poof

y=(1+2 ln x/x^2)+1/2; y'=(x^2-2x^2y+2)/x^3; y(1)=3/2

what am I doing wrong that I am getting the extra x^2 on the right hand side?

thanks!

I am so confused about how to figure this out but I believe the correct answer is A.

I am certain that A and C do not apply but unsure about B? Thanks!

The correct answer is step 3 but if N2O2 and N2O are both intermediates and step 3 is the slow step, wouldn't the rate law be rate=k[H2]^2, (since rate is determined by the slow step and the previous step) which would only be 2nd order overall not 4th? I am so confused !

I have been following all of my examples from class as much as possible but we have never done an example quite like this and I am so confused! thank you for your help in advance :)

I have been following all of my examples from class as much as possible but we have never done an example quite like this and I am so confused! thank you for your help!

Please let me know if what I have written is not correct! I kind of just taught this to myself and want to make sure I understand it correctly! thank you!

So I understand that adding CO2 will shift it towards reagent and adding CaCO3 will shift it to product, but how do you determine the shift caused by either lowering or increasing the volume if both sides of the equation are equivalent in volume? As in both sides have 3 moles. if one side had less volume, then decreasing the volume would shift it towards that side but how do you determine that for this example? thank you!

I know exothermic means the products have less energy than the reactants, hence the reaction coordinate I drew, but I am struggling to determine the rate law because I am unsure how to figure out the order of reactions for both the reactants when only given one step from the reaction?