I believe that the choices for this question are:
C2H4O2, C4H8O4 CH2O, C6H12O6 C3H6O3, C6H12O6 C2H4O2, C6H12O6
The answer to this based on the molar masses given is:
C2H4O2, C6H12O6
To prove calculate the molar mass:
C2H4O2 = 2*12 + 4*1 + 2*16 = 60
C6H12O6 = 6*12 + 12*1 + 6*16 = 180
Answer:
Second step: 4-bromo-1-methyl-2-nitrobenzene.
Third step: 1.5-dibromo-2-methyl-3-nitrobenzene.
Explanation:
To solve this exercise I will use the concepts of electrophilic substitution. In these reactions, a functional group is displaced by an electrophile. In the attached image are the two main products.
You’ve not put anything below so I’m not sure what the options are but the relative charge of a proton is +1
Flourine.
You are very welcome
The central iodine atom in triiodide has sp3d hybridization.In triiodide anion, the central iodine atom has three equatorial lone pairs of electrons and the terminal iodines are bonded axially in a linear shape. Electrons in sp3d hybridization are arranged in trigonal bipyramidal symmetry.
Explanation:
When you draw the Lewis structure of this particle, you'll realize that the central I atom has a pair of bonds and three individual pairs of electrons. as a result of there are five things around that central I atom, it's<span> sp3d hybridized.
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The bonds during a gas<span> (CH4) molecule </span>are fashioned<span> by four separate </span>however<span> equivalent orbitals; </span>one<span> 2s and </span>3<span> 2p orbitals of the carbon </span>interbreed<span> into four sp3 orbitals. </span>within the<span> ammonia molecule (NH3), 2s and 2p orbitals </span>produce<span> four sp3hybrid orbitals, </span>one among that<span> is occupied by a lone </span>try<span> of electrons.</span><span>
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