/*! 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 14 Which of the following nuclei ha... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 14

Which of the following nuclei has 3 more neutrons than protons? (Remember: The number before the symbol indicates atomic mass.) (A) 11 \(\mathrm{B}\) (B) 37 \(\mathrm{Cl}\) (C) 24 \(\mathrm{Mg}\) (D) 70 \(\mathrm{Ga}\)

Short Answer

Expert verified
The nucleus which has 3 more neutrons than protons is \(37Cl\) - option (B).

Step by step solution

01

Calculate Neutrons in each Nucleus

For each of the nuclei given in the options, subtract the number of protons from the total atomic mass to get the number of neutrons. The atomic number (which is also the number of protons) of each element is: \(B\) - 5, \(Cl\) - 17, \(Mg\) - 12, \(Ga\) - 31. So number of neutrons for each are: \(11B\) - 6, \(37Cl\) - 20, \(24Mg\) - 12, \(70Ga\) - 39.
02

Calculate Difference between Neutrons and Protons

Subtract the number of protons from the calculated number of neutrons for each element to confirm which has exactly 3 more neutrons than protons. The difference for each are: \(11B\) - 1, \(37Cl\) - 3, \(24Mg\) - 0, \(70Ga\) - 8.
03

Choose the Correct Nucleus

The nucleus which has exactly 3 more neutrons than protons is \(37Cl\). Therefore, the correct choice is option (B).

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 91影视!

Key Concepts

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

Atomic Structure
Understanding atomic structure is crucial in chemistry. It forms the basis for everything, from the elements on the periodic table to complex chemical reactions. Atoms are made up of three major parts: protons, neutrons, and electrons.
  • Protons are positively charged particles found in the nucleus. The number of protons defines the element and is represented by the atomic number.
  • Neutrons are neutral particles, meaning they have no charge. They also reside in the nucleus and contribute to the atom's mass.
  • Electrons are negatively charged and orbit the nucleus in electron clouds or shells.
The atomic mass of an atom is the sum of its protons and neutrons. Electrons have a negligible mass and do not significantly affect the atomic mass. This composition of the atom is fundamental in determining the chemical behavior of an element, as well as its stability and isotope formation.
Neutrons and Protons Calculation
To solve problems involving the calculation of neutrons and protons, start by understanding the terms clearly:- **Protons:** Represented by the atomic number of the element, found on the periodic table.- **Neutrons:** Calculated by subtracting the number of protons from the atomic mass.Let's consider an example using chlorine, \(^{37}Cl\). The atomic number (number of protons) for chlorine is 17. Given its atomic mass as 37, we find the number of neutrons as follows:\[\text{Number of neutrons} = \text{atomic mass} - \text{number of protons} = 37 - 17 = 20.\]In practice:
  • The atomic mass number tells you the sum of protons and neutrons.
  • To find the neutrons, use the atomic number to determine the protons, then subtract from the atomic mass.
Estimating this helps you solve exercises related to the structure and characteristics of atoms. It's important to distinguish between the two types of particles, as both play critical roles in the identity and mass of an element.
Isotopes
Isotopes are different forms of the same element. They have the same number of protons but differ in neutrons. This difference in neutron count can affect the atomic mass and sometimes the stability of the element.
  • Isotopes retain the element's chemical identity as their proton number remains unchanged.
  • The atomic mass varies due to the additional or fewer neutrons.
For example, Carbon has isotopes like \(^{12}C\) and \(^{14}C\), where both have 6 protons but different numbers of neutrons, 6 and 8 respectively. These isotopes can create variations in weight and help in processes like carbon dating.In the exercise, assessing isotopes involves comparing the neutron-to-proton difference. Chlorine was identified with the nucleus \(^{37}Cl\) having 3 more neutrons than protons. This characteristic is typical of isotopic variations where the neutron count affects mass and potentially stability but not the chemical properties.

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

Starting with a stock solution of 18.0 \(\mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) , what is the proper procedure to create a 1.00 \(\mathrm{L}\) sample of a 3.0 \(\mathrm{M}\) solution of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) in a volumetric flask? (A) Add 167 mL of the stock solution to the flask, then fill the flask the rest of the way with distilled water while swirling the solution. (B) Add 600 mL of the stock solution to the flask, then fill the flask the rest of the way with distilled water while swirling the solution. (C) Fill the flask partway with water, then add 167 mL of the stock solution, swirling to mix it. Last, fill the flask the rest of the way with distilled water. (D) Fill the flask partway with water, then add 600 mL of the stock solution, swirling to mix it. Last, fill the flask the rest of the way with distilled water.

