> (a) with hcn the hcn adds across the α c=o group to form a cyanohydrin Underbrace (ch_3cocooh)_color (red) (pyruvic acid) + hcn →. 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. Conjugates are basically the other term 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 sodium ions remain in solution as spectator ions
If xs sodium hydroxide is added the precipitate redissolves to give the soluble plumbate (ii) ion A simple way of writing this is Since water is in excess, 67.7 g mgo are needed to produce 98.0 g mg(oh)_2 Balanced equation mgo(s) + h_2o(l)rarrmg(oh)_2(s) moles magnesium hydroxide start with the given mass of mg(oh)_2 and convert it to moles by dividing by its molar mass (58.319 g/mol) Since molar mass is a fraction, g/mol, we can divide by multiplying by the reciprocal of the molar mass, mol/g. As you know, ammonia is a weak base, which means that it does not ionize completely in aqueous solution
The correct answer is a) hydroxide, carbonate, and hydrogen carbonate Alkalinity is another word for basicity (the concentration of hydroxide ions) A) hydroxide ions oh⁻ are the strongest base you can have in water Carbonate ions and hydrogen carbonate ions also react with water to form hydroxide ions Co₃²⁻ + h₂o ⇌ hco₃⁻ + oh⁻ hco₃⁻ + h₂o ⇌ h₂co₃ + oh⁻ b. Answer is 286g (3s.f) concept required
Mole calculation first start off by finding the number of moles for both compounds Pbcl (oh)=0.185/ (207.2+35.5+16+1)*1000 (1kg=1000g) = 0.712 (to 3s.f) pb2cl2co3=0.185/ (207.2*2+35.5*2+12+16*3)*1000 = 0.339 (to 3s.f) next, find the moles of pb from each compound 1 therefore, moles of pb in pbcl (oh) is 0.712 pb
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