The concentration of Ba(103)2 in H20 at 35°C (Ksp = 3.00 x 10- Ba(103)2 (s) O Ba2+ + 2103 7.32 x 10-4 M 2.37 x 10-4 M 5.23 x 10-4 M 6.69 x 10-4 M 9.09 x 10-4 M
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- 1/T (K-¹) 2 N₂O5(g) → 4NO2(g) + O₂(g) K(S-¹) 4.87 x 10-3 1.50 x 10-³ 4.98 x 10-4 1.35 x 10-4 3.46 x 10-5 7.87 x 10-7 T (K) 338 0.0030 328 0.0030 318 0.0031 308 0.0032 298 0.0034 273 6.0037 The data in the table above provide the temperature dependence of the rate constant for the reaction shown. Complete the table with required information to 2 significant figures each. - a=-12571.42 In K, no units 5.3 - 6.5 7.6 8.9 Carefully construct a neat and well labelled graph/plot of In K (y-axis) versus 1/T (x-axis) in the appropriate quadrant (s), using EXCEL or the template below; then determine the slope and activation energy, Ea for the given reaction. Recall that the Arrhenius Equation is represented in a more useful form as In K= In A- (Ea/RT) & R= 8.31 x 10-³ kJ/mol-K. Determine by calculation, the value of Ea in kj/mole given that, slope = -E/R slope: - Ca = slope & - Ga= €ag Ea=-(-12571.42) x (8.3) x 10 Rg/mol-K) Ea=104.47 kj/md 10.3 - 14.1 Y = (-14.1)-(-5.3) = -8.8 x = (0.0037) -…14. A 50.0 g sample of SAE 15W-40 motor oil is sent for testing. It was found to contain 13.4 µg of lead. What is the concentration (in M) in the oil? (doil = 0.87 g mL-¹) - A: 1.1 x 10-6The Ksp of magnesium hydroxide is 2.06 x 10-13. What is its solubility in g/100 g H20? Density of water is 1.00 g/mL. O 3.72 x 10-4 g/100 g H20 2.17 x 10-6 g/100 g H20 O 2.17 x 10-4 g/100 g H20 O 3.72 x 10-6 g/100 g H2O
- The Ksp of CAF2 is 1.5 x 10-10, If CaF2 (s) is dissolved in pure water, what is the solubility in grams per liter (g/L)? О 2.6 х 10:2 O 6.8 x 10-4 О 3.3 х 10:2 O 4.1 x 10-2 О 3.3 х 10-41) Listen The Ksp value for calcium sulfate [CaSO4] is 2.40 X 10-5 so if a professor made 1825 mL of a CaSO4(ag) solution then how many grams of calcium (Ca2+) ion would be present in this solution. 0.3948 gram Ca2+ ion 0.4965 gram Ca2+ ion 0.2527 gram Ca2+ ion 0.4747 gram Ca2+ ion 0.3396 gram Ca2+ ion 0.2943 gram Ca2+ ion 1591.docx LAB EXP. #3 -.docx Show AllThe Ksp value for calcium sulfate [CaSO4] is 2.40 X 10-5 so if a professor made 1825 mL of a CaSO4lag) solution then how many grams of sulfate (SO42) ion would be present in this solution. 0.8345 gram SO42- ion 0.8588 gram SO42- ion 0.9331 gram SO42- ion 1.0036 gram SO42- ion 0.8921 gram SO42- ion 2.. 591..docx W- LAB EXP. #3 -.docx Show All
- The Ksp of Fe(OH)3(s) is 3 × 10-39. What concentration of Fe3+ can exist in solution at pH 3.0?10-27. A 10.231-g sample of window cleaner containing ammonia was diluted with 39.466 g of water. Then 4.373 g of solution were titrated with 14.22 mL of 0.106 3 M HCl to reach a bromocresol green end point. (a) What fraction of the 10.231-g sample of window cleaner is contained in the 4.373 g that were analyzed? (b) How many grams of NH3 (FM 17.031) were in the 4.373-g sample? (c) Find the weight percent of NH3 in the cleaner.• How many grans and how many mL of 40.0 wt-% urea solution (density 1.111 g/mL) are required to react with 4.00 mol of Fe3+ in the following reactions? • (H2N)2CO + 3 H2O → CO2 + 2 NH4+ + 2 OH- Fe3+ +3 OH- + (x-1)H2O - FEOOH x x H2O(s)
- What is the solubility of Ca3(PO4)2 Ksp = 1.3 x 10-32 in a 0.048 M Ca(NO3)2 solution? 5.4 x 10-15M , 1.1 x 10-14 M, 2.6 x 10-16 M, 2.9 x 10-29 MSalts containing the phosphate ion are added to municipal water supplies to prevent the corrosion of lead pipes. Based on the pKa values for phosphoric acid (pKa1 = 7.5 * 10-3, pKa2 = 6.2 * 10-8, pKa3 = 4.2 * 10-13) what is the Kbvalue for the PO43- ion?PbI2 has a Ksp equal to 1.4 x 10-8. A 1.0 L saturated solution of PbI2 contains a concentration of I- equal to 3.0 x 10-3 M. If 500 mL of water is evaporated from the solution, what will be the concentration of Pb2+ after equilibrium is reestablished? [Pb2+] = 1.5 x 10-3 M [Pb2+] = 7.5 x 10-4 M [Pb2+] = 3.0 x 10-3 M [Pb2+] = 0 M (no ions will be in solution)