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](https://tex.z-dn.net/?f=C_p%3DC_v%2BR)
We can determine
from above if we make
the subject of the formula as:
![C_v](https://tex.z-dn.net/?f=C_v)
![=C_p-R](https://tex.z-dn.net/?f=%3DC_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](https://tex.z-dn.net/?f=C_V%20%3D%2029.42-%282.7%2A10%5E%7B-3%7D%29T%2B%285.82%2A10%5E%7B-7%7D%29T2-%281.305%2A10%5E%7B-8%7D%29T3-%288.23%2A10%5E%7B-12%7D%29T4-8.314)
![C_V = 21.106-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4](https://tex.z-dn.net/?f=C_V%20%3D%2021.106-%282.7%2A10%5E%7B-3%7D%29T%2B%285.82%2A10%5E%7B-7%7D%29T2-%281.305%2A10%5E%7B-8%7D%29T3-%288.23%2A10%5E%7B-12%7D%29T4)
A).
The formula for calculating change in internal energy is given as:
![dU=C_vdT](https://tex.z-dn.net/?f=dU%3DC_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](https://tex.z-dn.net/?f=U2-U1%3D%5Cint%5Climits%5E%7B500%7D_%7B450%7D%2821.106-%282.7%2A10%5E%7B-3%7D%29T%2B%285.82%2A10%5E%7B-7%7D%29T2-%281.305%2A10%5E%7B-8%7D%29T3-%288.23%2A10%5E%7B-12%7D%29T4%5C%2C%29%20du)
![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\,)](https://tex.z-dn.net/?f=U2-U1%3D%5Cint%5Climits%5E%7B500%7D_%7B450%7D%2821.106-%282.7%2A10%5E%7B-3%7D%29T%2F1%2B%285.82%2A10%5E%7B-7%7D%29T2%2F2-%281.305%2A10%5E%7B-8%7D%29T3%2F3-%288.23%2A10%5E%7B-12%7D%29T4%2F4%5C%2C%29)
![U2-U1= 1059J/mol](https://tex.z-dn.net/?f=U2-U1%3D%201059J%2Fmol)
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\,)](https://tex.z-dn.net/?f=U2-U1%3D%5Cint%5Climits%5E%7B500%7D_%7B450%7D%2821.106-%282.7%2A10%5E%7B-3%7D%29T%2F1%2B%285.82%2A10%5E%7B-7%7D%29T2%2F2-%281.305%2A10%5E%7B-8%7D%29T3%2F3-%288.23%2A10%5E%7B-12%7D%29T4%2F4%5C%2C%29)
![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\,)](https://tex.z-dn.net/?f=U2-U1%3D%5Cint%5Climits%5E%7B500%7D_%7B450%7D%2821.106-%282.7%2A10%5E%7B-3%7D%29273%2F1%2B%285.82%2A10%5E%7B-7%7D%291073%2F2-%281.305%2A10%5E%7B-8%7D%29T3%2F3-%288.23%2A10%5E%7B-12%7D%29T4%2F4%5C%2C%29)
![U2-U1= 17,920 J/mol](https://tex.z-dn.net/?f=U2-U1%3D%2017%2C920%20J%2Fmol)
Answer:
<h3>The answer is 3.66 mL</h3>
Explanation:
The volume of a substance when given the density and mass can be found by using the formula
![volume = \frac{mass}{density} \\](https://tex.z-dn.net/?f=volume%20%3D%20%20%5Cfrac%7Bmass%7D%7Bdensity%7D%20%5C%5C)
From the question
mass = 38.4 g
density = 10.5 g/cm³
We have
![volume = \frac{38.4}{10.5} \\ = 3.6571428...](https://tex.z-dn.net/?f=volume%20%3D%20%20%5Cfrac%7B38.4%7D%7B10.5%7D%20%20%5C%5C%20%20%3D%203.6571428...)
We have the final answer as
<h3>3.66 mL</h3>
Hope this helps you
Answer:
Molecular composition of iodine heptafluoride is one iodine atom and seven fluorine atoms.
Explanation:
The molecular formula of iodine heptafluoride is
and it is also known as Iodine (VII)fluoride.
According to the VSEPR theory,The geometry of the iodine heptafluoride is "petagonal bipyrmidal."
From the molecular formula of iodine heptafluoride has seven atoms of fluorine and only one atom of iodine.
Therefore, the molecular composition of iodine heptafluoride is one iodine atom and seven fluorine atoms.
Determining the mass of a compound or any substance from its number of moles, you would need data on its molar mass since it allows the conversion from moles to grams and grams to moles. For aluminum, it has an atomic mass of 26.98 g/mol.
3.57 mol Al ( 26.98 g / mol ) = 96.32 g Al