The net ionic reaction for the balanced equation below is 2AgNO_3(aq) + K_2SO_4(

Question

The net ionic reaction for the balanced equation below is 2AgNO_3(aq) + K_2SO_4(aq) rightarrow 2KNO_3(aq) + Ag2SO_4(s) A) Ag^+(aq) + NO_3^-(aq) rightarrow AgNO_3(aq). B) 2K^+(aq) + SO_4^_2(aq) rightarrow K_2SO_4(aq). C) 2Ag^+(aq) + SO_4^-2(aq) rightarrow Ag_2SO_4(s). D) K^+(aq) + NO_3^-(aq) rightarrow KNO_3(aq). What is the concentration, in mass percent, of a solution prepared from 10.0 g NaCl and 150.0 g of water? 6.25% 33.3% 40.0% 25.0% A patient needs to receive 20 grams of glucose every 12 hours. What volume of a 5.0% (m/v) glucose solution needs to be administered to the patient every 12 hours? 1700 mL 400 mL 6000 mL 17 mL What is the molarity of a solution that contains 3.25 moles of NaNO_3 in 250. mL of solution? 3.25 M 6.50 M 0.0130 M 13.0 M

Explanation / Answer

30. 2AgNO3(aq) + K2SO4(aq) -------> 2KNO3(aq) + Ag2SO4(s)

    2Ag+(aq) +2NO3-(aq) + 2K+(aq)+SO42-(aq) -------> 2K+(aq) +2NO3-(aq) + Ag2SO4(s)

2Ag+(aq) +SO42-(aq) -------> Ag2SO4(s) net ionic equation

answer C

31. percentage = mass of NaCl*100/total mass of solution

                          = 10*100/10+150

                        = 10*100/160   = 6.25%

answer A

32.

33. Molarity = no of moles/volume in L

                  = 3.25/0.25 = 13M

answer D

  

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