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### Normality-Molarity Calculator Tool

Normality-Molarity Calculator Tool

# Normality-Molarity Calculator Tool

## You can directly watch on YouTube by click on following linkhttps://youtu.be/0h0LTaf2-EI

Are you struggling to solve Equilibrium exercise problems from the NCERT textbook for Class 11? Look no further than our comprehensive solution guide, "Class 11 Unit 7 Equilibrium Exercise Solution 7.1 to 7.10 NCERT Solution 2023," available on YouTube.

This educational video is designed specifically to help students of Class 11 understand and solve Equilibrium exercise problems. Our experienced tutor begins by introducing the importance of Equilibrium in chemistry, and then proceeds to explain concepts such as Le Chatelier's Principle, Equilibrium Constants, and Equilibrium Expressions.

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So, if you're a Class 11 student looking to excel in Equilibrium exercises, make sure to check out our video, "Class 11 Unit 7 Equilibrium Exercise Solution 7.1 to 7.10 NCERT Solution 2023," and start solving problems with confidence!

7.1 A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased.

a) What is the initial effect of the change on vapour pressure?

b) How do rates of evaporation and condensation change initially?

c) What happens when equilibrium is restored finally and what will be the final vapour pressure?

7.2 What is Kc for the following equilibrium when the equilibrium concentration of each substance is: [SO2] = 0.60M, [O2] = 0.82M and [SO3] = 1.90M ?

2SO2(g) + O2(g) 2SO3(g)

7.3 At a certain temperature and total pressure of 105Pa, iodine vapour contains 40% by volume of I atoms

I2(g) 2I(g)

Calculate Kp for the equilibrium.

7.4 Write the expression for the equilibrium constant, Kc for each of the following reactions:

(i) 2NOCl(g) 2NO(g) + Cl2(g)

(ii) 2Cu(NO3)2(s) 2CuO(s) + 4NO2(g) + O2(g)

(iii) CH3COOC2H5(aq) + H2O(l) CH3COOH(aq) + C2H5OH(aq)

(iv) Fe3+(aq) + 3OH–(aq) Fe(OH)3(s)

(v) I2(s) + 5F2 2IF5

7.5 Find out the value of Kc for each of the following equilibria from the value of Kp:

(i) 2NOCl(g) 2NO(g) + Cl2(g); Kp = 1.8 × 10–2 at 500 K

(ii) CaCO3(s) CaO(s) + CO2(g); Kp= 167 at 1073 K

7.6 For the following equilibrium, Kc= 6.3 × 1014 at 1000 K

NO(g) + O3(g) NO2(g) + O2(g)

Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is Kc, for the reverse reaction?

7.7 Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression?

7.8 Reaction between N2 and O2 takes place as follows:

2N2(g) + O2(g) 2N2O(g)

If a mixture of 0.482 mol N2 and 0.933 mol of O2 is placed in a 10 L reaction vessel and allowed to form N2O at a temperature for which Kc= 2.0 × 10–37, determine the composition of equilibrium mixture.

7.9 Nitric oxide reacts with Br2 and gives nitrosyl bromide as per reaction given below:

2NO(g) + Br2(g) 2NOBr(g)

When 0.087 mol of NO and 0.0437 mol of Br2 are mixed in a closed container at constant temperature, 0.0518 mol of NOBr is obtained at equilibrium. Calculate equilibrium amount of NO and Br2.

7.10 At 450K, Kp = 2.0 × 1010/bar for the given reaction at equilibrium.

2SO2(g) + O2(g) 2SO3(g)

What is Kc at this temperature?

### કલિલ અવસ્થા : TY BSc, SEM 6, PHYSICAL CHEMISTRY US04CCHE23, UNIT 4

નીચે આપેલ ચિત્ર પર ક્લિક કરીને સંપૂર્ણ વિડીયો જોઈ શકો છો.

TY BSc, SEM 6, PHYSICAL CHEMISTRY US04CCHE23, UNIT 4

### Accuracy vs. Precision: Understanding the Difference - Class 11 Chemistry

Hi guys, This Dr. Nileshkumar Vala from My Smart Class, and in this video I am going to teach you all about Accuracy and Precision.

In the field of science, it is essential to understand and distinguish between accuracy and precision. Accuracy refers to how close a measurement is to the true or accepted value, while precision refers to how consistent a set of measurements are when repeated under the same conditions.

In this video, we will delve into the concepts of accuracy and precision in the context of Class 11 Chemistry. We will explore the differences between these two terms and understand why they are important in scientific experiments.

First, we will define accuracy and discuss how it is calculated. We will look at examples of accurate measurements and understand the impact of systematic and random errors on accuracy. Next, we will define precision and discuss how it is calculated. We will look at examples of precise measurements and understand the impact of random errors on precision.

We will then explore the relationship between accuracy and precision and understand how they are related but different concepts. We will also discuss the concept of significant figures and how it relates to accuracy and precision.

