19. Combining aqueous solutions of Barium lodide and Sodium Sulfate produces a p

Question

19. Combining aqueous solutions of Barium lodide and Sodium Sulfate produces a precipitate of Barium Sulfate. Which ion(s) is/are spectator ions in the reaction? B) Na only C) Na+ and D) Ba2 only E) SO. and 20. How many grams of RbOH are there in 400 mL of a 0.175 M RbOH solution? A) 7.17 B) 114 C) 13.99 D) 3.58 E) 1.399 21. The concentration (M) of an aqueous methanol solution produced when 880mL of a 0250 M solution was diluted to 9.50L is A) 0.800 B) 0.00020 C) 0.500 D) 0.0232 E) 0.00232 22. A 127.5 mL sample of HCI (aq) of unknown concentration was titrated with 0.226 M NaOH (aq). It took 75 mL of the base to reach the endpoint of the titration. The concentration (M) of the acid was A) 0.1329 B) 0.0907 C) 0.1815 D) 0.113 E) 0.227

Explanation / Answer

19) C. Na+ , I-

BaI2(aq) + Na2SO4(aq) --------> BaSO4(s) + 2NaI(aq)

NaI2(aq) ---------> Na+(aq) + I-(aq)

20) A. 7.17 g

Molarity = ( mass / molar mass ) x 1/V(L)

Molarity = 0.175 M

molar mass = 102.5 g/mol

Volume of solution = 400 mL = 0.400 L

=> 0.175 = (mass / 102.5 ) x 1/0.400

=> mass = 7.17 g

21) D. 0.0232

M1 = 0.250 M

V1 = 880 mL = 0.880 L

M2 = ?

V2 = 9.50 L

=> 0.250 x 0.880 = M2 x 9.50

=> M2 = 0.0232 M

22) A. 0.1329

NaOH + HCl ------> NaCl + H2O

M1V1/n1 = M2V2/n2

M1 x 127.5 / 1 = 0.226 x 75 / 1

=> M1 = [HCl] = 0.1329 M

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