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Chapter 18: Problem 7

Water has a mass per mole of \(18.0 \mathrm{g} / \mathrm{mol},\) and euch water molecule (H, O ) has 10 clectrons. (a) How many clectrons are there in onc 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
(a) \(3.34 \times 10^{26}\) electrons, (b) \(-5.34 \times 10^{7} \text{ C}\).

Step by step solution

01

Find the number of moles in 1 liter of water

The density of water is approximately \(1 \text{ g/mL}\), which means \(1\text{ L}\) of water has a mass of \(1000 \text{ g}\). We use the formula for moles \( n = \frac{\text{mass}}{\text{molar mass}} \). Here the molar mass is \(18.0 \text{g/mol}\). Substitute the mass of water:\[ n = \frac{1000 \text{ g}}{18.0 \text{ g/mol}} \approx 55.56 \text{ mol} \]
02

Calculate the number of water molecules

Avogadro's number, \( 6.022 \times 10^{23} \text{ molecules/mol} \), gives the number of molecules per mole. Multiply the moles of water by Avogadro's number:\[ \text{No. of molecules} = 55.56 \text{ mol} \times 6.022 \times 10^{23} \text{ molecules/mol} \approx 3.34 \times 10^{25} \text{ molecules} \]
03

Determine the total number of electrons

Each water molecule has 10 electrons. Therefore, multiply the number of molecules by the number of electrons per molecule:\[ \text{Total electrons} = 3.34 \times 10^{25} \text{ molecules} \times 10 \text{ electrons/molecule} \approx 3.34 \times 10^{26} \text{ electrons} \]
04

Calculate the net charge of the electrons

The charge of a single electron is \(-1.60 \times 10^{-19} \text{ C/electron}\). Multiply the total number of electrons by this charge:\[ \text{Net charge} = 3.34 \times 10^{26} \text{ electrons} \times (-1.60 \times 10^{-19} \text{ C/electron}) \approx -5.34 \times 10^{7} \text{ C} \]

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Key Concepts

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

Density of Water
Water is a unique substance with a consistent density, which is approximately 1 gram per milliliter (g/mL). This means that in a standard atmospheric condition, each liter of water weighs approximately 1000 grams. This property is pivotal for various calculations in chemistry, particularly when converting between volume and mass.
  • This density value allows for straightforward conversion: 1 liter of water equals 1000 grams.
  • It is a fundamental constant used in scientific calculations involving water.
  • Understanding this concept is crucial for determining how much matter is present in a given volume of water.
The consistency of this value makes water an excellent standard in scientific experiments and calculations, simplifying the process of calculating moles and molecules in water-based solutions.
Avogadro's Number
Avogadro's Number, represented as \(6.022 \times 10^{23}\), is an essential constant in chemistry. It denotes the number of particles, typically atoms or molecules, in one mole of a substance.
  • This number serves as a bridge between the macroscopic scale, which we can measure, and the microscopic scale of atoms and molecules.
  • In practical terms, it allows us to quantify atoms or molecules by measuring mass, turning grams into moles.
  • In exercises like the one provided, Avogadro's number is used to convert the number of moles of water into the actual number of molecules.
Essentially, it provides a way to count things that are far too small to individually count in a lab setting, making calculations of molecular quantities manageable.
Electron Charge
Electrons are subatomic particles with a negative electric charge, typically measured in Coulombs (C). Each electron carries a charge of \(-1.60 \times 10^{-19}\) C. This charge is small but fundamental to understanding atomic interactions and chemical reactions.
  • In a single water molecule, there are ten electrons, each contributing to the molecule's overall electronic characteristics.
  • The net charge of a group of electrons is calculated by multiplying the number of electrons by the charge per electron.
  • Understanding electron charge is crucial when exploring electrical properties of substances and the flow of electrons in electric currents.
This fundamental constant helps quantify the effects of electron movement in physics and chemistry, particularly in problems involving electricity or charge distribution.
Molar Mass Calculation
Molar mass is a substance's mass per mole, commonly expressed in grams per mole (g/mol). For water, the molar mass is calculated as 18.0 g/mol.
  • The molar mass is derived from the individual atomic masses of hydrogen and oxygen; here, hydrogen contributes about 1.0 g/mol, and oxygen contributes about 16.0 g/mol.
  • To find the number of moles in a substance, you divide the mass of the substance by its molar mass using the formula: \( n = \frac{\text{mass}}{\text{molar mass}} \).
  • This calculation is crucial for converting mass into a usable form for further calculations like the number of molecules or atoms.
Knowing how to calculate and use molar mass allows chemists to prepare solutions and understand proportions of components in chemical reactions effectively.

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Most popular questions from this chapter

Four identical metal spheres have charges of \(q_{\mathrm{A}}=-8.0 \mu \mathrm{C}\) \(q_{\mathrm{B}}=-2.0 \mu \mathrm{C}, q_{C}=+5.0 \mu \mathrm{C},\) and \(q_{\mathrm{D}}=+12.0 \mu \mathrm{C}\) (a) Two of the spheres are brought together so they touch, and then they are separated. spheres are brought together so they touch, and then they are separated. Which spheres are they, if the final charge on each one is \(+5.0 \mu \mathrm{C} ?\) (b) In a similar manncr, which three spheres are brought together and then separated, if the final charge on each of the three is \(+3.0 \mu \mathrm{C} ?\) (c) The final charge on each of the three separated spheres in part (b) is \(+3.0 \mu \mathrm{C}\) . How many clectrons would have to be addcd to onc of these spheres to make it electrically neutral?

Two tiny conducting spheres are identical and carry charges of \(-20.0 \mu \mathrm{C}\) and \(+50.0 \mu \mathrm{C}\) . They are separated by a distance of 2.50 \(\mathrm{cm}\) . (a) What is the magnitude of the force that each sphere experiences, and is the force attractive or repulsive? (b) The spheres are brought into contact and then separated to a distance of 2.50 cm. Determine the magnitude of the force that each sphere now experiences, and state whether the force is attractive or repulsive.

When point chargcs \(q_{1}=+8.4 \mu \mathrm{C}\) and \(q_{2}=+5.6 \mu \mathrm{C}\) are brought near each other, each experiences a repulsive force of magnitude 0.66 \(\mathrm{N}\) . Determine the distance between the charges.

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 small object has a mass of \(3.0 \times 10^{-3} \mathrm{kg}\) and a charge of \(-34 \mu \mathrm{C}\) . It is placcd at a certain spot where there is an clectric ficld. When released, the object experiences an acceleration of \(2.5 \times 10^{3} \mathrm{m} / \mathrm{s}^{2} in the direction of the \)+x$ axis. Determine the magnitude and direction of the electric field.
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