H3O+(aq) + OH-(aq) --> 2H2O (l)
NaHCO3(s) --> NaH 2+ (aq) + CO3 2- (aq)
NaH 2+ (aq) + H2O (l) --> Na+ (aq) + H3O+ (aq)
H2O (l) + CO3 2- (aq) --> OH- (aq) + HCO3- (aq)
(I'm not completely sure if I did the third question right) I'm sorry if I got it wrong
Answer:
They are 1.204×10^24 atoms of hydrogen present in 18 grams of water. In order to calculate this,it is necessary to compute the number of hydrogen moles present in the sample.
You mix 35mL of 0.54M NaI with 14mL of 0.84M Pb(NO3)2.
A is the answer to the question you asked.
