Longer, this is because the H in HNO2 is bonded with an oxygen, no longer allowing this structure to have a resonance structure.
NO2 on the other hand has one double bond and one single bond, so it has a resonance structure. And resonance structures are actually one structure so there isn't really a single and double bond, it's actually a 1 and 1/2 bond that calls for a higher bond order.
And I higher bond order will result in a shorter lengths!
I hope this helps out!!! And just out of curiosity, is this off of an AP FRQ packet??
Answer:
6,8 g
Explanation:
c = 4.18 J/(g * °C) = 4180 J / (kg * °C)
= 25 °C
= 36,4 °C
Q = 325 J
The formula is: Q = c * m * (
)
m = 
Calculating:
m = 325 / 4180 * (36,4 - 25) ≈ 0,0068 kg = 6,8 g
To calculate the <span>δ h, we must balance first the reaction:
NO + 0.5O2 -----> NO2
Then we write all the reactions,
2O3 -----> 3O2 </span><span>δ h = -426 kj eq. (1)
O2 -----> 2O </span><span>δ h = 490 kj eq. (2)
NO + O3 -----> NO2 + O2 </span><span>δ h = -200 kj eq. (3)
We divide eq. (1) by 2, we get
</span>O3 -----> 1.5O2 δ h = -213 kj eq. (4)
Then, we subtract eq. (3) by eq. (4)
NO + O3 -----> NO2 + O2 δ h = -200 kj
- (O3 -----> 1.5 O2 δ h = -213 kj)
NO -----> NO2 - 0.5O2 δ h = 13 kj eq. (5)
eq. (2) divided by -2. (Note: Dividing or multiplying by negative number reverses the reaction)
O -----> 0.5O2 <span>δ h = -245 kj eq. (6)
</span>
Add eq. (6) to eq. (5), we get
NO -----> NO2 - 0.5O2 δ h = 13 kj
+ O -----> 0.5O2 δ h = -245 kj
NO + O ----> NO2 δ h = -232 kj
<em>ANSWER:</em> <em>NO + O ----> NO2 δ h = -232 kj</em>
Answer:
The temperature to the nearest 0.5°C is 98.5°C
