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??
I know for number 4 the answer is c, sorry I can't help with the others.
Answer:
Explanation:
You need to remember that the oxidation number of H is +1, except when it is in a metal hydrites like NaH, where its oxidation number is -1. Then, the oxidation number of O is -2, but in peroxides is -1. So with these rules you just have to multiply the ox. number with the name of atoms and all the elements in the reaction must sum 0.
moles NaOH = c · V = 0.1973 mmol/mL · 29.43 mL = 5.806539 mmol
moles H2SO4 = 5.806539 mmol NaOH · 1 mmol H2SO4 / 2 mmol NaOH = 2.9032695 mmol
Hence
[H2SO4]= n/V = 2.9032695 mmol / 32.42 mL = 0.08955 M
The answer to this question is [H2SO4] = 0.08955 M