Concept explainers
(a)
Interpretation:
The physical state of air in the room is to be determined.
Concept introduction:
Anything that has mass and volume is defined as a matter. The matter is classified as solids, liquids, and gases on the basis of a state that depends on the physical form of matter.
In solids, the atoms and molecules have fixed positions and are closely packed to each other. The atoms and molecules in the solid state only vibrate and do not move over each other. Therefore, a solid has a rigid shape and fixed volume. The examples of matter that are solid are ice and diamond.
In liquids, atoms and molecules are also closely packed to each other but they can move over each other. Thus, liquids have fixed volume but they do not have a fixed shape. Liquids occupy the shape of the container. The examples of matter that are liquid are water and alcohol.
In gases, the atoms and molecules have space between them and can easily move over each other hence gases are compressible. Gases neither have fixed shape nor volume. It occupies the shape and volume of the container. The examples of matter that are gases are nitrogen and carbon dioxide.
(b)
Interpretation:
The physical state of tablets in a bottle of vitamins is to be determined.
Concept introduction:
Anything that has mass and volume is defined as a matter. The matter is classified as solids, liquids, and gases on the basis of a state that depends on the physical form of matter.
In solids, the atoms and molecules have fixed positions and are closely packed to each other. The atoms and molecules in the solid state only vibrate and do not move over each other. Therefore, a solid has a rigid shape and fixed volume. The examples of matter that are solid are ice and diamond.
In liquids, atoms and molecules are also closely packed to each other but they can move over each other. Thus, liquids have fixed volume but they do not have a fixed shape. Liquids occupy the shape of the container. The examples of matter that are liquid are water and alcohol.
In gases, the atoms and molecules have space between them and can easily move over each other hence gases are compressible. Gases neither have fixed shape nor volume. It occupies the shape and volume of the container. The examples of matter that are gases are nitrogen and carbon dioxide.
(c)
Interpretation:
The physical state of sugar in a packet is to be determined.
Concept introduction:
Anything that has mass and volume is defined as a matter. The matter is classified as solids, liquids, and gases on the basis of a state that depends on the physical form of matter.
In solids, the atoms and molecules have fixed positions and are closely packed to each other. The atoms and molecules in the solid state only vibrate and do not move over each other. Therefore, a solid has a rigid shape and fixed volume. The examples of matter that are solid are ice and diamond.
In liquids, atoms and molecules are also closely packed to each other but they can move over each other. Thus, liquids have fixed volume but they do not have a fixed shape. Liquids occupy the shape of the container. The examples of matter that are liquid are water and alcohol.
In gases, the atoms and molecules have space between them and can easily move over each other hence gases are compressible. Gases neither have fixed shape nor volume. It occupies the shape and volume of the container. The examples of matter that are gases are nitrogen and carbon dioxide.
Want to see the full answer?
Check out a sample textbook solutionChapter 1 Solutions
CHEMISTRY: MOLECULAR...(LLF) W/CONNECT
- Show work with explanation needed. don't give Ai generated solutionarrow_forwardShow work with explanation needed. Don't give Ai generated solutionarrow_forward7. 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_forward
- A 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_forward2. Write a complete mechanism for the reaction shown below. NaOCH LOCH₁ O₂N NO2 CH₂OH, 20 °C O₂N NO2arrow_forward
- 4. 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_forwardPlease correct answer and don't use hand rating and don't use Ai solutionarrow_forward
- 1. 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_forwardSelect all possible products from the following reaction: NaOH H₂O a) b) ОН HO O HO HO e) ОН f) O HO g) h) + OHarrow_forward
- ChemistryChemistryISBN:9781305957404Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCostePublisher:Cengage LearningChemistryChemistryISBN:9781259911156Author:Raymond Chang Dr., Jason Overby ProfessorPublisher:McGraw-Hill EducationPrinciples of Instrumental AnalysisChemistryISBN:9781305577213Author:Douglas A. Skoog, F. James Holler, Stanley R. CrouchPublisher:Cengage Learning
- Organic ChemistryChemistryISBN:9780078021558Author:Janice Gorzynski Smith Dr.Publisher:McGraw-Hill EducationChemistry: Principles and ReactionsChemistryISBN:9781305079373Author:William L. Masterton, Cecile N. HurleyPublisher:Cengage LearningElementary Principles of Chemical Processes, Bind...ChemistryISBN:9781118431221Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. BullardPublisher:WILEY