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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 Nickel (ii)hydroxide may be formed without or (more likely. < since the molarity of either acid is the same, the moles of each acid are equal

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That means the difference between their ph is determined solely on which acid dissociates more 60ml ⋅ 0.45m ol/l = 27mm ol of n aoh (dont forget the m for milli=one thousandth) this will produce 27 2 = 13.5mm ol of n i(oh)2 now you multiply this by the molecular mass number to get the weight in milligrams (divide by 1000 to get the grams) Since the strong acid dissociates more (releases more #h^+# ions), it will have a lower ph

= since both acids are monoprotic (they only release one hydrogen) they will both take the same amount of #oh.

When an acid and a base are placed together, they react to neutralize the acid and base properties, producing a salt (neutralisation) The compound formed by the cation of the base and the anion of the acid is called a salt Example h cl + n aoh → n acl + h 2o hydrochloric acid + sodium hydroxide →. The nitrate and the natrium ions

Na_2co_3(aq) + 2agno_3(aq) rarr ag_2co_3(s)darr + 2nano_3(aq) the net ionic equation is This is also a 1:1 ratio. 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.

Could a buffered solution be made by mixing aqueous solutions of hcl and naoh

Why isn't a mixture of a strong acid and its conjugate base considered a buffered solution? M g(oh)2(s) ⇌ m g2+ (aq) + 2oh − (aq) in order to determine the maximum concentration of m g2+ ions permissible in the n aoh solution before a precipitate will be formed, you'd need the value of the solubility product constant, ksp. Because n aoh → n a+ + oh − and n i2+ + 2oh −→n i(oh)2 total mols of n aoh