The fomula is NH4 (1+)
There are only two elements N and H.
As per oxidation state rules, the most electronegative element will have a negative oxidation state and the other element will have a positive oxidation state.
N is more electronative than H, so H will have a positive oxidation state and nitrogen will have a negative oxidation state.
You can also use the rule that states the hydrogen mostly has 1+ oxidation state,except when it is bonded to metals.
In conclusion the oxidation state of H in NH4 (1+) is 1+.
Now you must know that the sum of the oxidations states equals the charge of the ion, which in this case is 1+.
That implies that 4* (1+) + x = 1+
=> x = (1+) - 4(+) = 3-
Answer: the oxidation state of N is 3-, that is the option b.
Answer:
Kc = [H₂S]² . [CH₄] / [ H₂O]⁴ . [CS₂]
Explanation:
The equilibrium constant indicates the % of the yield reaction and can shows where the reaction is going to be equilibrated.
It works with molar concentrations on the equilibrium and it does not consider the solids compounds
Kc also can be modified by the time of the reaction.
This reaction is:
CS₂ (g) + 4 H₂O(g) ⇌ CH₄ (g) + 2H₂S (g)
Kc = [H₂S]² . [CH₄] / [ H₂O]⁴ . [CS₂]
Answer : The activation energy of the reaction is, 
Solution :
The relation between the rate constant the activation energy is,
![\log \frac{K_2}{K_1}=\frac{Ea}{2.303\times R}\times [\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Clog%20%5Cfrac%7BK_2%7D%7BK_1%7D%3D%5Cfrac%7BEa%7D%7B2.303%5Ctimes%20R%7D%5Ctimes%20%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= initial rate constant = 
= final rate constant = 
= initial temperature = 
= final temperature = 
R = gas constant = 8.314 kJ/moleK
Ea = activation energy
Now put all the given values in the above formula, we get the activation energy.
![\log \frac{8.75\times 10^{-3}L/mole\text{ s}}{4.55\times 10^{-5}L/mole\text{ s}}=\frac{Ea}{2.303\times (8.314kJ/moleK)}\times [\frac{1}{468K}-\frac{1}{531K}]](https://tex.z-dn.net/?f=%5Clog%20%5Cfrac%7B8.75%5Ctimes%2010%5E%7B-3%7DL%2Fmole%5Ctext%7B%20s%7D%7D%7B4.55%5Ctimes%2010%5E%7B-5%7DL%2Fmole%5Ctext%7B%20s%7D%7D%3D%5Cfrac%7BEa%7D%7B2.303%5Ctimes%20%288.314kJ%2FmoleK%29%7D%5Ctimes%20%5B%5Cfrac%7B1%7D%7B468K%7D-%5Cfrac%7B1%7D%7B531K%7D%5D)
Therefore, the activation energy of the reaction is,
Answer:
The correct formula for tetraphosphorous octaoxide is P₂O₅
