Concept explainers
Draw a “formula” for each of the following molecules using circular symbols of your choice to represent atoms:
a. A diatomic molecule of an element
b. A diatomic molecule of a compound
c. A triatomic molecule of an element
d. A molecule of a compound containing one atom of one element and four atoms of another element
(a)
Interpretation:
The formula for the diatomic molecule of an element by using circular symbols to represent atoms is to be drawn.
Concept introduction:
A molecular formula represents the number of atoms of each element present in a molecule of a compound.
The number of atoms present in molecule is determined by the subscript written below the normal line in the molecular formula.
Answer to Problem 2.1E
The formula for the diatomic molecule of an element by using circular symbols to represent atoms is shown below.
Explanation of Solution
It is given that the molecule of an element is a diatomic which means that the compound consists of two atoms with similar identity. This can be explained with the help of one example. Considering a diatomic molecule of an element that is chlorine gas. The chemical formula of chlorine gas is
Therefore, the formula for the diatomic molecule of an element by using circular symbols to represent atoms is shown below.
Figure 1
In the given figure, white circles represent the chlorine atoms.
The formula for the diatomic molecule of an element by using circular symbols to represent atoms is shown in figure 1.
(b)
Interpretation:
The formula for the diatomic molecule of a compound by using circular symbols to represent atoms is to be drawn.
Concept introduction:
A molecular formula represents the number of atoms of each element present in a molecule of a compound.
The number of atoms present in molecule is determined by the subscript written below the normal line in the molecular formula.
Answer to Problem 2.1E
The formula for the diatomic molecule of a compound by using circular symbols to represent atoms is shown below.
Explanation of Solution
It is given that the molecule of a compound is diatomic which means that the compound consists of two different atoms. This can be explained with the help of one example. Considering an example that is hydrogen fluoride. The chemical formula of hydrogen fluoride is
Therefore, the formula for the diatomic molecule of a compound by using circular symbols to represent atoms is shown below.
Figure 2
In the given figure, grey circle represent the fluorine atom, whereas white circle represent the hydrogen atoms.
The formula for the diatomic molecule of a compound by using circular symbols to represent atoms is shown in figure 2.
(c)
Interpretation:
The formula for the triatomic molecule of an element by using circular symbols to represent atoms is to be drawn.
Concept introduction:
A molecular formula represents the number of atoms of each element present in a molecule of a compound.
The number of atoms present in molecule is determined by the subscript written below the normal line in the molecular formula.
Answer to Problem 2.1E
The formula for the triatomic molecule of an element by using circular symbols to represent atoms is shown below.
Explanation of Solution
It is given that the molecule of an element is a triatomic which means that the compound consists of three atoms with similar identity. This can be explained with the help of one example. Considering an example that is ozone. The chemical formula of ozone is
Therefore, the formula for the triatomic molecule of an element by using circular symbols to represent atoms is shown below.
Figure 3
In the given diagram, white circles represent oxygen atoms.
The formula for the triatomic molecule of an element by using circular symbols to represent atoms is shown in figure 3.
(d)
Interpretation:
The formula for a molecule of a compound containing one atom of one element and four atoms of another element by using circular symbols to represent atoms is to be drawn.
Concept introduction:
A molecular formula represents the number of atoms of each element present in a molecule of a compound.
The number of atoms present in molecule is determined by the subscript written below the normal line in the molecular formula.
Answer to Problem 2.1E
The formula for a molecule of a compound containing one atom of one element and four atoms of another element by using circular symbols to represent atoms is shown below.
Explanation of Solution
It is given that the molecule contains one atom of one element and four atoms of another element. This can be explained with the help of one example. Considering an example that is carbon tetrachloride. The chemical formula of carbon tetrachloride is
Therefore, the formula for a molecule of a compound containing one atom of one element and four atoms of another element by using circular symbols to represent atoms is shown below.
Figure 4
In the given diagram, white circle represent carbon atom, whereas grey circle represent chlorine atoms.
The formula for a molecule of a compound containing one atom of one element and four atoms of another element by using circular symbols to represent atoms is shown in figure 4.
Want to see more full solutions like this?
