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kondor19780726 [428]
3 years ago
8

Using your value of Ksp, and starting with an equilibrium system consisting of a saturated solution of calcium hydroxide, predic

t whether the solubility will increase or decrease it you add a. A. 5.00 mL of 0.100 M HNO3 Ksp = 2.966610-5 b. 5.00 mL of 0.100 M NaOH C. 5.00 mL of 0.100 M Ca(NO3)2
Chemistry
1 answer:
Ugo [173]3 years ago
6 0

Answer:

(A) HNO_{3} Solubility will increase.

(B) NaOH Solubility will decrease.

(C) Ca(NO_{3} )_{2} Solubility will decrease.

Explanation:

a) According to Le Chatelier's Principle when a change is introduced in a reaction system at equilibrium, the system responds by the reaction shifting in the direction to counter the change introduced.

When the change introduced is addition of acid, some OH^{-} are consumed due to it and the system would respond by reaction shifting towards creation of more OH^{-} in the system by increased dissolution. Thus solubility increase.

(b) When OH^{-} are added, the reaction shifts to counter the change introduced of increased OH^{-} by shifting to side that decreases the increased OH^{-}. That happens by reaction shifting towards some Ca (OH)_{2} precipitating back. Hence solubility decreases.

(c) The change introduced is addition of Ca^{2+} which will be countered by reaction shifting to side that decreases increased Ca ^{2+}, which happens by some Ca(OH)_{2} precipitating back. Hence solubility decreases.

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Which reactant will be used up first if 78.1g of o2 is reacted with 62.4g of c4h10?
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Reagent O₂ will be consumed first.

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Then, by reaction stoichiometry, the following amounts of reactants and products participate in the reaction:

  • C₄H₁₀: 2 moles
  • O₂: 13 moles
  • CO₂: 8 moles
  • H₂O: 10 moles

Being:

  • C: 12 g/mole
  • H: 1 g/mole
  • O: 16 g/mole

The molar mass of the compounds that participate in the reaction is:

  • C₄H₁₀: 4*12 g/mole + 10*1 g/mole= 58 g/mole
  • O₂: 2*16 g/mole= 32 g/mole
  • CO₂: 12 g/mole + 2*16 g/mole= 44 g/mole
  • H₂O: 2*1 g/mole + 16 g/mole= 18 g/mole

Then, by reaction stoichiometry, the following mass quantities of reactants and products participate in the reaction:

  • C₄H₁₀: 2 moles* 58 g/mole= 116 g
  • O₂: 13 moles* 32 g/mole= 416 g
  • CO₂: 8 moles* 44 g/mole= 352 g
  • H₂O: 10 moles* 18 g/mole= 180 g

If 78.1 g of O₂ react, it is possible to apply the following rule of three: if by stoichiometry 416 g of O₂ react with 116 g of C₄H₁₀, 62.4 g of C₄H₁₀ with how much mass of O₂ do they react?

mass of O_{2} =\frac{416grams of O_{2}*62.4 grams ofC_{4}H_{10}   }{116 grams of C_{4}H_{10}}

mass of O₂= 223.78 grams

But 21.78 grams of O₂ are not available, 78.1 grams are available. Since you have less mass than you need to react with 62.4 g of C₄H₁₀, <u><em>reagent O₂ will be consumed first.</em></u>

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