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Bishop - An Intro to Chemistry - Chemistry First (Homework)

James Finch

Chemistry - College, section 1, Fall 2019

Instructor: Dr. Friendly

Current Score : 12 / 13

Due : Saturday, August 17, 2019 08:00 EDT

Last Saved : n/a Saving...  ()

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  • Instructions

    An Introduction to Chemistry - Chemistry First 2/e by Mark Bishop, is published by Chiral Publishing. An Introduction to Chemistry includes over 1500 assignable online homework problems from the text and 900 test bank questions. Every problem includes a link to the appropriate section of the eBook, as both pdf and audio files. Where appropriate, problems also link to additional resources, such as author-created animations based on key illustrations in the textbook. Click here for a list of all of the questions coded in WebAssign. This demo assignment, using a range of problems from An Introduction to Chemistry, uses many of WebAssign's hallmark features, including our chemPad answer entry palette, which provides intelligent feedback based on specific student answers and MarvinSketch, an assessable molecular drawing tool. Most problems include randomized and chemically parameterized values.

    This demo allows many submissions and allows you to try another version of the same question for practice wherever the problem has randomized values.

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1. –/4 points PrepChemCF1 2.1.C.012. My Notes
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The average distance between particles in a gas is about (No Response) seenKey

ten times

 the diameter of each particle. This leads to the gas particles themselves taking up only about (No Response) seenKey

0.1%

 of the total volume. The other (No Response) seenKey

99.9%

 of the total volume is empty space. In contrast, the particles of a liquid fill about (No Response) seenKey

70%

 of the liquid's total volume.

Solution or Explanation
The average distance between particles in a gas is about ten times the diameter of each particle. This leads to the gas particles themselves taking up only about 0.1% of the total volume. The other 99.9% of the total volume is empty space. In contrast, the particles of a liquid fill about 70% of the liquid's total volume.

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2. –/3 points PrepChemCF1 3.1.P.044. My Notes
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–/3
 
Write the chemical formula for each of the following compounds. List the symbols for the elements in the order that the elements are mentioned in the description.
(a) a compound that contains one aluminum atom for every three chlorine atoms

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AlCl3


(b) a compound that contains two lithium atoms and one carbon atom for every three oxygen atoms

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Li2CO3


(c) a compound with molecules that consist of two nitrogen atoms and three oxygen atoms

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N2O3


Solution or Explanation
(a) A compound that contains one aluminum atom for every three chlorine atoms contains aluminum (Al) and chlorine (Cl) in a ratio of 1:3. The elements are written as their symbols and the ratios of the elements are written as subscripts after the element, omitting the subscript when the ratio is one (1), to give AlCl3.

(b) A compound that contains two lithium atoms and one carbon atom for every three oxygen atoms contains lithium (Li), carbon (C), and oxygen (O) in a ratio of 2:1:3. The elements are written as their symbols and the ratios of the elements are written as subscripts after the element, omitting the subscript when the ratio is one (1), to give Li2CO3.

(c) A compound with molecules that consist of two nitrogen atoms and three oxygen atoms contains nitrogen (N) and oxygen (O) in a ratio of 2:3. The elements are written as their symbols and the ratios of the elements are written as subscripts after the elementto give N2O3.

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3. –/3 points PrepChemCF1 3.3.P.064. My Notes
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Draw a Lewis structure for each of the following formulas. (Assign lone pairs, radical electrons, and atomic charges where appropriate.)
(a) oxygen difluoride, OF2 (an unstable, colorless gas)
(No Response)
Marvin JS

(b) bromoform, CHBr3 (used as a sedative)
(No Response)
Marvin JS

(c) phosphorus triiodide, PI3 (used to make organic compounds)
(No Response)
Marvin JS


Solution or Explanation
When working with Lewis structures carefully consider the following principles outlined in the text:
  • The molecular formula often reflects the molecular structure.
  • Try to arrange the atoms to yield the most typical number of bonds for each atom.
  • Apply the following guidelines in deciding what element belongs in the center of your structure.
    • Hydrogen and fluorine atoms are never in the center.
    • Oxygen atoms are rarely in the center.
    • The element with the fewest atoms in the formula is often in the center.
    • The atom that is capable of making the most bonds is often in the center.
(a) oxygen difluoride, OF2 (an unstable, colorless gas)
(1) Determine the total number of valence shell electrons from the formula: 1 × 6 (from O) + 2 × 7 (from F) = 20
(2) Determine the number of valence electron pairs that can be created from the structure: 20 ÷ 2 = 10
(3) Identify the central atom(s): O
(4) Identify the ligand atom(s): F
(5) Place the central atom(s) approximately at the center of the drawing area:
(6) Place the ligand atom(s) approximately equally spaced around the central atom:
(7) Place an electron pair, i.e. a single bond, between the central atom(s) and each of the connecting atoms:
(8) Determine how many electron pairs remain to be distributed: 10 − 2 = 8
(9) Place electron pairs around each ligand atom until it has an octet of electrons, except for hydrogen which can only have one shared electron pair (a duet):
(10) Determine how many electron pairs remain to be distributed: 10 − 2 − (2 × 3 ) = 2
(11) Place any remaining electrons on the central atom(s) to obtain the intermediate Lewis structure:
(12) If necessary, rearrange electron pairs in your structure to give each atom its most common bonding pattern, and obtain the final Lewis structure:

