A solution of calcium chlorate was poured into a sodium fluoride solution. Would you expect a precipitate to form if 255.0 mL of
the calcium chlorate solution (2.0x10-5 M) was mixed with 300.0 mL of a 2.5x10-3 M sodium fluoride solution? A. No, because Qsp < Ksp for calcium fluoride B. Yes, because Qsp < Ksp for calcium fluoride C. Yes, because Qsp = Ksp for sodium chlorate D. Yes, because Qsp = Ksp for calcium fluoride E. No, because Qsp = Ksp for sodium chlorate F. No, because Qsp < Ksp for sodium chlorate G. No, because Qsp > Ksp for calcium fluoride H. No, because Qsp > Ksp for sodium chlorate I. Yes, because Qsp < Ksp for sodium chlorate J. No, because Qsp = Ksp for calcium fluoride K. Yes, because Qsp > Ksp for sodium chlorate L. Yes, because Qsp > Ksp for calcium fluoride
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14.292 grams of Fe2O3 is formed when 10 gram of iron metal is burned.
Explanation:
The balanced equation for the reaction is to be known so that number of moles taking part can be known.
The balanced chemical equation is
4Fe + 3⇒ 2
From the given weight of iron to be used for the production of , number of moles of Fe taking part in the reaction can be known by the formula:
Number of moles= mass ÷ Atomic mass of one mole of the element.
(Atomic weight of Fe is 55.845 gm/mole)
Putting the values in equation
Number of moles = 10 gm ÷ 55.845 gm/mole
= 0.179 moles
Applying the stoichiometry concept
4 moles of Fe gives 2 Moles of Fe2O3
0.179 moles will produce x moles of Fe2O3
So, 2÷ 4 = x ÷ 0.179
2/4 = x/ 0.179
2 × 0.179 = 4x
2 × 0.179 / 4 = x
x = 0.0895 moles
So from 10 grams of iron metal 0.0895 moles of Fe2O3 is formed.
Now the formula used above will give the weight of Fe2O3
weight = atomic weight × number of moles
= 159.69 grams × 0.0895
= 14.292 grams of Fe2O3 formed.