Question

In: Chemistry

How many moles and numbers of ions of each type are present in the following aqueous solution?

How many moles and numbers of ions of each type are present in the following aqueous solution? 9.91 L of a solution containing 1.05 × 1020 FU cesium nitrate/L: ________mol of cesium _______× 10__ cesium ions _______mol of nitrate ______× 10__ nitrate ions

Solutions

Expert Solution

9.91 L of a solution containing 1.05 × 1020 FU cesium nitrate/L    = 9.91L * 1.05*10^20 FU cesium nitrate/L

                                                                                                         = 9.91*1.05*10^20 Fu cesium nitrate

                                                                                                           = 10.4055*10^20 FU cesium nitrate

1 mole of cesium nitrate = 6.023*10^23 FU cesium nitrate

6.023*10^23 FU cesium nitrate = 1 mole of Cesium nitrate

10.4055*10^20 FU cesium nitrate = 1 mole * 10.4055*10^20FU/6.023*10^23FU    = 0.00173 moles of Cesium nitrate

CsNO3 ---------------------> Cs^+(aq)              +   NO3^-

1.73*10^-3 moles              1.73*10^-3 moles   1.73*10^-3 moles

1.73*10^-3 moles of cesium ion >>>>.answer

1.73*10^-3 moles of nitrate ions >>>>>>answer


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