You may want to reference (Page) Section 12.4 while completing this problem. Enter the balanced chemical equation for the complete combustion of each of the following hydrocarbons found in gasoline used to start a fire:. Part A. liquid nonane
Express your answer as a chemical equation including phases.
Part B. liquid isopentane (2-methylbutane)
Express your answer as a chemical equation including phases.
Part C. liquid 3-ethyltoluene
Express your answer as a chemical equation including phases.

See explanation

Explanation:

In this question, we have to remember that in all combustion reactions we will have Oxygen as reactive () and the products are Carbon dioxide () and Water (). Additionally, for the states, we will have (l) for the liquid state and (g) for the gas state. So, we can analyze each reaction:

Part A.

The formula for nonane is , with this in mind we can write the combustion reaction:

When we balance the reaction we will obtain:

Part B.

The formula for 2-methylbutane is , with this in mind we can write the combustion reaction:

When we balance the reaction we will obtain:

Part C.

The formula for 3-ethyltoluene is , with this in mind we can write the combustion reaction:

When we balance the reaction we will obtain:

See figure 1 for further explanations.

I hope it helps!

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole