Answer:
Mostly Para
Explanation:
First, let's assume that the molecule is the toluene (A benzene with a methyl group as radical).
Now the nitration reaction is a reaction in which the nitric acid in presence of sulfuric acid, react with the benzene molecule, to introduce the nitro group into the molecule. The nitro group is a relative strong deactiviting group and is metha director, so, further reactions that occur will be in the metha position.
Now, in this case, the methyl group is a weak activating group in the molecule of benzene, and is always ortho and para director for the simple fact that this molecule (The methyl group) is a donor of electrons instead of atracting group of electrons. Therefore for these two reasons, when the nitration occurs,it will go to the ortho or para position.
Now which position will prefer to go? it's true it can go either ortho or para, however, let's use the steric hindrance principle. Although the methyl group it's not a very voluminous and big molecule, it still exerts a little steric hindrance, and the nitro group would rather go to a position where no molecule is present so it can attach easily. It's like you have two doors that lead to the same place, but in one door you have a kid in the middle and the other door is free to go, you'll rather pass by the door which is free instead of the door with the kid in the middle even though you can pass for that door too. Same thing happens here. Therefore the correct option will be mostly para.
Nitrogen is a diatomic molecule in the VA family on the periodic table. Nitrogen has five valence electrons, so it needs three more valence electrons to complete its octet. A nitrogen atom can fill its octet by sharing three electrons with another nitrogen atom, forming three covalent bonds, a so-called triple bond.
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1:2
The ratio of carbon, hydrogen, and oxygen in most carbohydrates is 1:2:1. This means for every one carbon atom there are two hydrogen atoms and one...
Theses can include the power supply circuit a joule meter to measure the energy transferred which makes the calculations a lot easier.
Answer:
Explanation:
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In this case, since the combustion of butane is:
Thus, since the molar mass of butane is 58.12 g/mol and that of carbon dioxide is 44.01 g/mol, we obtain the following mass of CO2 product:
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