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.
Answer:
False
Explanation:
On the left side of the equation (Li + O2), there is 1 Li atom and 2 O atoms.
but on thw right side of the equation (Li2O,) there is 2 Li atoms and 1 O atom
I am not sure about the first question but the temperature has an important role in this situation because as the temp goes up particles moves at a faster speed and spread out every where.
Answer:
The hydrogen ion concentration in a solution, [H+], in mol L-1, can be calculated if the pH of the solution is known.
pH is defined as the negative logarithm (to base 10) of the hydrogen ion concentration in mol L-1 pH = -log10[H+] ...
[H+] in mol L-1 can be calculated using the equation (formula): [H+] = 10-pH
<u>Answer:</u> The equilibrium concentration of water is 0.597 M
<u>Explanation:</u>
Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as 
For a general chemical reaction:
The expression for 
is written as:
![K_{c}=\frac{[C]^c[D]^d}{[A]^a[B]^b}](https://tex.z-dn.net/?f=K_%7Bc%7D%3D%5Cfrac%7B%5BC%5D%5Ec%5BD%5D%5Ed%7D%7B%5BA%5D%5Ea%5BB%5D%5Eb%7D)
The concentration of pure solids and pure liquids are taken as 1 in the expression.
For the given chemical reaction:
The expression of 
for above equation is:
![K_c=\frac{[H_2O]^2}{[H_2S]^2\times [O_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BH_2O%5D%5E2%7D%7B%5BH_2S%5D%5E2%5Ctimes%20%5BO_2%5D%7D)
We are given:
![[H_2S]_{eq}=0.671M](https://tex.z-dn.net/?f=%5BH_2S%5D_%7Beq%7D%3D0.671M)
![[O_2]_{eq}=0.587M](https://tex.z-dn.net/?f=%5BO_2%5D_%7Beq%7D%3D0.587M)
Putting values in above expression, we get:
![1.35=\frac{[H_2O]^2}{(0.671)^2\times 0.587}](https://tex.z-dn.net/?f=1.35%3D%5Cfrac%7B%5BH_2O%5D%5E2%7D%7B%280.671%29%5E2%5Ctimes%200.587%7D)
![[H_2O]=\sqrt{(1.35\times 0.671\times 0.671\times 0.587)}=0.597M](https://tex.z-dn.net/?f=%5BH_2O%5D%3D%5Csqrt%7B%281.35%5Ctimes%200.671%5Ctimes%200.671%5Ctimes%200.587%29%7D%3D0.597M)
Hence, the equilibrium concentration of water is 0.597 M
