Answer:
cellular respiration
Explanation:
All exergonic processes produced in the cell, through which substances oxidize and chemical energy is released, are grouped under the name of cellular respiration, but to break down an organic molecule the cells employ, mainly dehydrogenations that can be carried carried out in the presence or absence of atmospheric O2 oxygen. There are therefore two types of breathing: aerobic respiration and anaerobic respiration. The latter also called fermentation.
Aerobic respiration (oxidative phosphorylation)
- Use molecular O2.
- It degrades glucose to CO2 and H2O
- Exergonic
- Recovers about 50% of chemical energy
- Present in most organisms.
- It uses enzymes located in the mitochondria.
You have to calculate the oxidation estates of the atoms in each compound.
I will start with K2Cr2O7 because I believe that Cr is the best candidate to reduce its oxidation number in 3 units.
In K2Cr2O7:
- K has oxidation state of 1+, then K2 has a charge of 2* (1+) = 2+.
- O has oxidation state of 2*, then O7 has a charge of 7* (2-) = 14-.
That makes that Cr2 has charge of 14 - 2 = +12, so each Cr has +12/2 = +6 oxidation state.
In Cr2O3:
- O has oxidation state of 2-, then O3 has charge 3 * (2-) = - 6
- Then, Cr2 has charge 6+, and each Cr has charge 6+ / 2 = 3+.
So, we have seen that Cr reduced its oxidation state in 3 units, from 6+ to 3+.
Answer: Cr has a change in oxidation number of - 3.
The number of molecules decrease
Answer:
Molarity of NaOH = 0.025 M
Explanation:
Given data:
Molarity of HCl = C₁ = 0.05 M
Volume of HCl = V₁= 50 mL
Molarity of NaOH = C₂=?
Volume of NaOH =V₂= 100 mL
Solution:
Formula:
C₁V₁ = C₂V₂
C₁ = Molarity of HCl
V₁ = Volume of HCl
C₂ = Molarity of NaOH
V₂ = Volume of NaOH
Now we will put the values:
C₁V₁ = C₂V₂
0.05 M × 50 mL = C₂ × 100 mL
2.5 M.mL =C₂ × 100 mL
C₂ = 2.5 M.mL /100 mL
C₂ = 0.025 M
