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Licemer1 [7]
3 years ago
7

Write a nuclear reaction for the neutron-induced fission of u−235 to form xe−144 and sr−90.

Chemistry
1 answer:
rusak2 [61]3 years ago
3 0

Answer:

235/92U+10n→144/54Xe+90/38Sr+2/10n

Explanation:

  • The nuclear reaction for the neutron-induced fission of u−235 to form xe−144 and sr−90 is represented by;

235/92U+10n→144/54Xe+90/38Sr+2/10n

  • In nuclear fission reactions a heavy nuclide is split into two light nuclides and is coupled by the release of energy.
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he rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy . If the rate c
Leya [2.2K]

The question is incomplete, here is the complete question:

The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy Ea = 71.0 kJ/mol . If the rate constant of this reaction is 6.7 M^(-1)*s^(-1) at 244.0 degrees Celsius, what will the rate constant be at 324.0 degrees Celsius?

<u>Answer:</u> The rate constant at 324°C is 61.29M^{-1}s^{-1}

<u>Explanation:</u>

To calculate rate constant at two different temperatures of the reaction, we use Arrhenius equation, which is:

\ln(\frac{K_{324^oC}}{K_{244^oC}})=\frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_{244^oC} = equilibrium constant at 244°C = 6.7M^{-1}s^{-1}

K_{324^oC} = equilibrium constant at 324°C = ?

E_a = Activation energy = 71.0 kJ/mol = 71000 J/mol   (Conversion factor:  1 kJ = 1000 J)

R = Gas constant = 8.314 J/mol K

T_1 = initial temperature = 244^oC=[273+244]K=517K

T_2 = final temperature = 324^oC=[273+324]K=597K

Putting values in above equation, we get:

\ln(\frac{K_{324^oC}}{6.7})=\frac{71000J}{8.314J/mol.K}[\frac{1}{517}-\frac{1}{597}]\\\\K_{324^oC}=61.29M^{-1}s^{-1}

Hence, the rate constant at 324°C is 61.29M^{-1}s^{-1}

8 0
3 years ago
Please help me vote you brainiest
Viktor [21]

Answer:

Explanation:

A equal

B opposite

C equal

D opposite

6 0
3 years ago
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Janet runs 6.4 miles in one hour. Is it speed or velocity or acceleration
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Answer:

acceleration

Explanation:

6 0
2 years ago
How much pressure does an elephant with a mass of 2200 Kg and a total footprint area of 4500 cm2 exert on the ground?
Debora [2.8K]

Answer:

47911.1 pa

Explanation:

The SI base unit of pressure is pascal, which is N/m^2.

2200 kg is 2200*9.8=21560 N, and 4500 cm^2=4500/10000=0.45 m^2.

So the total pressure exerted on the ground (!!) is 21560/0.45= 47911.1 Pa.

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3 years ago
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A gold nugget has a volume of 20.0 cm3 and its density is 19.3 g/cm3. What is the mass of the gold nugget?
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Density * Volume = Mass

Now we substitute the values in.

19.3 g/cm^3 + 20 cm^3 = 386 g

Mass = 386 g
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