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lawyer [7]
3 years ago
11

A polycondensation reaction takes place between 1.2 moles of a dicarboxylic acid, 0.4 moles of glycerol (a triol) and 0.6 moles

of ethylene glycol (a diol). A.Calculate the critical extents of reaction for gelation using (i) the statistical theory of Flory and (ii) the Carothers theory.B.Comment on the observation that the measured value of the critical extent of reaction is 0.866.
Physics
1 answer:
katrin2010 [14]3 years ago
3 0

Answer:

A) i) using statistical theory of floxy

(Pa)c = 0.816

(Pb)c = 0.816

ii) using Carothers theory

( Pc ) = 0.917

B) To Obtain the measured value of critical extent of reaction ( 0.866) 1 mol of Glycerol  will react with 1 mol of dicarboxylic acid, but the same can not be applied to our obtained value because our stoichiometry is different

Explanation:

Given data :

Polycondensation reaction takes place between : 1.2 moles of dicarboxylic acid , 0.4 moles of glycerol and 0.6 moles of ethylene glycol

A) Calculate the critical extents of reaction for gelation

i) using statistical theory of floxy

(Pa)c = 0.816

(Pb)c = 0.816

ii) using Carothers theory

( Pc ) = 0.917

attached below is the detailed solution

B) To Obtain the measured value of critical extent of reaction ( 0.866) 1 mol of Glycerol  will react with 1 mol of dicarboxylic acid, but the same can not be applied to our obtained value because our stoichiometry is different

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Three persons wants to push a wheel cart in the direction marked x in Fig. The two person push with horizontal forces F1 and F2
Svetllana [295]

Answer:

<u>I had to search the Figure on Google to solve this question.</u>

a) The magnitude of the force F₃ is:

F_{3} = 87.47 N

And the direction of F₃:

\alpha = 79.04 ^{\circ}  (with respect to the y-direction, in the third quadrant)

b) P = 4.22 N  

Explanation:

<u>I had to search the Figure on Google to solve this question.</u>

a) We can find the force of the third person as follows:

\Sigma F_{x} = F_{1x} + F_{2x} + F_{3x} = 0

\Sigma F_{y} = F_{1y} - F_{2y} + F_{3y} = 0

So, in x-direction we have:

\Sigma F_{x} = 45 N*cos(70) + 75 N*cos(20) + F_{3x} = 0

F_{3x} = -85.87 N

In y-direction we have:

\Sigma F_{y} = 45 N*sin(70) - 75 N*sin(20) + F_{3y} = 0

F_{3y} = -16.63 N

The magnitude of the force F₃ is:

F_{3} = \sqrt{F_{3x}^{2} + F_{3y}^{2}} = \sqrt{(-85.87 N)^{2} + (-16.63 N)^{2}} = 87.47 N

To find the direction of F₃ we need to calculate its angle with respect to the y-direction (in the third quadrant):

tan(\alpha) = \frac{|F_{3x}|}{|F_{3y}|} = \frac{85.87 N}{16.63 N}

\alpha = 79.04 ^{\circ}

<em>b) If the third person exerts the force found in part (a) the car will stop, so the only way for the cart to accelerate at 200 m/s² is that the third person does not exert the force found in a. </em>      

<u>To find the weight of the cart​ when it accelerates at 200 m/s², we need to consider: F₃ = 0</u>.  

First, we need to find the cart's mass. Since the car is moving in the x-direction we have:

\Sigma F_{x} = F_{1x} + F_{2x} = ma

45 N*cos(70) + 75 N*cos(20) = m*200 m/s^{2}

m = \frac{45 N*cos(70) + 75 N*cos(20)}{200 m/s^{2}} = 0.43 kg

Now, the weight of the cart​ is:

P = mg = 0.43 kg*9.81 m/s^{2} = 4.22 N

I hope it helps you!                                                                                    

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