27 I think beacuse if you subtract
The bond angles a and b are 120° respectively. The bond angle c is 111.4° .while the bond angle d is 120°. The bond angles e and f are 120° respectively.
In the carbonate ion, all the bond angles and bond lengths are equal hence three equivalent resonance structures can be drawn for the ion. All the bond angles, ( a and b) in carbonate ion all have bond angle of 120°.
The bond angle marked c in OCCl2 has a bond angle 111.4°, the bond angle marked d in the compound has the bond angle, 120°.
There are three bond angles present in the nitrate (NO3-) ion. Three resonance structures contribute to this bond. Based on these structures, the bond angles e and f in the molecule is 120°.
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Answer:
o.251 prduces 45.7L of oxogen
Explanation:
hope this helps
The question is incomplete, here is the complete question:
At elevated temperature, nitrogen dioxide decomposes to nitrogen oxide and oxygen gas
The reaction is second order for 
with a rate constant of 
at 300°C. If the initial [NO₂] is 0.260 M, it will take ________ s for the concentration to drop to 0.150 M
a) 1.01 b) 5.19 c) 0.299 d) 0.0880 e) 3.34
<u>Answer:</u> The time taken is 5.19 seconds
<u>Explanation:</u>
The integrated rate law equation for second order reaction follows:
![k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B1%7D%7Bt%7D%5Cleft%20%28%5Cfrac%7B1%7D%7B%5BA%5D%7D-%5Cfrac%7B1%7D%7B%5BA%5D_o%7D%5Cright%29)
where,
k = rate constant =
t = time taken = ?
[A] = concentration of substance after time 't' = 0.150 M
![[A]_o](https://tex.z-dn.net/?f=%5BA%5D_o)
= Initial concentration = 0.260 M
Putting values in above equation, we get:
Hence, the time taken is 5.19 seconds
Answer:
C3H8 +5O2 arrow 3CO2 +4H2O
