3. Which of the following represents the balanced equation for the reaction described in #2? * a. 2HCl + NaOH = 2H2O + NaCl b. HCl + NaOH = Na2O + OHCl c. 2HCl + 2NaOH = 2H2O + 2NaCl d. HCl + NaOH = H2O + NaCl4. What is the pH of the solution after the titration is completed? * a. 0.947 b. 4.00 c. 6.17 d. 7.005. If solution A was titrated with 0.80M Ba(OH)2, what volume would be required? * a. 1.77mL b. 3.53mL c. 7.06mL d. 25.0mL6. Which of the following represents the balanced equation for the reaction described in #5? * a. HCl + Ba(OH)2 = BaCl2 + H2O b. 2HCl + Ba(OH)2 = 2BaCl + H2O c. HCl + Ba(OH)2 = BaH2 + OHCl d. 2HCl + Ba(OH)2 = BaCl2 + 2H2OUse the following for #7 – 8HS- + NH4+ = H2S + NH37. What is the Bronsted-Lowry acid? * a. HS- b. NH4+ c. H2S d. NH38. What is the Bronsted-Lowry base? * a. HS- b. NH4+ c. H2S d. NH39. In a reaction which would be the conjugate acid for HTe-? * a. Te2- b. H2Te c. HTe- d. H3Te+10. What is the pH of a 6.2 x 10-3M H2SO4 solution? * a. 3.21 b. 2.21 c. 11.79 d. 10.7911. What is the pOH of a solution with a pH of 12.75? * a. 2.25 b. 1.25 c. 12.75 d. 1412. What is the [OH] of a solution with pOH equal to 4.8? * a. 6.31×10-10 b. 6.31×10-9 c. 1.58×10-4 d. 1.58×10-5Solution B – Use for #13-1510.00mL of 0.222M KOH is added to a flask13. How many milliliters of a 2.5M HCl solution is required to neutralize Solution B? * a. 0.889mL b. 112mL c. 10.0mL d. Not enough information14. Which of the following represents the balanced equation for the reaction explained above? * a. 2HCl + KOH = 2H2O + KCl2 b. HCl + KOH = H2O + KCl c. HCl + KOH = K2O + OHCl d. 2HCl + 2KOH = 2H2O + 2KCl15. What is the pH of the solution after the titration is completed? * a. 14.00 b. 10.00 c. 8.00
https://nursingessaytutors.com/wp-content/uploads/2019/11/Untitled-1.png 0 0 admin https://nursingessaytutors.com/wp-content/uploads/2019/11/Untitled-1.png admin2020-11-21 20:55:532020-11-21 20:55:533. Which of the following represents the balanced equation for the reaction described in #2?
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