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P4O10H N heatHsOtNHHeat1. KOH; Hzo2. HtCEN...

Question

P4O10H N heatHsOtNHHeat1. KOH; Hzo2. HtCEN

P4O10 H N heat HsOt NH Heat 1. KOH; Hzo 2. Ht CEN



Answers

(a) heat $\mathrm{H}_{2} / \mathrm{Pd}$ (b) (i) $\mathrm{HCl}$ (c) (i) $\mathrm{SOCl}_{2} / \mathrm{Et}_{3} \mathrm{~N}$ (d) (i) $\mathrm{Br}_{2} / \mathrm{hv}$ (ii) KOH/EtOH/heat (ii) $\mathrm{KOH} / \mathrm{EtOH} /$ heat (ii) $\mathrm{KOH} / \mathrm{EtOH} /$ heat

Here we are balancing are some chemical equations. So this is where we have the same number and type of atoms on the left hand side, as we do on the right hand side. So fastly we have two and a at two That yields NH 202. Following on from this, we've got P four add 5 to. That's an equilibrium with P 4010 as our products. Following on from this, you've got CAHC- 032 at N A. To C 03 sodium carbonate. That's an equilibrium with CACO. three, calcium carbonate, had to lots of N A H C 03. And so an extra we have is following well and H three at 5- five moles of oxygen. That's an equilibrium with four moles of nitrogen monoxide at six H 20. And so our final example is to an A two H 20. That's an equilibrium of two, lots of N H O H two moles of sodium hydroxide and H two

So this problem tells us we're burning ten grams of phosphorus p for ah in oh, two to form P for Oh, ten solid. And so we're gonna call this p four since we have a P for over here. So the equation is balanced, and the first thing we can do is find the heat. Since we're told that this is enough heat Ah, for a certain mass of water to change a certain temperature, we confined the amount of heat that is. And so we have our massive water. Twenty nine fifty grams times are specific heat of water four point one eight four Jules per gram degrees Celsius. And then our delta T is thirty eight point zero degree Celcius minus eighteen point zero degrees Celsius. And so that gives us a heat of two, four, six, eight, five, six Jules or two four, six point eight five, six. Kill it, Jules. And now Ah, we know that this is an extra thermic reaction, so we know our heat is going to be a negative value, and that is just a sign convention. And so now that we have our heat and kill a Jules, we confined our amount of moles in order to determine our final answer. And so we have ten grams of P for we want to divide that by the Moler massive p for one twenty three point nine zero grams per mole. And since we want P for oh, ten, we just take one mall of P four is one mall of P for oh ten, since that is our one to one ratio. And then we get our final answer in moles of P for oh ten, which is equal to zero point zero eight zero seven moles of P for Oh ten. And so now we just take our negative two forty six x point eight five six Killa jewels and divided by our zero point zero eight zero seven moles on We get our final answer to be negative three thousand zero fifty eight point five. Kill a Jules Permal, and that is our final answer.

In this problem, the reaction can be done something like this. Just look at it carefully. So I'm dish drying the Reactant 1st. Here it is. Beyond it is in pigeons. Oh, K O H he H plus S two will give the product ID this component, which I am at a thing here just located carefully. He heritage, Ohh Group. And in this side it is CH two CS three group according to the option. In this problem of some age, correct. They eat good day.

Ammonium by government only hitting the Great, I guess the same gas will be opened by running back room where the Liberator, nitrogen gas and still gas will be delivered on the Hitting off and it's four and and as for one or two when he did, then it will get into. Plus, that's true, and ammonium diagram it also gives, and to the correct option will be first. Okay, Thank you. Hope you will understand my explanation.


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