(b) bromoform, CHBr3 (used as a sedative)
(1) Determine the total number of valence shell electrons from the formula: 1 × 4 (from C) + 1 × 1 (from H) + 3 × 7 (from Br) = 26
(2) Determine the number of valence electron pairs that can be created from the structure: 26 ÷ 2 = 13
(3) Identify the central atom(s): C
(4) Identify the ligand atom(s): H, Br
(5) Place the central atom(s) approximately at the center of the drawing area:
(6) Place the ligand atom(s) approximately equally spaced around the central atom:
(7) Place an electron pair, i.e. a single bond, between the central atom(s) and each of the connecting atoms:
(8) Determine how many electron pairs remain to be distributed: 13 − 4 = 9
(9) Place electron pairs around each ligand atom until it has an octet of electrons, except for hydrogen which can only have one shared electron pair (a duet):
(10) Determine how many electron pairs remain to be distributed: 13 − 4 − (3 × 3) = 0
(11) Place any remaining electrons on the central atom(s) to obtain the intermediate Lewis structure:
(12) If necessary, rearrange electron pairs in your structure to give each atom its most common bonding pattern, and obtain the final Lewis structure:

(c) phosphorus triiodide, PI3 (used to make organic compounds)
(1) Determine the total number of valence shell electrons from the formula: 1 × 5 (from P) + 3 × 7 (from I) = 26
(2) Determine the number of valence electron pairs that can be created from the structure: 26 ÷ 2 = 13
(3) Identify the central atom(s): P
(4) Identify the ligand atom(s): I
(5) Place the central atom(s) approximately at the center of the drawing area:
(6) Place the ligand atom(s) approximately equally spaced around the central atom:
(7) Place an electron pair, i.e. a single bond, between the central atom(s) and each of the connecting atoms:
(8) Determine how many electron pairs remain to be distributed: 13 − 3 = 10
(9) Place electron pairs around each ligand atom until it has an octet of electrons, except for hydrogen which can only have one shared electron pair (a duet):
(10) Determine how many electron pairs remain to be distributed: 13 − 3 − (3 × 3) = 1
(11) Place any remaining electrons on the central atom(s) to obtain the intermediate Lewis structure:
(12) If necessary, rearrange electron pairs in your structure to give each atom its most common bonding pattern, and obtain the final Lewis structure:

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4. –/1 points PrepChemCF1 4.2.P.051. My Notes
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Chromium(III) phosphate is a paint pigment that is made in a precipitation reaction between water solutions of chromium(III) chloride and sodium phosphate. Write the complete equation for this reaction. (Use the lowest possible whole number coefficients. Include states-of-matter in your answer.)

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CrCl3(aq) + Na3PO4(aq)  →  CrPO4(s) + 3 NaCl(aq)


Solution or Explanation
To solve, determine the formulas for the possible products using the general double-displacement equation, AB + CD → AD + CB: Chromium(III) chloride and sodium phosphate yields chromium(III) phosphate and sodium chloride. Now, we convert the names to formulas. Chromium(III) chloride is an ionic compound, so we write the name of the cation followed by the name of the anion. The charge of the cation, Cr, is +3, and the charge of the anion, Cl, is −1. Thus, the formula for chromium(III) chloride is: CrCl3. Sodium phosphate is an ionic compound. The charge of the cation, Na, is +1, and the charge of the anion, PO4, is −3. Thus, the formula for sodium phosphate is: Na3PO4. Chromium phosphate is also an ionic compound. The charge of the cation, Cr, is +3, and the charge of the anion, PO4, is −3. Thus, the formula for chromium phosphate is: CrPO4. Sodium chloride is an ionic compound with the formula: NaCl. The unbalanced equation then is: CrCl3 + Na3PO4 → CrPO4 + NaCl.

Next, add in the state-of-matter for each formula. The reactants are followed by (aq). The first product is CrPO4. By definition, compounds containing phosphate ions are insoluble in water except with group 1 metallic ions and ammonium ions. Chromium is not a group 1 metallic ion. Thus, this product is insoluble and is followed by (s). The second product is NaCl. By definition, compounds containing chloride ions are water-soluble except with silver ions and lead(II) ions. Thus, this product is soluble and is followed by (aq). The unbalanced equation with states-of-matter is: CrCl3(aq) + Na3PO4(aq) → CrPO4(s) + NaCl(aq).

Finally, balance the equation with the lowest possible whole number coefficients. Cr is the first element in the first formula. The moles of Cr appear to be balanced at this stage. The next element in the formula is Cl. There are 3 moles of Cl listed on the reactant side of the unbalanced equation, but there is only 1 mole of Cl listed on the product side. We can balance them with 3 moles of NaCl, which corresponds to 3 ✕ 1 = 3 moles of Cl.
CrCl3(aq) + Na3PO4(aq) → CrPO4(s) + 3 NaCl(aq)
The next element in the equation is Na, which appears to be balanced at this stage with 3 moles of Na on each side of the unbalanced equation. Since PO4 appears in the same form on both sides of the chemical equation, it can be balanced as though it was a single atom. The moles of PO4 appear to be balanced at this stage with 1 mole of PO4 on both sides of the unbalanced equation. The complete equation, then, is: CrCl3(aq) + Na3PO4(aq) → CrPO4(s) + 3 NaCl(aq).

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5. –/1 points PrepChemCF1 18.1.P.076. My Notes
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Marie Curie won the Nobel Prize for physics in 1903 for her study of radioactive nuclides, including polonium-218 (which was named after her native country, Poland). Polonium-218 undergoes alpha emission. Write the nuclear equation for this change. (Enter all particles in the form
A X
Z
.)

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218
84
Po  →  
214
82
Pb + 
4
2
He


Solution or Explanation
Polonium-218 is written as 21884Po. When it undergoes alpha emission, it loses two protons and two neutrons and produces a helium nucleus (42He) to yield 21482Pb.
21884Po → 21482Pb + 42He

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6. –/1 points PrepChemCF1 13.TB.1.010. My Notes
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Which of the following statements is true about gases?
    

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