Answer:
The ethanol has 21 vibrational modes.
Explanation:
A molecule can show 3 types of motions: one external called translational and two internal called rotational and vibrational.
In order to calculate the vibrational modes of a molecule we need to know the degrees of freedom of this molecule, it means the number of variables that are involved in the movement of this particle.
If we know that atoms are three dimensional we will know that they have 3 coordinates expressed as 3N. But the atoms are bonded together so they can move not only in translational but also rotational and vibrational. So, the rotational move can be described in 3 axes and the other vibrational move can be described as
3N-5 for linear molecules
3N-6 For nonlinear molecules like ethanol
So using the formula for nonlinear molecules where N is the amount of atoms in the chemical formula, so ethanol has 9 atoms
3(9)-6= 21
Thus, ethanol has 21 vibrational modes.
<u>Charge:</u>
An electron has a negative charge and a <em>n</em>eutron has a <em>n</em>eutral charge.
<u>Size:</u>
Electrons have a really small mass whereas the neutron has a mass of about 1 amu.
<u>Location:</u>
Neutrons are found in center of an atom, but electrons are around it.
<u>Number:</u>
The number of electrons and neutrons in atom varies.
Answer : The final temperature of the mixture is, ![22.14^oC](https://tex.z-dn.net/?f=22.14%5EoC)
Explanation :
First we have to calculate the mass of ethanol and water.
![\text{Mass of ethanol}=\text{Density of ethanol}\times \text{Volume of ethanol}=0.789g/mL\times 45.0mL=35.5g](https://tex.z-dn.net/?f=%5Ctext%7BMass%20of%20ethanol%7D%3D%5Ctext%7BDensity%20of%20ethanol%7D%5Ctimes%20%5Ctext%7BVolume%20of%20ethanol%7D%3D0.789g%2FmL%5Ctimes%2045.0mL%3D35.5g)
and,
![\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1.0g/mL\times 45.0mL=45.0g](https://tex.z-dn.net/?f=%5Ctext%7BMass%20of%20water%7D%3D%5Ctext%7BDensity%20of%20water%7D%5Ctimes%20%5Ctext%7BVolume%20of%20water%7D%3D1.0g%2FmL%5Ctimes%2045.0mL%3D45.0g)
Now we have to calculate the final temperature of the mixture.
In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.
![q_1=-q_2](https://tex.z-dn.net/?f=q_1%3D-q_2)
![m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)](https://tex.z-dn.net/?f=m_1%5Ctimes%20c_1%5Ctimes%20%28T_f-T_1%29%3D-m_2%5Ctimes%20c_2%5Ctimes%20%28T_f-T_2%29)
where,
= specific heat of ethanol = ![2.42J/g^oC](https://tex.z-dn.net/?f=2.42J%2Fg%5EoC)
= specific heat of water = ![4.18J/g^oC](https://tex.z-dn.net/?f=4.18J%2Fg%5EoC)
= mass of ethanol = 35.5 g
= mass of water = 45.0 g
= final temperature of mixture = ?
= initial temperature of ethanol = ![8.0^oC](https://tex.z-dn.net/?f=8.0%5EoC)
= initial temperature of water = ![28.6^oC](https://tex.z-dn.net/?f=28.6%5EoC)
Now put all the given values in the above formula, we get:
![35.5g\times 2.42J/g^oC\times (T_f-8.0)^oC=-45.0g\times 4.18J/g^oC\times (T_f-28.6)^oC](https://tex.z-dn.net/?f=35.5g%5Ctimes%202.42J%2Fg%5EoC%5Ctimes%20%28T_f-8.0%29%5EoC%3D-45.0g%5Ctimes%204.18J%2Fg%5EoC%5Ctimes%20%28T_f-28.6%29%5EoC)
![T_f=22.14^oC](https://tex.z-dn.net/?f=T_f%3D22.14%5EoC)
Therefore, the final temperature of the mixture is, ![22.14^oC](https://tex.z-dn.net/?f=22.14%5EoC)
Answer:
When a substance is heated ,the kinetic energy of its molecules also increase.
Explanation:
K.E is directly proportional to T
Capture all of the smoke and weight it. it will weigh exactly the same before and after you burn it but will just be CO2 and H2O gas.