Answer:
Both contain a hydroxyl group on the second carbon.
Neither contain a hydroxyl group on an end chain carbon.
Explanation:
2,2-butanediol Formula: C4H10O2
2,3,4-pentanetriol Formula: C5H12O3
Both contain a hydroxyl group on the second carbon.
Neither contain a hydroxyl group on an end chain carbon.
<span>The metal that would more easily lose an electron would be potassium. It is more reactive than sodium. Also, looking on the periodic table, </span><span>from top to bottom for groups 1 and 2, reactivity increases. So, it should be potassium. Hope this answers the question. Have a nice day.</span>
Kekulé/Vant Hoff and Le Bel?
<span>The number of valence electrons from the electron configuration,write 1s², 2s² 2p⁶, 3s² 3p⁴.
The "big" numbers are the principal quantum numbers (n) which correspond to the major energy levels, and sub-scripted numbers are the number of electrons in the sub-energy level.
</span><span>For sulfur, the highest principal quantum number is 3, and there are 6 electrons in the 3s² and 3p⁴ sublevels. </span>
First, we have to get the initial [C6H8O6] = mass/molar mass
when the molar mass of C6H8O6 = 176.12 g/mol
∴[C6H8O6] = 0.25 g / 176.12 g/mol
= 0.00142 M
when
C6H8O6 ⇄ H+ + C6H7O6-
intial 0.00142 M 0 0
change -X +X +X
Equ (0.00142-X) X X
so, Ka = [H+][C6H7O6-] / [C6H8O6]
by substitution:
8 x 10^-5 = X * X / (0.00142-X) by solving this equation for X
∴ X = 0.000299
∴[H+] = 0.000299
∴PH = -㏒[H+]
= -㏒ 0.000299
= 3.52
