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alex41 [277]
2 years ago
10

What can be said about a reaction with AH= 62.4 kJ/mol and AS= 0.145

Chemistry
1 answer:
seropon [69]2 years ago
6 0

Answer:

<em>B.</em><em> </em><em>It </em><em>is </em><em>spontaneous </em><em>at </em><em>5</em><em>0</em><em>0</em><em>k</em><em> </em>

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Technetium (Tc; Z = 43) is a synthetic element used as a radioactive tracer in medical studies. A Tc atom emits a beta particle
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Question in incomplete, complete question is:

Technetium (Tc; Z = 43) is a synthetic element used as a radioactive tracer in medical studies. A Tc atom emits a beta particle (electron) with a kinetic energy (Ek) of 4.71\times 10^{-15}J . What is the de Broglie wavelength of this electron (Ek = ½mv²)?

Answer:

6.762\times 10^{-12} m is the de Broglie wavelength of this electron.

Explanation:

To calculate the wavelength of a particle, we use the equation given by De-Broglie's wavelength, which is:

\lambda=\frac{h}{\sqrt{2mE_k}}

where,

= De-Broglie's wavelength = ?

h = Planck's constant = 6.624\times 10^{-34}Js

m = mass of beta particle = 9.1094\times 10^{-31} kg

E_k = kinetic energy of the particle = 4.71\times 10^{-15}J

Putting values in above equation, we get:

\lambda =\frac{6.624\times 10^{-34}Js}{\sqrt{2\times 9.1094\times 10^{-31} kg\times 4.71\times 10^{-15}J}}

\lambda = 6.762\times 10^{-12} m

6.762\times 10^{-12} m is the de Broglie wavelength of this electron.

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