Answer:
Explanation:
The acid level has changed
Let's eliminate these one by one.
The first pair would not be the same, as X would most likely be in group IA, and Y would be in group VIIA, because of their tendency to gain and lose electrons.
The second pair would also violate the same rule, but X would most likely be in group IIA, and Y would most likely be in group VIA.
The third pair would not be the same, as X is most likely in group VIIA, and since Y has eight valence electrons, it is most likely a noble gas.
The final pair has X with atomic number 15, making it phosphorous. Phosphorous wants to gain 3 electrons to have a full octet of 8 outer "valence" electrons, and Y would also like to gain 3 electrons. This means it is possible that the final pair would be in the same group.
4.2 hours, do 315 miles / 75 miles per hour
According to this formula :
㏑[A] /[Ao] = - Kt
when we have Ao = 0.3 m
and K =0.46 s^-1
t = 20min = 0.2 x 60 =12 s
So by substitution :
㏑[A] / 0.3 = - 0.46 * 12
㏑[A] / 0.3 = - 5.52
by taking e^x for both side of the equation we can get [A]
∴[A] = 0.0012 mol dm^-3
Answer:
The equilibrium concentrations are:
[SO2]=[NO2] = 0.563 M
[SO3]=[NO] = 1.04 M
Explanation:
<u>Given:</u>
Equilibrium constant K = 3.39
[SO2] = [NO2] = [SO3] = [NO] = 0.800 M
<u>To determine:</u>
The equilibrium concentrations of the above gases
Calculation:
Set-up an ICE table for the given reaction
I 0.800 0.800 0.800 0.800
C -x -x +x +x
E (0.800-x) (0.800-x) (0.800+x) (0.800+x)
The equilibrium constant is given as:
![Keq = \frac{[SO3][NO]}{[SO2][NO2]}=\frac{(0.800+x)^{2}}{(0.800-x)^{2}}](https://tex.z-dn.net/?f=Keq%20%3D%20%5Cfrac%7B%5BSO3%5D%5BNO%5D%7D%7B%5BSO2%5D%5BNO2%5D%7D%3D%5Cfrac%7B%280.800%2Bx%29%5E%7B2%7D%7D%7B%280.800-x%29%5E%7B2%7D%7D)
x = 0.2368 M
[SO2]=[NO2] = 0.800 -x = 0.800 - 0.2368 = 0.5632 M
[SO3]=[NO] = 0.800 +x = 0.800 + 0.2368 = 1.037 M
