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3 years ago

What is the iupac name of the following compound? 2,2-dimethyl-4-ethylheptane

Chemistry

1 answer:

Lady bird [3.3K]3 years ago

4 0

Answer:

The IUPAC name of the compound has already been given which is 2,2-dimethyl-4-ethylheptane.

Explanation:

The IUPAC (International Union of Pure and Applied Chemistry) is an authority in chemistry that provides a guideline and standardized methods in the naming of compounds formed from the periodic table.

In order the give an IUPAC name to a compound, certain steps needs to be followed, these includes:

--> Identify the functional group in the compound as this will form the suffix. For example if the functional group is an alkane the suffix will be -ane.

--> Identify the longest carbon chain (it may not be a straight chain) that contains the functional group. This forms the prefix. Example: if the longest carbon chain is 7 carbon atoms then the prefix will be hept-

--> All the carbons of the longest chain should be numbered

--> Identify branched groups on the chain and name them according to the number of carbon atoms. They usually end with -yl.

--> Finally, combine the elements of the name is a single word.

The structural formula of the IUPAC compound can be found in the attached file for a better understanding. The branched groups are circled.

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3 years ago

A mixture of CS2(g) and excess O2(g) is placed in a 10 L reaction vessel at 100.0 ∘C and a pressure of 3.10 atm . A spark causes

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Answer:

PCO2 = 0.6 25 atm

PSO2 = 1.2 75 atm

PO2 = 0.6 atm

Explanation:

Step 1: Data given

Volume = 10.0 L

Temperature = 100.0 °C

Pressure = 3.10 °C

After reaction, the temperature returns to 100.0 ∘C, and the mixture of product gases (CO2, SO2, and unreacted O2) is found to have a pressure of 2.50 atm

Step 2: The balanced equation

CS2(g)+3O2(g)→CO2(g)+2SO2(g)

Step 3: Name the reactants and products

a = CS2

b = O2 before reaction

c = CO2

d = SO2

e = nS O2 after reaction with n = the number of moles

Step 4: Calculate moles before reaction

PV = nRT

n = PV/(RT)

(na + nb) = (3.10atm) * (10.0L) / ((0.08206 Latm/moleK) * (373.15K))

(na + nb) = 1.0124

Step 5: Calculate moles after reaction

PV = nRT

n = PV/(RT)

nc + nd + ne) = PV/(RT) = (2.50 atm)*(10.0L) / ((0.08206 Latm/moleK)*(373.15K))

(nc + nd + ne) = 0.816 moles

Step 6: Calculate mol fraction

For 1 mole CS2 we need 3 moles O2 to produce 1 mole of CO2 and 2 moles of SO2

moles O2 remaining = ne = nb - 3na

moles CO2 produced = nc = na

moles SO2 producted = nd = 2na

(nc + nd + ne) = 0.816 moles = nb - 3na + na + 2na = 0.816

nb = 0.816

. (na + nb) = 1.0124

na = 1.0124 moles - 0.816 moles = 0.208

which leads to

nc = na = 0.208

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ne = 0.816 - 3*0.208 = 0.192

mole fraction CO2 = 0.208 / (0.208 + 0.416 + 0.192) = 0.25

mole fraction SO2 = 0.416 / (0.208 + 0.416 + 0.192) = 0.5 1

mole fraction O2 = 0.192 /(0.208 + 0.416 + 0.192) = 0.24

Step 6: Calculate partial pressure

PCO2 = 0.25 * 2.50 atm = 0.6 25 atm

PSO2 = 0.51 * 2.50 atm = 1.2 75 atm

PO2 = 0.24 * 2.50 atm = 0.6 atm

Step 7: Control results

now let's verify a couple of things

PV = nRT

P = nRT/V

before rxn

P = (0.208 + 0.816) * (0.08206 L*atm/mole*K) * (373.15K) / (10.0L) ≈ 3.10 atm

after rxn

P = ((0.208 +0.416+0.192) * (0.08206 L*atm/mole*K) * (373.15K) / (10.0L) ≈ 2.50 atm

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Explanation:

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