Chapter 2 Solutions
Chemistry For Today: General, Organic, And Biochemistry, Loose-leaf Version
Additional Science Textbook Solutions
Microbiology Fundamentals: A Clinical Approach
Campbell Essential Biology with Physiology (5th Edition)
Human Biology: Concepts and Current Issues (8th Edition)
Fundamentals Of Thermodynamics
Physical Science
Organic Chemistry
- 7. Calculate the following for a 1.50 M Ca(OH)2 solution. a. The concentration of hydroxide, [OH-] b. The concentration of hydronium, [H3O+] c. The pOH d. The pHarrow_forwardA first order reaction is 46.0% complete at the end of 59.0 minutes. What is the value of k? What is the half-life for this reaction? HOW DO WE GET THERE? The integrated rate law will be used to determine the value of k. In [A] [A]。 = = -kt What is the value of [A] [A]。 when the reaction is 46.0% complete?arrow_forward3. Provide the missing compounds or reagents. 1. H,NNH КОН 4 EN MN. 1. HBUCK = 8 хно Panely prowseful kanti-chuprccant fad, winddively, can lead to the crading of deduc din-willed, tica, The that chemooices in redimi Грин. " like (for alongan Ridovi MN نيا . 2. Cl -BuO 1. NUH 2.A A -BuOK THE CF,00,H Ex 5)arrow_forward
- 2. Write a complete mechanism for the reaction shown below. NaOCH LOCH₁ O₂N NO2 CH₂OH, 20 °C O₂N NO2arrow_forward4. Propose a synthesis of the target molecules from the respective starting materials. a) b) LUCH C Br OHarrow_forwardThe following mechanism for the gas phase reaction of H2 and ICI that is consistent with the observed rate law is: step 1 step 2 slow: H2(g) +ICI(g) → HCl(g) + HI(g) fast: ICI(g) + HI(g) → HCl(g) + |2(g) (1) What is the equation for the overall reaction? Use the smallest integer coefficients possible. If a box is not needed, leave it blank. + → + (2) Which species acts as a catalyst? Enter formula. If none, leave box blank: (3) Which species acts as a reaction intermediate? Enter formula. If none, leave box blank: (4) Complete the rate law for the overall reaction that is consistent with this mechanism. (Use the form k[A][B]"..., where '1' is understood (so don't write it) for m, n etc.) Rate =arrow_forward
- Please correct answer and don't use hand rating and don't use Ai solutionarrow_forward1. For each of the following statements, indicate whether they are true of false. ⚫ the terms primary, secondary and tertiary have different meanings when applied to amines than they do when applied to alcohols. • a tertiary amine is one that is bonded to a tertiary carbon atom (one with three C atoms bonded to it). • simple five-membered heteroaromatic compounds (e.g. pyrrole) are typically more electron rich than benzene. ⚫ simple six-membered heteroaromatic compounds (e.g. pyridine) are typically more electron rich than benzene. • pyrrole is very weakly basic because protonation anywhere on the ring disrupts the aromaticity. • thiophene is more reactive than benzene toward electrophilic aromatic substitution. • pyridine is more reactive than nitrobenzene toward electrophilic aromatic substitution. • the lone pair on the nitrogen atom of pyridine is part of the pi system.arrow_forwardThe following reactions are NOT ordered in the way in which they occur. Reaction 1 PhO-OPh Reaction 2 Ph-O -CH₂ heat 2 *OPh Pho -CH2 Reaction 3 Ph-O ⚫OPh + -CH₂ Reaction 4 Pho Pho + H₂C OPh + CHOPh H₂C -CH₂ Reactions 1 and 3 Reaction 2 O Reaction 3 ○ Reactions 3 and 4 ○ Reactions 1 and 2 Reaction 4 ○ Reaction 1arrow_forward
- Select all possible products from the following reaction: NaOH H₂O a) b) ОН HO O HO HO e) ОН f) O HO g) h) + OHarrow_forward3. Draw diagrams to represent the conjugation in these molecules. Draw two types of diagram: a. Show curly arrows linking at least two different ways of representing the molecule b. Indicate with dotted lines and partial charges (where necessary) the partial double bond (and charge) distribution H₂N* H₂N -NH2arrow_forwardQuestion 2 of 25 point Question Attempt 3 of Ulimited Draw the structure for 3-chloro-4-ethylheptane. Part 2 of 3 Click and drag to start drawing a structure. Draw the structure for 1-chloro-4-ethyl-3-lodooctane. Click and drag to start drawing a structure. X G X B c Part 3 of 30 Draw the structure for (R)-2-chlorobutane. Include the stereochemistry at all stereogenic centers. Check Click and drag to start drawing a structure. G X A 。 MacBook Pro G P Save For Later Submit Assignment Privacyarrow_forward
- Introductory Chemistry: An Active Learning Approa...ChemistryISBN:9781305079250Author:Mark S. Cracolice, Ed PetersPublisher:Cengage LearningChemistry for Today: General, Organic, and Bioche...ChemistryISBN:9781305960060Author:Spencer L. Seager, Michael R. Slabaugh, Maren S. HansenPublisher:Cengage Learning
- Chemistry: Matter and ChangeChemistryISBN:9780078746376Author:Dinah Zike, Laurel Dingrando, Nicholas Hainen, Cheryl WistromPublisher:Glencoe/McGraw-Hill School Pub CoWorld of Chemistry, 3rd editionChemistryISBN:9781133109655Author:Steven S. Zumdahl, Susan L. Zumdahl, Donald J. DeCostePublisher:Brooks / Cole / Cengage Learning