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SashulF [63]
4 years ago
7

A civil engineering student is considering buying an electric car. However, the conscious student wants to make sure her carbon

footprint will decrease with her new investment. Her current car runs on gasoline (C7H16) and has quite good fuel efficiency at 36.5mi/gal. The electric vehicle she is looking to buy can perform at 5.2mi/kWh. The student will recharge her car battery at school and the energy there comes from an 85% efficient power plant that runs on natural gas (CH4). Help the student make her decision by comparing the volume of CO2 (in gal) produced
Chemistry
1 answer:
Ganezh [65]4 years ago
7 0

Answer: Vₙ / Vₓ = 11039.78 gal / 2046.04 gal = 5.3

Therefore, gasoline-powered vehicle produces 5.3 times as much CO₂ as the electric vehicle.

Note: The question is incomplete. The complete question is given below:

A civil engineering student is considering buying an electric car. However, the conscious student wants to make sure her carbon footprint will decrease with her new investment. Her current car runs on gasoline (C7H16) and has quite good fuel efficiency at 36.5mi/gal. The electric vehicle she is looking to buy can perform at 5.2mi/kWh. The student will recharge her car battery at school and the energy there comes from an 85% efficient power plant that runs on natural gas (CH4). Help the student make her decision by comparing the volume of CO2 (in gal) produced in a month by both the gasoline and the electric cars. Her commute is 16 miles per day and she drives to school 5 days of the week.

Explanation:

Distance travelled in an average month =  16 * 5 * 52wks/12mos = 346.7 mi

Using a gasoline powered vehicle Vₙ

Volume (in liters) of gasoline used = 346.7mi / (36.5mi/gal) * 3.785 = 35.95 L

Mass of gasoline used = density * volume = 679.5g/L * 35.95 L = 24428.03g

Number of moles of gasoline used = mass/molar mass

n = 24428.03 g / 100g/mol = 244.28 mol

Equation for combustion  : C₇H₁₆ + O₂ ⟶ 7CO₂ + 8H₂O  

Moles of CO₂ formed = 244.28 * 7 = 1710 mol

Volume (in gal) of CO₂ formed at 25 °C and 1 atm

V = {nRT/P} / 3.785

V = {1710 * 0.082 *298 / 1} / 3.785 = 11039.78 gal

Using an electric vehicle Vₓ

<em>Theoretical energy (in KJ) used</em> = 346.7mi / 5.2mi/ KWh * 3600KJ/KWh

= 240023.08 KJ

<em>Actual energy used </em> since the power station is only 85 % efficient =

240023.08 /0.85 = 282380.09 KJ

Equation for the combustion of CH₄ : CH₄ + 2O₂ ⟶ CO₂ +2 H₂O

<em>The heat of combustion of methane at 25 °C and 1 atm is -891.0 kJ·mol⁻¹ </em>

Number of moles of CO₂ produced since mole ratio to methane is 1:1 is given as

n = 282380.09 KJ / 891.0 kJ·mol⁻¹ of methane * (1 mol CO₂/1 mol of methane) = 316.92 moles of CO₂

Volume of CO₂ (in gal) = {nRT/P} / 3.785

V = {316.92 *0.082 * 298/1} /3.785 = 2046.04 gal

Comparing the two:

Vₙ / Vₓ = 11039.78 gal / 2046.04 gal = 5.3

Therefore, gasoline-powered vehicle produces 5.3 times as much CO₂ as the electric vehicle.

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GenaCL600 [577]

Answer:

P_A=4.20atm\\\\P_B=17.1atm

Explanation:

Hello!

In this case, since the equation for the ideal gas is:

PV=nRT

For each gas, given the total volume, temperature (28.1+273.15=301.25K) and moles, we can easily compute the partial pressure as shown below:

P_A=\frac{n_ART}{V} =\frac{1.21mol*0.082\frac{atm*L}{mol*K}*301.25K}{7.12L} \\\\P_A=4.20atm\\\\P_B=\frac{n_BRT}{V} =\frac{4.94mol*0.082\frac{atm*L}{mol*K}*301.25K}{7.12L} \\\\P_B=17.1atm

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A first-order reaction has a half-life of 16.7 s . How long does it take for the concentration of the reactant in the reaction t
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A <em>half-life</em> is the time it takes for the concentration to fall to half its original value.

Assume the initial concentration is 1.00 mol/L. Then,

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