gram molecular weight of Ca(NO3)2 = 40+ 14*2 + 16*6 = 164g/mole no of moles of C
ID: 699982 • Letter: G
Question
gram molecular weight of Ca(NO3)2 = 40+ 14*2 + 16*6 = 164g/mole
no of moles of Ca(NO3)2 = W/G.M.Wt
= 2.11*10^-4/164 = 1.286*10^-6 moles of Ca(NO3)2
Ca(NO3)2 -------------------------> Ca^2+ (aq) + 2NO3^-
1.286*10^-6 mole 1.286*10^-6 2*1.286*10^-6
no of moles of Ca^2+ ions = 1.286*10^-6 moles
no of moles of NO3^- ions = 2.572*10^-6 moles
Total no of moles of ions = 1.286*10^-6 + 2.572*10^-6
= (1.286+2.572)*10^-6
= 3.858*10^-6 moles of ions
Explanation / Answer
Be sure to answer all parts. How many total moles of ions are released when the following sample dissolves completely in water? Enter your answer in scientific notation. 2.11 × 104 g of Ca(NO3)2: ___× 10__ mol
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