Answer:
- <em>During the polymerization of a 20 monomer-long cellulose molecule,</em> <u>19 molecules of water are released.</u>
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Explanation:
In simple terms, <em>cellulose </em>is the biopolymer formed by many glucose units. This is cellulose is the polymer and glucose is the monomer.
To have a <em>20 monomer-long cellulose molecule</em>, 20 monomers have been chemically bonded by reacting 19 times, as it is explained in the next paragrpahs, and so 19 molecules of water have been released.
You can imaging the polymerization process as a step-by-step reaction in which the first step is the condensation reaction of one glucose molelecule to produce a 2 monomer-long chain, with the release of one molecule of water: the second step would be the condensation reaction between the 2 monomer-long chain with another glucose molecule, with the release of an additional molecule of water, and so on, until 19 condensation reactions happen, to obtain the 20 monomer-long cellulose molecule.
Condensation is the loss of water in a chemical reaction.
When two glucose molecules react together, condensation occurs. One OH group from each glucose molecule come together, the OH from one glucose molecule combines with the H part of the OH from the other glucose molecule, to form H₂O (water that is released).
The two glucose molecules (monomers) will form one bigger molecule where the two glucose monomers are bonded through the oxygen atom that did not form part of the water molecule released.
Then, a 20-monomer chain means 19 condenstation reactions, with the release of 19 molecules of water.
I think it would be homogeneous and heterogeneous
Answer : The concentration of NOBr after 95 s is, 0.013 M
Explanation :
The integrated rate law equation for second order reaction follows:
![k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B1%7D%7Bt%7D%5Cleft%20%28%5Cfrac%7B1%7D%7B%5BA%5D%7D-%5Cfrac%7B1%7D%7B%5BA%5D_o%7D%5Cright%29)
where,
k = rate constant = 
t = time taken = 95 s
[A] = concentration of substance after time 't' = ?
![[A]_o](https://tex.z-dn.net/?f=%5BA%5D_o)
= Initial concentration = 0.86 M
Now put all the given values in above equation, we get:
![0.80=\frac{1}{95}\left (\frac{1}{[A]}-\frac{1}{(0.86)}\right)](https://tex.z-dn.net/?f=0.80%3D%5Cfrac%7B1%7D%7B95%7D%5Cleft%20%28%5Cfrac%7B1%7D%7B%5BA%5D%7D-%5Cfrac%7B1%7D%7B%280.86%29%7D%5Cright%29)
[A] = 0.013 M
Hence, the concentration of NOBr after 95 s is, 0.013 M
Answer:
1.9 L
Explanation:
Step 1: Given data
- Initial pressure (P₁): 1.5 atm
- Initial volume (V₁): 3.0 L
- Initial temperature (T₁): 293 K
- Final pressure (P₂): 2.5 atm
- Final temperature (T₂): 303 K
Step 2: Calculate the final volume of the gas
If we assume ideal behavior, we can calculate the final volume of the gas using the combined gas law.
P₁ × V₁ / T₁ = P₂ × V₂ / T₂
V₂ = P₁ × V₁ × T₂ / T₁ × P₂
V₂ = 1.5 atm × 3.0 L × 303 K / 293 K × 2.5 atm = 1.9 L
