POH of a 0.0072 M=-lg(0.0072) = 2.1426675
Its hybridization would be sp because Be only has 2 covalent bonds with Cl
Answer:
pH = 5.7
Explanation:
pH = -log[H^+]
For HCl pH = -log[HCl] = - log [2.10 x 10⁻⁶ ] = 5.7
You have to figure out a way to write the two unknown abundances in terms of one variable.
The total abundance is 1 (or 100%). So if you say the abundance for the first one is X then the abundance for the second one has to be 1-X (where X is the decimal of the percentage so say 0.8 for 80%).
203(X) + 205(1-X) = 204.4
Then you just solve for X to get the percentage for TI-203.
And then solve for 1-X to get the percentage for TI-205.
After that the higher percentage would be the most abundant.
203x + 205 - 205x = 204.4
-2x + 205 = 204.4
-2x = -0.6
x = 0.3
1-x = 0.7
Then the TI-205 would have the highest percentage and would be the most abundant.
By use IUPAC nomenclature rule compound N2O3 is named as
Dinitrogen trioxide ( answer d)
Nitrogen (N2) is named before oxygen(O3) since they are arranged alphabetically. In addition the prefix Di is used infront of nitrogen since they are two nitrogen atoms while prefix tri is used infront of oxide since they are 3 oxygen atoms
