Answer:
The answer to this is
Unsaturated solution
Explanation:
An unsaturated solution has the property of having a solute concentration lower than the the solubility at equilibrium at a given temperature hence it has the capacity to dissolve more solutes. Is is a solution containing a lower amount of solute than a saturated solution
The two processes that occur on dissolving a solute in a solvent are dissolution and crystallization and in an unsaturated solution the rate of dissolution is greater than the rate of crystallization
To tract genetic inheritance ( phenotype or genotype) from one generation to another generation, PEDIGREE CHART is used. Pedigree Chart is a diagram shows the occurrence and appearance of a particular genes of an individual organism from its ancestor down to the generations it follows. Relationship in a pedigree is shown as a series of linear graph. Parents are connected to a horizontal lines while vertical lines will lead to the offspring.
Answer:
pH = 11.95≈12
Explanation:
Remember the reaction among aqueous acetic acid (
) and aqueous sodium hydroxide (NaOH)
First step. Need to know how much moles of the substances are present
![0.1 \frac{mol NaOH}{L} * 0.030 L = 0.003 mol NaOH[tex]0.1 \frac{mol CH_3COOH}{L} * 0.025 L= 0.0025 mol CH_3COOHSecond step. Know wich substance is in excess.0.0025 mol CH_3COOH * [tex]1 mol NaOH/ 1 mol CH_3COOH](https://tex.z-dn.net/?f=0.1%20%5Cfrac%7Bmol%20NaOH%7D%7BL%7D%20%2A%200.030%20L%20%3D%200.003%20mol%20NaOH%3C%2Fp%3E%3Cp%3E%5Btex%5D0.1%20%5Cfrac%7Bmol%20CH_3COOH%7D%7BL%7D%20%2A%200.025%20L%3D%200.0025%20mol%20CH_3COOH%3C%2Fp%3E%3Cp%3E%3C%2Fp%3E%3Cp%3ESecond%20step.%20Know%20wich%20substance%20is%20in%20excess.%3C%2Fp%3E%3Cp%3E0.0025%20mol%20CH_3COOH%20%2A%20%5Btex%5D1%20mol%20NaOH%2F%201%20mol%20CH_3COOH)
= 0.0025 mol NaOH
0.003 mol NaOH * 
/ 1 mol NaOH = 0.003 mol CH_3COOH[/tex]
NaOH is in excess. Now, how much?
0.003 mole NaOH - 0.0025 mole NaOH = 0.0005 mole NaOH
Then, that amount in excess would be responsable for the pH.
Third step. Know the pH
Remember that pH= -log[H+]
According to the dissociation of water equilibrium
Kw=[H+]*[OH-]= 10^(-14)
The dissociation of NaOH is
NaOH -> 
Now, concentration of OH^{-}[/tex] would be given for the excess of NaOH.
[OH-]= 0.0005 mole / 0.055 L = 0.00909 M
Careful: we have to use the total volumen
Les us to calculate pH
![pH= -log [H+]\\pH= -log \frac{K_w}{[OH-]} \\pH= 11.95](https://tex.z-dn.net/?f=pH%3D%20-log%20%5BH%2B%5D%5C%5CpH%3D%20-log%20%5Cfrac%7BK_w%7D%7B%5BOH-%5D%7D%20%5C%5CpH%3D%2011.95)
C. Diffusion is not a phase change.
