Osmosis and diffusion are related processes that display similarities. Both osmosis and diffusion equalize the concentration of two solutions. Both diffusion and osmosis are passive transport processes, which means they do not require any input of extra energy to occur. In both diffusion and osmosis, particles move from an area of higher concentration to one of lower concentration. Osmosis and facilitated diffusion both account for movement of molecules from a region of high concentration to a region of low concentration.
Answer:
Axial
Explanation:
In the most stable conformation of Cis-3-tert-Butylcyclohexanol, the tert-butyl group is at equatorial position and the alcohol group is in the axial position.
If the tert-butyl group is placed in equatorial position, repulsions are minimized. The bulkier the group, the greater the energy difference between the axial and equatorial conformers. Hence for a ring having a bulky substituent, such bulky substituent is better placed in the equatorial position.
The energy difference between the conformers of Cis-3-tert-Butylcyclohexanol is so high that the compound is almost "frozen" in a conformation where the tert-butyl groups are equatorial and the -OH groups are axial. This conformer is more stable by 24 KJ/mol.
K₃PO₄ → 3K⁺ (aq) + PO₄³⁻(aq)
One mole of PO₄³⁻ ion gets dissociated from one mole of K₃PO₄
As per the definition of Avogadro's number, 1 mole = 6.022 x 10²³ ions
One mole of PO₄³⁻ ions x (6.022 x 10²³ ions/ 1 mole of PO₄³⁻ ions )
= 6.022 x 10²³ ions
Therefore , there are 6.022 x 10²³ PO₄³⁻ ions in a mole of K₃PO₄.
Answer:
Explanation:
<u>1) Data:</u>
a) Hypochlorous acid = HClO
b) [HClO} = 0.015
c) pH = 4.64
d) pKa = ?
<u>2) Strategy:</u>
With the pH calculate [H₃O⁺], then use the equilibrium equation to calculate the equilibrium constant, Ka, and finally calculate pKa from the definition.
<u>3) Solution:</u>
a) pH
b) Equilibrium equation: HClO (aq) ⇄ ClO⁻ (aq) + H₃O⁺ (aq)
c) Equilibrium constant: Ka = [ClO⁻] [H₃O⁺] / [HClO]
d) From the stoichiometry: [CLO⁻] = [H₃O⁺] = 2.29 × 10 ⁻⁵ M
e) By substitution: Ka = (2.29 × 10 ⁻⁵ M)² / 0.015M = 3.50 × 10⁻⁸ M
f) By definition: pKa = - log Ka = - log (3.50 × 10 ⁻⁸) = 7.46
