Answer:
Mixtures
Explanation:
Matter can be classified as a compound and a mixture.
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)
Aluminum has three oxidation states. The most common one is +3. The other two are +1 and +2. One +3 oxidation state for Aluminum can be found in the compound aluminum oxide, Al2O3.
In a water wave all particles travel in clockwise circles.
Answer:
The pressures will remain at the same value.
Explanation:
A catalyst is a substance that alter the rate of a chemical reaction. It either speeds up the or slows down the rate of a chemical reaction.
While a catalyst affects the rate, it is noteworthy that it has no effect on the equilibrium position of the chemical reaction. A catalyst works by creating an alternative pathway for the reaction to proceed. Most times, it decreases the activation energy needed to kickstart the chemical reaction.
Hence, we know that it has no effect on the equilibrium position. Factors affecting equilibrium position includes, temperature and concentration of reactants and products( pressure in terms of gases).
The reactants and the products here are gaseous, and as such pressure affects the equilibrium position. Now, we have established that the equilibrium position is unaffected. And as such the pressure affecting it does not change.
Thus, we have established that the pressure of the products and reactants are unaffected and as such they remain at their value unaffected.
