So I believe you are supposed to take notes based on the guiding questions (the questions on the side).
Explanation:
The solution of the lactic acd and sodium lactate is referred to as a buffer solution.
A buffer solution is an aqueous solution consisting of a mixture of a weak acid and its conjugate base, or vice versa. In this case, the weak acid is the lactic acid and the conjugate base is the sodium lactate.
Buffer solutions are generally known to resist change in pH values.
When a strong base (in this case, NaOH) is added to the buffer, the lactic acid will give up its H+ in order to transform the base (OH-) into water (H2O) and the conjugate base, so we have:
HA + OH- → A- + H2O.
Since the added OH- is consumed by this reaction, the pH will change only slightly.
The NaOH reacts with the weak acid present in the buffer sollution.
WHAT THE HECK!?!?!?! AM I SUPPOSE TO KNOW WHATEVER LANGUAGE THAT IS?!?!
Answer:
The equilibrium concentrations are:
[SO2]=[NO2] = 0.563 M
[SO3]=[NO] = 1.04 M
Explanation:
<u>Given:</u>
Equilibrium constant K = 3.39
[SO2] = [NO2] = [SO3] = [NO] = 0.800 M
<u>To determine:</u>
The equilibrium concentrations of the above gases
Calculation:
Set-up an ICE table for the given reaction

I 0.800 0.800 0.800 0.800
C -x -x +x +x
E (0.800-x) (0.800-x) (0.800+x) (0.800+x)
The equilibrium constant is given as:
![Keq = \frac{[SO3][NO]}{[SO2][NO2]}=\frac{(0.800+x)^{2}}{(0.800-x)^{2}}](https://tex.z-dn.net/?f=Keq%20%3D%20%5Cfrac%7B%5BSO3%5D%5BNO%5D%7D%7B%5BSO2%5D%5BNO2%5D%7D%3D%5Cfrac%7B%280.800%2Bx%29%5E%7B2%7D%7D%7B%280.800-x%29%5E%7B2%7D%7D)

x = 0.2368 M
[SO2]=[NO2] = 0.800 -x = 0.800 - 0.2368 = 0.5632 M
[SO3]=[NO] = 0.800 +x = 0.800 + 0.2368 = 1.037 M
Answer:
191.11 grams of oxygen gas should be produced.
Explanation:
The balanced reaction is:
2 Al₂O₃ → 4 Al + 3 O₂
By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
- Al₂O₃: 2 moles
- Al: 4 moles
- O₂: 3 moles
Being the molar mass of each compound:
- Al₂O₃: 102 g/mole
- Al: 27 g/mole
- O₂: 32 g/mole
By reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
- Al₂O₃: 2 moles* 102 g/mole= 204 grams
- Al: 4 moles* 27 g/mole= 108 grams
- O₂: 3 moles* 32 g/mole= 96 grams
Then you can apply the following rule of three: if by stoichiometry 108 grams of aluminum are produced along with 96 grams of oxygen, 215 grams of aluminum are produced along with how much mass of oxygen?

mass of oxygen= 191.11 grams
<u><em>191.11 grams of oxygen gas should be produced.</em></u>