Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Calculate the mass ratio of a bromine atom to an atom of (a) neon (b) calcium (c) helium

Short Answer

Expert verified
Answer: (a) The mass ratio of a bromine atom to an atom of neon is 3.96, (b) the mass ratio of a bromine atom to an atom of calcium is 1.99, and (c) the mass ratio of a bromine atom to an atom of helium is 19.98.

Step by step solution

01

Find the atomic masses of bromine, neon, calcium, and helium

You can find the atomic masses of these elements in the periodic table. You will find that: - Bromine (Br) has an atomic mass of approximately 79.9 - Neon (Ne) has an atomic mass of approximately 20.2 - Calcium (Ca) has an atomic mass of approximately 40.1 - Helium (He) has an atomic mass of approximately 4.0
02

Calculate the mass ratio of bromine to neon

To find the mass ratio of bromine to neon, divide the atomic mass of bromine by the atomic mass of neon: Mass ratio = (Mass of Br)/(Mass of Ne) = (79.9)/(20.2) = 3.96
03

Calculate the mass ratio of bromine to calcium

To find the mass ratio of bromine to calcium, divide the atomic mass of bromine by the atomic mass of calcium: Mass ratio = (Mass of Br)/(Mass of Ca) = (79.9)/(40.1) = 1.99
04

Calculate the mass ratio of bromine to helium

To find the mass ratio of bromine to helium, divide the atomic mass of bromine by the atomic mass of helium: Mass ratio = (Mass of Br)/(Mass of He) = (79.9)/(4.0) = 19.98 In conclusion, (a) The mass ratio of a bromine atom to an atom of neon is 3.96 (b) The mass ratio of a bromine atom to an atom of calcium is 1.99 (c) The mass ratio of a bromine atom to an atom of helium is 19.98

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 Vaia!

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

Consider the following nuclear symbols. How many protons, neutrons, and electrons does each element have? What elements do \(\mathrm{R}, \mathrm{T},\) and \(\mathrm{X}\) represent? (a) \({ }_{14}^{30} \mathrm{R}\) (b) \({ }_{39}^{89} \mathrm{~T}\) (c) \({ }_{55}^{133} \mathrm{X}\)

Consider the following nuclear symbols. How many protons, neutrons, and electrons does each element have? What elements do \(\mathrm{A}, \mathrm{L},\) and \(\mathrm{Z}\) represent? (a) \(\frac{75}{33} \mathrm{~A}\) (b) \({ }_{23}^{51} \mathrm{~L}\) (c) \({ }_{54}^{131} \mathrm{Z}\)

Nuclei with the same mass number but different atomic numbers are called isobars. Consider Ca-40, Ca-41, K-41 and \(\mathrm{Ar}-41\) (a) Which of these are isobars? Which are isotopes? (b) What do Ca-40 and Ca-41 have in common? (c) Correct the statement (if it is incorrect): Atoms of \(\mathrm{Ca}-41, \mathrm{~K}-41,\) and \(\mathrm{Ar}-41\) have the same number of neutrons.

Give the number of protons and electrons in the following: (a) \(\mathrm{S}_{8}\) molecule. (b) \(\mathrm{SO}_{4}^{2-}\) ion. (c) \(\mathrm{H}_{2} \mathrm{~S}\) molecule. (d) \(S^{2-}\) ion.

A student saw the following nuclear symbol for an unknown element: \({ }_{11}^{23} \mathrm{X}\). Which of the following statements about \(\mathrm{X}\) and \({ }_{11}^{23} \mathrm{X}\) are true? (a) \(\mathrm{X}\) is sodium. (b) \(\mathrm{X}\) is vanadium. (c) X has 23 neutrons in its nucleus. (d) \(\mathrm{X}^{2+}\) has 13 electrons. (e) \({ }_{11}^{23} \mathrm{X}\) has a proton/neutron ratio of about 1.1 .

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.

Sign-up for free