.2\times (0.50)^2\times (0.075)^1\\\\\text{Rate}=2.25\times 10^{-2}Ms^{-1}

Hence, the initial rate for the formation of C at 25°C is 2.25\times 10^{-2}Ms^{-1}

8 0
3 years ago
Based on the article "Will the real atomic model please stand up?,” describe one major change that occurred in the development o
nydimaria [60]
I don't know this article, but I do know some major changes: first, the change from the plum pudding model (no nucleus, just electrons) to the gold foil experiment, which had Rutherford shoot alpha particles at a sheet of gold only to find them rebounding, proving the existence of a positively charged mass, i.e a nucleus, in the atom. However, this changed again when Bohr realized that the negatively charged electrons should be attracted to the positively charged center, so that there must be something else inside the nucleus. 
3 0
2 years ago
Read 2 more answers
0.446 g of hydrogen gas fills a 5.0 L bag determine the density of hydrogen
Oduvanchick [21]

The density of hydrogen : ρ = 0.0892 g/L

<h3>Further explanation</h3>

Given

mass of Hydrogen : 0.446 g

Volume = 5 L

Required

The density

Solution

Density is a quantity derived from the mass and volume  

Density is the ratio of mass per unit volume  

The unit of density can be expressed in g/cm³, kg/m³, or g/L

Density formula:  

\large {\boxed {\bold {\rho ~ = ~ \frac {m} {V}}}}

Input the value :

ρ = m : V

ρ = 0.446 g : 5 L

ρ = 0.0892 g/L

5 0
3 years ago
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