/*! 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 7 Water has a mass per mole of \(1... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Water has a mass per mole of \(18.0 \mathrm{g} / \mathrm{mol}\), and each water molecule \(\left(\mathrm{H}_{2} \mathrm{O}\right)\) has 10 electrons. (a) How many electrons are there in one liter \(\left(1.00 \times 10^{-3} \mathrm{m}^{3}\right)\) of water? (b) What is the net charge of all these electrons?

Short Answer

Expert verified
There are approximately \(3.34 \times 10^{26}\) electrons with a net charge of \(-5.34 \times 10^{7}\) C.

Step by step solution

01

Determine the Mass of Water in One Liter

Since the density of water is approximately 1 g/mL, the mass of 1 liter (which is 1000 mL) of water is approximately 1000 grams.
02

Calculate the Number of Moles of Water

Using the mass of water per mole given as 18.0 g/mol, calculate the number of moles of water in 1000 grams:\[ \text{Number of moles} = \frac{1000 \text{ g}}{18.0 \text{ g/mol}} \approx 55.56 \text{ moles} \]
03

Determine the Number of Water Molecules

Using Avogadro's number (\(6.022 \times 10^{23}\) molecules/mol), calculate the number of water molecules in 55.56 moles:\[ \text{Number of molecules} = 55.56 \times 6.022 \times 10^{23} \approx 3.34 \times 10^{25} \text{ molecules} \]
04

Calculate the Total Number of Electrons

Since each water molecule has 10 electrons, multiply the number of molecules by the number of electrons per molecule:\[ \text{Number of electrons} = 3.34 \times 10^{25} \times 10 = 3.34 \times 10^{26} \text{ electrons} \]
05

Compute the Net Charge of the Electrons

Each electron has a charge of \(-1.6 \times 10^{-19}\) coulombs. So, the total charge is:\[ \text{Net charge} = 3.34 \times 10^{26} \times (-1.6 \times 10^{-19}) \approx -5.34 \times 10^{7} \text{ C} \]

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.

Mole Calculations
Solving problems in chemistry often requires us to determine the number of moles of a substance. The mole is a unit that helps chemists count particles at the atomic scale, like molecules or atoms. It's used because dealing with such tiny units one by one would be impractical. To perform mole calculations, we use the formula:
  • Number of moles = \( \frac{\text{mass of substance (g)}}{\text{molar mass (g/mol)}} \)
This formula allows us to find how many moles are present in a given mass of a substance. For water, with a molar mass of 18.0 g/mol, you can find the number of moles in a specified mass by dividing the total mass of water by its molar mass. If you have 1000 grams of water, you calculate the moles by dividing 1000 by 18. This approach makes it easier to move forward with further calculations involving the substance.
Avogadro's Number
Avogadro's number, \(6.022 \times 10^{23}\), is a huge number that represents the quantity of atoms, molecules, or particles in one mole of any substance. It's a fundamental constant in chemistry, allowing chemists to transition from the microscopic scale to the macroscopic scale.
Whenever you need to find out how many molecules or atoms are in a given amount of moles, you multiply by Avogadro's number:
  • Number of molecules = (number of moles) \( \times \) Avogadro's number
This means if you have approximately 55.56 moles of water, you multiply that by \(6.022 \times 10^{23}\) to find the number of molecules. Understanding Avogadro's number helps comprehend the immense number of particles contained in even a small amount of material.
Density of Water
The density of a substance is essential in converting between volume and mass. For water, the density is usually 1 g/mL, meaning that one milliliter of water has a mass of one gram. Knowing this, you can easily determine the mass of a certain volume of water. For example, 1 liter—which is 1000 milliliters—has a mass of 1000 grams.
  • Mass of water (g) = Volume (mL) \( \times \) Density (g/mL)
This relationship is particularly useful in experiments and calculations where the conversion between mass and volume is required. Consistently using density in such calculations helps ensure that your results for matters like chemical reactions or molecular computations are accurate.
Electric Charge of Electrons
Electrons carry a fundamental unit of charge, known as the elementary charge, which is quantitatively \(-1.6 \times 10^{-19}\) coulombs. In many calculations, particularly those involving chemistry or physics, understanding this value is crucial.
If you know the number of electrons, you can calculate the total charge by multiplying the number of electrons by the charge of one electron:
  • Total charge \(= \, \text{Number of electrons} \times -1.6 \times 10^{-19} \, \text{Coulombs/electron}\)
In the problem given, after finding the total number of electrons in a certain volume of water, this formula allows the calculation of their net charge. Understanding electron charge is vital for grasping electrochemical reactions and the behavior of substances in electric fields.

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

The membrane surrounding a living cell consists of an inner and an outer wall that are separated by a small space. Assume that the membrane acts like a parallel plate capacitor in which the effective charge density on the inner and outer walls has a magnitude of \(7.1 \times 10^{-6} \mathrm{C} / \mathrm{m}^{2} .\) (a) What is the magnitude of the electric field within the cell membrane? (b) Find the magnitude of the electric force that would be exerted on a potassium ion \(\left(\mathrm{K}^{+} ;\right.\) charge \(\left.=+e\right)\) placed inside the membrane.

A charge \(+q\) is located at the origin, while an identical charge is located on the \(x\) axis at \(x=+0.50 \mathrm{m}\). A third charge of \(+2 q\) is located on the \(x\) axis at such a place that the net electrostatic force on the charge at the origin doubles, its direction remaining unchanged. Where should the third charge be located?

ssm Consider three identical metal spheres, A, B, and C. Sphere A carries a charge of \(+5 q .\) Sphere \(\mathrm{B}\) carries a charge of \(-q .\) Sphere \(\mathrm{C}\) carries no net charge. Spheres \(A\) and \(B\) are touched together and then separated. Sphere \(C\) is then touched to sphere \(A\) and separated from it. Last, sphere \(C\) is touched to sphere \(\mathrm{B}\) and separated from it. (a) How much charge ends up on sphere \(\mathrm{C} ?\) What is the total charge on the three spheres (b) before they are allowed to touch each other and (c) after they have touched?

A circular surface with a radius of \(0.057 \mathrm{m}\) is exposed to a uniform external electric field of magnitude \(1.44 \times 10^{4} \mathrm{N} / \mathrm{C}\). The magnitude of the electric flux through the surface is \(78 \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}\). What is the angle (less than \(90^{\circ}\) ) between the direction of the electric field and the normal to the surface?

A particle of charge \(+12 \mu C\) and \(\operatorname{mass} 3.8 \times 10^{-5} \mathrm{kg}\) is released from rest in a region where there is a constant electric field of \(+480 \mathrm{N} / \mathrm{C}\). What is the displacement of the particle after a time of \(1.6 \times 10^{-2} \mathrm{s} ?\)

See all solutions

Recommended explanations on Physics 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.