/*! 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} QCYL Round the following to the indic... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: QCYL (page 717)

Round the following to the indicated number of significant figures:

(a) 0.424 (to two significant figures)

(b) 0.0038661 (to three significant figures)

(c) 421.25 (to four significant figures)

(d) 28,683.5 (to five significant figures)

Short Answer

Expert verified

a) 0.424 rounds down to 0.42.

b) 0.0038661 rounds up to 0.00387.

c) 421.25 rounds down to 421.2.

d) 28,683.5 rounds up to 28,684.

Step by step solution

01

Significant figures

All the digits in measurement, including the uncertain last digit, are called significant figures or significant digits.

02

Rules used to determine the significant figures in a number

  • Non-zero digits are always significant.
  • Any zero between two significant digits is significant.

a) 0.424 rounds "down" to 0.42 (The removed digit, 4, is less than 5).

b) 0.0038661 rounds "up" to 0.00387 (The removed digit, 6, and the retained digit are odd).

c) 421.25 rounds "down" to 421.2 (The removed digit, 5, and the retained digit are even).

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 91影视!

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

Complete the changes in concentrations (or pressure, if requested) for each of the following reactions.

What property of a reaction can we use to predict the effect of a change in temperature on the value of an equilibrium constant?

The following reaction has \({K_P} = 4.50 \times {10^{ - 5}}\) at \(720\;{\rm{K}}\).

\({{\rm{N}}_2}(g) + 3{{\rm{H}}_2}(g) \rightleftharpoons 2{\rm{N}}{{\rm{H}}_3}(g)\)

If a reaction vessel is filled with each gas to the partial pressures listed, in which direction will it shift to reach equilibrium?

\(P\left( {{\rm{N}}{{\rm{H}}_3}} \right) = 93\;{\rm{atm}},\;P\left( {\;{{\rm{N}}_2}} \right) = 48\;{\rm{atm}},\;{\rm{and}}\;P\left( {{{\rm{H}}_2}} \right) = 52\)

Question: Consider the equilibrium

4NO2(g) + 6H2 O(g) 鈬� 4NH3(g) + 7O2(g)

(a) What is the expression for the equilibrium constant (Kc) of the reaction?

(b) How must the concentration of NH3 change to reach equilibrium if the reaction quotient is less than the equilibrium constant?

(c) If the reaction were at equilibrium, how would a decrease in pressure (from an increase in the volume of the reaction vessel) affect the pressure of NO2?

(d) If the change in the pressure of NO2 is 28 torr as a mixture of the four gases reaches equilibrium, how much will the pressure of O2 change?

Question:What is the value of the equilibrium constant at \(50{0^o}C\) for the formation of \(N{H_3}\)according to the following equation? N2(g) + 3H2(g) 鈬� 2NH3(g)

An equilibrium mixture of \(N{H_3}(g)\) \({H_2}(g)\) and \({N_2}(g)\) at \(50{0^o}C\) was found to contain\(1.35M{H_2},1.15M{N_2}\)and \(4.12\)\( \times 1{0^{ - 1}}MN{H_3}\)

See all solutions

Recommended explanations on Chemistry Textbooks

Making Measurements

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Kinetics

Read Explanation

Physical Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Chemical Reactions

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.