The first dissociation for H2X:
H2X +H2O ↔ HX + H3O
initial 0.15 0 0
change -X +X +X
at equlibrium 0.15-X X X
because Ka1 is small we can assume neglect x in H2X concentration
Ka1 = [HX][H3O]/[H2X]
4.5x10^-6 =( X )(X) / (0.15)
X = √(4.5x10^-6*0.15)
∴X = 8.2 x 10-4 m
∴[HX] & [H3O] = 8.2x10^-4
the second dissociation of H2X
HX + H2O↔ X^2 + H3O
8.2x10^-4 Y 8.2x10^-4
Ka2 for Hx = 1.2x10^-11
Ka2 = [X2][H3O]/[HX]
1.2x10^-11= y (8.2x10^-4)*(8.2x10^-4)
∴y = 1.78x10^-5
∴[X^2] = 1.78x10^-5 m
Answer:
MECHANISM:
1) The lone pair on oxygen attacks the H-Br molecule forming a hydronium ion.
2) Formation of carbocation.
3) Attack of Nucleophile Br − .
Explanation:
Answer:
1s² 2s²2p³
Explanation:
If the atom has seven electrons, it is Element 7 (nitrogen).
In the Periodic Table, you count the electrons in all the subshells up to
No. 7.
In the first Period, you have filled the 1s level (2 electrons).
In the second Period, you have filled the 2s subshell (2 electrons) and put three electrons in the 2p subshell.
Thus, the electron configuration is
1s² 2s²2p³
Note how the superscripts tell you the number of electrons in each subshell: <em>2 + 2 + 3 = 7</em>.
Yes, because it comes from a one thing and spreads throughout the entire space. Similar to dripping foot coloring into a glass of water, or spraying air freshener.
