As in math or
?
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Explanation:
a)phenotype= 100% big
b) genotype=1:2:1 (FF-25%, Ff-50%, ff-25%)
phenotype=3:1 (big=75%, small= 25%)
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.
First write all of the compounds/atoms in either side then fill in existing values and balance
Na- 1
Br- 1
Ca- 1
Cl- 2
Na- 1
Cl- 1
Ca-1
Br-2
Balance to get
2NaBr+CaCl2=2NaCl+CaBr2
Consider the isomerization of butane with equilibrium constant is 2.5 .The system is originally at equilibrium with :
[butane]=1.0 M , [isobutane]=2.5 M
If 0.50 mol/L of butane is added to the original equilibrium mixture and the system shifts to a new equilibrium position, what is the equilibrium concentration of each gas?
Answer:
The equilibrium concentration of each gas:
[Butane] = 1.14 M
[isobutane] = 2.86 M
Explanation:
Butane ⇄ Isobutane
At equilibrium
1.0 M 2.5 M
After addition of 0.50 M of butane:
(1.0 + 0.50) M -
After equilibrium reestablishes:
(1.50-x)M (2.5+x)
The equilibrium expression will wriiten as:
![K_c=\frac{[Isobutane]}{[Butane]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BIsobutane%5D%7D%7B%5BButane%5D%7D)

x = 0.36 M
The equilibrium concentration of each gas:
[Butane]= (1.50-x) = 1.50 M - 0.36M = 1.14 M
[isobutane]= (2.5+x) = 2.50 M + 0.36 M = 2.86 M