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Valence elections, electron affinity, electronegativity, atomic radius
The reaction that should be followed is
Na2SO4 + C<span>a(NO3)2 --> CaSO4 + 2NaNO3</span>
first calculate the limiting reactant
mol Na2SO4 = 0.075 L (<span>1.54×10−2 mol / L) = 1.155x10-3 mol
mol Ca(NO3)2 = 0.075 L (</span><span>1.22×10−2 mol / L) = 9.15x10-4 mol
so the limiting reactant is the Ca(NO3)2
so all of the Ca2+ will be precipitated, percentage unprecipitated = 0.00 % </span>
{ 12 mol * 27 g / mol * 0.902 j/gdeg C* ( 658 -72 ) deg C + 12 * 27 * 3.95 }
quantity of energy required to heat 12 mol of aluminium for 72 deg C to its melting point 658 C
so energy applied is 1451 057 . 33 J
The sample contains 1.00 mol H₂O.
Moles of water = 18.0 g H₂O × 1 mol H₂O/18.0 g H₂O = 1.00 mol H₂O
