Hi there!
Although there are only 20 amino acids, these amino acids can combine into an innumerable amount of combinations to form different and unique proteins.
In case that doesn't make sense to you, I'll provide you with an analogy. You could be provided with 20 different LEGO bricks to work with. While there may only be 20 bricks, these bricks can combine into a vast amount of different formations, structures, etc. Amino Acids work in the same way.
1s to the second power, 2s to the second power, 2p to the 6th power, 3s to the second power and 3p to the third power.
Answer : The equilibrium concentration of 
in the solution is, 
Explanation :
The dissociation of acid reaction is:
Initial conc. c 0 0
At eqm. c-x x x
Given:
c = 
The expression of dissociation constant of acid is:
![K_a=\frac{[H_3O^+][C_6H_5COO^-]}{[C_6H_5COOH]}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BH_3O%5E%2B%5D%5BC_6H_5COO%5E-%5D%7D%7B%5BC_6H_5COOH%5D%7D)
Now put all the given values in this expression, we get:
![6.3\times 10^{-5}=\frac{(x)\times (x)}{[(7.0\times 10^{-2})-x]}](https://tex.z-dn.net/?f=6.3%5Ctimes%2010%5E%7B-5%7D%3D%5Cfrac%7B%28x%29%5Ctimes%20%28x%29%7D%7B%5B%287.0%5Ctimes%2010%5E%7B-2%7D%29-x%5D%7D)
Thus, the equilibrium concentration of in the solution is,
Answer:
it's to blurry you can't see anything
<h3>Answer:</h3>
Limiting reactant is Lithium
<h3>
Explanation:</h3>
<u>We are given;</u>
- Mass of Lithium as 1.50 g
- Mass of nitrogen is 1.50 g
We are required to determine the rate limiting reagent.
- First, we write the balanced equation for the reaction
6Li(s) + N₂(g) → 2Li₃N
From the equation, 6 moles of Lithium reacts with 1 mole of nitrogen.
- Second, we determine moles of Lithium and nitrogen given.
Moles = Mass ÷ Molar mass
Moles of Lithium
Molar mass of Li = 6.941 g/mol
Moles of Li = 1.50 g ÷ 6.941 g/mol
= 0.216 moles
Moles of nitrogen gas
Molar mass of Nitrogen gas is 28.0 g/mol
Moles of nitrogen gas = 1.50 g ÷ 28.0 g/mol
= 0.054 moles
- According to the equation, 6 moles of Lithium reacts with 1 mole of nitrogen.
- Therefore, 0.216 moles of lithium will require 0.036 moles (0.216 moles ÷6) of nitrogen gas.
- On the other hand, 0.054 moles of nitrogen, would require 0.324 moles of Lithium.
Thus, Lithium is the limiting reagent while nitrogen is in excess.
