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Mazyrski [523]
3 years ago
15

You are given that kc = 10-1 kg eq-1 min-1, ku = 10-3 kg2 eq-2 min-1 and [A]0 = 10 eq kg-1, where kc is the rate constant for a

catalyzed step-growth polymerization, ku is the rate constant for an uncatalyzed step-growth polymerization, and [A]0 is the initial concentration of monomer A in the step-growth polymerization. Assume that the stoichiometry of monomer A to monomer B is 1:1, that the reactive groups on A only react with the reactive groups on B, and that the reactivity of A and B is independent of the chain length.
A. Calculate the time in minutes for conversion (p) to reach values of 0.2, 0.4, 0.6, 0.8, and 0.99 for both catalyzed and uncatalyzed polymerizations.B. Next, calculate the number-average degree of polymerization at each of these times.C. Assume that you can recover all of the catalyst (i.e., the cost of the catalyst is not a factor), which polymerization route is more economical from an industrial perspective?
Engineering
1 answer:
victus00 [196]3 years ago
7 0

Answer:

-ajkdbvójadh`toug511

Explanation:

b

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What is the least count of screw gauge?<br> (a) 0.01 cm<br> (b) 0.001 cm<br> (c) 0.1 cm<br> (d) 1 mm
Nonamiya [84]
Its 0.001

0.01 x100 = 1mm
0.001x100=0.1mm
0.1=10mm
1m
3 0
2 years ago
How would you describe what would happen to methane if the primary bonds were to break?
erastova [34]

Answer:

All the bonds in methane (CH4CH4) are equivalent, and all have the same dissociation energy.

The product of the dissociation is methyl radical (CH3CH3). All the bonds in methyl radical are equivalent, and all have the same dissociation energy.

The product of that dissociation is methylene (CH2CH2). All the bonds in methylene are equivalent, and all have the same dissociation energy.

The product of that dissociation is methyne (CHCH) .

The C-H bonds in methane do not have the same dissociation energy as C-H bonds in methyl radical, which in turn do not have the same dissociation energy as the C-H bonds in methylene, which are again different from the C-H bond in methyne.

If (by some miracle) you were able to get all four bonds in methane to dissociate absolutely simultaneously, they would all show the same dissociation energy… but that energy, per bond broken, would be different than the energy required to break just one C-H bond in methane, because the products are different.

(In this case, it’s CH4→C+4HCH4→C+4H versus CH4→CH3+HCH4→CH3+H.)

To alter hydrocarbons you add enough energy to break a C-H bond. Why does only one bond break? What concentrates the energy on one C-H bond?

the weakest CH bond is the one that breaks. in plain alkanes it has to do with the molecular orbital interactions between neighboring carbon atoms. look at propane for example. the middle carbon has two C-C bonds, and each of those C-C bonds is strengthened by slight electron delocalization from the C-H bonds overlapping with the antibonding orbitals of the adjacent carbons.

since the C-H bonds on the middle carbon donate electron density to both of its neighbors, those two are weakest.

one of them will break preferentially.

which one actually breaks depends on the reaction conditions (kinetics). frankly it's whichever one ramdomly approaches a nucleophile first. when the nucleophile pulls of one of the H's, the other C-H bonds start to share (delocalize) the negative charge across the whole molecule. so while the middle C feels the majority of the negative charge character, the other two C's take on a fair amount as well...

by the way, alkanes don't really like to break and form anions like that.

a better example would be something like isopropyl iodide, where the C-I bond breaks and the I carries away the electron pair, forming a carbocation (also not particularly stable, but more so than the carbanion).

7 0
2 years ago
Identify three material considerations an engineer would need to consider when working on a design process.
Anika [276]

Answer:

Three material considerations are;

1. Identify and appraise the attainment of the goal of the with the design specification

2. Ascertain the required load the product being designed will experience and the suitability of the design specification to that load

3. Review the producibility of the design to ensure that it can be produced with the available technology

Explanation:

1. The appraisal of the design includes the consideration of the factors of the design and the building of reliability and efficiency into the design from the beginning

2. Ascertain if the product will require toughness, elasticity, and if will be subject to sudden or repeated loading conditions

3. Ensure that the design can be readily produced with the accessible manufacturing equipment during the conceptualization stage of the design.

4 0
3 years ago
If gas costs $3.50 per gallon, how much would it cost to drive 500 miles in a city in a car that is 58.3 km/L
Akimi4 [234]
1 liter = .264 gallon
1 km = .621 mile

this means that 58.3km/L is equal to 137.13mpg

so

500/137.13 = 3.65 gallons of gas

3.65 x 3.5 = $12.78
5 0
2 years ago
A basketball has a 300-mm outer diameter and a 3-mm wall thickness. Determine the normal stress in the wall when the basketball
faltersainse [42]

Answer:

2.65 MPa

Explanation:

To find the normal stress (σ) in the wall of the basketball we need to use the following equation:

\sigma = \frac{p*r}{2t}

<u>Where:</u>

p: is the gage pressure = 108 kPa

r: is the inner radius of the ball

t: is the thickness = 3 mm  

Hence, we need to find r, as follows:

r_{inner} = r_{outer} - t    

r_{inner} = \frac{d}{2} - t

<u>Where:</u>

d: is the outer diameter = 300 mm

r_{inner} = \frac{300 mm}{2} - 3 mm = 147 mm

Now, we can find the normal stress (σ) in the wall of the basketball:

\sigma = \frac{p*r}{2t} = \frac{108 kPa*147 mm}{2*3 mm} = 2646 kPa = 2.65 MPa

Therefore, the normal stress is 2.65 MPa.

I hope it helps you!

3 0
3 years ago
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