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Why Is the Solubility of Oxygen Lower in the Pond Water Rather Than Tap Water?

Who says the solubility of O2 is lower in pond water? There are several reasons why that *might* be the case, but I do not think it is universally true. The factors with the biggest influence on O2 solubility are (1) temperature and (2) dissolved solids concentration. Pond water *might* be warmer than tap water and therefore might have lower O2 solubility. (But in winter, pond water might be much colder.) Pond water *might* have higher dissolved solids concentration than tap water and this might reduce O2 solubility. (But, just as likely, the chlorination of tapwater might increase the dissolved solids content and reduce solubility of oxygen in tapwater.)

1. The solubility product constant of calcium sulfate, CaSO4, is 7.10Ã—10âˆ’5. Its molar mass is 136.1 g/mol....?

Calcium sulfate dissociates in water forming a saturated solution according to the following equation: CaSO4(s) Ca^2(aq) SO4^2-(aq) The equilibrium constant [solubility product] expression is in the form of the following equation: Ksp = [Ca^2][SO4^2-] According to the above equation, one mole of CaSO4 yields one mole of Ca^2 and one mole of SO4^2-. Therefore, the molar concentration of dissolved calcium sulfate is the same as the molar concentration of either calcium ion or sulfate ion. Moreover, the molar concentration of dissolved calcium ion equals the molar concentration of dissolved sulfate ion. Since the value of the solubility product constant of calcium sulfate is given, we can calculate the molar concentration of either calcium ion or sulfate ion, since they are equal to each other: Let x = [Ca^2] = [SO4^2-] Ksp = [Ca^2][SO4^2-] 7.10 10^-5 = [x][x] 7.10 10^-5 = x^2 71.0 10^-6 = x^2 8.4261497 10^-3 M = x = [Ca^2] = [SO4^2-] = [CaSO Now that molar concentration of calcium sulfate in a saturated solution of this salt is known, we can calculate the number of moles of calcium sulfate in 71.0 L of water. Note that the solubility product constant for calcium sulfate is a very small number, therefore its saturated solution in water is virtually mostly water and the change in volume is insignificant. Molarity CaSO4 = (moles CaSO4) / (Liters of solution) or moles CaSO4 = (Molarity CaSO4)(Liters of solution) moles CaSO4 = (8.4261497 10^-3 M)(71.0 L) moles CaSO4 = 598.25662 10^-3 mol = 0.598 mol To find the mass of 0.598 mol of calcium sulfate, we need to multiply the number of moles by the mole mass of calcium sulfate given in the problem: (0.598 mol)(136.1 g/mol) = 81.422723 g or 81.4 g Answer: About 81.4 g of calcium sulfate can be dissolved in 71.0 L of pure water.

2. Chemistry: rule for solubility and non-polar molecules?

what did you say??????

3. Solubility of a molecular compound?

Decomposition of the reaction mixture with acid will allow some of the product to go into the acidic solution by the conversion to the pyridinium salt. Instead, use NaOH to dissolve all the aluminum salts as Al(OH)4-. Then add some satd NaCl to salt out any solubles. keeping the mixture alkaline will also keep any unreacted nicotinic acid in the aqueous phase. Make the second washof the aqueous phase with with CH2Cl2

4. chemistry (saturated solution, unsaturated solution, solubility question)?

The solution is satured because it has the maximum amount dissolved, and any left over accumulates at the bottom as a crystal or on the sides

5. Which of the following are arranged in increasing order of their solubility in water?

I think no 3, because potassium ion is the largest and hence has a smallest attraction and polarizing effect on Co3 2(-)ion compared to Na and Ca 2. Ca 2 has a greater polarising effect because of its charge (2) compared to Na(1) hence calcium hydrogen carbonate should have the lowest solubility ... you can say calcium is strongly attracted to and attracting the carbonate ion hence it wo not easily let go. I hope I could be of help to you

6. Solubility of Organic Compound confusion

There are two regions of bond energy to look at here:The bonding between water molecules is strong enough to resist separation by methane molecules because the gain could be no more than the stabilization in methane (\$pu0.223 kcal/mol\$), but the loss would be as much as \$pu1.435 kcal/mol\$. The first sentence in the question refers to this region of bond energies, not to the much higher energy region ( \$pu100 kcal/mol\$) that would be extensively covered in a chapter on bond enthalpy.

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