In electromagnetic waves, energy is transferred through vibrations of electric and magnetic fields. In sound waves, energy is transferred through vibration of air particles or particles of a solid through which the sound travels.
CaCO₃ partially dissociates in water as Ca²⁺ and CO₃²⁻. The balanced equation is,
CaCO₃(s) ⇄ Ca²⁺(aq) + CO₃²⁻(aq)
Initial Y - -
Change -X +X +X
Equilibrium Y-X X X
Ksp for the CaCO₃(s) is 3.36 x 10⁻⁹ M²
Ksp = [Ca²⁺(aq)][CO₃²⁻(aq)]
3.36 x 10⁻⁹ M² = X * X
3.36 x 10⁻⁹ M² = X²
X = 5.79 x 10⁻⁵ M
Hence the solubility of CaCO₃(s) = 5.79 x 10⁻⁵ M
= 5.79 x 10⁻⁵ mol/L
Molar mass of CaCO₃ = 100 g mol⁻¹
Hence the solubility of CaCO₃ = 5.79 x 10⁻⁵ mol/L x 100 g mol⁻¹
= 5.79 x 10⁻³ g/L
Answer:
Vapour pressure of cyclohexane at 50°C is 490torr
Vapour pressure of benzene at 50°C is 90torr
Explanation:
Using Raoult's law, pressure of a solution is defined by the sum of the product sbetween mole fraction of both solvents and pressure of pure solvents.

In the first solution:


<em>(1)</em>
For the second equation:


<em>(2)</em>
Replacing (2) in (1):


-122.5torr = -0.250P°A

<em>Vapour pressure of cyclohexane at 50°C is 490torr</em>
And for benzene:


<em>Vapour pressure of benzene at 50°C is 90torr</em>
I think the answer is D no change. Though you add more CO2, but the pressure is not mentioned. If the pressure is constant and the reaction is already balanced, the H2O is also saturation and can not absorb more CO2.