A 5.000g sample of pesticide was decomposed with metallic sodium in alcohol and

Question

A 5.000g sample of pesticide was decomposed with metallic sodium in alcohol and the liberated chloride ion was precipitated as AgCl. Express the results of this analysis as percent DDT, C14H9Cl5, 354.5 g/mol based on 0.1606 g AgCl recovered. 100.00 ml of 0.1000 M AgNO3 is mixed with 37.00 ml of 0.1000 M HCl. What are the final concentrations of all the ions in this solution? 100.00 ml of 0.1000 M Cd(NO3)2 is mixed with 100.00 ml of 0.1000 M HNO3 and 100.0 ml of 0.1000 M Na2S. Calculate the final molar concentrations of all the ions in this solution.

Explanation / Answer

For each start by converting mL to L. so 100mL is .1L

27mL is .037L

then M*L = mol

2.) .1mol*.1L = .01mol Ag+, .01mol NO3-

.1M*.037L = .0037mol H+, .0037mol Cl-

3.) .1L*.1M Cd(NO3)2 = .01mol Cd, .02NO3- (There are 2X as many NO3 ions as Cd ions)

.1*.1 = .01mol H+, .01mol NO3-

.1*.1 = .02 Na, .01S

Total: .01mol Cd, .03mol NO3-, .01mol H, .02mol Na, .01mol S

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