1) Reaction
<span>NH4Cl(s) ---> NH3(g) + HCl(g)
2) equilibrium equation, Kc
Kc = [NH3] * [HCl]
3) Table of equilibrium formation
step concentrations
</span>
<span> NH4Cl(s) NH3(g) HCl(g)
start 1.000 mole 0 0
react - x
produce +x + x
------------------ ---------- -----------
end 1 - x +x +x
1 - x = 0.3 => x = 1 - 0.3 = 0.7
[NH3] = [HCl] = 0.7/0.5 liter = 1.4 (I used 0.500 dm^3 = 0.5 liter)
4) Equilibrium equation:
Kc = [NH3] [HCl] = (1.4)^2 = 1.96
Which is the number that you were looking for.
Answer: Kc = 1.96
</span>
there's no picture here but I guess the answer would be:
considering the constant temperature, if you double the volume, the pressure would be halved.
like: volume is 2, pressure is 4
if 2×2, then:4÷2
Answer : The cell potential for this reaction is 0.50 V
Explanation :
The given cell reactions is:
The half-cell reactions are:
Oxidation half reaction (anode): 
Reduction half reaction (cathode): 
First we have to calculate the cell potential for this reaction.
Using Nernest equation :
![E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Zn^{2+}]}{[Pb^{2+}]}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3DE%5Eo_%7Bcell%7D-%5Cfrac%7B2.303RT%7D%7BnF%7D%5Clog%20%5Cfrac%7B%5BZn%5E%7B2%2B%7D%5D%7D%7B%5BPb%5E%7B2%2B%7D%5D%7D)
where,
F = Faraday constant = 96500 C
R = gas constant = 8.314 J/mol.K
T = room temperature = 
n = number of electrons in oxidation-reduction reaction = 2
= standard electrode potential of the cell = +0.63 V
= cell potential for the reaction = ?
![[Zn^{2+}]](https://tex.z-dn.net/?f=%5BZn%5E%7B2%2B%7D%5D)
= 3.5 M
![[Pb^{2+}]](https://tex.z-dn.net/?f=%5BPb%5E%7B2%2B%7D%5D)
= 
Now put all the given values in the above equation, we get:
Therefore, the cell potential for this reaction is 0.50 V
Ok thanks for the valuble info.
Answer:
18 oxygen atoms
Explanation:
in order to from the 6 molecules carbon dioxide and 6 molecules of water you will have a total of 18 oxygen atoms
