An ornament weighs 36 g in air and 34 g in water. Assuming the...

Loss of weight of the ornament in water $= 36 - 34 = 2 \text{ g}$

Since the density of water is $1 \text{ g/cm}^3$, the volume of the displaced water is $2 \text{ cm}^3$. Therefore, the total volume of the ornament is $V = 2 \text{ cm}^3$.

Let the mass of copper in the ornament be $m$. Then the mass of gold is $36 - m$.

The volume of copper is $V_c = \dfrac{m}{8.9}$ and the volume of gold is $V_g = \dfrac{36 - m}{19.3}$.

The total volume of the ornament is the sum of the volumes of copper and gold:

$\dfrac{m}{8.9} + \dfrac{36 - m}{19.3} = 2$

$\dfrac{m}{8.9} - \dfrac{m}{19.3} = 2 - \dfrac{36}{19.3}$

$m \left( \dfrac{19.3 - 8.9}{8.9 \times 19.3} \right) = \dfrac{38.6 - 36}{19.3}$

$m \left( \dfrac{10.4}{8.9 \times 19.3} \right) = \dfrac{2.6}{19.3}$

$m = \dfrac{2.6 \times 8.9}{10.4}$

$m = \dfrac{8.9}{4} = 2.225 \text{ g}$

The amount of copper present is approximately $2.2 \text{ g}$.