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Over [174]
2 years ago
12

Why is the combined gas law the most practiced gas law

Chemistry
2 answers:
vfiekz [6]2 years ago
8 0
The combined gas law allows you to derive any of the relationships needed by combining all of the changeable peices in the ideal gas law: namely pressure, temperature and volume
rjkz [21]2 years ago
3 0

The combined gas law allows you to derive any of the relationships needed by combining all of the changeable peices in the ideal gas law: namely pressure, temperature and volume.
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How would you abbreviate you as an element?​
qaws [65]

Answer:

i am unique and have a specific number in the periodic table

5 0
3 years ago
The ideal gas heat capacity of nitrogen varies with temperature. It is given by:
hammer [34]

Answer:

A)  1059 J/mol

B)  17,920 J/mol

Explanation:

Given that:

Cp = 29.42 - (2.170*10^-3 ) T + (0.0582*10^-5 ) T2 + (1.305*10^-8 ) T3 – (0.823*10^-11) T4

R (constant) = 8.314

We know that:

C_p=C_v+R

We can determine C_v from above if we make C_v the subject of the formula as:

C_v=C_p-R

C_V = 29.42-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4-8.314

C_V = 21.106-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4

A).

The formula for calculating change in internal energy is given as:

dU=C_vdT

If we integrate above data into the equation; it implies that:

U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4\,) du

U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T/1+(5.82*10^{-7})T2/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)

U2-U1= 1059J/mol

Hence, the internal energy that must be added to nitrogen in order to increase its temperature from 450 to 500 K = 1059 J/mol.

B).

If we repeat part A for an initial temperature of 273 K and final temperature of 1073 K.

then T = 273 K & T2 = 1073 K

∴

U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T/1+(5.82*10^{-7})T2/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)

U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})273/1+(5.82*10^{-7})1073/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)

U2-U1= 17,920 J/mol

3 0
3 years ago
How do I do these problems
horsena [70]

you calculate all the letters together and it would be Zn2

8 0
3 years ago
A composition of reflections over parallel lines is the same as a __________.
8090 [49]

Answer : Option A) Translation


Explanation : A composition of reflections over parallel lines is the same as a <u>Translation.</u>


To identify if the composition of reflections over parallel lines are same as translation or not?


We can check using a picture of some shape in the plane. Place the picture on the right side of two vertical parallel. Now, we can see the reflected the shape over the nearest parallel line, then check the reflection over the other parallel line. We see that the shape winds up in the same orientation, like it was just shifted over to the right. Hence, it is translation.

6 0
3 years ago
Read 2 more answers
Duncan knows that it takes 36400 cal of energy to heat a pint of water from room temperature to boiling. However, Duncan has pre
antiseptic1488 [7]
However, Duncan has prepared ramen noodles so many times he does not need to measure the water carefully. If he happens to heat 0.850 ...
7 0
3 years ago
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