/*! 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 24 If the volume of a proton is sim... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

If the volume of a proton is similar to the volume of an electron, how will the densities of these two particles compare to each other?

Short Answer

Expert verified
The density ratio of a proton to an electron can be determined by taking the ratio of their masses, since their volumes are assumed to be similar. The mass of a proton is about 1.0073 atomic mass units (u), and the mass of an electron is about 0.0005485799 u. Converting to kilograms and finding the ratio, the density of a proton is approximately 1836 times greater than the density of an electron.

Step by step solution

01

1. Finding the masses of proton and electron

We need to know the masses of a proton and an electron to compare their densities. From the atomic mass unit (u), we have the following values: Mass of proton = 1.0073 u Mass of electron = 0.0005485799 u
02

2. Converting the atomic mass unit (u) to kilograms

Since density is generally given in kilograms per cubic meter (kg/m³), we need to convert the mass of proton and electron from atomic mass units to kilograms. The conversion factor is 1 u = \(1.66054 × 10^{-27}\) kg. Thus, the mass of a proton in kilograms is: \(1.0073 \times 1.66054 × 10^{-27}\) kg And the mass of an electron in kilograms is: \(0.0005485799 \times 1.66054 × 10^{-27}\) kg
03

3. Finding the density ratio

Since the volumes of the proton and electron are similar, their ratios of densities can be found by taking the ratio of their masses: Density ratio (proton/electron) = \(\frac{Mass \thinspace of \thinspace proton}{Mass \thinspace of \thinspace electron}\) Plugging in their masses in kilograms from step 2: Density ratio (proton/electron) = \(\frac{1.0073 \times 1.66054 × 10^{-27}}{0.0005485799 \times 1.66054 × 10^{-27}}\) Notice that the factor \(1.66054 × 10^{-27}\) is present in both the numerator and the denominator, so we can cancel it out. Density ratio (proton/electron) = \(\frac{1.0073}{0.0005485799}\)
04

4. Calculating the density ratio

Density ratio (proton/electron) = \(1836.152\) This means that the density of a proton is approximately 1836 times greater than the density of an electron.

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!

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 formulas and common names for several substances are given below. Give the systematic names for these substances. a. sugar of lead \(\mathrm{Pb}\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)_{2}\) \(\begin{array}{ll}\text { b. blue vitrol } & \mathrm{CuSO}_{4}\end{array}\) c. quicklime \(\mathrm{CaO}\) d. Epsom salts \(\mathrm{MgSO}_{.}\) e. milk of magnesia \(\quad \mathrm{Mg}(\mathrm{OH})_{2}\) f. gypsum \(\mathrm{CaSO}_{4}\) \(\mathrm{g}\). laughing gas \(\mathrm{N}_{2} \mathrm{O}\)

Section \(2.3\) describes the postulates of Dalton's atomic theory With some modifications, these postulates hold up very well regarding how we view elements, compounds, and chemical reactions today. Answer the following questions concerning Dalton's atomic theory and the modifications made today. a. The atom can be broken down into smaller parts. What are the smaller parts? b. How are atoms of hydrogen identical to each other and how can they be different from each other? c. How are atoms of hydrogen different from atoms of helium? How can \(\mathrm{H}\) atoms be similar to \(\mathrm{He}\) atoms? d. How is water different from hydrogen peroxide \(\left(\mathrm{H}_{2} \mathrm{O}_{2}\right)\) even though both compounds are composed of only hydrogen and oxygen? e. What happens in a chemical reaction and why is mass conserved in a chemical reaction?

The isotope of an unknown element, \(X\), has a mass number of 79 . The most stable ion of the isotope has 36 electrons and forms a binary compound with sodium having a formula of \(\mathrm{Na}_{2} \mathrm{X}\). Which of the following statements is(are) true? For the false statements, correct them. a. The binary compound formed between \(\mathrm{X}\) and fluorine will be a covalent compound. b. The isotope of \(\mathrm{X}\) contains 38 protons. c. The isotope of \(X\) contains 41 neutrons. d. The identity of \(X\) is strontium, \(S r\).

Knowing the names of similar chlorine oxyanions and acids, deduce the names of the following: \(\mathrm{IO}^{-}, \mathrm{IO}_{2}^{-}, \mathrm{IO}_{3}^{-}, \mathrm{IO}_{4}^{-}, \mathrm{HIO}\) \(\mathrm{HIO}_{2}, \mathrm{HIO}_{3}, \mathrm{HIO}_{4}\)

Give the names of the metals that correspond to the following symbols: \(\mathrm{Sn}, \mathrm{Pt}, \mathrm{Hg}, \mathrm{Mg}, \mathrm{K}, \mathrm{Ag}\).

See all solutions

Recommended explanations on Chemistry 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.