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This is also a 1:1 ratio. For every acid, you have a conjugate base (that no longer has that extra h^+ ion), and for every base, you have a conjugate acid (that has an extra h^+ ion). 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.
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 Conjugates are basically the other term They will be completely consumed by the reaction.
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. 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.
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