Molten \(\mathrm{AlCl}_{3}\) is electrolyzed with a constant current of 5.00 amperes over a period of 600.0 seconds. Which of the following expressions is equal to the maximum mass of Al(s) that plates out? (1 faraday = 96,500 coulombs) (A) \(\frac{(600)(5.00)}{(96,500)(3)(27.0)}\) grams (B) \(\frac{(600)(5.00)(3)(27.0)}{(96,500)}\) grams (C) \(\frac{(600)(5.00)(27.0)}{(96,500)(3)}\) grams (D) \(\frac{(96,500)(3)(27.0)}{(600)(5.00)}\) grams

\(2 \mathrm{HI}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{HCl}(g)+\mathrm{I}_{2}(g)+\) energy A gaseous reaction occurs and comes to equilibrium, as shown above. Which of the following changes to the system will serve to increase the number of moles of \(\mathrm{I}_{2}\) present at equilibrium? (A) Increasing the volume at constant temperature (B) Decreasing the volume at constant temperature (C) Increasing the temperature at constant volume (D) Decreasing the temperature at constant volume

Use the following information to answer questions 1-5. \(\begin{array}{ll}{\text { Reaction } 1 : \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{H}_{2}(g) \rightarrow 2 \mathrm{NH}_{3}(g)} & {\Delta H=?} \\ {\text { Reaction } 2 : \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{CH}_{4} \mathrm{O}(l) \rightarrow \mathrm{CH}_{2} \mathrm{O}(g)+\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g)} & {\Delta H=-37 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}} \\ {\text { Reaction } 3 : \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightarrow 2 \mathrm{NH}_{3}(g)} & {\Delta H=-46 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}} \\ {\text { Reaction } 4 : \mathrm{CH}_{4} \mathrm{O}(l) \rightarrow \mathrm{CH}_{2} \mathrm{O}(g)+\mathrm{H}_{2}(g)} & {\Delta H=-65 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}}\end{array}\) What is the enthalpy change for reaction 1\(?\) (A) \(-148 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}\) (B) \(-56 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}\) (C) \(-18 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}\) (D) \(+148 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}\)

Use the following information to answer questions 25-28. A voltaic cell is created using the following half-cells: \(\begin{array}{ll}{\mathrm{Cr}^{3+}+3 e \rightarrow \mathrm{Cr}(s)} & {E^{\circ}=-0.41 \mathrm{V}} \\ {\mathrm{Pb}^{2+}+2 e \rightarrow \mathrm{Pb}(s)} & {E^{\circ}=-0.12 \mathrm{V}}\end{array}\) The concentrations of the solutions in each half-cell are 1.0 M. Which of the following best describes the activity in the salt bridge as the reaction progresses? (A) Electrons flow through the salt bridge from the \(\mathrm{Pb} / \mathrm{Pb}^{2+}\) half-cell to the \(\mathrm{Cr} / \mathrm{Cr}^{3+}\) half-cell. (B) \(\quad \mathrm{Pb}^{2+}\) flows to the \(\mathrm{Cr} / \mathrm{Cr}^{3+}\) flows to the \(\mathrm{Pb} / \mathrm{Pb}^{2+}\) half-cell. (C) \(\mathrm{Na}^{+}\) flows to the \(\mathrm{Cr} / \mathrm{Cr}^{3+}\) half-cell, and \(\mathrm{Cl}^{-}\) flows to the \(\mathrm{Pb} / \mathrm{Pb}^{2+}\) half- cell. (D) \(\quad \mathrm{Na}^{+}\) flows to the \(\mathrm{Pb} / \mathrm{Pb}^{2}\) half- cell, and \(\mathrm{Cl}^{-}\) flows to the \(\mathrm{Cr} / \mathrm{Cr}^{3+}\) half- cell.
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Physical Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Nuclear Chemistry

Read Explanation

Organic Chemistry

Read Explanation
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.