Category: Chapter 8 assessment chemistry

Chapter 8 assessment chemistry

A number of phenomena, both physical as well as biological, are concerned with redox reactions.

chapter 8 assessment chemistry

These reactions find extensive use in pharmaceutical, biological, industrial, metallurgical and agricultural areas. After studying this unit you will be able to identify redox reactions as a class of reactions in which oxidation and reduction reactions occur simultaneously; define the terms oxidation, reduction, oxidant oxidising agent and reductant reducing agent ; explain mechanism of redox reactions by electron transfer process; use the concept of oxidation number to identify oxidant and reductant in a reaction; classify redox reaction into combination synthesisdecomposition, displacement and disproportionation reactions; suggest a comparative order among various reductants and oxidants; balance chemical equations using i oxidation number ii half reaction method; learn the concept of redox reactions in terms of electrode processes.

Unit 8, Unit 9, Unit 10 and Unit 11 hold a total weightage of 16 marks in the final examination. The topics and sub-topics in Chapter 8 Redox Reaction are given below.

Other than given exercises, you should also practice all the solved examples given in the book to clear your concepts on Redox Reactions.And it includes the short questions of exercise and more short questions made from within the chapter reading section.

chapter 8 assessment chemistry

By studying these notes you can prepare your whole 2nd year eight chapter easily. The post is tagged and categorized under in 12th chemistry12th chemistry notesEducation NewsNotes Tags. For more content related to this post you can click on labels link.

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And to help the students in their studies. Find me on Facebook: Maher Afrasiab. Chapter 5 Notes Here. Chapter 6 Notes Here. Chapter 7 Notes Here. You may also like: Newer Post Older Post. Next Prev Post.

Previous Next Post. Anonymous 8 April at Anonymous 21 July at Subscribe to: Post Comments Atom. If you are a student of matric class 9 then you should get a new syllabus book of Pak studies edition.

Here we have shared and made Looking for the 11th class chapter 4 of chemistry notes? If you are searching for the text book of chemistry of class 10 Matric in pdf format then you are at right page because here we have sharMake sure that your printout includes all content from the page. If it doesn't, try opening this guide in a different browser and printing from there sometimes Internet Explorer works better, sometimes Chrome, sometimes Firefox, etc.

Diamond is the hardest natural material known on Earth. Yet diamond is just pure carbon. What is special about this element that makes diamond so hard? In a perfect diamond crystal, each C atom makes four connections—bonds—to four other C atoms in a three-dimensional matrix. Four is the greatest number of bonds that is commonly made by atoms, so C atoms maximize their interactions with other atoms.

This three-dimensional array of connections extends throughout the diamond crystal, making it essentially one large molecule.

Breaking a diamond means breaking every bond at once. Also, the bonds are moderately strong. There are stronger interactions known, but the carbon-carbon connection is fairly strong itself. Not only does a person have to break many connections at once, but also the bonds are strong connections from the start. There are other substances that have similar bonding arrangements as diamond does. Silicon dioxide and boron nitride have some similarities, but neither of them comes close to the ultimate hardness of diamond.

How do atoms make compounds? Typically they join together in such a way that they lose their identities as elements and adopt a new identity as a compound.

CHE 105/110 - Introduction to Chemistry - Textbook

These joins are called chemical bonds. But how do atoms join together? Ultimately, it all comes down to electrons. Before we discuss how electrons interact, we need to introduce a tool to simply illustrate electrons in an atom. In almost all cases, chemical bonds are formed by interactions of valence electrons in atoms.

To facilitate our understanding of how valence electrons interact, a simple way of representing those valence electrons would be useful.

Chemistry Test Chapter 8 9

A Lewis electron dot diagram or electron dot diagram or a Lewis diagram or a Lewis structure is a representation of the valence electrons of an atom that uses dots around the symbol of the element. The number of dots equals the number of valence electrons in the atom. These dots are arranged to the right and left and above and below the symbol, with no more than two dots on a side. It does not matter what order the positions are used. For example, the Lewis electron dot diagram for hydrogen is simply.

Because the side is not important, the Lewis electron dot diagram could also be drawn as follows:. The electron dot diagram for helium, with two valence electrons, is as follows:.

By putting the two electrons together on the same side, we emphasize the fact that these two electrons are both in the 1 s subshell; this is the common convention we will adopt, although there will be exceptions later. The next atom, lithium, has an electron configuration of 1 s 2 2 s 1so it has only one electron in its valence shell. Its electron dot diagram resembles that of hydrogen, except the symbol for lithium is used:.

