1. three.

ÝÝ (1) all 5 carbons line up linearly.

ÝÝ (2) one carbon is center and other four sround it.(the center one connected with 4 carbons)

ÝÝ (3) four carbons form linear and the other bond with the second/third one.

1)Ý The are three isomers of the hydrocarbon containing five carbons and twelve hydrogens.Ý The first is a continuous chain of five carbons.Ý The first is connected to the second.Ý The second is connected to the first and third.Ý The third is connected to the second and fourth etc.Ý Three hydrogens are bonded to each of the end carbons and two hydrogens are bonded to each of the three remaining carbons.Ý The second isomer consists of four carbons in a row with a fifth carbon bonded to the second carbon in the chain.Ý The second carbon also has one hydrogen bonded to it.Ý The two end carbons and the one bonded to the second carbon each has three hydrogens.Ý The remaining two are bonded to two hydrogens each.Ý The third isomer consists of a central carbon bonded to four other carbons.Ý The central carbon is bonded to no hydrogens.Ý Each of the other four is bonded to four hydrogens.

Carbon attached to nitrogen would come somewhere between carbon-carbon and carbon-oxygen, but probably closer to carbon-oxygen.Ý Because the electronegativity of carbon nitrogen is quite high.Ý THe difference in electronegativity causes dipole moments between atoms.Ý THe higher the electronegativity the more drastic the dipole moments.Ý In a sense the longer an atom spends around the negatively charged nucleus the longer the dipole moments.Ý Therefore nitrogen which has a high electronegativity would spend a lot of time around the nucleus which would cause a long dipole moment, whereas carbon is not that electronegative and therefore it would not spend quite as much time around the nucleus.

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2)the signal would come at about 40 ppm because the electronegativity of nitrogen is less than that of oxygen which gives a signal at 50 ppm and the electronegativity of nitrogen is much greater than that of carbon which gives a signal at 15 ppm

2)Ý The ppm values for silicon bonded to carbon, carbon bonded to carbon, and oxygen bonded to carbon were plotted on a graph with their respective electronegativities.Ý These three points were connected with a smooth, curved line to produce a graph.Ý This graph was then used with nitrogen's electronegativity to estimate what the ppm value for nitrogen bonded to carbon would be.Ý The result suggested that a nitrogen bonded to a carbon would have a ppm value near 27.5.

3)ÝÝ Knowing both the empirical formula and the mass to charge ratio of the charged cation (taken to be the molecular mass) the molecular formula can be determined by dividing the molecular mass by the total mass of the empirical formula.Ý This result gives the factor that the subscripts that the empirical formula must be multiplied by to yield the molecular formula.Ý In this case, the factor was 4, and the resulting molecular formula was determined to have four carbons and eight oxygens.Ý This was choice "C."