Answer:
K loses one electron to CI
Explanation:
The lewis electron dot notation shows only the chemical symbol of the element surrounded by dots to represent the valence electrons.
We have atom of K with one valence electrons
Cl with 7 valence electrons
For an electrostatic attraction to occur, both particles must be charged. To do this, one of the species must lose an electron, and the other gains it.
This will make both species attain a stable octet;
Hence, K will lose 1 electron and Cl will gain the electrons.
Ionization energy increases from left to right in the row and from bottom to top in a column. Also as we get closer to the nucleus it would be harder to take electrons out. B (atomic #5) has 2 layers of electron 2 and 3 atom in each layer. P has 15 so it would be 2,8 and 5 respectively. Ca is 20 so 2,8,8,2 and Zn is 30 and it would be 2,18,8,2.
For energy between second and third ionization we are looking at taking out the 3rd electron. B already has 3 electron in the first layer so its easy to take them all. P has 5 in the last layer so again easy. But when we look at Ca and Zn after the 2nd electron (in the last layer) we should change the layer go one layer inside. So this needs more energy. To pick between Zn and Ca (they are in the same row) I mentioned earlier that in one row as we go to the right ionization energy increases so the answer is Zn.
Dmitri Mendeleev was the creator and therefore organizer of the periodic table.
Answer:
Explanation:
From the information given:
oxidation of oxidized solution takes place at 950° C
For wet oxidation:
The linear and parabolic coefficient can be computed as:
![\dfrac{B}{B/A} = D_o \ exp \Big [\dfrac{-\varepsilon a}{k_BT} \Big]](https://tex.z-dn.net/?f=%5Cdfrac%7BB%7D%7BB%2FA%7D%20%3D%20D_o%20%5C%20exp%20%5CBig%20%5B%5Cdfrac%7B-%5Cvarepsilon%20a%7D%7Bk_BT%7D%20%5CBig%5D)
Using 
and 
values obtained from the graph:
Thus;
![\dfrac{B}{A} = 1.63 \times 10^8 exp \Big [ \dfrac{-2.05}{8.617 \times 10^{_-5}\times 1173}\Big] \\ \\ = 0.2535 \ \ \mu m/hr](https://tex.z-dn.net/?f=%5Cdfrac%7BB%7D%7BA%7D%20%3D%201.63%20%5Ctimes%2010%5E8%20exp%20%5CBig%20%5B%20%5Cdfrac%7B-2.05%7D%7B8.617%20%5Ctimes%2010%5E%7B_-5%7D%5Ctimes%201173%7D%5CBig%5D%20%5C%5C%20%5C%5C%20%3D%200.2535%20%5C%20%5C%20%5Cmu%20m%2Fhr)
![B= 386 \ exp \Big [-\dfrac{0.78}{8.617 \times 10^{-3} \times 1173} \Big] \\ \\ = 0.1719 \ \mu m^2/hr](https://tex.z-dn.net/?f=B%3D%20386%20%5C%20%20exp%20%5CBig%20%5B-%5Cdfrac%7B0.78%7D%7B8.617%20%5Ctimes%2010%5E%7B-3%7D%20%5Ctimes%201173%7D%20%5CBig%5D%20%5C%5C%20%5C%5C%20%20%3D%200.1719%20%5C%20%5Cmu%20m%5E2%2Fhr)
So, the initial time required to grow oxidation is expressed as:
∴
NOW;
Thus; since we will consider the positive sign, the initial thickness 
is ;
≅ 0.261 μm
