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In the laboratory a student combines 36.8 mL of a 0.252 M sodium nitrate solution with 14.4 mL of a 0.533 M copper(II) nitrat
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Both sodium nitrate (NaNO₂) and copp- er (11) ritrate [Cu(NO3)2 are stroong electrolyte No. of moles of sodium nitrate in 36.Copper (11) nitrate dissociates as below - C (NO3) È Curt + 2NO3- Thus from 7-67 x 10-3 mol of C (NO3)2 we will get 2x 7.67 xNow, total volume of the solution - (36.8 + 14.4) mL = 51.2 ml = 51.2 x 103 o final concentration of nitrate anion = 24.6 x 1I have given the answer in three significant figures.....if the answer does not match with the computer database due to significant figures.....then please comment below.......

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