This is how I did it: mass=3.25x10 -3 kg =3.25g gold amu= 196.97g 3.25g/196.97am

Question

This is how I did it: mass=3.25x10-3kg =3.25g gold amu= 196.97g 3.25g/196.97amu=.0165molx6.022x1023=9.94x1021atoms because each gold atom has 79 electrons, so9.94x1021x79=7.85x1023 electrons Q=(#e-)q=7.85x1023x1.6x10-19= 125594.6C 2.78hour=10008second I=Q/t=125594.6/10008=12.55A This is how I did it: mass=3.25x10-3kg =3.25g gold amu= 196.97g 3.25g/196.97amu=.0165molx6.022x1023=9.94x1021atoms because each gold atom has 79 electrons, so9.94x1021x79=7.85x1023 electrons Q=(#e-)q=7.85x1023x1.6x10-19= 125594.6C 2.78hour=10008second I=Q/t=125594.6/10008=12.55A

Explanation / Answer

   if NA is the avagadro number, then    NA = 6.023 * 1023    molecular weigth (M) of gold = 197g/mol    now the mass (mg) of the gold atom will be    mg = M / NA          = ...........kg    as the gold ions charge,    the numbe (n) charge carriers moving to negativeelectrode will be    n = 3.75 10-3 kg  /mg    now the current will be as follows    if Q is the charge that passes through thecross section of the conductor in time t, then    from the theory the electric current I in aconductor is defined as    I = Q / t    if e is the charge on electron, then    Q = (n)(e)    solve for I    hope u got it

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