Question

Chemistry is an experimental science in which a consistent system of measurements is used. Scientists throughout the world use the International System of Units (Système International or SI), which is based on the metric system. However, the metric system in not used on an everyday basis in the United States. It is often necessary to convert between the English system, which is used in the United States, and the SI system. One problem with the SI system is the size of the units sometimes turns out to be inconveniently large or small. Therefore it is often necessary to make conversions within the SI system. For this reason, SI units are modified through the use of prefixes when they refer to either larger or smaller quantities.
Part A

In Europe, gasoline efficiency is measured in km/L. If your car's gas mileage is 22.0mi/gal , how many liters of gasoline would you need to buy to complete a 142-km trip in Europe? Use the following conversions:1km=0.6214mi and 1gal=3.78L.

Part B

While in Europe, if you drive 115km per day, how much money would you spend on gas in one week if gas costs 1.10 euros per liter and your car's gas mileage is 30.0mi/gal ? Assume that 1euro=1.26dollars.

Part C

A sprinter set a high school record in track and field, running 200.0 m in 20.9s . What is the average speed of the sprinter in kilometers per hour?

Part D

A specific brand of gourmet chocolate candy contains 7.00 g of dietary fat in each 22.7-g piece. How many kilograms of dietary fat are in a box containing 4.00lb of candy?

Answer 1

Answer:

(a) 142 km trip requires 15.16 Liters of Gasoline

(b) The money spent is 69.34 euros or $87.37.

(c) The average speed of sprinter is 34.45 km/h.

(d) 4 lb of candy contains 0.559 kg of dietary fat.

Step-by-step explanation:

(a)

Given that;

Mileage = 22 mi/gal

Converting it to km/L

Mileage = (22 mi/gal)(1 gal/3.78 L)(1 km/0.6214 mi)

Mileage = 9.36 km/L

Now, the gasoline required for 142 km trip:

Gasoline Required = (Length of trip)/(Mileage)

Gasoline Required = (142 km)/(9.36 km/L)

Gasoline Required = 15.16 L

(b)

Given that;

Mileage = 30 mi/gal

Converting it to km/L

Mileage = (30 mi/gal)(1 gal/3.78 L)(1 km/0.6214 mi)

Mileage = 12.77 km/L

Now, the gasoline required for 142 km trip:

Gasoline Required = (Length of trip)/(Mileage)

Gasoline Required = (115 km/day)(7 days/week)/(12.77 km/L)

Gasoline Required = 63 L/week

Now, we find the cost:

Weekly Cost = (Gasoline Required)(Unit Cost)

Weekly Cost = (63 L/week)(1.1 euros/L)

Weekly Cost = 69.34 euros/week = $87.37

since, 1 euro = $1.26

(c)

Average Speed = (Distance Travelled)/(Time Taken)

Average Speed = 200 m/20.9 s

Average Speed = (9.57 m/s)(3600 s/1 h)(1  km/ 1000 m)

Average Speed = 34.45 km/h

(d)

22.7 g piece contains = 7 g dietary fat

(22.7 g)(1 lb/453.592 g) piece contains = (7 g)(1 kg/1000 g) dietary fat

0.05 lb piece contains = 0.007 kg dietary fat

1 lb piece contains = (0.007/0.05) kg dietary fat

4 lb piece contains = 4(0.007/0.05) kg dietary fat

4 lb piece contains = 0.559 kg dietary fat