Concept explainers
Interpretation:
The reason due to which one cannot pass hand through a solid object made up of atom needs to be explained.
Concept introduction:
Atom is the unit of the matter. It composed of protons, electrons, and neutrons. The central region of an atom is the nucleus, the protons, and neutrons present in the nucleus of the atom. The Outer sphere-like structure is a shell where the electrons are moving around the nucleus.

Answer to Problem 112A
The atoms having electron clouds around the nucleus make it just like a solid object, due to the presence of electron cloud one cannot pass hand or any object through it.
Explanation of Solution
The atom consist three main subatomic particles. The revolving of an electron is due to the
In nature, the intermolecular forces are very strong in the solid particle and the intermolecular space is negligible so the particles are tightly joined together. In the solid-state, the arrangement of particles very close to each other, and two atoms are joined in a fixed and complex manner, there is no space to pass any object through it for example your hand or any other object.
The electrons are repelled to each other and the nucleus attracted them with equal force of attraction. So a cloudy wall is created due to center force of attraction and one cannot pass hand through it.
Chapter 4 Solutions
Chemistry: Matter and Change
Additional Science Textbook Solutions
Concepts of Genetics (12th Edition)
Campbell Biology (11th Edition)
Campbell Biology: Concepts & Connections (9th Edition)
Microbiology with Diseases by Body System (5th Edition)
Microbiology: An Introduction
Physics for Scientists and Engineers: A Strategic Approach, Vol. 1 (Chs 1-21) (4th Edition)
- a) A favorable entropy change occurs when ΔS is positive. Does the order of the system increase or decrease when ΔS is positive? (b) A favorable enthalpy change occurs when ΔH is negative. Does the system absorb heat or give off heat when ΔH is negative? (c) Write the relation between ΔG, ΔH, and ΔS. Use the results of parts (a) and (b) to state whether ΔG must be positive or negative for a spontaneous change. For the reaction, ΔG is 59.0 kJ/mol at 298.15 K. Find the value of K for the reaction.arrow_forwardA sample of hydrated magnesium sulfate (MgSO4⋅xH2O) is analyzed using thermogravimetric analysis (TGA). The sample weighs 2.50 g initially and is heated in a controlled atmosphere. As the temperature increases, the water of hydration is released in two stages: (a) The first mass loss of 0.72 g occurs at 150°C, corresponding to the loss of a certain number of water molecules. (b) The second mass loss of 0.90 g occurs at 250°C, corresponding to the loss of the remaining water molecules. The residue is identified as anhydrous magnesium sulfate (MgSO4) Questions: (i) Determine the value of x (the total number of water molecules in MgSO4⋅xH2O) (ii) Calculate the percentage of water in the original sample. Write down the applications of TGA.arrow_forwardThe solubility product of iron(III) hydroxide (Fe(OH)3) is 6.3×10−38. If 50 mL of a 0.001 M FeCl3 solution is mixed with 50 mL of a 0.005 M NaOH solution, will Fe(OH)3 precipitate? Show all step-by-step calculations. To evaluate the equilibrium constant, we must express concentrations of solutes in mol/L, gases in bars, and omit solids, liquids, and solvents. Explain why.arrow_forward
- Predict the major products of this organic reaction.arrow_forward2. Provide the structure of the major organic product in the following reaction. Pay particular attention to the regio- and stereochemistry of your product. H3CO + H CN Aarrow_forwardPredict the major products of the following organic reaction.arrow_forward
- 1) The isoamyl acetate report requires eight paragraphs - four for comparison of isoamyl alcohol and isoamyl acetate (one paragraph each devoted to MS, HNMR, CNMR and IR) and four for comparison of acetic acid and isoamyl acetate ((one paragraph each devoted to MS, HNMR, CNMR and IR. 2) For MS, the differing masses of molecular ions are a popular starting point. Including a unique fragmentation is important, too. 3) For HNMR, CNMR and IR state the peaks that are different and what makes them different (usually the presence or absence of certain groups). See if you can find two differences (in each set of IR, HNMR and CNMR spectra) due to the presence or absence of a functional group. Include peak locations. Alternatively, you can state a shift of a peak due to a change near a given functional group. Including peak locations for shifted peaks, as well as what these peaks are due to. Ideally, your focus should be on not just identifying the differences but explaining them in terms of…arrow_forwardWhat steps might you take to produce the following product from the given starting material? CI Br Он до NH2 NH2arrow_forward1) The isoamyl acetate report requires eight paragraphs - four for comparison of isoamyl alcohol and isoamyl acetate (one paragraph each devoted to MS, HNMR, CNMR and IR) and four for comparison of acetic acid and isoamyl acetate ((one paragraph each devoted to MS, HNMR, CNMR and IR. 2) For MS, the differing masses of molecular ions are a popular starting point. Including a unique fragmentation is important, too. 3) For HNMR, CNMR and IR state the peaks that are different and what makes them different (usually the presence or absence of certain groups). See if you can find two differences (in each set of IR, HNMR and CNMR spectra) due to the presence or absence of a functional group. Include peak locations. Alternatively, you can state a shift of a peak due to a change near a given functional group. Including peak locations for shifted peaks, as well as what these peaks are due to. Ideally, your focus should be on not just identifying the differences but explaining them in terms of…arrow_forward
- №3 Fill in the below boxes. HN 1. LAH 2. H3O+ NH2arrow_forwardFor the photochemical halogenation reaction below, draw both propagation steps and include the mechanism arrows for each step. H CH ot CH3 CI-CI MM hv of CH H-CI CH3 2nd attempt See Periodic Table See Hint Draw only radical electrons; do not add lone pair electrons. Note that arrows cannot meet in "space," and must end at either bonds or at atoms. 1 i Add the missing curved arrow notation to this propagation step. 20 H ن S F P H CI Br 品arrow_forwardThe radical below can be stabilized by resonance. 4th attempt Draw the resulting resonance structure. DOCEarrow_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





