<span>False,
This is because when you can easily ionize and atom or the chances of it being ionizable are quite high, it means that that particular atom have very low ionization potential that is the reason why it was easily ionizable
An atom with a high ionization power and a firmly negative electron fondness will both pull in electrons from different particles and oppose having its electrons taken away; it will be an exceedingly electronegative molecule.</span>
Has 9 protons and 9 electrons in per atom.........
Answer:
<u><em>Arrhenius Acid:</em></u>
According to Arrhenius concept, Acids are proton donors.
Since H₂SO₄ have a proton (H⁺ ion) and it can donate it to be made a sulphate ion, So it is an Arrhenius acid.
See the following reaction =>
<u><em>H₂SO₄ + H₂O => HSO₄ + H₃O⁺</em></u>
<u><em>Arrhenius Base:</em></u>
An Arrhenius base is a a proton acceptor.
KOH accepts the proton to to made to KOH₂ and a proton acceptor.
See the following reaction =>
<u><em>KOH + H₂o => KOH₂ + OH⁻</em></u>
<u><em></em></u>
You start numbering from <span>the side with the methyl group and find the longest chain, which is seven = hept. The triple bond is -yne at C3. The methyl group is on C2.
This means that the best name for this molecule is </span><span>2-methyl-3-heptyne.</span>
Complete Question
The complete question is shown on the first uploaded image
Answer:
The pressure is ![[O_2] = 4.8 *10^{-5} \ atm](https://tex.z-dn.net/?f=%5BO_2%5D%20%3D%20%204.8%20%2A10%5E%7B-5%7D%20%5C%20atm)
Explanation:
From the question we are told that
The pressure of 
is ![[SO_3 ] = 0.63 \ atm](https://tex.z-dn.net/?f=%5BSO_3%20%5D%20%3D%20%200.63%20%5C%20atm)
The pressure of 
is ![[SO_ 2] = 0.30 \ atm](https://tex.z-dn.net/?f=%5BSO_%202%5D%20%20%3D%20%200.30%20%5C%20atm)
The equilibrium constant is 
The reaction is
⇔ 
Generally the equilibrium constant is mathematically represented as
=> ![[O_2] = \frac{k_p * [SO_3] ^2 }{[SO_2]^2}](https://tex.z-dn.net/?f=%5BO_2%5D%20%3D%20%20%5Cfrac%7Bk_p%20%2A%20%5BSO_3%5D%20%5E2%20%7D%7B%5BSO_2%5D%5E2%7D)
substituting values
![[O_2] = \frac{1.2 *10^{-5} * 0.60 ^2 }{0.30^2}](https://tex.z-dn.net/?f=%5BO_2%5D%20%3D%20%20%5Cfrac%7B1.2%20%2A10%5E%7B-5%7D%20%2A%200.60%20%5E2%20%7D%7B0.30%5E2%7D)
