Answer:
The number of neutrons present in one atom of isotope of Silicon of mass 28 amu is<u> 14 neutrons</u>
Explanation:
Symbol of Si isotope
![_{14}^{28}\textrm{Si}](https://tex.z-dn.net/?f=_%7B14%7D%5E%7B28%7D%5Ctextrm%7BSi%7D)
<u>Number of Neutron = Mass number - Atomic Number</u>
Mass number = Total number of protons and neutrons present in the nucleus of the atom.For Si = 28 amu
Atomic Number = Total number of Protons present in the nucleus.
Si = 14
Number of neutron = 24 - 14
= 14
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
. What is the de Broglie wavelength of this electron (Ek = ½mv²)?
Answer:
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}}](https://tex.z-dn.net/?f=%5Clambda%3D%5Cfrac%7Bh%7D%7B%5Csqrt%7B2mE_k%7D%7D)
where,
= De-Broglie's wavelength = ?
h = Planck's constant = ![6.624\times 10^{-34}Js](https://tex.z-dn.net/?f=6.624%5Ctimes%2010%5E%7B-34%7DJs)
m = mass of beta particle = ![9.1094\times 10^{-31} kg](https://tex.z-dn.net/?f=%209.1094%5Ctimes%2010%5E%7B-31%7D%20kg)
= kinetic energy of the particle = ![4.71\times 10^{-15}J](https://tex.z-dn.net/?f=4.71%5Ctimes%2010%5E%7B-15%7DJ)
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}}](https://tex.z-dn.net/?f=%5Clambda%20%3D%5Cfrac%7B6.624%5Ctimes%2010%5E%7B-34%7DJs%7D%7B%5Csqrt%7B2%5Ctimes%209.1094%5Ctimes%2010%5E%7B-31%7D%20kg%5Ctimes%204.71%5Ctimes%2010%5E%7B-15%7DJ%7D%7D)
![\lambda = 6.762\times 10^{-12} m](https://tex.z-dn.net/?f=%5Clambda%20%3D%206.762%5Ctimes%2010%5E%7B-12%7D%20m)
is the de Broglie wavelength of this electron.
Answer:
I'm converting this if I could remember how
2.882568
2 110321/ 125000
T-T sorry if I'm wrong I have bad memory
so I recommend not using my answer at all,
if that is even how y'all write it.
Answer:
1.427x10^-3mol per L
Explanation:
![Y(IO_{3} )_{3} ---- Y^{3+} +IO_{3} ^{3-}](https://tex.z-dn.net/?f=Y%28IO_%7B3%7D%20%29_%7B3%7D%20----%20Y%5E%7B3%2B%7D%20%2BIO_%7B3%7D%20%5E%7B3-%7D)
I could use ⇌ in the math editor so I used ----
from the question each mole of Y(IO3)3 is dissolved and this is giving us a mole of Y3+ and a mole of IO3^3-
Ksp = [Y^3+][IO3-]^3
So that,
1.12x10^-10 = [S][3S]^3
such that
1.12x10^-10 = 27S^4
the value of s is 0.001427mol per L
= 1.427x10^-3mol per L
so in conclusion
the molar solubility is therefore 1.427x10^-3mol per L