/*! 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 3 BIO Estimate how many electrons ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 21: Problem 3

BIO Estimate how many electrons there are in your body. Make any assumptions you feel are necessary, but clearly state what they are. (Hint: Most of the atoms in your body have equal numbers of electrons, protons, and neutrons.) What is the combined charge of all these electrons?

Short Answer

Expert verified
The body has approximately \(1.4 \times 10^{28}\) electrons with a total charge of \(-2.24 \times 10^{9}\) C.

Step by step solution

01

Estimate the Average Mass of a Human Body

Let's assume the average mass of a human body is about 70 kg. This is a commonly used average for calculations and simplifies the problem.
02

Estimate the Number of Atoms in the Body

Assume that the body is mostly composed of water (about 60% by weight), and the remaining is mostly proteins and similar organic molecules. Water's approximate molecular weight is 18 g/mol, so there are approximately \( \frac{1000 \text{ g}}{18 \text{ g/mol}} \approx 55.5 \text{ mol/L} \) in pure water. Converting 70 kg of body weight to water weight (70 kg \(\times\) 0.6 = 42 kg, or 42,000 g), we get \( \frac{42,000 \text{ g}}{18 \text{ g/mol}} \approx 2333 \text{ mol} \).
03

Calculate the Number of Water Molecules

Since each mole contains Avogadro's number \( (6.022 \times 10^{23}) \) of molecules, the number of water molecules is \( 2333 \text{ mol} \times 6.022 \times 10^{23} \text{ molecules/mol} \approx 1.4 \times 10^{27} \text{ molecules} \).
04

Determine the Number of Electrons per Water Molecule

Each water (H2O) molecule contains 10 electrons: two from each hydrogen atom and eight from the oxygen atom. Multiplying the number of water molecules by the number of electrons gives: \( 1.4 \times 10^{27} \text{ molecules} \times 10 \text{ electrons/molecule} \approx 1.4 \times 10^{28} \text{ electrons} \).
05

Calculate Total Charge of Electrons

Each electron has a charge of \(-1.6 \times 10^{-19} \) coulombs. Thus, the total charge is \( 1.4 \times 10^{28} \text{ electrons} \times (-1.6 \times 10^{-19} \text{ C/electron}) \approx -2.24 \times 10^{9} \text{ C} \). This is a large negative charge, as expected due to the vast number of electrons.

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!

Key Concepts

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

Electron Calculation
To estimate the number of electrons in the human body, we begin by considering the composition of the body. The hint from our exercise suggests most atoms have equal numbers of electrons, protons, and neutrons, meaning a neutral charge overall. By focusing on the most abundant molecule in the body, which is water ( H2O), we can simplify our calculations significantly. First, we assume the average mass of a human is 70 kg, of which about 60% is water. This translates to roughly 42 kg of water in the body. By converting this mass to moles using the molecular weight of water (18 g/mol), we calculate there are around 2333 moles of water in the body. As each molecule of water contains 10 electrons (2 from each hydrogen and 8 from oxygen), multiplying the number of water molecules by 10 gives us approximately and impressively 1.4 x 10^28 electrons in a typical human body.
Molecular Weight
Understanding molecular weight is vital to estimating components of the human body. Molecular weight (or molecular mass) refers to the mass of a molecule, calculated from the sum of the atomic masses of all atoms present.In our specific calculation about the human body, we use the molecular weight of water, H2O, which is approximately 18 g/mol (grams per mole). This value stems from the atomic masses:- Hydrogen has an atomic mass of approximately 1 g/mol.- Oxygen has an atomic mass of approximately 16 g/mol.Together, the molecular weight of water becomes \( 2 \times 1 + 16 = 18 \) g/mol. This enables us to convert grams to moles, a critical conversion when calculating the number of molecules and subsequently the number of electrons in the body.
Avogadro's Number
Avogadro's number is a cornerstone in chemistry as it relates to the amount of substance one mole corresponds to. This number, 6.022 x 10^23, represents the number of atoms, ions, or molecules in one mole of a substance. Using Avogadro's number, we can determine the number of molecules from moles, a key step in calculating components of substances. For example, when we say that water has a molecular weight of 18 g/mol, using Avogadro's number lets us know that 18 grams of water will have approximately 6.022 x 10^23 molecules. In our exercise, knowing there are roughly 2333 moles of water, applying Avogadro's number provides the total number of molecules in the human body as around 1.4 x 10^27. From this, we effectively determine how many electrons those molecules account for.
Charge of an Electron
Every electron carries a specific amount of electric charge, which is a fundamental property of matter. The standard value for the charge of an electron is \(-1.6 \times 10^{-19} \) coulombs. This negative charge plays a crucial role in chemical reactions and interactions within biological and physical systems.In the electron calculation for the human body, the task is to find the total charge by multiplying the number of electrons by the charge of a single electron. From the exercise, the total number of electrons in the body is approximately 1.4 x 10^28. Multiplying this by \(-1.6 \times 10^{-19} \, \text{C/ electron}\), results in a total charge of about \(-2.24 \times 10^9 \, \text{C}\).This calculation encapsulates an immense amount of total charge, although in practice, it's balanced by positive charges in the body, maintaining neutrality.

