C.
centi- is essentially 10^2 of one meter.
If you had 100m, multiplying 100 by 10^2 (or 100) would give you 10000 cm.
Answer:
option C= hydrolysis and break down
Explanation:
All other three pairs are correct coupling of each others.
Option A= dehydration synthesis and hydrolysis
Dehydration synthesis:
In dehydration synthesis monomers combine through the covalent bonds and form large molecules. The large molecules are called polymers. The water as a byproduct also released when monomers joints together.
Hydrolysis:
In hydrolysis the polymers are break down into monomers by using water molecules. The catalysts are also required in this process.
Option B= Catabolic and Anabolic
Anabolic:
In this process smaller molecules combine to gather to form large complex molecules by using energy.
For example simple glucose molecules join together to form large disaccharides.
Catabolic:
It is the break down of large complex molecules to the smaller molecules.
For example during cellular respiration sugar molecules break down and generate energy.
Option D= Break down and synthesis
The break down and synthesis are also reverse pair of each others. The synthesis involve the formation of molecules form smaller component while the break down involve destruction of molecules into smaller units.
Answer:
This is a typical stoichiometry question.To answer this question you want to get a relationship between
N
a
2
O and NaOH.
So you can get a relationship between the moles of
N
a
2
O
and moles of NaOH by the concept of stoichiometry.
N
a
2
O +
H
2
O ----------------> 2 NaOH.
According to above balanced equation we can have the stoichiometry relationship between
N
a
2
O and NaOH. as 1:2
It means 1 moles of
N
a
2
O is required to react with one mol of
H
2
O to produce 2 moles of NaOH.
in terms of mass 1 mole of
N
a
2
O has mass 62 g on reaction with water produces 2 moles of NaOH or 80 g of NaOH.
62 g of
N
a
2
O produces 80 g of NaOH.
1g of NaOH is produced from 62/80 g of
N
a
2
O
1.6 x
10
2
g of NaOH will require 62 x 1.6 x
10
2
g / 80 of
N
a
2
O
124g of
N
a
2
O.
Explanation:
Answer:
![[I_2]=[Br]=0.31M](https://tex.z-dn.net/?f=%5BI_2%5D%3D%5BBr%5D%3D0.31M)
Explanation:
Hello there!
In this case, according to the given information, it is possible for us to set up the following chemical equation at equilibrium:
Now, we can set up the equilibrium expression in terms of x (reaction extent) whereas the initial concentration of both iodine and bromine is 0.5mol/0.250L=2.0M:
![K=\frac{[IBr]^2}{[I_2][Br_2]} \\\\1.2x10^2=\frac{(2x)^2}{(2.0-x)^2}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BIBr%5D%5E2%7D%7B%5BI_2%5D%5BBr_2%5D%7D%20%5C%5C%5C%5C1.2x10%5E2%3D%5Cfrac%7B%282x%29%5E2%7D%7B%282.0-x%29%5E2%7D)
Thus, we solve for x as show below:
Therefore, the concentrations of both bromine and iodine are:
![[I_2]=[Br]=2.0M-1.69M=0.31M](https://tex.z-dn.net/?f=%5BI_2%5D%3D%5BBr%5D%3D2.0M-1.69M%3D0.31M)
Regards!
