A precipitate forms when mixing solutions of sodium fluoride (NaF) and lead II nitrate (Pb(NO3)2). Complete and balance the net
1 answer:
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Answer:- 27.7 grams of are produced.
Solution:- The balanced equation is:
let's convert the grams of each reactant to moles and calculate the grams of the product and see which one gives least amount of the product. This least amount would be the answer as the least amount we get is from the limiting reactant.
Molar mass of = 207.2+2(126.90) = 461 gram per mol
let's do the calculations for the grams of the product for the given grams of each of the reactant:
=
=
From above calculations, NaI gives least amount of , so the answer is, 27.7 g of
are produced.
Answer : The molal freezing point depression constant of X is
Explanation : Given,
Mass of urea (solute) = 5.90 g
Mass of X liquid (solvent) = 450.0 g
Molar mass of urea = 60 g/mole
Formula used :
where,
= change in freezing point
= freezing point of solution =
= freezing point of liquid X=
i = Van't Hoff factor = 1 (for non-electrolyte)
= molal freezing point depression constant of X = ?
m = molality
Now put all the given values in this formula, we get
Therefore, the molal freezing point depression constant of X is