First, we can neglect the Ka2 value and use Ka1:
according to this equation and by using the ICE table:
H2Se ⇄ H+ +HSe-
initial 0.4 M 0 0
change -X +X +X
Equ (0.4-X) X X
so,
Ka1 = [H+][HSe-] / [H2Se]
so by substitution:
1.3 x 10^-4 = X*X / (0.4 -X) by solving for X
∴X = 0.00715
∴[H+] = 0.00715 m
∴PH = -㏒[H+]
= -㏒ 0.00715
= 2.15
What elem eat is this? That has three orbitals? Usually the first orbital will contain only two electrons. The second orbital would contain four, the third would contain 16
Answer:
it would be a cation ... i believe
Explanation:
To become a cation, a positively charged ion, an atom would have to lose at least two electrons
Answer : The reaction 2 is spontaneous.
Explanation :
As we know that:
= +ve, reaction is non spontaneous
= -ve, reaction is spontaneous
= 0, reaction is in equilibrium
For the reaction to be spontaneous, the Gibbs free energy of the reaction must come out to be negative.
Reaction 1:
Glucose + Pi 
glucose-6-phosphate + H₂O, ΔG = +13.8 kJ/mol
Reaction 2:
ATP + H₂O ADP + Pi, ΔG = -30.5 kJ/mol
From this we conclude that the value of ΔG is negative. So, reaction 2 is a spontaneous reaction.
