Answer:
Meiosis
Explanation:
Meiosis is the formation of a new species from an existing species and can occur in two phases.
Hope it helps.
D) a change in protein synthesis
One major reason for exploring extreme environments is to find new resources. An example of a resource is the search for a new antibiotic. Extreme environments have rare species endemic to the regions which can offer different insights from the normative environment we are used to.
No, I would not. This is because the risk of death is higher in an extreme environment. Example of these risks are deadly diseases or harsh climate. The body may be unable to adjust to the extreme environment hence may be unable to restore homoeostasis; such as in restoring body heat in extremely cold environments hence leading to hypothermia.
Answer:
two sets of magnifying lenses
Explanation:
This is because a compound microscope is a high magnifying microscope that have multiple lenses or two magnifying lenses which help to see or observe or magnify image very well compare to single lense. It is a high power microscope and it has high level of magnification because of the
two magnifying lenses. It produces two dimensional image. The eyepieces and the objective lenses give a very high magnification, a condenser that is below the stage focus the light into the specimen to be viewed.
Answer:The ratio of the concentrations of 
and 
when the buffer has a pH of 7.02 is 0.69
Explanation:
The dissociation constant for formic acid = 
Concentration of HA= 0.5 mM
Concentration of = 0.1 mM
pH = 6.16
First we have to calculate the value of 
.
Using Henderson Hesselbach equation :
![pH=pK_a+\log \frac{[A^-}{[HA]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BA%5E-%7D%7B%5BHA%5D%7D)
Now put all the given values in this expression, we get:
![6.16=pK_a+\log (\frac{[0.1]}{0.5})](https://tex.z-dn.net/?f=6.16%3DpK_a%2B%5Clog%20%28%5Cfrac%7B%5B0.1%5D%7D%7B0.5%7D%29)
To calculate the ratio of the concentrations of and when the buffer has a pH of 7.02.
Using Henderson Hesselbach equation :
![7.02=6.86+\log \frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=7.02%3D6.86%2B%5Clog%20%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
![\frac{[A^-]}{[HA]}=1.44](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D%3D1.44)
![\frac{[HA]}{[A^-]}=0.69](https://tex.z-dn.net/?f=%5Cfrac%7B%5BHA%5D%7D%7B%5BA%5E-%5D%7D%3D0.69)
Thus the ratio of the concentrations of and when the buffer has a pH of 7.02 is 0.69
