Good idea!maybe I should try that
Using the Henderson-Hasselbalch equation on the solution before HCl addition: pH = pKa + log([A-]/[HA]) 8.0 = 7.4 + log([A-]/[HA]); [A-]/[HA] = 4.0. (equation 1) Also, 0.1 L * 1.0 mol/L = 0.1 moles total of the compound. Therefore, [A-] + [HA] = 0.1 (equation 2) Solving the simultaneous equations 1 and 2 gives: A- = 0.08 moles AH = 0.02 moles Adding strong acid reduces A- and increases AH by the same amount. 0.03 L * 1 mol/L = 0.03 moles HCl will be added, soA- = 0.08 - 0.03 = 0.05 moles AH = 0.02 + 0.03 = 0.05 moles Therefore, after HCl addition, [A-]/[HA] = 0.05 / 0.05 = 1.0 Resubstituting into the Henderson-Hasselbalch equation: pH = 7.4 + log(1.0) = 7.4, the final pH.
Answer: Silicon the first two electrons will go in the 1s orbital. Since 1s can only hold two electrons the next 2 electrons for Silicon go in the 2s orbital. The next six electrons will go in the 2p orbital. The p orbital can hold up to six electrons.
Hope this helps! :)
Given that
1 skvarnick = 45 quibs
3 quibs = 7 sleps
Now if we have 45 quibs it means we have you have one skvarnick
three quibs = 7 sleps
so one quib = 7/3 sleps
so 45 quibs = 7 X 45 / 3 = 105 sleps
