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IetHnalRnnl alnenlanaealru naiye) WlDacaltoMerdHp?...

Question

IetHnalRnnl alnenlanaealru naiye) WlDacaltoMerdHp?

i et HnalRnnl alnenlanae alru naiye) WlDacalto MerdHp?



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The sccond ionizarion cncrgics of the $\mathrm{C}, \mathrm{N}, \mathrm{O}$ and $\mathrm{F}$ aroms are such rhar (a) $\mathrm{C}>>\mathrm{N}>\mathrm{F}>\mathrm{C}$ (b) $\mathrm{F}>\mathrm{O}>\mathrm{N}>\mathrm{C}$ (c) $C>\mathrm{O}>\mathrm{N}>\mathrm{T}$ (d) $\mathrm{O}>\mathrm{F}>\mathrm{N}>\mathrm{C}$

So for this problem because of the fact that we know that, uh, Jay, why is congruent to J X? We're going to be able to fill that out right away. Um, as one of her options and then by the triangle mid segment here. We know that, um, since l is going to be the midpoint of side y Z in K is the midpoint of side XY. That means that the segment between L and K is going to be parallel, do you side x y, which means that KL is exactly half of side ex wife. And so since 1/2 of x Y is going to be, uh, either X J or J Y, that means that KL is also going to be king. Grew TJ Why m. J.

In this problem, I can write the reaction something like this. Just look at it carefully. Here it is. C. O. Be it. Here it is B. A. This in presence of that and it to all As well as CH two as village as three oh positive will give the product ID this component to give the productive this component here it is C. O. At Heritage. Okay, so according to the optioning this problem, option C H. Obstinacy each but it then said for this problem.

As we all know that L three plus and BI three plus and Creasy tweeted as their IDOC side as they are, I drew oxides, L o H all three and B i o edge all three with ammonium with ammonium I dockside in the presence of in the questions of ammonium chloride. So according to the option, option B each correct.

Hello there. Okay so here we have this venture this in in C. Three and we need to find the country, get the real part imaginary part and the norm of these. Very so the country, it is really easy. We only need to change the sign of those that are that have the I the imaginary part. That's the the only thing that we need to do when we talk about the country gate. We need to change the sign of the imaginary court. So we did the following factor too. Plus I Lux minus here buying this four. I I'm here -1 1 -9 So as you can see we only change the sign of the imaginary. Then we need to get the real part Basically is all the numbers that are not together with the iPod. So 201 imaginary port instead is those that have the I. So minus one 41 and finally the normal to the norm in the complex space is fine as the other problem of you with you in the complex the square root of the product of the vector with itself. But this is the square root of you don't product you congregate, and this result into the square root of four plus one plus four. I was 16 here and last one plus. Okay, so the result of this is the square root of 23.


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