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Elina [12.6K]
2 years ago
7

It takes 330 joules of energy to raise the temperature of 24.6 gbenzene from 21 degrees Celsius to 28.7 degrees Celsius at const

antpressure. What is the molar hear capacity of benzene at constantpressure?
Chemistry
1 answer:
lilavasa [31]2 years ago
8 0

Given :

Energy , E = 330 J .

Initial temperature , T_i=21^oC .

Final temperature , T_f=24.6^oC .

Mass of benzene , m = 24.6 g .

To Find :

The molar hear capacity of benzene at constant pressure .

Solution :

Molecular mass of benzene , M = 78 g/mol .

Number of moles of benzene :

n=\dfrac{24.6}{78} \ mol\\\\n=0.32 \ mol

Energy required is given by :

q=nC_p\Delta T\\\\330=0.32\times C_p\times (28.7-21)\\\\C_p=\dfrac{330}{0.32\times 7.7}\ J\ mol^{-1}^oC^{-1} \\\\C_p=133.9\ J\ mol^{-1}^oC^{-1}

Hence , this is the required solution .

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Answer:

Explination:

Given Data:

                  Trail 1          Trial 2         Trial 3

Student A 448.0 cm 485.6 cm 463.4 cm

Student B 450.5 cm 441.3 cm         446.8 cm

Student C 422.6 cm 445.2 cm 432.7 cm

Accepted Value = 435.0 cm

Required:

A: Accurate measurement =?

B: Reason for the answer =?

C: Precise measurement =?

D: Reason for the answer =?

Solution:

Student A:

Trail 1: 435.0 cm – 448.0 cm = (13.0 cm greater than accepted value)

Trail 2: 435.0 cm – 485.6 cm = (50.6 cm greater than accepted value)

Trial 3: 435.0 cm – 463.4 cm = (28.4 cm greater than accepted value)

Student B:

Trail 1: 435.0 cm – 450.5 cm = (14.5 cm greater than accepted value)

Trail 2: 435.0 cm – 441.3 cm = (6.3 cm greater than accepted value)

Trial 3: 435.0 cm – 446.8 cm = (11.8 cm greater than accepted value)

Student C:

Trial 1: 435.0 cm – 422.6 cm = (12.4 cm less than accepted value)

Trial 2: 435.0 cm – 445.2 cm = (10.2 cm greater than accepted value)

Trial 3: 435.0 cm – 432.7 cm = (1.3 cm less than accepted value)

A: The 3rd trial of the student C is accurate measurement = 432.7 cm.

B: The 3rd value of students’ C measurement is accurate because it is quite near to the accepted value, i.e.  435.0 cm.

As we know that Accuracy refers to the closeness of a measured value to a standard or known value.

This value has only the difference of 1.3 cm.

435.0 cm – 432.7 cm = (1.3 cm less than accepted value)

All the other have larger difference which is as above.

___________________

Student A:

1. 448.0 cm – 485.6 cm = (37.6 cm far)

2. 485.6 cm – 463.4cm = (22.2 cm far)

Student B:

1. 450.5 cm – 441.3 cm = (9.2 cm far)

2. 441.3 cm – 446.8 cm = (5.5 cm far)

Student C:

1. 422.6 cm – 445.2 cm = (22.6 cm far)

2. 445.2 cm – 432.7cm = (12.5 cm far)

So,

C: The values of Student B are more precise.

D: As we know that Precision refers to the closeness of two or more measurements to each other.

The measurements of student C are more close to each other. The values are only 9.2 cm and 5.5 cm far from each other.

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Read 2 more answers
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