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Which of the following is adsorbed in maximum amount by activated charcoal?a. $mathrm{N}_{2}$b. $mathrm{CO}$c. $mathrm{HCl}$d. $mathrm{CO}_{2}$...

Question

Which of the following is adsorbed in maximum amount by activated charcoal?a. $mathrm{N}_{2}$b. $mathrm{CO}$c. $mathrm{HCl}$d. $mathrm{CO}_{2}$

Which of the following is adsorbed in maximum amount by activated charcoal? a. $mathrm{N}_{2}$ b. $mathrm{CO}$ c. $mathrm{HCl}$ d. $mathrm{CO}_{2}$



Answers

Give the oxidation state for carbon in:
a.$\mathrm {co}$ $\quad$ b. $\mathrm{co}_{2} \quad$ c. $\mathrm{c}_{3} \mathrm{O}_{2}$

So this question is about Graham's law of diffusion and were asked in two in two pairs of gases. Which of them is going to refuse faster and also are asked how much faster eso the first step that the principle behind it is that the smaller molecule is going to refuse faster, which everyone has a lower mass is going to if use faster than the one that has a higher mass, and we'll look at the equation in a moment. But the first thing I'd like to dio is come up with the molar masses for each of these. So if we look at our periodic table, the molar mass for Krypton is 84 the molar mass for oxygen, a single oxygen atom is 16. But of course this is 02 so it's gonna be 32 moving down to the second part for nitrogen. A single nitrogen is 14 but this is N two, so it's gonna be 28 and a settling here, So carbon is 12 but there are two of them's that's 24 and hydrogen is one. But there two of them. So that's to say 24 plus two is 26. So in both pairs, we are going to I want the smaller of the two. So between Krypton and oxygen, oxygen is considerably smaller than the Krypton, considerably lighter. So we know that that one is going to be the faster one. And then between the nitrogen and the A, settling the A settling is slightly smaller, so it's gonna be slightly faster. Now. Let's look at the equation. So Grams Lava Fusion says that the ratio of the rates of effusion is equal to the square root of the ratio of the masses, and you'll notice that we inverted it. So we have our one over R to believe them to over him one And again. That's because as we have smaller particles as mass goes down, the rate goes up. So we need to swap those because it's an inverse relationship in inverse square relationship here. And so now if we go ahead and we do the math, we're trying to figure out how much faster the oxygen is compared to the Krypton. We comm plug in our masses there, so we have are larger, one on top and we take the square root. So 84 Divide by 32 is about 2.6, and if we take the square root of that, we get 1.62 So the oxygen is about 1.62 times faster, then the Krypton is. We do the same thing for the other one. We have 28 over 26 is about 1.7 on. And if we take the square root of that, we get 1.377 So let's call it one point four. So the settling then is gonna be 1.4 times faster than the nitrogen at defusing. And that's how you use Graham's law of diffusion.

Ls for this question that we are asked to find solid ability of a certain substance. I recommend falling along this video with the rules for soluble ity near you. So that way you are able to compare answers. Our first compound is called zinc sulfide. No, most sulfides, except for the ones with the exceptions, um are insoluble. Therefore, we can determine that zinc sulfide is insoluble in the water. Next, we had a compound gold carbonate a you to C 03 Now this is ah in. So this is insoluble in water because most, except for the exceptions, most things with carbonate are insoluble. So we also have an insoluble compound here. Next we have P b. See how to no this This is also insoluble because chlorine is normally the chlorine ion is normally soluble but ah, lead is its exception. So it's also insoluble. Finally, we have, um ano too. No, no. M N 02 is an intermediate, so it isn't a salt, it's assault. It's a collide. All ended acts as an intermediate, so it's partially soluble from colloidal form to colloidal form. Therefore, we say it's partially

So now let's look at another Valence electron counting problem in a compound. So we need to remind ourselves that the Valence Electron number of an atom is the same as its group number. So this is the rule that we're going to be working with here and now. I would like to show in an example. So we've got ch three seat out, and that is Tector. Huge off shape. And so let's address all of the group numbers that we're working with hair. So Carbon is in group for hydrogen is in Group one. Glory in is in group seven. And so we have one carpets. That's fine. We can leave its full have we've got three hydrogen. So we need to multiply this by three. And then we've got one chlorine, so we can leave that seven. And so simply all we need to do is add those up and we're left with 14 valence electrons for this compound structure.

In order for the bond order to increase with the loss of two electrons. The two electrons that air lost must have been in anti bonding molecular orbital's. So if we look at A, which is B two, and look at the molecular orbital diagram for B two in figure 5 50 we will see that the two highest energy electrons are in bonding Molecular Orbital's. So the removal of these will not increase bond order. It will decrease bond order. Same thing with C two. They're both the highest energy. Electrons are in pie bonding molecular orbital's. We then go to en to and we see that its highest energy electrons are in Sigma two p bonding Molecular orbital's the only one that is going to result in an increase in bond order by removal of electrons ISO to going toe to two plus. Because the two electrons that air removed from 02 to form two plus comes from pie two p anti bonding molecular orbital's. So with the removal of those electrons, this bond order of 022 plus is greater than 02


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