9.91 L of a solution containing 1.05 × 1020 FU cesium nitrate/L = 9.91L * 1.05*1

Question

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

Explanation / Answer

Be sure to answer all parts. 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

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