Question

Two small aluminum spheres, each of mass 0.0250 kilograms, areseparated by 80.0 centimeters.

Answer 1

Answer:

Total number of electrons

$N = 7.25 \times 10^{24}$

electrons removed from each sphere

$N = 5.27 \times 10^{15}$

Fraction of electrons transferred is given as

$f = 7.27 \times 10^{-10}$

Explanation:

As we know that moles is defined as

$n =\frac{mass}{molar mass}$

$n = \frac{0.0250}{0.026982}$

$n = 0.93$

so number of atoms of Al in each sphere is given as

$N = 0.93(6.02 \times 10^{23})$

$N = 5.58 \times 10^{23}$

Now number of electrons in each atom is given as

atomic number = number of electrons in each atom = 13

total number of electrons in each sphere is

$N = 13 \times (5.58 \times 10^{23})$

$N = 7.25 \times 10^{24}$

Also we know that force of attraction between them is given as

$F= \frac{kq_1q_2}{r^2}$

$1.00 \times 10^4 = \frac{(9\times 10^9)q^2}{0.80^2}$

$q = 8.4 \times 10^{-4} C$

now we have

$q = Ne$

$8.4 \times 10^{-4} = N(1.6 \times 10^{-19}$

$N = \frac{(8.4 \times 10^{-4})}{1.6 \times 10^{-19}}$

$N = 5.27 \times 10^{15}$

Fraction of electrons transferred is given as

$f = \frac{5.27 \times 10^{15}}{7.25 \times 10^{24}}$

$f = 7.27 \times 10^{-10}$