Answer:
D i think, sorry if its wrong
Explanation:
HA ⇄ H⁺ + A⁻
so:
![\frac{[H^+][A^-]}{[HA]} = 1.5 x 10^{-5}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5BH%5E%2B%5D%5BA%5E-%5D%7D%7B%5BHA%5D%7D%20%3D%201.5%20x%2010%5E%7B-5%7D%20%20)
and now:
= 1.5 x 10⁻⁵
x is considered very small compared to 0.15
x² = 2.25 x 10⁻⁶
x = 1.5 x 10⁻³
So [H⁺] = 1.5 x 10⁻³
pH = - log [H⁺] = - log (1.5 x 10⁻³) = 2.83
The partial stress of H2 is 737.47 mmHg Let's observe the Ideal Gas Law to find out the whole mols.
We count on that the closed vessel has 1L of volume
- P.V=n.R.T
- We must convert mmHg to atm. 760 mmHg.
- 1 atm
- 755 mmHg (755/760) = 0.993 atm
- 0.993 m.1L=n.0.082 L.atm/mol.K .
- 293 K(0.993 atm 1.1L)/(0.082mol.K /L.atm).
- 293K = n
- 0.0413mols = n
These are the whole moles. Now we are able to know the moles of water vapor, to discover the molar fraction of it.
- P.V=n.R.T
- 760 mmHg. 1 atm
- 17.5 mmHg (17.5 mmHg / 760 mmHg)=0.0230 atm
- 0.0230 m.1L=n.0.082 L.atm/mol.K.293 K(0.0230atm.1L)/(0.082mol.K/L.atm .293K)=n 9.58 × 10 ^ 4 mols = n.
- Molar fraction = mols )f gas/general mols.
- Molar fraction water vapor =9.58×10^ -four mols / 0.0413 mols
- Sum of molar fraction =1
- 1 - 9.58 × 10 ^ 4 × mols / 0.0413 ×mols = molar fraction H2
- 0.9767 = molar fraction H2
- H2 pressure / Total pressure =molar fraction H2
- H2 pressure / 55mmHg = =0.9767 0.9767 = h2 pressure =755 mmHg.
- 737,47 mmHg.
<h3>What is a mole fraction?</h3>
Mole fraction is a unit of concentration, described to be identical to the variety of moles of an issue divided through the whole variety of moles of a solution. Because it's miles a ratio, mole fraction is a unitless expression.
Thus it is clear that the partial pressure of H2 is 737,47 mmHg.
To learn more about partial pressure refer to the link :
brainly.com/question/19813237
<h3 />
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
