In an ionic solution, a current consists of Ca2+Ca2+ ions (of charge +2e+2e) and Cl−Cl− ions (of charge −e−e) traveling in opposite directio

Question

In an ionic solution, a current consists of Ca2+Ca2+ ions (of charge +2e+2e) and Cl−Cl− ions (of charge −e−e) traveling in opposite directions. Part A If 5.02×1018 Cl−Cl− ions go from A to B every 0.620 minmin , while 3.25×1018 Ca2+Ca2+ ions move from B to A, what is the current (in mAmA) through this solution? Part B In which direction (from A to B or from B to A) is the current in part A going?

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Đan Thu 4 years 2021-08-03T12:16:23+00:00 1 Answers 53 views 0

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  1. diemthu
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    2021-08-03T12:17:29+00:00
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    Answer:

    A

    The current through the solution is   I =35.61mA

    B

    The current is moving from B to A

    Explanation:

    From the question we are told that

          The number of  Cl^{-} that move from A to B is N_{cl} = 5.02 *10^{18} ions

          The time taken to move from A to B   t_{cl} = 0.620

                                                                              =37.2 s

    Since the value of 1 charge is  q =1.602 *10^{-19} C

    The quantity of charge Q that flow from A to B is mathematically given as

                             Q_{cl} = 5.02 *10^{18} * 1.602*10^{-19}

                                    =0.804C

       The number of  Ca^{+} that move from A to B is N_{ca} = 3.25*10^{18} ions

          Since time taken to move from A to B is equal to time taken to move from B to A   t =t_{cl} =t_{ca}= 0.620=37.2s

     

    The quantity of charge Q that flow from B to A is mathematically given as

                             Q_{ca} = 3.25 *10^{18} * 1.602*10^{-19}

                                    =0.5207C

    The total quantity of charge is

                                    Q_{tot}=Q_{cl} + Q_{ca}

                                    Q_{tot}= 0.804 + 0.5207

                                           = 1.325C

    The current flowing through the solution is

                             I =\frac{Q_{tot}}{t}

                                I = \frac{1.325}{37.2}

                                    = 0.03561A

                                I =35.61mA

    The flow is from B to A cause current  flow from the positive terminal to negative terminal

                   

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