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GarryVolchara [31]
2 years ago
11

One way to identify the type of radioactive decay produced in a reaction is to pass the emission through an electric field. Desc

ribe the type of radioactive emission produced from the decay of uranium-
238 to thorium-234 and its reaction to the electric field
Chemistry
1 answer:
Sergeeva-Olga [200]2 years ago
4 0

Decay of Uranium-238 produces gamma radiaton whereas decay of thorium-234 releases beta radiation.

<h3>What type of radiation is produced?</h3>

In the decay of U-238, two gamma rays of different energies are emitted in addition to the alpha particle while on the other hand, in the decay of thorium-234 , beta rays are emitted.

So we can conclude that decay of Uranium-238 produces gamma radiaton whereas decay of thorium-234 releases beta radiation.

Learn more about decay here: brainly.com/question/25537936

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Alecsey [184]
Elements can be defined by their unique properties and atomic.
3 0
3 years ago
The equilibrium constant for the formation of ammonia from nitrogen and hydrogen is 1.6 × 102. what is the form of the equilibri
Nimfa-mama [501]

Answer: The expression for equilibrium constant is \frac{[NH_3]^2}{[H_2]^3[N_2]}

Explanation: Equilibrium constant is the expression which relates the concentration of products and reactants preset at equilibrium at constant temperature. It is represented as k_c

For a general reaction:

aA+bB\rightleftharpoons cC+dD

The equilibrium constant is written as:

k_c=\frac{[C]^c[D]^d}{[A]^a[B]^b}

Chemical reaction for the formation of ammonia is:

N_2+3H_3\rightleftharpoons 2NH_3

k_c=1.6\times 10^2

Expression for k_c is:

k_c=\frac{[NH_3]^2}{[H_2]^3[N_2]}

1.6\times 10^2=\frac{[NH_3]^2}{[H_2]^3[N_2]}

8 0
3 years ago
In the rock cycle, the process of weathering and erosion happens between:
Natalija [7]
Sedimentary and metamorphic :)


i hope this helps!!
8 0
2 years ago
What is Rutherford atomic modle
Ksju [112]

Answer:

Rutherford's atomic model explained how the electrons surrounded the nucleus of protons and neutrons. His model showed how J. J. Thomson's Plum Pudding model was incorrect.

4 0
3 years ago
Read 2 more answers
How long does it take for a 12.62g sample of ammonia to heat from 209K to 367K if heated at a constant rate of 6.0kj/min? The me
Georgia [21]
First, consider the steps to heat the sample from 209 K to 367K.

1) Heating in liquid state from 209 K to 239.82 K

2) Vaporaizing at 239.82 K

3) Heating in gaseous state from 239.82 K to 367 K.


Second, calculate the amount of heat required for each step.

1) Liquid heating

Ammonia = NH3 => molar mass = 14.0 g/mol + 3*1g/mol = 17g/mol

=> number of moles = 12.62 g / 17 g/mol = 0.742 mol

Heat1 = #moles * heat capacity * ΔT

Heat1 = 0.742 mol * 80.8 J/mol*K * (239.82K - 209K) = 1,847.77 J

2) Vaporization

Heat2 = # moles * H vap

Heat2 = 0.742 mol * 23.33 kJ/mol = 17.31 kJ = 17310 J

3) Vapor heating

Heat3 = #moles * heat capacity * ΔT

Heat3 = 0.742 mol * 35.06 J / (mol*K) * (367K - 239.82K) = 3,308.53 J

Third, add up the heats for every steps:

Total heat = 1,847.77 J + 17,310 J + 3,308.53 J = 22,466.3 J

Fourth, divide the total heat by the heat rate:

Time = 22,466.3 J / (6000.0 J/min) = 3.7 min

Answer: 3.7 min


3 0
3 years ago
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