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

A point charge is at the origin. With this point charge as the source point, what is the unit vector \(r\) in the direction of (a) the field point at \(x=0, \quad y=-1.35 \mathrm{m}\) ; (b) the field point at \(x=12.0 \mathrm{cm}, y=12.0 \mathrm{cm} ;(\mathrm{c})\) the field point at \(x=-1.10 \mathrm{m},\) \(y=2.60 \mathrm{m} ?\) Express your results in terms of the unit vectors \(\hat{\boldsymbol{\imath}}\) and \(\hat{\boldsymbol{J.}}\)

Three point charges are placed on the \(y\) -axis: a charge \(q\) at \(y=a,\) a charge \(-2 q\) at the origin, and a charge \(q\) at \(y=-a\) . Such an arrangement is called an electric quadrupole. (a) Find the magnitude and direction of the electric field at points on the positive \(x\) -axis. (b) Use the binomial expansion to find an approximate expression for the electric field valid for \(x \gg a\) . Contrast this behavior to that of the electric field of a point charge and that of the electric field of a dipole.

Three identical point charges \(q\) are placed at each of three corners of a square of side \(L .\) Find the magnitude and direction of the net force on a point charge \(-3 q\) placed (a) at the center of the square and \((b)\) at the vacant corner of the square. In each case, draw a free-body diagram showing the forces exerted on the \(-3 q\) charge by each of the other three charges.

Point charge \(q_{1}=-5.00 \mathrm{nC}\) is at the origin and point charge \(q_{2}=+3.00 \mathrm{nC}\) is on the \(x\) -axis at \(x=3.00 \mathrm{cm} .\) Point \(P\) is on the \(y\) -axis at \(y=4.00 \mathrm{cm} .\) (a) Calculate the electric fields \(\vec{\boldsymbol{E}}_{1}\) and \(\vec{\boldsymbol{E}}_{2}\) at point \(P\) due to the charges \(q_{1}\) and \(q_{2} .\) Express your results in terms of unit vectors (see Example 21.6 ). (b) Use the results of part (a) to obtain the resultant field at \(P\) , expressed in unit vector form.

A negative point charge \(q_{1}=-4.00 \mathrm{nC}\) is on the \(x\) -axis at \(x=0.60 \mathrm{m} .\) A second point charge \(q_{2}\) is on the \(x\) -axis at \(x=-1.20 \mathrm{m} .\) What must the sign and magnitude of \(q_{2}\) be for the net electric field at the origin to be (a) 50.0 \(\mathrm{N} / \mathrm{C}\) in the \(+x\) -direction and \((\mathrm{b}) 50.0 \mathrm{N} / \mathrm{C}\) in the \(-x\) -x-direction?
See all solutions

Recommended explanations on Physics Textbooks

Fields in Physics

Read Explanation

Classical Mechanics

Read Explanation

Rotational Dynamics

Read Explanation

Physical Quantities And Units

Read Explanation

Astrophysics

Read Explanation

Electric Charge Field and Potential

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.