Finally, we will discuss the importance of accuracy and precision in scientific experiments. We will understand how they impact the validity and reliability of results and why it is essential to achieve both accuracy and precision in scientific measurements.

By the end of this video, viewers will have a clear understanding of accuracy and precision and their importance in scientific experiments. They will be able to distinguish between these two terms and apply them to real-world situations. This video is a valuable resource for Class 11 Chemistry students and anyone interested in the field of science.

This video is perfect for Class 11 Chemistry students who want to learn more about accuracy and precision. It is also useful for anyone who wants to gain a deeper understanding of these fundamental concepts in science. By watching this video, you will learn how to calculate accuracy and precision, understand their relationship, and appreciate their importance in scientific experiments.

In addition, this video is designed to be engaging and easy to follow. We use clear and concise language, real-world examples, and interactive visuals to help you understand these concepts better. Whether you are a visual, auditory, or kinesthetic learner, you will find this video helpful and informative.

So, if you want to take your understanding of accuracy and precision to the next level, be sure to watch this video. It will provide you with a solid foundation for further study and help you achieve success in your academic and professional pursuits. Don't miss out!

Hope you like it :) Thanks For Watching :)

## Class 12 Chemistry Chapter 6 NCERT Solutions – Free PDF Download

The most significant practice materials for the CBSE Class 12 Chemistry test and competitive exams are the NCERT Solutions for Class 12 Chemistry Chapter 6 General Concepts and Methods of Isolation of Elements. Students must take Class 12 Chemistry carefully whether they plan to continue their studies or if they want to get ready for a career. The NCERT Answers for Class 12 Chemistry include both exceptional and significant questions from past years' test questions in addition to answers to textbook questions.

Any questions and concerns about any chapter of CBSE Class 12 Chemistry can be answered using the NCERT Answers for Class 12 Chemistry. The solutions are clear and effectively organized in simple words for learning. The associated link provided below allows students to download the NCERT Answers for Class 12 Chemistry Chapter 6.

### Important Questions from Class 12 Chemistry (General Principles and Processes of Isolation of Elements) NCERT Solutions

00:20 Numerical 6.1 Copper can be extracted by hydrometallurgy but not zinc. Explain.

00:20 Numerical 6.2 What is the role of depressant in froth floatation process?

00:20 Numerical 6.3 Why is the extraction of copper from pyrites more difficult than that from its oxide are through reduction?

00:20 Numerical 6.4 Explain: (i) Zone refining (ii) Column chromatography.

00:20 Numerical 6.5 Out of C and CO, which is a belter reducing agent at 673 K?

00:20 Numerical 6.6 Name the common elements present in the anode mud in electrolytic refining of copper. Why are they so present?

00:20 Numerical 6.7 Write down the reactions taking place in different zones in the blast furnace during the extraction of iron.

00:20 Numerical 6.8 Write chemical reactions taking place in the extraction of zinc from zinc blende.

00:20 Numerical 6.9 State the role of silica in the metallurgy of copper.

00:20 Numerical 6.10 What is meant by the term "chromatography”?

00:20 Numerical 6.11 What criterion is followed for the selection of the stationary phase in chromatography?

00:20 Numerical 6.12 Describe a method for refining nickel.

00:20 Numerical 6.13 How can you separate alumina from silica in a bauxite ore associated with silica? Give equations. if any.

00:20 Numerical 6.14 Giving examples. Differentiate between 'roasting' and 'calcination'.

00:20 Numerical 6.15 How is 'cast iron' different from “pig iron”?

00:20 Numerical 6.16 Differentiate between “minerals” and "ores".

00:20 Numerical 6.17 Why copper matte is put in silica lined converter?

00:20 Numerical 6.18 What is the role of cryolite in the metallurgy of aluminium?

00:20 Numerical 6.19 How is leaching carried out in case of low grade copper ores?

00:20 Numerical 6.20 Why is zinc not extracted from zinc oxide through reduction using CO?

00:20 Numerical 6.21 The value of for formation of Cr2O3 is - 540 kJmol-1 and that of Al2O3 is - 827 kJmol-1. Is the reduction of Cr2O3 possible with Al ?

00:20 Numerical 6.22 Out of C and CO, which is a better reducing agent for ZnO ?

00:20 Numerical 6.23 The choice of a reducing agent in a particular case depends on thermodynamic factor. How far do you agree with this statement? Support your opinion with two examples.

00:20 Numerical 6.24 Name the processes from which chlorine is obtained as a by-product. What will happen if an aqueous solution of NaCl is subjected to electrolysis?

00:20 Numerical 6.25 What is the role of graphite rod in the electrometallurgy of aluminium?

00:20 Numerical 6.26 Outline the principles of refining of metals by the following methods:

(i) Zone refining

(ii) Electrolytic refining

(iii) Vapour phase refining

00:20 Numerical 6.27 Predict conditions under which Al might be expected to reduce MgO.

(Hint: See Intext question 6.4)