If the [H3O+] of a solution is 1x 10-8 mol/L the [OH-] is

(H30+)= 1x10^-6 M

Explanation:

Both pH and pOH have a relationship to belonging to the same aqueous solution: the expression of the Kwater (ionic product of the water Kw) is used:

1x 10-8 mol/L equals to1x10-8 M

(H3O+) x (OH-) = 1x10^-14

(H30+)x 1x 10^-8 =1x10^-14

(H30+)= 1x10^-14/1x 10^-8

(H30+)= 1x10^-6 M

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

14,16 18 reacts with group