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frutty [35]
3 years ago
12

The British gold sovereign coin is an alloy of gold and copper having a total mass of 7.988 g, and is 22-karat gold.

Chemistry
2 answers:
Luden [163]3 years ago
8 0

Answer:

a

The mass of gold is  L  =  7.322 *10^{-3} \ kg

b

     The volumes of gold and copper is  V_g   = 3.794 *10^{-7} \ m^3  , V_c = 7.426 *10^{-8} \ m^3

c

      The density of the British sovereign coin

          \rho = 17.593*10^{3} \ kg/m^3

Explanation:

From the question we are told that

    The total mass of the gold is  K  = 7.988 \ g =  7.988 * 10^{-3} \ kg

      The karat of the British gold sovereign is  z = 22

     

Let the mass of gold in the alloy be  L  

Now we are told that

       z =  24 * \frac{L}{K}

substituting value  

        22  =  24 * \frac{L}{7.988 * 10^{-3}}

  So  L  =  \frac{22}{24} * 7.899*10^{-3}

        L  =  7.322 *10^{-3} \ kg

The volume of the gold coin is  mathematically represented as

         V_g   = \frac{L}{\rho_g }

Where   \rho_g is the density of the gold which a constant with value  

     \rho_g =  19.3 *10^{3} \  kg /m^3

So

         V_g   = \frac{7.322 *10^{-3}}{19.3 *10^{3} }

         V_g   = 3.794 *10^{-7} \ m^3

The mass of copper is mathematically evaluated as

        m_c =  K -  L

        m_c =  7.988*10^{-3} - 7.322 *10^{-3}

        m_c =  6.657 *10^{-4} \ kg

Volume of the copper is  

          V_c =  \frac{m_c}{\rho_c}

Where   \rho_c is the density of the copper which a constant with value  

        \rho_c =  8.92 * 10^{3} \ kg/m^3

So

       V_c =  \frac{6.657 *10^{-4}}{8.92 *10^{3}}

       V_c = 7.426 *10^{-8} \ m^3

The total volume of the British gold sovereign coin is  \

        V  =  V_g + V_c

substituting values

        V  =  3.7939 *10^{-7} +  7.4626 *10^{-7}

        V  =  4.54 *10^{-7} \ m^3

The density of the British gold sovereign coin is

      \rho =  \frac{K}{V}

substituting values

         \rho =  \frac{7.988 *10^{-3}}{4.54 *10^{-7}}

         \rho = 17.593*10^{3} \ kg/m^3

azamat3 years ago
3 0

Answer:

(a) m_{gold}=7.322g

(b)

V_{gold}=0.379cm^3

V_{copper}=0.122cm^3

(c) \rho _{coin}=15.94g/cm^3

Explanation:

Hello,

(a) In this case, with the given formula we easily compute the mass of gold contained in the sovereign  as shown below:

m_{gold}=\frac{m_{tota}*karats}{24}=\frac{7.988g*22}{24}=7.322g

(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:

V_{gold}=\frac{m_{gold}}{\rho_{gold}} =\frac{7.322g}{19.32g/cm^3}=0.379cm^3

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

(c) Finally, the volume is computed by dividing the total mass over the total volume containing both gold and copper:

\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

Best regards.

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