Chapter 2: Problem 81
A student loses \(3.3 \mathrm{lb}\) in one month. How many grams did he lose?
Short Answer
Expert verified
The student lost approximately 1497 grams.
Step by step solution
01
Understanding the conversion factors
To convert pounds (lb) to grams (g), you need to know the conversion factor. There are approximately 453.59237 grams in a pound.
02
Set up the conversion equation
You can set up a conversion equation to change pounds to grams. If 1 lb equals to approximately 453.59237 g, then for any number of pounds, you multiply the number of pounds by 453.59237 to find out how many grams it is.
03
Perform the conversion
Multiply the number of pounds lost by the student, 3.3 lb, by the conversion factor, 453.59237 g/lb. The calculation will look like this: 3.3 lb * 453.59237 g/lb.
04
Calculate the result
Performing the multiplication gives you the weight loss in grams: 3.3 * 453.59237 = 1496.864821 g.
05
Round the answer
If necessary, you may want to round the result to a more manageable number, depending on the level of precision required. Rounding to the nearest gram, the student lost approximately 1497 grams.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Pounds to Grams Conversion
Understanding how to convert units from pounds to grams is a valuable skill in chemistry and everyday life. It’s a simple process that requires you to multiply the weight in pounds by a conversion factor to obtain the weight in grams. In practical terms, if a student loses 3.3 pounds over a month, to find out how much that is in grams, we use the conversion factor that 1 pound is approximately equal to 453.59237 grams.
Here’s how it works: take the number of pounds lost (\(3.3 \text{ lb}\)) and multiply by the conversion factor (\(453.59237 \text{ g/lb}\)). The equation would be \(3.3 \text{ lb} \times 453.59237 \text{ g/lb}\). After performing the multiplication, we find that the student lost approximately 1496.864821 grams. For convenience and ease of understanding, this number can be rounded to the nearest gram, resulting in 1497 grams lost.
Here’s how it works: take the number of pounds lost (\(3.3 \text{ lb}\)) and multiply by the conversion factor (\(453.59237 \text{ g/lb}\)). The equation would be \(3.3 \text{ lb} \times 453.59237 \text{ g/lb}\). After performing the multiplication, we find that the student lost approximately 1496.864821 grams. For convenience and ease of understanding, this number can be rounded to the nearest gram, resulting in 1497 grams lost.
Conversion Factors
Conversion factors play a crucial role in chemistry calculations. They are essentially ratios that let you express the same quantity in different units, allowing for the conversion from one unit to another. These factors are based on equivalences, such as 1 pound being equal to 453.59237 grams.
In the given example, the conversion factor for pounds to grams is integral to solving the problem. Memorizing common conversion factors or knowing where to find them can greatly simplify many chemistry problems. Besides the aforementioned conversion, others like converting inches to centimeters or liters to gallons are just as important in various scenarios. Being adept with these factors ensures accuracy in calculations, which is imperative in any scientific or mathematical context.
In the given example, the conversion factor for pounds to grams is integral to solving the problem. Memorizing common conversion factors or knowing where to find them can greatly simplify many chemistry problems. Besides the aforementioned conversion, others like converting inches to centimeters or liters to gallons are just as important in various scenarios. Being adept with these factors ensures accuracy in calculations, which is imperative in any scientific or mathematical context.
Chemistry Calculations
Chemistry calculations often require converting measurements into different units to solve problems accurately. These problems can range from simple tasks like converting pounds to grams, as seen in our example, to more complex stoichiometric calculations in chemical reactions.
To tackle these problems effectively, understanding the concept of moles, molar mass, concentration, and gas laws is crucial. The process usually starts with writing down the information provided, understanding the conversion needed, setting up an equation with appropriate conversion factors, and then calculating the result.
For the student's weight loss example, the initial step was to multiply the pounds by the gram conversion factor. The final answer was then rounded to the nearest whole number for ease of communication, showing how rounding can sometimes be necessary to make the result usable.
To tackle these problems effectively, understanding the concept of moles, molar mass, concentration, and gas laws is crucial. The process usually starts with writing down the information provided, understanding the conversion needed, setting up an equation with appropriate conversion factors, and then calculating the result.
For the student's weight loss example, the initial step was to multiply the pounds by the gram conversion factor. The final answer was then rounded to the nearest whole number for ease of communication, showing how rounding can sometimes be necessary to make the result usable.