Answer:
Electronic configuration of Ga³⁺:
Ga = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s⁰ 3d¹⁰ 4p⁰
Explanation:
Gallium is present in group 13.
It has 3 valance electrons.
Electronic configuration:
Ga = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p¹
There are two possibilities of gallium to form cation.
By removing one 4p electron to form Ga +1 cation.
By removing two 4s electrons and one 4p to form Ga +3 cation.
Gallium is preset in group thirteen that's why common oxidation state is +3.
Electronic configuration of Ga⁺:
Ga = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁰
Electronic configuration of Ga³⁺:
Ga = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s⁰ 3d¹⁰ 4p⁰
Answer:
German chemist G.E. Stahl
I would say carbon monoxide
Firstly because the image shows two different atoms bonded so it cannot be nitrogen or krypton
Secondly sulphur dioxide has 3 atoms bonded (two oxygens and a sulphur atom) so it can’t be that
Finally it can’t be hydrogen chloride because chloride is significantly larger than hydrogen
Thus it must be carbon monoxide as carbon and oxygen are bonded (CO) and are both relatively similar in size
it should be True unless if I’m wrong
Answer:
pH at equivalence point is 8.52
Explanation:
1 mol of HCOOH reacts with 1 mol of NaOH to produce 1 mol of 
So, moles of NaOH used to reach equivalence point equal to number of moles produced at equivalence point.
As density of water is 1g/mL, therefore molarity is equal to molality of an aqueous solution.
So, moles of produced = 
Total volume of solution at equivalence point = (25+29.80) mL = 54.80 mL
So, at equivalence point concentration of = 
At equivalence point, pH depends upon hydrolysis of . So, we have to construct an ICE table.
I: 0.1940 0 0
C: -x +x +x
E: 0.1940-x x x
So, ![\frac{[HCOOH][OH^{-}]}{[HCOO^{-}]}=K_{b}(HCOO^{-})=\frac{10^{-14}}{Ka(HCOOH)}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BHCOOH%5D%5BOH%5E%7B-%7D%5D%7D%7B%5BHCOO%5E%7B-%7D%5D%7D%3DK_%7Bb%7D%28HCOO%5E%7B-%7D%29%3D%5Cfrac%7B10%5E%7B-14%7D%7D%7BKa%28HCOOH%29%7D)
species inside third bracket represent equilibrium concentrations
So, 
or,
So, 
So, 
So, ![pH=14-pOH=14+log[OH^{-}]=14+logx=14+log(3.285\times 10^{-6})=8.52](https://tex.z-dn.net/?f=pH%3D14-pOH%3D14%2Blog%5BOH%5E%7B-%7D%5D%3D14%2Blogx%3D14%2Blog%283.285%5Ctimes%2010%5E%7B-6%7D%29%3D8.52)
