Covalent bond , hopefully this help :) explanation : .....
The question requires us to explain the differences in radii of neutral atoms, cations and anions.
To answer this question, we need to keep in mind that a neutral atom presents the same number of protons (positive particles) and electrons (negative particles). Another important information is that the protons are located in the nucleus of the atom, while the electrons are around the nucleus. Also, there is an electrostatic force between protons and electrons, which means that they the protons tend to attract the electrons to the nucleus.
While a neutral atom presents the same number of protons and electrons, a cation is an ion with positive charge, which means it has lost one or more electrons. In a cation, the balance between protons and electrons doesn't exist anymore: now, there is more positive than negative charge (more protons than electrons), and the overall attractive force that the protons have for the electrons is increased. As a result, the electrons stay closer to the nucleus and the radius of a cation is smaller than the neutral atom from which it was derived.
On the other side, anions present negative charge, which means they have received electrons. Similarly to cations, the balance between protons and electrons doesn't exist anymore, but in this case, there are more electrons than protons. In an anion, the overall attractive force that the protons have for the electrons is decreased. As a result, the electrons are "more free" to move and, as they are not so attracted to the nucleus, they tend to stay farther from the positive nucleus compared to the neutral atom - because of this, the radius of an anion is larger than the neutral atom from which it was derived.
The formula is:
frequency * h (Planck's constant) = Energy
So, to find frequency you need just divide energy by the constant:
frequency = (8 * 10^-15 J) / (6.63 * 10^-34 J*s) = 1.2 * 10^19 1/s or 1.2 * 10^19 Hz
Answer:
M
Explanation:
Concentration of ![Cu^{2+} = [Cu(NO_3)_2]](https://tex.z-dn.net/?f=Cu%5E%7B2%2B%7D%20%3D%20%5BCu%28NO_3%29_2%5D)
= 0.020 M
Constructing an ICE table;we have:
![Cu^{2+}+4NH_3_{aq} \rightleftharpoons [Cu(NH_3)_4]^{2+}_{(aq)}](https://tex.z-dn.net/?f=Cu%5E%7B2%2B%7D%2B4NH_3_%7Baq%7D%20%5Crightleftharpoons%20%5BCu%28NH_3%29_4%5D%5E%7B2%2B%7D_%7B%28aq%29%7D)
Initial (M) 0.020 0.40 0
Change (M) - x - 4 x x
Equilibrium (M) 0.020 -x 0.40 - 4 x x
Given that: 
![K_f } = \frac{[Cu(NH_3)_4]^{2+}}{[Cu^{2+}][NH_3]^4}](https://tex.z-dn.net/?f=K_f%20%7D%20%3D%20%5Cfrac%7B%5BCu%28NH_3%29_4%5D%5E%7B2%2B%7D%7D%7B%5BCu%5E%7B2%2B%7D%5D%5BNH_3%5D%5E4%7D)
Since x is so small; 0.40 -4x = 0.40
Then:
M
