Calculate the molar concentration of H3O ions and the pH of the following solutions: (a) 25.0 cm3 of 0.144 m HCl(aq) was added to 25.0 cm3 of 0.125 m NaOH(aq), (b) 25.0 cm3 of 0.15 m HCl(aq) was added to 35.0 cm3 of 0.15 m KOH(aq), and (c) 21.2 cm3 of 0.22 m HNO3(aq) was added to 10.0 cm3 of 0.30 m NaOH(aq).

elements

explanation:

yes the other ones are right just show me a pic of ur face cuz u look hot

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g

the atomic mass is the bottom number on the element.

-steel jelly