The number-average molecular weight of a polypropylene is 663419 g/mol. Compute the degree of polymerization (Key in your answer as the whole number only. Avoiding round-off error by using all the numbers during intermediate steps.

Answers

Answer 1
Answer:

Answer:

15766

Explanation:

The average molecular weight of the polypropylene = 663419 g/mol.

The mass of the repeating unit , monomer which is propylene = 42.08 g/mol

The degree of polymerization is:

DP_n=\frac {Total\ molecular\ weight\ of\ the\ polymer}{Molecular\ weight\ of\ the\ monomer}

DP_n=\frac {663419\ g/mol}{42.08\ g/mol}

Degree of polymerization = 15766


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If the molecule C6H12 does not contain a double bond, and there are no branches in it, what will its structure look like?

Answers

I have attached a photo of the structure. 
You can get better at solving problems like this by practicing a lot! 




Which of the following is an example of physical change?

Answers

Answer:

The glass cup falling from the counter

Explanation:

the glass isn't changing in any chemical way. it's still made of the same material, just broken apart.

Final answer:

Physical changes involve the alteration of the state or appearance of matter, without changing the composition. An example is solid wax turning into liquid wax when heated, or steam condensing inside a cooking pot.

Explanation:

The question asks for an example of a physical change. Physical changes involve alterations in the state or appearance of matter, without changing its composition. For example, solid wax turning into liquid wax when heated is a physical change. The wax is still the same substance, it's just in a different state. Similarly, steam condensing inside a cooking pot is also a physical change. The water vapor turns back into liquid water, but it's still water. These are distinguished from chemical changes, which transform one substance into a different substance.

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How much heat is required to change the temperature of two cups of water (500 g) from room temperature (25◦C) to boiling? Specific heat of water is c=4.184 J/(g oC) a 78.5 kJ b 15.7 kJ c 157 kJ d 1.57 kJ

Answers

The heat that is required to change the temperature of two cups of water (500 g) from room temperature (25◦C) to boiling

C) 157 kJ

Heat

Heat required= Mass of water x specific heat capacity of water x change in temperature of water required

Q=m* c* delta T

M = 500g

C = 4.184 g°C

Delta T = 100 - 25(room temp) = 75°C

Heat = 500 x 4.184 x 75

Heat = 156900 J

Heat = 156.9 KJ

Heat ~ 157.0 KJ (3.D.P)

Thus, the correct answer is C.

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Answer:

C

Explanation:

Heat required= Mass of water x specific heat capacity of water x change in temperature of water required

q \:  = m  * c  * delta \: t

M = 500g

C = 4.184 g°C

Delta T = 100 - 25(room temp) = 75°C

Heat needed= 500 x 4.184 x 75

= 156900 J

= 156.9 KJ

~ 157.0 KJ (3.D.P)

PLEASE HELP FAST!!If metal X is lower than metal Y on the activity series, then:
A. X will react in water, but only if the temperature is low enough
B. Y will form oxides of X, but only indirectly
C. X will replace ions of Y in a solution
D. Y will replace ions of X in a solution

Answers

Answer:

D. Y will replace ions of X in a solution

Explanation:

If metal X is lower than metal Y on the activity series, then Y will replace ions of X in a solution.

This is the crux of single displacement reactions.

  • In a single displacement reaction, a metallic ion in solution is replaced by a metal higher in the activity series than the metal in solution.
  • On the activity series, metals higher are more reactive and will displace the lower and less reactive ones.
  • Reactivity increases up the group.

The boiling point of an aqueous 1.83 m (nh4)2so4 (molar mass = 132.15 g/mol) solution is 102.5°c. determine the value of the van't hoff factor for this solute if the kb for water is 0.512°c/m. 2.7 1.8 2.3 3.0 3.6

Answers

The value of the van't Hoff factor for the given solute is 2.7

Option (A) 2.7 is correct.

What is van't Hoff factor?

The Van't Hoff factor is a measure of a solute's effect on colligative qualities such boiling point elevation, osmotic pressure, and relative vapor pressure reduction.

Given,

Molar mass is 132.15 g/mol

Boiling point elevation is 102.5 °C

kb for water is  0.512°c/m

Molarity of solution is 1.83 m

The value of van't Hoff factor is: 2.66 or 2.7 (approx)

Now, from the solution of colligative properties to calculate elevation in boiling point.

\bold{\Delta Tb = ikbm}

where Δ = elevation of boiling point (102.5)

\bold{102.5* i* 0.512 * 1.83  }

\bold{i = 2.66 = 2.7 }

Thus, The value is 2.7 option (A) is correct.

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We need to know the value of van't hoff factor.

The van't hoff factor is: 2.66 or 2.7 (approximately)

(NH₄)₂SO₄ is an ionic compound, so it dissociates in solution and produces 3 ionic species. Therefore van't hoff factor is more than one.

From the equation: ΔT_(b)=i K_(b).m, where ΔT_(b)= elevation of boiling point=102.5 - 100=2.5°C.

m=molality of solute=1.83 m (Given)

K_(b)= Ebullioscopic constant or Boiling point elevation constant= 0.512°C/m (Given)

i= Van't Hoff factor

So, 2.5= i X 0.512 X 1.83

i=(2.5)/(0.512 X 1.83)

i=2.66= 2.7 (approx.)


Calculate the mass of a sample of lead (cPb = 0.16 J/g℃) when it loses 200 J cooling from 75.0℃ to 42.0℃.

Answers

looses 299 please collins 75.0