Answer:
There are no options but the question can be answered based on general understanding of experimentation.
Group A: Control group
Group B: Experimental group
Explanation:
In an experiment, there are two groups viz: the control group and the experimental group. A control group is the group that does not receive any experimental treatment. It is considered the normal group while the experimental group is the group that is subjected to experimental treatment.
N.B: The experimental treatment in this case is the "Oxygen level"
According to the question, out of 20 petri-dishes used to grow bacteria, half (10) of the petri-dishes were placed in a container with normal atmosphere while the other 10 petri-dishes were placed in another container that have double the normal oxygen level. The first group of bacteria labelled GROUP A are the CONTROL GROUP because they were not given any treatment i.e placed under normal conditions while GROUP B are the EXPERIMENTAL GROUP because they are the group that received the experimental treatment (increased oxygen level).
Answer: less oxygen available at higher altitudes
Explanation:
The higher in elevation you get, the lower the air pressure, allowing oxygen atoms to spread out farther, therefore making available oxygen less.
Answer:
lesser the molar mass of the gas higher the no. of moles included in a certain mass sample. ie at STP more volume is required for the gas having less molar mass.
He has the smallest molar mass.
Therefore bag of He is the biggest.
Answer: The equilibrium constant is ![3.3\times 10^{-4}](https://tex.z-dn.net/?f=3.3%5Ctimes%2010%5E%7B-4%7D)
Explanation:
Initial concentration of
= 0.095 M
The given balanced equilibrium reaction is,
![I_2(g)\rightleftharpoons 2I(g)](https://tex.z-dn.net/?f=I_2%28g%29%5Crightleftharpoons%202I%28g%29)
Initial conc. 0.095 M 0 M
At eqm. conc. (0.095-x) M (2x) M
Given : 2x = 0.0055
x = 0.00275
The expression for equilibrium constant for this reaction will be,
Now put all the given values in this expression, we get :
![K_c=\frac{(0.0055)^2}{(0.095-0.00275)}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%280.0055%29%5E2%7D%7B%280.095-0.00275%29%7D)
![K_c=\frac{(0.0055)^2}{0.09225}=0.00033](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%280.0055%29%5E2%7D%7B0.09225%7D%3D0.00033)
Thus the equilibrium constant is ![3.3\times 10^{-4}](https://tex.z-dn.net/?f=3.3%5Ctimes%2010%5E%7B-4%7D)