PKa = -log (Ka) = log [HPO4(2-)] - log[H+]^2 = - log(4.2×10^-13)
pH = - log [H+]
- log [H+]^2 = - 2 log [H+]
2pH = - log (4.2×10^-13) - log [HPO4(2-)]
2pH = - log (4.2×10^-13) - log (0.550)
pH = 6.32
To add tartness / good luck and will you mark me brainliest please
Percentage yield= actual yield/theoretical yield x100
So you would have to do-
15/22 x 100. Hope this helps!!
Answer:
Because CLEARLY, each mole of glucose, C6H12O6 contains 6⋅mol oxygen atoms.
