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Answer: The solubility of this compound in pure water is 0.012 M
Explanation:
Solubility product is defined as the equilibrium constant in which a solid ionic compound is dissolved to produce its ions in solution. It is represented as 
The equation for the ionization of the is given as:
By stoichiometry of the reaction:
1 mole of 
gives 1 mole of 
and 2 mole of 
When the solubility of is S moles/liter, then the solubility of will be S moles\liter and solubility of will be 2S moles/liter.
![K_{sp}=[Ca^{2+}][OH^{-}]^2](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BCa%5E%7B2%2B%7D%5D%5BOH%5E%7B-%7D%5D%5E2)
![6.5\times 10^{-6}=[S][2S]^2](https://tex.z-dn.net/?f=6.5%5Ctimes%2010%5E%7B-6%7D%3D%5BS%5D%5B2S%5D%5E2)
Thus solubility of this compound in pure water is 0.012 M
Answer:
Where is the results and what is the question or is there a picture
Explanation:
Answer:
6.93
Explanation:
Step 1: Given data
- Standard Gibbs free energy (∆G°): -5.20 kJ
- Equilibrium constant (K): ?
Step 2: Convert the temperature to the Kelvin scale
We will use the following expression.
K = °C + 273.15
K = 50°C + 273.15
K = 323 K
Step 3: Calculate K
We will use the following expression.
∆G° = -R × T × ln K
-5.20 × 10³ J = -(8.314 J/mol.K) × 323 K × ln K
K = 6.93
