Answer: CO is a limiting reagent with regards to the Fe production.
Explanation:

Moles of CO = 
moles of 
According to reaction , 3 mole of CO reacts with 1 mole of
then , 0.4714 moles of CO will react with :
moles of
that is 0.1571 moles.
0.4714 moles of CO will react with 0.1571 moles of
which means that CO is present in limited amount acting as limiting reagent.
Mole remaining of
= 0.2673 mol - 0.1571 mol = 0.1102 mol
Hence, CO is a limiting reagent and
is an excessive reagent.
Construction of dam : scarcity of water for animals
Answer : The pH of the solution is, 2.67
Explanation :
The equilibrium chemical reaction is:

Initial conc. 0.450 0 0
At eqm. (0.450-x) x x
As we are given:

The expression for equilibrium constant is:

Now put all the given values in this expression, we get:


The concentration of
= x = 0.00212 M
Now we have to calculate the pH of solution.
![pH=-\log [H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%5BH%5E%2B%5D)


Therefore, the pH of the solution is, 2.67
1. Phosphate Group,Pentose sugar, and Nitrogen base
2. Not sure
3.These are found in RNA cytosine, guanine, adenine and uracil.