Beryllium has two valence electrons in its 2 s shell, so its electron dot diagram is like that of helium:. The next atom is boron. Its valence electron shell is 2 s 2 2 p 1so it has three valence electrons. The third electron will go on another side of the symbol:. Again, it does not matter on which sides of the symbol the electron dots are positioned.NCERT Solutions for Class 11 Chemistry Chapter 8 Redox Reactions includes all the important topics with detailed explanation that aims to help students to understand the concepts better.

Going through the solutions provided on this page will help you to know how to approach and solve the problems. Question 1. Let the oxidation no. Question 2. What are the oxidation number of the underlined elements in each of the following and how do you rationalise your results?

The oxidation number of two iodine atoms forming the I2 molecule is zero while that of iodine ion forming the coordinate bond is Thus, the O.

Chemistry Chapter 8 Test

Question 3. Question 4. Question 5. Suggest structure of these compounds. Count for the fallacy. Question 6. Question 7. Question 8. While sulphur dioxide and hydrogen peroxide can act as oxidising as well as reducing agents in their reactions, ozone and nitric acid act only as oxidants.

Therefore, sulphur dioxide acts as a reducing agent as well as oxidising agent. Similarly, the oxidation state of oxygen in hydrogen peroxide is It can be oxidised to O 2 zero oxidation state or reduced to H 2 O or OH — -2 oxidation state and therefore, acts as reducing as well as oxidising agents. However, both ozone and nitric acid can only decrease their oxidation number and therefore, act only as oxidising agents.

Question 9. Question The compound AgF 2 is unstable compound. However, if formed, the compound acts as a very strong oxidising agent. Whenever a reaction between an oxidising agent and a reducing agent is carried out, a compound of lower oxidation state is formed if the reducing agent is in excess and a compound of higher oxidation state is formed if the oxidising agent is in excess.

Justify this statement giving three illustrations. How do you count for the following observations?

Write a balanced redox equation for the reaction. Identify the substance oxidised, reduced, oxidising agent and reducing agent for each of the following reactions: Solution. Consider the reactions: Why does the same reductant, thiosulphate react differently with iodine and bromine?

Justify giving reactions that among halogens, fluorine is the best oxidant and among hydrohalic compounds, hydroiodic acid is the best reductant. The halogens X 2 have strong electron accepting tendency and have positive standard oxidation potential values. They are therefore, powerful oxidising agents. The decreasing order of oxidising powers of halogens is :. Why does the following reaction occur? Consider the reactions :. Balance the following redox reactions by ion — electron method: Solution.Chemistry 10th Edition by Whitten Test Bank.

This new edition of CHEMISTRY continues to incorporate a strong molecular reasoning focus, amplified problem-solving exercises, a wide range of real-life examples and applications, and innovative technological resources.

With this text's focus on molecular reasoning, readers will learn to think at the molecular level and make connections between molecular structure and macroscopic properties.

The Tenth Edition has been revised throughout and now includes a reorganization of the descriptive chemistry chapters to improve the flow of topics, a new basic math skills Appendix, an updated art program with new "talking labels" that fully explain what is going on in the figure, and much more.

Chapter 1. The Foundations of Chemistry. Chapter 2. Chemical Formulas and Composition Stoichiometry. Chapter 3. Chemical Equations and Reaction Stoichiometry. Chapter 4. The Structure of Atoms. Chapter 5. Chemical Periodicity. Chapter 6. Some Types of Chemical Reactions. Chapter 7. Chemical Bonding. Chapter 8. Molecular Structure and Covalent Bonding Theories.

Chapter 9. Molecular Orbitals in Chemical Bonding. Chapter Gases and the Kinetic-Molecular Theory. Liquids and Solids. Chemical Thermodynamics. Chemical Kinetics. Chemical Equilibrium.Find Flashcards. Browse over 1 million classes created by top students, professors, publishers, and experts, spanning the world's body of "learnable" knowledge. A Level Exams. AP Exams.

chapter 8 assessment chemistry

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Joe Porrello (Cannery, GT Bookies Battle champion): 49ers, 28-21. A lot of people should be "Kaepernicking" after the game. Ken Miller (Cantor Gaming): 49ers: 27-21. The last two teams the Ravens played featured Manning and Brady, who are pocket QB passers and pose zero threat of running the ball.

chapter 8 assessment chemistry

Not so SB Sunday. Bob Scucci (Orleans): 49ers, 24-17. I felt the NFC was the tougher, more physical conference all year. While the Ravens have certainly shown they can beat anyone I think the 49ers come out on top.

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I noticed the Ravens pass defense, net yards, rushing defense, red zone offense and defense were all regressing. Niners are peaking at the right time. JT The Brick (FOX national radio host): 49ers, 27-23.

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