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=WC12 L(1 + (WCR)2) + CR2 _ 07 L+L_ = CRP 02C'...

Question

=WC12 L(1 + (WCR)2) + CR2 _ 07 L+L_ = CRP 02C'

=WC12 L(1 + (WCR)2) + CR2 _ 0 7 L+ L_ = CRP 02C'



Answers

The oxidarion srare of xenon arom in $\mathrm{X} \mathrm{c} \mathrm{F}_{s}$, $\mathrm{HX}_{\mathrm{c} \mathrm{O}}_{4}, \mathrm{Na}_{4} \mathrm{X} \mathrm{cO}_{6}$ are, rcspecrively (a) $+i,+6,+8$ (b) $+1,+6,+6$ (c) $+1,+6,+7$ (d) $+1,+5,+8$

We have given to have celebration uh in order for cl two plus two E two C L S uh 1.63 bolt. So here we consider this uh even uh E one In what? one is equal to 1.36 world for the reaction we have Yeah cl two in gaseous state. Bless two electrons. And to see el negative in egg was form. And next we have we have to calculate for you do in order to is equal to unknown. So the reaction foreigner to is cl negative. In this, he is prone to have cl two in gaseous form plus electron. So here we can use the equation by It do not equal to -11 not so by rearranging this equation, if one not is able to Even not is equal to -1.36 old. So the next half cell electrode electrode potential value for the and on external house, L is minus 1.36 fold. So here we have option B is the correct option minus 1.36 world

Let's complete the following nuclear equations. A atomic master means and changed, and this is going to be atomic number 51 which is the elements and Timoney in the B atomic masses to six 86 element atomic number 86 is read on and see the Atomic Mass is unchanged 36 element three sixes. Krypton, 37 again on Atomic Mass is unchanged, says on that 35 which

In this question, you're simply asked to complete the following nuclear reactions. To complete them, you need to make certain the some of the mass and atomic numbers are the same on both sides of the equation. If we have oxygen 16 and oxygen 16 going to something and an alpha particle than four Plus, Something Needs to Equal 32. That would be 28 and two plus something needs to equal 16. So that would be 14 Periodic table states that atomic number 14 is silicone For the next one. We have uranium 235 colliding with a neutron Creating s. r. 90 something and three neutrons. So we've got three plus something Plus 90, gives us 2 36. That something needs to be 1 43 and then zero plus 38 gives us 92. That something needs to be 54. And the periodic table tells us that 54 is en in And we've got carbon 13 plus an alpha particle Gives us oxygen 16 plus something. So something Plus 16 needs to be 17, so that'll be one and something plus eight gives us eight. So that will be zero. The only thing that has a massive one and no charges in neutron. And then we've got bismuth to 10 going to something and beta particle, so zero plus something gives us to 10. That'll be 2, 10 And then -1 plus something gives us 83. So that'll be 84 and 84 is polonium. And for the last one, we've got carbon 12 plus a proton goes to Gamma mission. So we've got 12 plus one is going to be 13 and six plus one is going to be seven. So that'll be nitrogen 13.

In this problem, Bandon we will reactivate cl CH two CH two c s t D in presence of LCL three to give the product AD this compound, which I am and I think he had plus this compound C s t d c o CST. Therefore, according to the option options, see it correct Option C it direct answer for this problem. I hope you understand the reaction of this problem.


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