It would be acidic rather than basic.
Beyond that, you'd need to provide additional details.
0.1125
60 g NaOH x 1 mole NaOH/40 g NaOH = 1.5 moles NaOH
The molar weight of NaOH is approx. 40 g/molSo 12.0 g of NaOH would be approx 12/40 = 0.3 mol NaOH
There is no NaCl2. It would simply be NaCl and the reactant would be HCl and NaOH. Thus,HCl + NaOH ==> NaCl + H2O
1280 grams
The alkali in NaOH would burn your skin.
0.1125
60 g NaOH x 1 mole NaOH/40 g NaOH = 1.5 moles NaOH
The molar weight of NaOH is approx. 40 g/molSo 12.0 g of NaOH would be approx 12/40 = 0.3 mol NaOH
There is no NaCl2. It would simply be NaCl and the reactant would be HCl and NaOH. Thus,HCl + NaOH ==> NaCl + H2O
1280 grams
As NaOH is a strong base I would not be surprised to see a 14 pH at least.
The premise of this question is incorrect. When NaOH is added to water the hydroxide concentration increases. NaOH is a base. If a substance decreases hydroxide concentration it would be an acid.
The product is sodium hydroxide. The reaction equation is Na2O + H2O -> 2 NaOH.
3.42 moles NaOH (39.998 grams/1 mole NaOH) = 137 grams NaOH
NaOH is required promote Aldol and dehydration at the end of the reaction. Moreover, the percentage of NaOH is less as greater %age would promote Cannizaro reaction.
Like other ionic compounds NaOH will not conduct electricity in its solid form, but will if dissolved in water or molten.