Oh− (aq) + h3o+ (aq) → 2h2o(l) so you can say that when you mix these two solutions, the hydronium cations present in the hydrochloric acid solution will be the limiting reagent, i.e They will be completely consumed by the reaction. > (a) with hcn the hcn adds across the α c=o group to form a cyanohydrin Underbrace (ch_3cocooh)_color (red) (pyruvic acid) + hcn →. So this is a propanol derivative Both names seem to be unambiguous.
6.3072 g >>molarity = moles of solute/volume of solution (in litres) 0.45 m = n/0.4 l n = 0.45 m × 0.4 l = 0.18 mol you need 0.18 mol of nh_4oh molar mass of nh_4oh is 35.04 g/mol mass of solute = 0.18 cancelmol × 35.04 g/cancelmol = 6.3072 g Start by writing the balanced chemical equation for this neutralization reaction The added water to reach 100.00 ml doesn't change the mols of hcl present, but it does decrease the concentration by a factor of 100//40 = 2.5 Regardless, what matters for neutralization is what amount of naoh you add to what number of mols of hcl I got ph's of 1.36, 1.51, 1.74, 2.54 You started with 0.1100 m hcl, but it was diluted from 40 ml to 100 ml
