Question

What mass of aluminum has a total nuclear charge of 3.0 C?Aluminum has atomic number 13. Suppose the aluminum is all of the isotope with 14 neutrons.

Answer 1

Answer:

The value is $m = 6.48*10^{-5} \ g$

Explanation:

From the question we are told that

The total nuclear charge is $Q_t = 3.0 \ C$

The atomic number is $n = 13$

The number of neutrons is $n = 14$

Generally the number of protons is mathematically represented as

$N_p = \frac{Q_t}{p}$

Here p is the charge on a proton with the value $p = 1.60 *10^{-19} \ C$

=> $N_p = \frac{3}{p}$

=> $N_p = \frac{3}{1.60 *10^{-19} }$

=> $N_p = 1.875*10^19 \ protons$

Generally the number of atoms present is mathematically represented as

$N_a = \frac{N_p}{n }$

=> $N_a = \frac{ 1.875*10^19}{13}$

=> $N_a = 1.44*10^{18} \ atoms$

Generally the number of moles of atom present is mathematically represented as

$N = \frac{N_a}{A_n }$

Here $A_n$ is the Avogadro's number with a constant value

$A_n = 6.023*10^{23}$

So

$N = \frac{1.44*10^{18}}{6.023*10^{23} }$

=> $N = 2.4*10^{-6} \ mol$

Generally the total mass is mathematically represented as

$m = N * M$

Here  M is the molar mass of   aluminum with

         $M = 27 \ g/mol$

So

        $m = 2.4*10^{-6} * 27$

         $m = 2.4*10^{-6} * 27$

         $m = 6.48*10^{-5} \ g$