L = an integer from 0 to n-1
<span>mL = an integer from -L to +L </span>
<span>ms = + or - 1/2 </span>
<span>the format is (n, L, mL, ms) </span>
<span>so your </span>
<span>n = 3... this is OK </span>
<span>L = 2.. also OK.. if n=3, L can = 0, 1, or 2 </span>
<span>mL = -3 is NOT ok... if L=2.. mL can only be -2, -1, 0, +1, or +2 </span>
<span>ms = +1/2 and that is ok
Hope this helps!</span>
Answer:
The answer to your question is: Flask X
Explanation:
Data
Flask X Flask B
Molar mass 30 g 60 g
mass 1.2g 1.2 g
Pressure 1 atm 0.5 atm
Formula PV = nRT
In the formula, we can notice that the number of moles (n)
is directly proportional to the pressure.
Then, let's calculate the number of moles
flask X flask Y
30 g --------------- 1 mol 60 g -------------- 1 mol
1.2 g ---------------- x 1.2 g ------------- x
x = (1.2 x 1) / 30 x = (1.2 x 1) / 60
x = 0.04 mol x = 0.02 mol
From the results, we conclude that the flask with the gas of molar mass 30g is the flask with pressure of 1 atm, because the higher the number of moles, the higher the pressure.
Answer:
The equilibrium constant for CO now
= 0.212 M
For H₂O
= 0.212 M
For CO₂ = x = 0.2880 M
For H₂ = x = 0.2880 M
Explanation:
The chemical equation for the reaction is:
CO(g) + H2O(g) ⇌ CO2(g) + H2(g)
The ICE Table for this reaction can be represented as follows:
CO(g) + H2O(g) ⇌ CO2(g) + H2(g)
Initial 0.5 0.5 - -
Change -x -x + x + x
Equilibrium 0.5 -x 0.5 - x
The equilibrium constant![K_c = \dfrac{[x][x]}{[0.5-x][0.5-x]}](https://tex.z-dn.net/?f=K_c%20%3D%20%5Cdfrac%7B%5Bx%5D%5Bx%5D%7D%7B%5B0.5-x%5D%5B0.5-x%5D%7D)
![K_c = \dfrac{[x]^2}{[0.5-x]^2}](https://tex.z-dn.net/?f=K_c%20%3D%20%5Cdfrac%7B%5Bx%5D%5E2%7D%7B%5B0.5-x%5D%5E2%7D)
where; 



1.3583 (0.5-x) = x
0.67915 - 1.3583x = x
0.67915 = x + 1.3583x
0.67915 = 2.3583x
x = 0.67915/2.3583
x = 0.2880
The equilibrium constant for CO now = 0.5 - x
= 0.5 - 0.2880
= 0.212 M
For H₂O = 0.5 - x
= 0.5 - 0.2880
= 0.212 M
For CO₂ = x = 0.2880 M
For H₂ = x = 0.2880 M
Answer:
Y is a 3-chloro-3-methylpentane.
The structure is shown in the figure attached.
Explanation:
The radical chlorination of 3-methylpentane can lead to a tertiary substituted carbon (Y) and to a secondary one (X).
The E2 elimination mechanism, as shown in the figure, will happen with a simulyaneous attack from the base and elimination of the chlorine. This means that primary and secondary substracts undergo the E2 mechanism faster than tertiary substracts.