I understand there is a 1:2 ratio and before doing the initial concentrations th
ID: 493306 • Letter: I
Question
I understand there is a 1:2 ratio and before doing the initial concentrations the M of [IO3-] was double the M of [Cu2+]. I think the copper is the limiting reagent, however I can't figure out how many times the iodate decreases by test tube test tube test tube test tube test tube 5 (std) 0.015 [Cu in mixture equilibrium decrease in i. Why was there a decrease in [Cu 1? What became of the ions in solution? from equali orum, ii. Write the net ionic equation for the precipitation reaction: test tube test tube Ttest tube test tube LIO3 in mixture before precipitation iii. The decrease in [IO3 s times the decrease in [Cu +1. decrease in [IO3 at equilibrium Show your calculations in the space below (or on a separate sheet of paper).Explanation / Answer
Answer:-
The iodate will decrease two times of the Cu2+ decrease because the mole ratio is 1:2 . Its means if one mole of Cu2+ decreases then the decrease in iodate will be double. i.e., two times