Answer: Every enzyme has a specific name that can give us insight into the specific reaction that that enzyme can catalyze. We divide them into six different categories.
1) Oxidoreductase - includes two different types of reactions by transferring electrons from either molecule A to B or vice versa. It is involved in oxidizing electrons away from a molecule.
2) Hydrolase - uses water to divide a molecule into two other molecules.
3) Transferase - you move some functional group X from molecule B to molecule A
4) Ligase - catalyzes reactions between two molecules, A and B, that are combining to form a complex between the two. (example: DNA replication)
5) Lyase - divides a molecule into two other molecules without using water and without reducing or oxidation
Answer:
C) acid-base neutralization
Explanation:
NaOH + CH₃COOH = CH₃COONa + H₂O
Break the solutions apart:
NaOH = Na⁺ + OH⁻
CH₃COOH = CH₃COO⁻ + H⁺
Combine the resulting solution after the reaction:
OH⁻ + H⁺ = H₂O
Answer:
1. 72.9 atm
2. 0.43937 K
Explanation:
1. Gray- lussacs law is p1/t1=p2/t2 so we use this formula to figure it out by filling in the variables and solving
p1=45.0 atm
t1=323 K
p2= ?
t2=523 K
Now we fill in this in the formula and solve - 45.0 atm/ 323 K = p2/ 523 K
and now we solve for p2 by multiplying 535k by each side to give us p2
2. Using the same formula we get 10.0atm/? = 12.0 atm/ 273.15 k and we divide both sides by 10.0 atm
<h3>
Answer:</h3>
Chlorine gas (Cl₂)
<h3>
Explanation:</h3>
- According to the Graham's law of diffusion, the diffusion rate of a gas is inversely proportional to the square root of its density or molar mass.
- Therefore, a lighter gas will diffuse faster at a given temperature compared to a heavy gas.
- Consequently, the heavier a gas is then the denser it is and the slower it diffuses at a given temperature and vice versa.
In this case we are given gases, CI₂
, H₂,He and Ne.
- We are required to identify the gas that will diffuse at the slowest rate.
- In other words we are required to determine the heaviest gas.
Looking at the molar mass of the gases given;
Cl₂- 70.91 g/mol
H₂- 2.02 g/mol
He - 4.00 g/mol
Ne- 20.18 g/mol
Therefore, chlorine gas is the heaviest and thus will diffuse at the slowest rate among the choices given.
Answer:
d. 0.121 M HC2H3O2 and 0.116 M NaC2H3O2
Explanation:
Hello,
In this case, since the pH variation is analyzed via the Henderson-Hasselbach equation:
![pH=pKa+log(\frac{[Base]}{[Acid]} )](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%28%5Cfrac%7B%5BBase%5D%7D%7B%5BAcid%5D%7D%20%29)
We can infer that the nearer to 1 the ratio of of the concentration of the base to the concentration of the acid the better the buffering capacity. In such a way, since the sodium acetate is acting as the base and the acetic acid as the acid, we have:
a. ![\frac{[Base]}{[Acid]}=\frac{0.497M}{0.365M}=1.36](https://tex.z-dn.net/?f=%5Cfrac%7B%5BBase%5D%7D%7B%5BAcid%5D%7D%3D%5Cfrac%7B0.497M%7D%7B0.365M%7D%3D1.36)
b. ![\frac{[Base]}{[Acid]}=\frac{0.217M}{0.521M}=0.417](https://tex.z-dn.net/?f=%5Cfrac%7B%5BBase%5D%7D%7B%5BAcid%5D%7D%3D%5Cfrac%7B0.217M%7D%7B0.521M%7D%3D0.417)
c. ![\frac{[Base]}{[Acid]}=\frac{0.713M}{0.821M}=0.868](https://tex.z-dn.net/?f=%5Cfrac%7B%5BBase%5D%7D%7B%5BAcid%5D%7D%3D%5Cfrac%7B0.713M%7D%7B0.821M%7D%3D0.868)
d. ![\frac{[Base]}{[Acid]}=\frac{0.116M}{0.121M}=0.959](https://tex.z-dn.net/?f=%5Cfrac%7B%5BBase%5D%7D%7B%5BAcid%5D%7D%3D%5Cfrac%7B0.116M%7D%7B0.121M%7D%3D0.959)
Therefore, the d. solution has the best buffering capacity.
Regards.
