An irregular chunk of iron has a density of 7.87 g/cm3. A 14.5 gram chunk is added to a graduated cylinder. If the final volume
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Answer:
Solution A that will form a precipitate with Ksp = 2.3 x 10−4
Explanation:
Li₃PO₄ ⇄ 3 Li⁺(aq) + PO₄³⁻(aq)
3S S
Where S = Solubility(mole/lit) and Ksp = Solubility product
⇒ Ksp = (3S)³ x (S)
⇒ 27S⁴ = 2.3x10−4
⇒ S = 0.05 mol/lit
Concentration of Li₃PO₄ precipitate = 0.05
<u>Solution A </u>
0.500 lit of a 0.3 molar LiNO₃ contains 0.5 x 0.3 = 0.15 mole
0.4 lit of a 0.2 molar Na₃PO₄ contains = 3 x 0.4 x 0.2 = 0.24 mole
3 LiNO₃ + Na₃PO₄ → 3 NaNO₃ + Li₃PO₄
(Mole/Stoichiometry)
= 0.05 = 0.24
Since from (Mole/Stoichiometry) ratio we can conclude that LiNO₃ is limiting reagent.
So concentration of Li₃PO₄ is equal to 0.05.
Answer:
Exothermic reaction for the HCl, endothermic reaction for the water
Explanation:
Heat was lost by HCl as its temperature lowered, so it was an exothermic reaction for the HCL.
Heat was gained by water as its temperature increased, so it was an endothermic reaction for the water.
m = Mass of water = 100 g
c = Specific heat of water =
= Change in temperature of water =
Heat is given by
Heat gained by water is .