An r-selected species reproduces much faster than K-selected species.
r-selected species focuses on maturing and reproducing quickly. r-selected species will probably reproduce when the water supply is there for the short period of time; thus, increasing the chance of the r-selected species of surviving.
K-selected species, on the other hand, focus on raising their young and reproduce later. Since the K-selected species take long to mature before reproducing, water supply may run out before they have a chance of fully maturing; thus, K-selected species have a lower chance of survival.
Hope this helps.
If you need anything more, feel free to comment! Have an awesome day! :)
~Collinjun0827, Junior Moderator
When we can get Pka for K2HPO4 =6.86 so we can determine the Ka :
when Pka = - ㏒ Ka
6.86 = -㏒ Ka
∴Ka = 1.38 x 10^-7
by using ICE table:
H2PO4- → H+ + HPO4
initial 0.4 m 0 0
change -X +X +X
Equ (0.4-X) X X
when Ka = [H+][HPO4] / [H2PO4-]
by substitution:
1.38 X 10^-7 = X^2 / (0.4-X) by solving for X
∴X = 2.3x 10^-4
∴[H+] = X = 2.3 x 10^-4
∴PH = -㏒[H+]
= -㏒ (2.3 x 10^-4)
∴PH = 3.6
I believe the correct answer is the first option. To increase the molar concentration of the product N2O4, you should increase the pressure of the system. You cannot determine the effect of changing the temperature since we cannot tell whether it is an endothermic or an exothermic reaction. Also, decreasing the number of NO2 would not increase the product rather it would shift the equilibrium to the left forming more reactants. The only parameter we can change would be the pressure. And, since NO2 takes up more space than the product increasing the pressure would allow the reactant to collide more forming the product.
Answer:
a) find attached image 1
b) find attached image 2
Explanation :
The more stable radical is formed by a reaction with smaller bond dissociation energy.
since the bond dissociation for cleavage of the bond to form primary free radical is higher, more energy must be added to form it. This makes primary free radical higher in energy and therefore less stable than secondary free radical.
