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:
Axis Labels
Explanation:
The axis labels are usually located on the x and y axis. This graph however is missing those.
hope this helps!
Answer:
it has 0g of sugar (hopes this helps)
Answer: Enthalpy of combustion (per mole) of 
is -2657.5 kJ
Explanation:
The chemical equation for the combustion of butane follows:
The equation for the enthalpy change of the above reaction is:
![\Delta H^o_{rxn}=[(8\times \Delta H^o_f_{CO_2(g)})+(10\times \Delta H^o_f_{H_2O(g)})]-[(1\times \Delta H^o_f_{C_4H_{10}(g)})+(4\times \Delta H^o_f_{O_2(g)})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%288%5Ctimes%20%5CDelta%20H%5Eo_f_%7BCO_2%28g%29%7D%29%2B%2810%5Ctimes%20%5CDelta%20H%5Eo_f_%7BH_2O%28g%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7BC_4H_%7B10%7D%28g%29%7D%29%2B%284%5Ctimes%20%5CDelta%20H%5Eo_f_%7BO_2%28g%29%7D%29%5D)
We are given:
Putting values in above equation, we get:
![\Delta H^o_{rxn}=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.6)+(4\times 0)]\\\\\Delta H^o_{rxn}=-5315kJ](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%288%5Ctimes%20-393.5%29%2B%2810%5Ctimes%20-241.82%29%5D-%5B%282%5Ctimes%20-125.6%29%2B%284%5Ctimes%200%29%5D%5C%5C%5C%5C%5CDelta%20H%5Eo_%7Brxn%7D%3D-5315kJ)
Enthalpy of combustion (per mole) of is -2657.5 kJ
Answer: 167 g
Explanation:
1) The depression of the freezing point of a solution is a colligative property ruled by this equation:
ΔTf = i × m × Kf
Where:
ΔTf is the decrease of the freezing point of the solvent due to the presence of the solute.
i is the Van't Hoof factor and is equal to the number of ions per each mole of solute. It is only valid for ionic compounds. Here the solute is not ionice, so you take i = 1
Kf is the molal freezing constant and is different for each solvent. For water it is 1.86 m/°C
2) Calculate the molality (m) of the solution
ΔTf = i × m × Kf ⇒ m = ΔTf / ( i × Kf) = 5.00°C / 1.86°C/m = 2.69 m
3) Calculate the number of moles from the molality definition
m = moles of solute / kg of solvent ⇒ moles of solute = m × kg of solvent
moles of solute = 2.69 m × 1.00 kg = 2.69 moles
4) Convert moles to grams using the molar mass
molar mass of C₂H₆O₂ = 62.07 g/mol
mass in grams = number of moles × molar mass = 2.69 moles × 62.07 g/mol = 166.97 g ≈ 167 g
