Skip to main content
Chemistry LibreTexts

Module 3 Practice

  • Page ID
    278643
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    Exercise \(\PageIndex{1}\)

    For each isotope shown, give the number of protons, neutrons, and electrons. 

    a)   199F1–

    b)  5224Cr3+

    c)  3115P3–

    Answer a

    9 protons, 10 neutrons, 10 electrons

    Answer b

    24 protons, 28 neutrons, 21 electrons

    Answer c

    15 protons, 16 neutrons, 18 electrons

     

     

    Exercise \(\PageIndex{1}\)

    Given the number of protons, neutrons, and electrons, what would be the mass number and the charge number of each isotope?  What would be the atomic number?  The identity of the element? 

    d)  14 protons, 15 neutrons, 14 electrons

    e)  3 protons, 4 neutrons, 2 electrons

    f)  30 protons, 36 neutrons, 28 electrons

    Answer d

    Mass number is 29, charge is 0.  Atomic number is 14, identity is silicon (Si).

    Answer e

    Mass number is 7, charge is +1.  Atomic number is 3, identity is lithium (Li).

    Answer f

    Mass number is 66, charge is +2.  Atomic number is 30, identity is zinc (Zn).

     

    Exercise \(\PageIndex{1}\)

    For each element shown, tell what charge it has when it forms an ion. 

    g)  S

    h)  Ag

    i)  Ca

    Answer g

    –2

    Answer h

    +1

    Answer i

    +2

     

    Exercise \(\PageIndex{1}\)

    Give the electron configuration for each of the following atoms.

    j)  Mg

    k)  Mn

    l)  Sr

    Answer j

    1s2 2s2 2p6 3s2     (total of 12 electrons)

    Answer k

    1s2 2s2 2p6 3s2 3p6 4s2 3d5  (total of 25 electrons)

    Answer l

    1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2  (total of 38 electrons)

     

    Exercise \(\PageIndex{1}\)

    Give the electron configuration for each of the following ions.

    m)  N3–

    n)  S2–

    o)  Rb1+

    Answer m

    1s2 2s2 2p6    (total of 10 electrons)

    Answer n

    1s2 2s2 2p6 3s2 3p6    (total of 18 electrons)

    Answer o

    1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6   (total of 36 electrons)

    Exercise \(\PageIndex{1}\)

              K                Ca

              Rb              Sr

    Out of the four elements listed above, choose the one that has

    j)  the largest radius

    k)  the least ionization energy

    l)  the most metallic character

    Answer j

    Rb

    Answer k

    Rb

    Answer l

    Rb

    Exercise \(\PageIndex{1}\)

    The element chlorine is found in nature as a mixture of 35Cl and 37Cl.  Given that the average atomic mass of chlorine is 35.45 amu, which isotope is more abundant?

    Answer

    35Cl is more abundant.  Since the average mass is closer to 35 amu, you have more of the isotope with a mass of 35 amu than you do of the isotope with a mass of 37 amu.

    Exercise \(\PageIndex{1}\)

    If there were chemists on another planet, their periodic table might be different because the elements might have different "natural" abundances.  If on the other planet, the element sulfur had the following natural abundances, calculate the average atomic mass that the chemists on that planet would use.

    68.44 % is 32S with mass of 31.9721 amu

    23.24 % is 33S with mass of 32.9715 amu

    8.32 % is 34S with mass of 33.9679 amu

    Answer

    Add up (uncertain digits are underlined):

    21.8817amu

      7.6626 amu

      2.8261 amu

    Total is 32.3704 amu which rounds to 32.37 amu


    Module 3 Practice is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

    • Was this article helpful?