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This is also a 1:1 ratio. You started with 0.1100 m hcl, but it was diluted from 40 ml to 100 ml 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. I got ph's of 1.36, 1.51, 1.74, 2.54 We want the standard enthalpy of formation for ca (oh)_2
Thus, our required equation is the equation where all the constituent elements combine to form the compound, i.e.
The acid in excess is then titrated with n aoh (aq) of known concentration.we can thus get back to the concentration or molar quantity of m (oh)2.as it stands the question (and answer) are hypothetical. 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 So this is a propanol derivative Both names seem to be unambiguous.
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
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