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Chapter 16: Problem 1

Find the total positive charge of all the protons in \(1.0 \mathrm{mol}\) of water.

Short Answer

Expert verified
Answer: The total positive charge of all the protons in 1.0 mol of water is approximately \(1.93\times10^{6}\) coulombs.

Step by step solution

01

Determine the number of protons in one water molecule

A water molecule has the chemical formula H₂O, which means it consists of two hydrogen atoms and one oxygen atom. Since each hydrogen atom contains 1 proton, there are 2 protons in a single water molecule.
02

Determine the number of molecules in 1.0 mol of water

One mole of any substance contains Avogadro's number (approximately \(6.022\times10^{23}\)) of particles (atoms, molecules, ions, etc.). Therefore, there are about \(6.022\times10^{23}\) water molecules in 1.0 mol of water.
03

Find the total number of protons in 1.0 mol of water

To determine the total number of protons in 1.0 mol of water, we need to multiply the number of protons in a single water molecule by the number of water molecules in 1.0 mol of water: Total protons \(= 2\) protons/water molecule \(\times 6.022\times10^{23}\) water molecules/mol \(= 1.2044\times10^{24}\) protons
04

Find the total positive charge of all the protons

The elementary charge of a single proton is approximately \(1.602\times10^{-19}\) coulombs. Therefore, to find the total positive charge of all the protons, we multiply the total number of protons by the elementary charge of a proton: Total positive charge \(= 1.2044\times10^{24}\) protons \(\times 1.602\times10^{-19}\) coulombs/proton \(\approx 1.93\times10^{6}\) coulombs So, the total positive charge of all the protons in 1.0 mol of water is about \(1.93\times10^{6}\) coulombs.

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Most popular questions from this chapter

Point charges are arranged on the vertices of a square with sides of $2.50 \mathrm{cm} .$ Starting at the upper left corner and going clockwise, we have charge A with a charge of \(0.200 \mu \mathrm{C}, \mathrm{B}\) with a charge of \(-0.150 \mu \mathrm{C}, \mathrm{C}\) with a charge of \(0.300 \mu \mathrm{C},\) and \(\mathrm{D}\) with a mass of \(2.00 \mathrm{g},\) but with an unknown charge. Charges \(\mathrm{A}, \mathrm{B},\) and \(\mathrm{C}\) are fixed in place, and \(\mathrm{D}\) is free to move. Particle D's instantaneous acceleration at point \(D\) is \(248 \mathrm{m} / \mathrm{s}^{2}\) in a direction \(30.8^{\circ}\) below the negative \(x\) -axis. What is the charge on D?
In a uniform electric field of magnitude \(E\), the field lines cross through a rectangle of area \(A\) at an angle of \(60.0^{\circ}\) with respect to the plane of the rectangle. What is the flux through the rectangle?
Use Gauss's law to derive an expression for the electric field outside the thin spherical shell of Conceptual Example 16.8
(a) Find the electric flux through each side of a cube of edge length \(a\) in a uniform electric field of magnitude \(E\) The field direction is perpendicular to two of the faces. (b) What is the total flux through the cube?
(a) What would the net charges on the Sun and Earth have to be if the electric force instead of the gravitational force were responsible for keeping Earth in its orbit? There are many possible answers, so restrict yourself to the case where the magnitude of the charges is proportional to the masses. (b) If the magnitude of the charges of the proton and electron were not exactly equal, astronomical bodies would have net charges that are approximately proportional to their masses. Could this possibly be an explanation for the Earth's orbit?
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