r The solubility product expression for silver(l) sulfate is 1) 2) For the react

Question

r The solubility product expression for silver(l) sulfate is 1) 2) For the reaction N2 (8)+ 2H2 (8)-N2H4 (8) (endothermic), what condition will increase the yield 2) of product? A) high temperature, low pressure C) low temperature, high pressure B) high temperature, high pressure D) low temperature, low pressure 3) For the reaction C2H2 (g)+ H20 (g) CH3CHO (g) (exothermic), what condition will increase the 3) yield of product? A) low temperature, high pressure C) high temperature, high pressure B) high temperature, low pressure D) low temperature, low pressure 4) If the solubility of calcium fluoride is 2.0 x 10-4 M, the solubility product value is. 4) A) 4.0 x 10-8 B) 3.2 x 10-11 C) 1.6x10-12 D) 8.0x 10-12 5) If the solubility of cupric sulfate is 4.9 x 10-3 M, the solubility product is B) 1.2 10-2 5) A) 8.7 10-8 C) 9.6 10-5 D) 2.4 x 10-5 6) Consider the equation 2NH3 (g) - N2 (g)+3H2 (g). Increasing the volume will 6) A) will have no effect B) shift the reaction to the right C) shift the reaction to the left D) One cannot say, since the temperature is unknown. 7) Consider the equation 2NH3 (g)- N2 (g)+3H2 (8). Decreasing the concentration of ammonia gas 7 will A) shift the reaction to the left B) shift the reaction to the right C) will have no effect D) One cannot say, since the temperature is unknown. 8) The equilibrium constant expression for the reaction 2 SO3 (g) + 2 Cl2 (g)-2502Cl2 (g) + O2 (g) is 8)--

Explanation / Answer

Please aks questions separately and not all at once

1)D [Ag+]2[SO42-] The coefficient comes as exponent on the outside .

2)B High temperature ,high pressure .

endothermic: temp decreases on product side .So increasing temperature will increase yield of product by pushing the reaction in the forward direction . Pressure is directly proportional to temperature .So high pressure will have same effect

3)D low temperature, low pressure. Exothermic: heat on product side. So opp conditions will force reaction in forward direction .

4)3.2*10^-11

the molecular weight =78

Given solubility of CaF2: 2.0*10^-4 M

Solubility equilibrium : CaF2--Ca2+ + 2F-

One mole of CaF2 gives one mole Ca2+ and 2 moles of F-

Hence [Ca2+]: 2.0*10^-4 mol/L

[F-] : 2* 2.0*10^-4= 4.0*10^-4

Using law of solubility product

Ksp: [Ca2+] [F-]2

Ksp=3.2*10^-11

Related Questions

Navigate

• Browse (All)
• Browse R
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.