(a)
Interpretation:
A balanced chemical equation for the equilibrium occurring when the given solute is added to water is to be stated and the
Concept Introduction:
In a saturated solution, the molar solubility of salt is used to determine the solubility product. The compounds which have high solubility product are more soluble in aqueous solution.
(b)
Interpretation:
A balanced chemical equation for the equilibrium occurring when the given solute is added to water is to be stated and the
Concept Introduction:
In a saturated solution, the molar solubility of salt is used to determine the solubility product. The compounds which have high solubility product are more soluble in aqueous solution.
(c)
Interpretation:
A balanced chemical equation for the equilibrium occurring when the given solute is added to water is to be stated and the
Concept Introduction:
In a saturated solution, the molar solubility of salt is used to determine the solubility product. The compounds which have high solubility product are more soluble in aqueous solution.
(d)
Interpretation:
A balanced chemical equation for the equilibrium occurring when the given solute is added to water is to be stated and the
Concept Introduction:
In a saturated solution, the molar solubility of salt is used to determine the solubility product. The compounds which have high solubility product are more soluble in aqueous solution.
(e)
Interpretation:
A balanced chemical equation for the equilibrium occurring when the given solute is added to water is to be stated and the
Concept Introduction:
In a saturated solution, the molar solubility of salt is used to determine the solubility product. The compounds which have high solubility product are more soluble in aqueous solution.
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Chemistry: The Molecular Science
- Hydrogen iodide gas decomposes to hydrogen gas and iodine gas: 2HI(g)H2(g)+I2(g)To determine the equilibrium constant of the system, identical one-liter glass bulbs are filled with 3.20 g of HI and maintained at a certain temperature. Each bulb is periodically opened and analyzed for iodine formation by titration with sodium thiosulfate, Na2S2O3. I2(aq)+2S2O32(aq)S4O62(aq)+2 I(aq)It is determined that when equilibrium is reached, 37.0 mL of 0.200 M Na2S2O3 is required to titrate the iodine. What is K at the temperature of the experiment?arrow_forwardWrite an equation for an equilibrium system that would lead to the following expressions (ac) for K. (a) K=(Pco)2 (PH2)5(PC2H6)(PH2O)2 (b) K=(PNH3)4 (PO2)5(PNO)4 (PH2O)6 (c) K=[ ClO3 ]2 [ Mn2+ ]2(Pcl2)[ MNO4 ]2 [ H+ ]4 ; liquid water is a productarrow_forwardA small quantity of a soluble salt is placed in water. Equilibrium between dissolved and undissolved salt may or may not be attained. Explain.arrow_forward
- Consider the system 4NH3(g)+3O2(g)2N2(g)+6H2O(l)H=1530.4kJ (a) How will the amount of ammonia at equilibrium be affected by 1. removing O2(g)? 2. adding N2(g)? 3. adding water? 4. expanding the container at constant pressure? 5. increasing the temperature? (b) Which of the above factors will increase the value of K? Which will decrease it?arrow_forwardWrite the expression for the equilibrium constant and calculate the partial pressure of CO2(g), given that Kp is 0.25 (at 427 C) for NaHCO3(s) NaOH(s) + CO2(g)arrow_forwardTo a beaker with 500 mL of water are added 95 mg of Ba(NO3)2, 95 mg of Ca(NO3)2, and 100.0 mg of Na2CO3. After equilibrium is established, will there be • no precipitate? • a precipitate of BaCO3 only? • a precipitate of CaCO3 only? • a precipitate of both CaCO3 and BaCO3? Assume that the volume of the solution is still 500.0 mL after the addition of the salts.arrow_forward
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