H20 = 2.741 x 10^23
C6H8 = 1.0823 x 10^23
Lets assume x volume of NaOH and x volume of HCl are added together.
NaOH ---> Na⁺ + OH⁻
NaOH is a strong base therefore it completely ionizes and releases OH⁻ ions into the medium
HCl ---> H⁺ + Cl⁻
HCl is a strong base and completely ionizes and releases H⁺ ions in to the medium. number of NaOH moles in 1 L - 0.1 mol
Therefore in x L - 0.1 /1 * x = 0.1x moles of NaOH present
Similarly in HCl x L contains - 0.1x moles of HCl
H⁺ + OH⁻ ---> H₂O
Due to complete ionisation, 0.1x moles of H⁺ ions and 0.1x moles of OH⁻ ions react to form 0.1x moles of H₂O. Therefore all H⁺ and OH⁻are completely used up and yield water molecules.
Then at this point the H⁺ and OH⁻ ions in the medium come from the weak dissociation of water. This is equivalent to 1 x 10⁻⁷M
pH = -log [H⁺]
pH = -log [10⁻⁷]
pH = 7
pH is therefore equals to 7 which means the solution is neutral
Al(NO3)3(aq) + 3NaOH(s) --> Al(OH)3 (s) + 3NaNO3 (aq)
The precipitate here is Al(OH)3 (s), since the solid reactant is the precipitate in the aqueous solution. Usually, it is okay to assume in basic chemistry that the transition metal is going to be part of the compound that is the precipitate, especially in an acidic salt and a strong base reaction that we have here.
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