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Particles in a Gold Ring. You have a pure ( 24 karal) gold ring with mass 17.7 \(\mathrm{g}\) . Gold has an atomic mass of 197 \(\mathrm{g} / \mathrm{mol}\) and an atomic number of \(79 .\) ( a) How many protons are in the ring, and what is their total positive charge? (b) If the ring carries no net charge, how many electrons are in it?

Short Answer

Expert verified
(a) The ring has approximately \(4.27 \times 10^{24}\) protons with a total positive charge of \(6.84 \times 10^{5}\) C. (b) It also has \(4.27 \times 10^{24}\) electrons since the ring is neutral.

Step by step solution

01

Calculate Moles of Gold

First, we need to determine the number of moles of gold in the ring. Use the formula:\[ \text{Moles of Gold} = \frac{\text{Mass of Gold}}{\text{Atomic Mass of Gold}} \]Given: Mass of Gold = 17.7 g, Atomic Mass of Gold = 197 g/mol. Thus, \[ \text{Moles of Gold} = \frac{17.7}{197} \approx 0.0898 \text{ moles} \].
02

Find Number of Atoms of Gold

Next, calculate the number of gold atoms using Avogadro's number, which is \(6.022 \times 10^{23}\) atoms/mol.\[ \text{Number of Atoms} = \text{Moles of Gold} \times 6.022 \times 10^{23} \]\[ \text{Number of Atoms} = 0.0898 \times 6.022 \times 10^{23} \approx 5.40 \times 10^{22} \text{ atoms} \].
03

Calculate Number of Protons

Since each gold atom has an atomic number of 79, each atom has 79 protons. The total number of protons is:\[ \text{Number of Protons} = 79 \times \text{Number of Atoms} \]\[ \text{Number of Protons} = 79 \times 5.40 \times 10^{22} \approx 4.27 \times 10^{24} \text{ protons} \].
04

Calculate Total Positive Charge

The charge of a proton is \(1.602 \times 10^{-19}\) Coulombs.The total positive charge is:\[ \text{Total Positive Charge} = \, \text{Number of Protons} \times \text{Charge of One Proton} \]\[ \text{Total Positive Charge} = \, 4.27 \times 10^{24} \times 1.602 \times 10^{-19} \approx 6.84 \times 10^{5} \text{ C} \].
05

Determine Number of Electrons

Since the gold ring carries no net charge, the number of electrons must equal the number of protons. Therefore, the number of electrons in the ring is also \(4.27 \times 10^{24}\).

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Key Concepts

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

Moles and Avogadro's Number
Understanding moles and Avogadro's number is key to tackling problems involving quantities of atoms in a substance. A "mole" is a unit used by chemists to count atoms, much like a "dozen" is used to count eggs. One mole contains exactly \(6.022 \times 10^{23}\) particles—whether they are atoms, molecules, or ions. This extremely large number is known as Avogadro's number. To find the number of moles in the gold ring, we divide its mass by the atomic mass of gold. For example:
  • Mass of the gold ring: 17.7 grams
  • Atomic mass of gold: 197 grams/mole
Hence, the moles of gold in the ring is \(\frac{17.7}{197} \approx 0.0898\) moles. Once moles are calculated, we use Avogadro's number to find the number of atoms. In this case, \(0.0898 \text{ moles} \times 6.022 \times 10^{23} \text{ atoms/mole} \approx 5.40 \times 10^{22} \text{ atoms}\). This calculation allows us to understand the scale of particles present even in a small jewelry item.
Protons and Atomic Number
Each element on the periodic table is defined by its atomic number, which tells us the number of protons in the nucleus of an atom. Protons are positively charged particles that reside in the nucleus. For gold, the atomic number is 79, indicating each gold atom has 79 protons. The atomic number is fundamental as it determines the element's identity. Considering the gold ring with \(5.40 \times 10^{22}\) atoms, the number of protons can be calculated as:
  • \(79 \times 5.40 \times 10^{22} \approx 4.27 \times 10^{24} \text{ protons}\)
The importance of protons extends beyond just counting particles; they contribute to the chemical properties and overall structure of the element. In this exercise, knowing the protons helps us calculate the total positive charge of the gold ring.
Electric Charge
The concept of electric charge is vital in understanding atomic interactions. Protons have a positive charge of \(1.602 \times 10^{-19}\) Coulombs each, which is fundamental to their role in atomic structure. In our example, once we calculate the number of protons in the gold ring, we can find the total positive charge by multiplying the number of protons by the charge of a single proton:
  • Total positive charge: \(4.27 \times 10^{24} \text{ protons} \times 1.602 \times 10^{-19} \text{ C/proton} \approx 6.84 \times 10^{5} \text{ C}\)
If the gold ring has no net charge, this implies the number of electrons equals the number of protons. Since electrons carry a negative charge equal in magnitude to that of protons, but negative, their presence balances out the total charge of the gold ring. Thus, we see that the number of electrons also amounts to \(4.27 \times 10^{24}\), ensuring electronic neutrality.

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Most popular questions from this chapter

Three point charges are arranged on a line. Charge \(q_{3}=+5.00 \mathrm{nC}\) and is at the origin. Charge \(q_{2}=-3.00 \mathrm{nC}\) and is at \(x=+4.00 \mathrm{cm} .\) Charge \(q_{1}\) is at \(x=+2.00 \mathrm{cm} .\) What is \(q_{1}\) (magnitude and sign) if the net force on \(q_{3}\) is zero?

A particle has charge \(-3.00 \mathrm{nC}\) (a) Find the magnitude and direction of the electric field due to this particle at a point 0.250 \(\mathrm{m}\) directly above it. (b) At what distance from this particle does its electric field have a magnitude of 12.0 \(\mathrm{N} / \mathrm{C} ?\)

A dipole consisting of charges \(\pm e, 220 \mathrm{nm}\) apart, is placed between two very large (essentially infinite) sheets carrying equal but opposite charge densities of 125\(\mu \mathrm{C} / \mathrm{m}^{2}\) (a) What is the maximum potential energy this dipole can have due to the sheets, and how should it be oriented relative to the sheets to attain this value? (b) What is the maximum torgue the sheets can exert on the dipole, and how should it be oriented relative to the sheets to attain this value? (c) What net force do the two sheets exert on the dipole?

(a) Sketch the electric field lines for an infinite line of charge. You may find it helpful to show the field lines in a plane containing the line of charge in one sketch and the field lines in a a plane perpendicular to the line of charge in a second sketch. (b) Explain how your sketches show (i) that the magnitude \(E\) of the electric field depends only on the distance \(r\) from the line of charge and (ii) that \(E\) decreases like \(1 / r .\)

Strength of the Electric Force. Imagine two \(1.0-\mathrm{g}\) bags of protons, one at the earth's north pole and the other at the south pole. (a) How many protons are in each bag? (b) Calculate the gravitational attraction and the electrical repulsion that each bag exerts on the other. (c) Are the forces in part (b) large enough for you to feel if you were holding one of the bags?

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