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- Calculate the pH of a 8.3 x 10-M solution of H2SO4 (K, = 1.2 x 10-2). pH =Calculate the pH at 25 °C of a 0.23 M solution of sodium hypochlorite (NaClO). Note that hypochlorous acid (HCIO) is a weak acid with a pK, of 7.50. a Round your answer to 1 decimal place. pH = 0 ?Given the following: [H3O+] = 5.112 x 10-13, please calculate the pOH.(Note please use 3 sig. figs for your answer.)
- 3. For the following Lewis acid-base reaction, identify the acid, the base, and the new covalent bond formed. 3NO2 (g) + H₂O (1) 2HNO3 (aq) + NO(g)Based on the Lewis structures of the reactants and product of the following equation, identify the Lewis acid and the Lewis base in it: CO, + OH → HCO, =c=0: + :0-H H-0- O: O Lewis acid: CO,; Lewis base: HCO, O Lewis acid: OH; Lewis base: CO, O Lewis acid: CO; Lewis base: OH-For the following acid-base reaction, (1) predict the products, showing both reactants and products complete Lewis structures and arrows showing electron flow; (2) label each structure with the lowing: Bronsted acid, Bronsted base, conjugate acid, conjugate base; (3) give a brief definition of a ronsted acid and Bronsted base; (4) predict the direction of the equilibrium and justify your answer. HC0OH + CH3 Nta PRん106Y pkb = 3.36
- Fill in the left side of this equilibrium constant equation for the reaction of diethylmethylamine (C3H13N), a weak base, with water. D= K, oloConsider the following data on some weak acids and weak bases: name K₂ formula HNO₂ 4.5 x 104 acetic acid HCH,CO₂ 1.8x105 acid nitrous acid solution 0.1 M HONHYBT 0.1 M KBr Use this data to rank the following solutions in order of increasing pH. In other words, select a 'l' next to the solution that will have the lowest pH, a '2' next to the solution that will have the next lowest pH, and so on. 0.1 M KCH₂CO₂ 0.1 M C₂H₂NHCI PH choose one choose one choose one base choose one K name formula hydroxylamine HONH₂ 1.1 x 107 pyridine CH₂N 1.7×10Calculate the pH and the pOH of these solutions: (a) 0.0017 M NaOH pH = pOH = (b) 0.158 M HCl pH = pOH = (c) 0.0277 M HC6H5O (Ka = 1.3 x 10-10) pH = pOH =
- 3. A 0.0560 g quantity of acetic acid is dissolved in enough water to make 50.0 mL of solution. Calculate the concentrations of H;O*, CH;COO and CH;COOH at equilibrium. What is the pH of the solution? (Ka = 1.8x10) CH;COOH(aq) = CH;COO (aq) + H;O* (aq) a) Calculate the initial concentration of CH;COOH. (C:12; H:1; 0:16) b) Calculate the concentration of CH;CoO (aq) and H;O* (aq) at equilibrium. c) Calculate pH of the solution.least acidic I < || < III most acidic least acidic II < I < III most acidic least acidic III < || < I most acidic least acidic III < | < || most acidicReaction 1: NaOH(s) -> Na+(AQ) + OH-(AQ) Reaction two: Na+(AQ) + OH-(AQ) + H+(AQ) + -> H2O(I) + Na+(AQ) Assuming reaction 1 has a enthalpy of 59.12 KJ/mol and reaction 2 has a enthalpy of 46.06 kJ/mol Calculate the change in enthalpy for the reaction shown below NaOH(s) + H+(aq) -> H2O(I) + Na+(aq) Delta H = ?