Explanation:
<h2>Melting point = -33.34°c</h2><h2>boiling point=77.73°c</h2>
Answer:
Would it be<em><u> 7.69 seconds</u></em>?
Explanation:
The maximum concentration (in m) of calcium ions[
] in the given solution is 6.2×![10^{-8}](https://tex.z-dn.net/?f=10%5E%7B-8%7D)
The balanced equillibrium equation is ,
(s) ↔
(aq) +
(aq)
From the reaction equation above, the formula for Ksp:
Ksp = [
][
] = 3.4×![10^{-9}](https://tex.z-dn.net/?f=10%5E%7B-9%7D)
We know the value for [
] , so we can solve for [
]as:
( 3.4×
) = [
] (0.055)
[
] = 61.8×![10^{-9}](https://tex.z-dn.net/?f=10%5E%7B-9%7D)
[
] = 6.2×![10^{-8}](https://tex.z-dn.net/?f=10%5E%7B-8%7D)
Therefore, the maximum concentration (in m) of calcium ions[
] in the given solution is 6.2×![10^{-8}](https://tex.z-dn.net/?f=10%5E%7B-8%7D)
Learn more about calcium ions[
] here:
brainly.com/question/19865131
#SPJ4
Explanation:
![E_n=-13.6\times \frac{Z^2}{n^2}eV](https://tex.z-dn.net/?f=E_n%3D-13.6%5Ctimes%20%5Cfrac%7BZ%5E2%7D%7Bn%5E2%7DeV)
Formula used for the radius of the
orbit will be,
(in pm)
where,
= energy of
orbit
= radius of
orbit
n = number of orbit
Z = atomic number
a) The Bohr radius of an electron in the n = 3 orbit of a hydrogen atom.
Z = 1
![r_3=476.1 pm](https://tex.z-dn.net/?f=r_3%3D476.1%20pm)
476.1 pm is the Bohr radius of an electron in the n = 3 orbit of a hydrogen atom.
b) The energy (in J) of the atom in part (a)
![E_n=-13.6\times \frac{Z^2}{n^2}eV](https://tex.z-dn.net/?f=E_n%3D-13.6%5Ctimes%20%5Cfrac%7BZ%5E2%7D%7Bn%5E2%7DeV)
![E_3=-13.6\times \frac{1^2}{3^2}eV=1.51 eV](https://tex.z-dn.net/?f=E_3%3D-13.6%5Ctimes%20%5Cfrac%7B1%5E2%7D%7B3%5E2%7DeV%3D1.51%20eV)
![1 eV=1.60218\times 10^{-19} Joules](https://tex.z-dn.net/?f=1%20eV%3D1.60218%5Ctimes%2010%5E%7B-19%7D%20Joules)
![1.51 eV=1.51\times 1.60218\times 10^{-19} Joules=2.4210\times 10^{-19} Joules](https://tex.z-dn.net/?f=1.51%20eV%3D1.51%5Ctimes%201.60218%5Ctimes%2010%5E%7B-19%7D%20Joules%3D2.4210%5Ctimes%2010%5E%7B-19%7D%20Joules)
is the energy of n = 3 orbit of a hydrogen atom.
c) The energy of an Li²⁺ ion when its electron is in the n = 3 orbit.
![E_n=-13.6\times \frac{Z^2}{n^2}eV](https://tex.z-dn.net/?f=E_n%3D-13.6%5Ctimes%20%5Cfrac%7BZ%5E2%7D%7Bn%5E2%7DeV)
n = 3, Z = 3
![E_3=-13.6\times \frac{3^2}{3^2}eV = -13.6 eV](https://tex.z-dn.net/?f=E_3%3D-13.6%5Ctimes%20%5Cfrac%7B3%5E2%7D%7B3%5E2%7DeV%20%3D%20-13.6%20eV)
![=-13.6eV = -13.6\times 1.60218\times 10^{-19} Joules=2.179\times 10^{-18} Joules](https://tex.z-dn.net/?f=%3D-13.6eV%20%3D%20-13.6%5Ctimes%201.60218%5Ctimes%2010%5E%7B-19%7D%20Joules%3D2.179%5Ctimes%2010%5E%7B-18%7D%20Joules)
is the energy of an Li²⁺ ion when its electron is in the n = 3 orbit.
d) The difference in answers is due to change value of Z in the formula which is am atomic number of the element..
![E_n=-13.6\times \frac{Z^2}{n^2}eV](https://tex.z-dn.net/?f=E_n%3D-13.6%5Ctimes%20%5Cfrac%7BZ%5E2%7D%7Bn%5E2%7DeV)