Answer:
(a) [tex]m_{gold}=7.322g[/tex]
(b)
[tex]V_{gold}=0.379cm^3[/tex]
[tex]V_{copper}=0.122cm^3[/tex]
(c) [tex]\rho _{coin}=15.94g/cm^3[/tex]
Explanation:
Hello,
(a) In this case, with the given formula we easily compute the mass of gold contained in the sovereign  as shown below:
[tex]m_{gold}=\frac{m_{tota}*karats}{24}=\frac{7.988g*22}{24}=7.322g[/tex]
(b) Now, by knowing the density of gold and copper, 19.32 and 8.94 g/cm³ respectively, we compute each volume, by also knowing that the rest of the coin contains copper:
[tex]V_{gold}=\frac{m_{gold}}{\rho_{gold}} =\frac{7.322g}{19.32g/cm^3}=0.379cm^3[/tex]
[tex]m_{copper}=7.988g-7.322g=1.09g\\V_{copper}=\frac{m_{copper}}{\rho_{copper}}=\frac{1.09g}{8.94g/cm^3} \\\\V_{copper}=0.122cm^3[/tex]
(c) Finally, the volume is computed by dividing the total mass over the total volume containing both gold and copper:
[tex]\rho _{coin}=\frac{m_{total}}{V_{gold}+V_{copper}}=\frac{7.988 g}{0.379cm^3+0.122cm^3}\\ \\\rho _{coin}=15.94g/cm^3[/tex]
Best regards.
Answer:
a
The mass of gold is  [tex]L = 7.322 *10^{-3} \ kg[/tex]
b
   The volumes of gold and copper is  [tex]V_g = 3.794 *10^{-7} \ m^3[/tex]  , [tex]V_c = 7.426 *10^{-8} \ m^3[/tex]
c
   The density of the British sovereign coin
     [tex]\rho = 17.593*10^{3} \ kg/m^3[/tex]
Explanation:
From the question we are told that
  The total mass of the gold is  [tex]K = 7.988 \ g = 7.988 * 10^{-3} \ kg[/tex]
   The karat of the British gold sovereign is  [tex]z = 22[/tex]
  Â
Let the mass of gold in the alloy be  L Â
Now we are told that
    [tex]z = 24 * \frac{L}{K}[/tex]
substituting value Â
    [tex]22 = 24 * \frac{L}{7.988 * 10^{-3}}[/tex]
 So  [tex]L = \frac{22}{24} * 7.899*10^{-3}[/tex]
    [tex]L = 7.322 *10^{-3} \ kg[/tex]
The volume of the gold coin is  mathematically represented as
     [tex]V_g = \frac{L}{\rho_g }[/tex]
Where  [tex]\rho_g[/tex] is the density of the gold which a constant with value Â
   [tex]\rho_g = 19.3 *10^{3} \ kg /m^3[/tex]
So
     [tex]V_g = \frac{7.322 *10^{-3}}{19.3 *10^{3} }[/tex]
     [tex]V_g = 3.794 *10^{-7} \ m^3[/tex]
The mass of copper is mathematically evaluated as
    [tex]m_c = K - L[/tex]
    [tex]m_c = 7.988*10^{-3} - 7.322 *10^{-3}[/tex]
    [tex]m_c = 6.657 *10^{-4} \ kg[/tex]
Volume of the copper is Â
     [tex]V_c = \frac{m_c}{\rho_c}[/tex]
Where  [tex]\rho_c[/tex] is the density of the copper which a constant with value Â
    [tex]\rho_c = 8.92 * 10^{3} \ kg/m^3[/tex]
So
    [tex]V_c = \frac{6.657 *10^{-4}}{8.92 *10^{3}}[/tex]
    [tex]V_c = 7.426 *10^{-8} \ m^3[/tex]
The total volume of the British gold sovereign coin is  \
    [tex]V = V_g + V_c[/tex]
substituting values
    [tex]V = 3.7939 *10^{-7} + 7.4626 *10^{-7}[/tex]
    [tex]V = 4.54 *10^{-7} \ m^3[/tex]
The density of the British gold sovereign coin is
   [tex]\rho = \frac{K}{V}[/tex]
substituting values
     [tex]\rho = \frac{7.988 *10^{-3}}{4.54 *10^{-7}}[/tex]
     [tex]\rho = 17.593*10^{3} \ kg/m^3[/tex]
