Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 1: Problem 87

In recent years, there has been a large push for an increase in the use of renewable resources to produce the energy we need to power our vehicles. One of the newer fuels that has become more widely available is a mixture of \(85 \%\) ethanol and \(15 \%\) gasoline, E85. Despite being more environmentally friendly, one of the potential drawbacks of E85 fuel is that it produces less energy than conventional gasoline. Assume a car gets \(28.0 \mathrm{mi} / \mathrm{gal}\) using gasoline at \(\$ 3.50 / \mathrm{gal}\) and \(22.5 \mathrm{mi} / \mathrm{gal}\) using \(\mathrm{E} 85\) at \(\$ 2.85 / \mathrm{gal}\). How much will it cost to drive 500 . miles using each fuel?

Short Answer

Expert verified
For gasoline, it will cost approximately \$62.51 to drive 500 miles, whereas for E85 fuel, it will cost approximately \$63.33. Hence, E85 fuel is slightly more expensive for driving the same distance despite its environmental benefits.

Step by step solution

01

Calculate the gallons of fuel needed for 500 miles for each type of fuel

To find out how many gallons of fuel are needed, divide the total distance (500 miles) by the miles per gallon for each fuel type. For gasoline: \(\frac{500 \ \mathrm{miles}}{28 \ \mathrm{mi/gal}}\) For E85: \(\frac{500 \ \mathrm{miles}}{22.5 \ \mathrm{mi/gal}}\)
02

Compute the gallons of fuel for each type

Now, let's do the calculations: For gasoline: \(\frac{500}{28} = 17.86 \ \mathrm{gal}\) (approx.) For E85: \(\frac{500}{22.5} = 22.22 \ \mathrm{gal}\) (approx.)
03

Calculate the cost of driving 500 miles using each type of fuel

To find the cost, multiply the total gallons of fuel needed by the cost per gallon for each fuel type. For gasoline: \(17.86 \ \mathrm{gal} \times \$3.50 /\mathrm{gal}\) For E85: \(22.22 \ \mathrm{gal} \times \$2.85 /\mathrm{gal}\)
04

Compute the cost for each type of fuel

Now, let's do the calculations: For gasoline: \(17.86 \times \$3.50 = \$62.51\) For E85: \(22.22 \times \$2.85 = \$63.33\)
05

Comparing the costs

After calculating the costs, we find out that it will cost: - \$62.51 to drive 500 miles using gasoline - \$63.33 to drive 500 miles using E85 fuel So, it will cost slightly more to drive 500 miles using E85 fuel, despite it being more environmentally friendly.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

How many significant figures are in each of the following? a. 100 e. \(0.0048\) b. \(1.0 \times 10^{2}\) f. \(0.00480\) c. \(1.00 \times 10^{3}\) g. \(4.80 \times 10^{-3}\) d. 100 . h. \(4.800 \times 10^{-3}\)

You have a \(1.0-\mathrm{cm}^{3}\) sample of lead and a \(1.0-\mathrm{cm}^{3}\) sample of glass. You drop each in separate beakers of water. How do the volumes of water displaced by each sample compare? Explain.

An experiment was performed in which an empty \(100-\mathrm{mL}\) graduated cylinder was weighed. It was weighed once again after it had been filled to the \(10.0-\mathrm{mL}\) mark with dry sand. A \(10-\mathrm{mL}\) pipet was used to transfer \(10.00 \mathrm{~mL}\) of methanol to the cylinder. The sand- methanol mixture was stirred until bubbles no longer emerged from the mixture and the sand looked uniformly wet. The cylinder was then weighed again. Use the data obtained from this experiment (and displayed at the end of this problem) to find the density of the dry sand, the density of methanol, and the density of sand particles. Does the bubbling that occurs when the methanol is added to the dry sand indicate that the sand and methanol are reacting? Mass of cylinder plus wet sand \(\quad 45.2613 \mathrm{~g}\) Mass of cylinder plus dry sand \(\quad 37.3488 \mathrm{~g}\) Mass of empty cylinder \(22.8317 \mathrm{~g}\) Volume of dry sand \(10.0 \mathrm{~mL}\) Volume of sand plus methanol \(\quad 17.6 \mathrm{~mL}\) Volume of methanol \(\quad 10.00 \mathrm{~mL}\)

Sterling silver is a solid solution of silver and copper. If a piece of a sterling silver necklace has a mass of \(105.0 \mathrm{~g}\) and a volume of \(10.12 \mathrm{~mL}\), calculate the mass percent of copper in the piece of necklace. Assume that the volume of silver present plus the volume of copper present equals the total volume. Refer to Table \(1.5\). Mass percent of copper \(=\frac{\text { mass of copper }}{\text { total mass }} \times 100\)

The density of pure silver is \(10.5 \mathrm{~g} / \mathrm{cm}^{3}\) at \(20^{\circ} \mathrm{C}\). If \(5.25 \mathrm{~g}\) of pure silver pellets is added to a graduated cylinder containing \(11.2 \mathrm{~mL}\) of water, to what volume level will the water in the cylinder rise?
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Physical Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Nuclear Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Inorganic Chemistry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free