/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 21 Perform the following calculatio... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Perform the following calculations; express each answer in exponential form and with the appropriate number of significant figures. (a) \(0.406 \times 0.0023=\) (b) \(0.1357 \times 16.80 \times 0.096=\) (c) \(0.458+0.12-0.037=\) (d) \(32.18+0.055-1.652=\)

Short Answer

Expert verified
(a) \(9.3 \times 10^{-4}\), (b) \(2.2 \times 10^{-1}\), (c) \(5.4 \times 10^{-1}\), (d) \(3.058 \times 10^{1}\)

Step by step solution

01

Solve (a) - Perform the Calculation

Perform multiplication for \(0.406 \times 0.0023\), which will give the result 0.000934.
02

Solve (a) - Determine Number of Significant Figures

The number of significant figures in the final calculation must be the same as the number with the least significance in the original numbers. Here, \(0.0023\) has two significant figures, hence the answer will also be in two significant figures, which results in 0.00093.
03

Solve (a) - Convert to Exponential Form

Write the answer in exponential form to give \(9.3 \times 10^{-4}\).
04

Solve (b) - Perform the Calculation

Perform multiplication for \(0.1357 \times 16.80 \times 0.096\), which will give the result 0.22058.
05

Solve (b) - Determine Number of Significant Figures

Based on the rule for multiplication and division, the number of significant figures in the final calculation must equal the smallest number of significant figures in the original numbers. Hence in this case, the answer will be in two significant figures 0.22.
06

Solve (b) - Convert to Exponential Form

Convert the decimal to exponential notation, which will yield \(2.2 \times 10^{-1}\).
07

Solve (c) - Perform the Calculation

Perform the operations in order \(0.458 + 0.12 - 0.037\), which will yield 0.541.
08

Solve (c) - Determine Number of Significant Figures

The answer for addition and subtraction should be reported to the lowest decimal place contained in the original numbers. Hence, the answer should be reported to the hundredth place to yield 0.54.
09

Solve (c) - Convert to Exponential Form

Write the calculation in exponential form, which will yield \(5.4 \times 10^{-1}\).
10

Solve (d) - Perform the Calculation

Perform the operations in order \(32.18+0.055-1.652\), which will yield 30.583.
11

Solve (d) - Determine Number of Significant Figures

Because we are adding and subtracting, the final answer should be reported to the hundredth decimal place, 30.58.
12

Solve (d) - Convert to Exponential Form

The final answer will be in exponential form, \(3.058 \times 10^{1}\).

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.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Exponential Notation
Exponential notation is a handy way to express very large or small numbers. It's often used in chemistry due to the vast range of measurements you encounter. By writing numbers in the form of \(a \times 10^n\), you simplify expressions and make them easier to manage.
This format, where \(a\) is a number called the coefficient, and \(n\) is an integer, helps in clearly showing the number of significant figures.
For example, the number 0.00093 is expressed as \(9.3 \times 10^{-4}\). This method ensures that each value is shown in its purest form without any unnecessary digits. Always remember that the exponent \(n\) indicates how many places the decimal point moves to convert the coefficient to the standard form.
  • Positive \(n\): Moves decimal to the right.
  • Negative \(n\): Moves decimal to the left.
Mastering exponential notation can make calculations less cumbersome and allow for consistent reporting of your final results.
Multiplication and Division in Chemistry
In chemistry, precision in reporting results is crucial, and this is where understanding significant figures comes into play, especially in multiplication and division. The rule is relatively simple: the number of significant figures in the result is determined by the original number with the fewest significant figures.
Consider the example of multiplying \(0.1357 \text{ and } 16.80 \text{ and } 0.096\). The smallest significant figure number here is from 0.096, which has two significant figures. Hence, any product from this operation should also be expressed with two significant figures, resulting in \(2.2 \times 10^{-1}\).
This ensures that precision is not overestimated in your results, maintaining the integrity of the data, especially when measurements have uncertainties inherent to the experiment. Being diligent in this practice is vital for accurate data reporting and interpretation.
Addition and Subtraction in Chemistry
When adding or subtracting numbers in chemistry, attention shifts slightly from significant figures to decimal places. The least precise measurement—the one with the fewest decimal places—determines the precision of the result.
Take for instance the operation \(32.18 + 0.055 - 1.652\). Here, the smallest decimal place is at the hundredth place, dictated by 32.18, which controls how many decimal places we report in our answer.
Therefore, your result should coincide with this precision level, yielding 30.58.
  • Always align decimals before performing operations.
  • Round the result to the precision of the least precise measure.
Understanding this concept helps ensure that the results you report are as precise as possible, without falsely increasing the apparent accuracy of your data.

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

Indicate whether each of the following is an exact number or a measured quantity subject to uncertainty. (a) the number of pages in this text (b) the number of days in the month of January (c) the area of a city lot (d) the distance between the centers of the atoms in a gold medal

Blood alcohol content (BAC) is sometimes reported in weight-volume percent and, when it is, a BAC of \(0.10 \%\) corresponds to \(0.10 \mathrm{g}\) ethyl alcohol per \(100 \mathrm{mL}\) of blood. In many jurisdictions, a person is considered legally intoxicated if his or her BAC is 0.10\%. Suppose that a 68 kg person has a total blood volume of 5.4 L and breaks down ethyl alcohol at a rate of 10.0 grams per hour. \(^{*}\) How many 145 mL glasses of wine, consumed over three hours, will produce a BAC of \(0.10 \%\) in this 68 kg person? Assume the wine has a density of \(1.01 \mathrm{g} / \mathrm{mL}\) and is \(11.5 \%\) ethyl alcohol by mass. (* The rate at which ethyl alcohol is broken down varies dramatically from person to person. The value given here for the rate is a realistic, but not necessarily accurate, value.)

Compared with its mass on Earth, the mass of the same object on the moon should be (a) less; (b) more; (c) the same; (d) nearly the same, but somewhat less.

The Greater Vancouver Regional District (GVRD) chlorinates the water supply of the region at the rate of 1 ppm, that is, 1 kilogram of chlorine per million kilograms of water. The chlorine is introduced in the form of sodium hypochlorite, which is \(47.62 \%\) chlorine. The population of the GVRD is 1.8 million persons. If each person uses 750 L of water per day, how many kilograms of sodium hypochlorite must be added to the water supply each week to produce the required chlorine level of 1 ppm?

A Boeing 767 due to fly from Montreal to Edmonton required refueling. Because the fuel gauge on the aircraft was not working, a mechanic used a dipstick to determine that 7682 L of fuel were left on the plane. The plane required \(22,300 \mathrm{kg}\) of fuel to make the trip. In order to determine the volume of fuel required, the pilot asked for the conversion factor needed to convert a volume of fuel to a mass of fuel. The mechanic gave the factor as \(1.77 .\) Assuming that this factor was in metric units (kg/L), the pilot calculated the volume to be added as 4916 L. This volume of fuel was added and the 767 subsequently ran out the fuel, but landed safely by gliding into Gimli Airport near Winnipeg. The error arose because the factor 1.77 was in units of pounds per liter. What volume of fuel should have been added?

See all solutions

Recommended explanations on Chemistry Textbooks

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.