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3. How many carbon atoms are present in the molecule shown?A) 6 8 8) 10 D) 11...

Question

3. How many carbon atoms are present in the molecule shown?A) 6 8 8) 10 D) 11

3. How many carbon atoms are present in the molecule shown? A) 6 8 8) 10 D) 11



Answers

How many carbon atoms are there in one molecule of each of the following compounds? (a) $\mathrm{CH}_{4} ;$ (b) $\mathrm{C}_{3} \mathrm{H}_{8} ;$ (c) $\mathrm{C}_{6} \mathrm{H}_{6}$ (d) $\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}$

Let us work on the problem from organic chemistry. This problem is based on the formula of organic compounds. Here we are given the number of carbon atoms in Elkins, we have to tell the number of hydrogen atoms that could be present in these Elkins. So we know that l gains are represented by the general formula. C N H two n plus two. We're and is the number of carbon atoms present in the alkaline. Now, here we are, given in the first one. There are eight carbon atoms. If we apply the value of N as eight in this formula, we come to the value that C eight edge to multiply eight plus two. So the formula comes out as C eight h 18. So in the Al Caine containing eight carbon atoms, the number of hydrogen atoms would be 18. The next one is 16 carbon atoms. On using the value of N S 16 in the formula For L. Kane, we can write C 16 h 16 multiplied to plus two on solving. We get the formula as C 16 h 34. So the number of hydrogen atoms in the Al Caine containing 16 carbon atoms are 34 hydrogen atoms. The next Elkin has one carbon atom, if any is one. Then on substituting the value one we can right c one h to multiply one plus two. So we get the formula C H four. So the number of hydrogen atoms in the Al Cain having one cabinet, um, are four in the last, we have four carbon atoms. When N is equal to four, we can grade C four h four multiplied two plus two on solving. We get the formula at C four h then so in the al cane with four cabinet terms, the number of hydrogen atoms are then.

Compound. We know that in order for the carbon to be neutral and have a full octet, it must have four bombs. So, lister, with this carbon, it is at the end of a chain. It is a carbon. It is attached to one oxygen. So it only has one bond. Meaning it must have three other bonds to be neutral. So this carbon has three. Hydrogen is attached to it. Move over to this Carmen. It currently has one, two, three, four bonds so it can't have any more bonds. It has zero. Hydrogen is attached to it. This carbon has one, two, three, four bonds. So it also doesn't have any Hydrogen is attached to it. This one has one, two, three bonds. It needs one more. Therefore, there must be one hydrogen coming off of this carbon. This carbon has one, two, three bonds. It must have one more to have a full octet and to be neutral. So it has one hydrogen this carbon for the same. Aziz too also has one hydrogen. This carbon has one, two, three, four bronze. It doesn't need anymore. Therefore, there are not any implied hydrogen zone that carbon this one has one to bonds. It needs two more. Therefore, there are two implied hydrogen is attached to that carbon. This one has one, two, three, four bonds. So there are zero hydrants attached to that carbon. Avery has all four bronze. This one has one to bonds. Meaning there. Two more hydrogen Sze. Same with this Kurban. This carbon in this carbon each only has two bonds. Therefore, it must have two more implied hydrogen is attached to it. This carbon has one, two, three bombs. So it needs one more. It must be to a hydrogen. Same with this carbon. It currently only has three bonds. The last bond is an implied hydrogen. This carbon has one, two, three bonds. It also has a hydrogen coming off of it. The central carbon. Here, this one only has one bond. It needs three more. So there must be three implied hydrogen is attached to it and saying with this method moving over to work Second compound. Let's start with this aromatic ring. This carbon has one, two, three bonds. It needs one more. So must be bombing one hydrogen. This carbon responded to one, two, three hike our carbons it needs one more bond to hydrogen. Same with this carbon This carbon This carbon This one This one This one This one in this one all have one hydrogen attached to it For the same reason They all have three bonds and need a fourth. If we look at these two carbons, they have the same bonding there. One, two, three, four bombs They do not need any more hydrogen Sze. We look at this carbon It has one, two, three, four bonds. It also doesn't have any hydrogen is attached to it. This carbon has one, two, three bonds. It needs one more to have a full architect This carbon has to bonds. Therefore it must have two more hydrogen Sze coming off of it in order to have a full octet. This carbon currently has two bonds. Therefore there are two implied hydrogen Sze Same with these two carbons. One to bonds means there's two extras one two means there must be two more bronze to hydrogen This carbon also on ly currently shows two bombs. Therefore it must have to bonds to hydrogen. Same with this one. And with this one one two bonds means there must be two more. This carbon has one two, three bonds, Meaning it's missing one. There's another implied hydrogen coming off this carbon. This carbon in the ring. She has one, two, three, four Bonds doesn't need any more this carbon, so it must have zero. Hydrogen is attached to it. So we should write that in here is well, zero zero and zero attached to these three carbons. We move up to this one, and it's one, two three. Bonds needs one more to hydrogen. Same with this one. Same with this one. And this one off for the same reason this carbon has one, two, three, four bonds, so it can't hold any more. Hydrogen zit has zero attached. This carbon has one, two three, four bonds coming off of it. Therefore, it also doesn't have any. Hydrogen is attached there. This carbon has one bond and therefore has three. Hydrogen is attached to it to get a full octet. Same with this method group. The last carbon here. It may help to actually draw out the functional group. So we're talking about a carb oxalic acid. I'm just going to call this are for the rest of this chain is the our group. So we're talking about this carbon. It has one, two, three, four bonds. Therefore there's zero. Hydrogen is attached to that carpet.

Hi everyone. So in this question does home training. Yeah. Of the negative items are represented by each chemical formula. So is carbon atoms in C. Two H five. Ceos he has three B option items since the Cielo for toys Then see hydrogen attempting an exploratory is HBO four. So relate this as a serial number Bang chemical formula. Give me color formula. Then the Merav indicated items. Oh, number of Mhm. Indicated victims. So, e chemical formulas C. H. For you. COCH three. So in this case, number of carbon items they ask. So a number of carbon items are no more. Oh. Mhm. Carbon atoms All right. Equal to two plus three plus 14 bam. Oh secondly CSL before toys. Yeah. C. A. Mhm. Mhm. Yeah. Fear or for toys? Mhm. But it's Yeah, I can see em part of clothing, correct? So this can be a vast number of auction items. So the mud off oxygen atoms are in quantity eight the next. Mm hmm. Seeing Yeah. Oh, an export toys. It's your fault. So the number of hydrogen items number of I brought your items? Yeah, I recall 2 9. No thanks.

We want to answer the question. How many atoms are in 0.59 animals. One mole contains 6.2 times 10 to the 23rd atoms we have options. A through F below A is 3.2 times some of the 14th B is 3.19 times 7 to 14. C is three times 10 to 14 days. 3.2 times 7 17. The is 3.1910, 7, 17 and Fs 3 10 7 17. We see that A through B and D through F are similar in that each of them has 10 to 14 and some of the 17 is our answer. And notably each of them also differs in a number of sick babies they use. So this answer is clearly testing to see whether or not we can direct the correct scientific notation 10 to the 14th and 17th and the correct number of sick things. Remember the results? Sick things have to be less than or equal to the initial six things. And that means both sets of six. So 6.2 times 10, 20 to 33.5 has won. Our results can only have one single thing, so next we simply have to multiply 10.5 animals by the number of moles or rather Adams normal to get our answer. So nano is 10 to the negative nine, so the number of atoms is five times that of the negative 10 moles counting 6.0 to 10, 7, 23rd out of normal or answer C three times 10 to the 14 atoms with the correct number of something.


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