1. Name the following alkanes.

a. 2-methyl-propane.

b. 2-methyl-butane.

c. 4-ethyl-heptane.

d. 4-methyl-3-ethyl-heptane.

e. 3,3,5-trimethyl-heptane.

f. 2-dimethyl-butane.

g. 2-dimethyl-propane.

h. 3-diethyl-pentane.

Explanation:

Hello, for IUPAC rules, take into account that for numbering the organic chain, we must start by the carbon havinf the closest radical to it. Moreover, for naming, we can name alphabetically or by complexity order as shown below:

a. In this case, we consider the middled methyl as a radical in the three-carbon chain, therefore, IUPAC name is 2-methyl-propane.

b. In this case, we consider the methyl at the second carbon in the four-carbon chain (longest one), therefore IUPAC name is 2-methyl-butane.

c. In this case, we consider the ethyl at the fourth carbon in the seven-carbon chain (longest one), therefore IUPAC name is 4-ethyl-heptane.

d. In this case, we consider the ethyl at the third carbon and the methyl at the fourth carbon in the seven-carbon chain (longest one), therefore IUPAC name is 4-methyl-3-ethyl-heptane.

e. In this case, we consider the three at the third and fifth carbons in the seven-carbon chain (longest one), therefore IUPAC name is 3,3,5-trimethyl-heptane.

f. In this case, we consider the two methyles at the second carbon in the four-carbon chain (longest one), therefore IUPAC name is 2-dimethyl-butane.

g. In this case, we consider the two methyles at the second carbon in the three-carbon chain (longest one), therefore IUPAC name is 2-dimethyl-propane.

h. In this case, we consider the two ethyles at the third carbon in the five-carbon chain (longest one), therefore IUPAC name is 3-diethyl-pentane.

Best regards.

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2