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Novosadov [1.4K]
2 years ago
8

21. aluminum sulfate + calcium hydroxide - aluminum hydroxide + calcium sulfate

Chemistry
1 answer:
Igoryamba2 years ago
3 0

Al2(SO4)3 + 3Ca(OH)2 -> 2Al(OH)3 + 3Ca(SO)4

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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
Sedaia [141]

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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