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 General Principles and Processes of Isolation of Elements NCERT Solutions

 

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)


Class 11 Unit 6 Thermodynamics Full Exercise Solution 6.1 to 6.22 NCERT Solution 2022

Class 11 Unit 6 Thermodynamics Full Exercise Solution 6.1 to 6.22 NCERT Solution 2022


Hi guys, This Falguni Vala from My Smart Class, in this video, I am going to teach you all about Class 11 Unit 6 Thermodynamics Full Exercise Solution 6.1 to 6.22 
Time Stamp of Numerical is given below in the description. 
6.1 Choose the correct answer. A thermodynamic state function is a quantity 
6.2 For the process to occur under adiabatic conditions, the correct condition is: 
6.3 The enthalpies of all elements in their standard states are: 
6.4 ∆U0of combustion of methane is – X kJ mol–1. The value of ∆H is 
6.5 The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, –890.3 kJ mol–1 –393.5 kJ mol–1, and –285.8 kJ mol–1 respectively. Enthalpy of formation of CH4(g) will be 
6.6 A reaction, A + B → C + D + q is found to have a positive entropy change. The reaction will be 
6.7 In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system. What is the change in internal energy for the process? 
6.8 The reaction of cyanamide, NH2CN (s), with dioxygen was carried out in a bomb calorimeter, and ∆U was found to be –742.7 kJ mol–1 at 298 K. Calculate enthalpy change for the reaction at 298 K. 
6.9 Calculate the number of kJ of heat necessary to raise the temperature of 60.0 g of aluminium from 35°C to 55°C. Molar heat capacity of Al is 24 J mol–1 K–1. 
6.10 Calculate the enthalpy change on freezing of 1.0 mol of water at10.0°C to ice at –10.0°C. 
6.11 Enthalpy of combustion of carbon to CO2 is –393.5 kJ mol–1. Calculate the heat released upon formation of 35.2 g of CO2 from carbon and dioxygen gas. 
6.12 Enthalpies of formation of CO(g), CO2(g), N2O(g) and N2O4(g) are –110, – 393, 81 and 9.7 kJ mol–1 respectively. Find the value of ∆rH for the reaction: 
6.13 Given N2(g) + 3H2(g) → 2NH3(g) ; ∆rH = –92.4 kJ mol–1 What is the standard enthalpy of formation of NH3 gas? 
6.14 Calculate the standard enthalpy of formation of CH3OH(l) from the following data: CH3OH (l) +3/2 O2(g) → CO2(g) + 2H2O(l) ; ∆rH = –726 kJ mol–1 C(graphite) + O2(g) → CO2(g) ; ∆cH = –393 kJ mol–1 H2(g) +1/2O2(g) → H2O(l) ; ∆f H = –286 kJ mol–1. 
6.15 Calculate the enthalpy change for the process CCl4(g) → C(g) + 4 Cl(g)and calculate bond enthalpy of C – Cl in CCl4(g).∆vapH(CCl4) = 30.5 kJ mol–1.∆fH (CCl4) = –135.5 kJ mol–1.∆Ha (C) = 715.0 kJ mol–1 , where ∆Hais enthalpy of atomisation∆Ha (Cl2) = 242 kJ mol–1 
6.16 For an isolated system, ∆U = 0, what will be ∆S ? 
6.17 For the reaction at 298 K, 2A + B → C ∆H = 400 kJ mol–1 and ∆S = 0.2 kJ K–1 mol–1 At what temperature will the reaction become spontaneous considering ∆H and ∆S to be constant over the temperature range. 
6.18 For the reaction, 2 Cl(g) → Cl2(g), what are the signs of ∆H and ∆S ? 
6.19 For the reaction 2 A(g) + B(g) → 2D(g) ∆U  = –10.5 kJ and ∆S = –44.1 JK–1. Calculate ∆G for the reaction, and predict whether the reaction may occur spontaneously. 
6.20 The equilibrium constant for a reaction is 10. What will be the value of ∆G ? R = 8.314 JK–1 mol–1, T = 300 K. 
6.21 Comment on the thermodynamic stability of NO(g), given 
6.22 Calculate the entropy change in surroundings when 1.00 mol of H2O(l) is formed under standard conditions. ∆f H = –286 kJ mol–1.

THERMODYNAMICS Exercise Solution 6.1 to 6.5, Class 11 Chemistry Chapter 6

 



6.1 Choose the correct answer. A thermodynamic state function is a quantity 
(i) used to determine heat changes
(ii) whose value is independent of path
(iii) used to determine pressure volume work
(iv) whose value depends on temperature only.

6.2 For the process to occur under adiabatic conditions, the correct condition is:
(i) ∆T = 0
(ii) ∆p = 0
(iii) q = 0
(iv) w = 0

6.3 The enthalpies of all elements in their standard states are:
(i) unity
(ii) zero
(iii)  0
(iv) different for each element

6.4 ∆U° of combustion of methane is – X kJ mol–1. The value of ∆H° is
(i) = ∆U°
(ii)  ∆U°
(iii)  ∆U°
(iv) = 0

6.5 The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, –890.3 kJ mol–1 –393.5 kJ mol–1, and –285.8 kJ mol–1 respectively. Enthalpy of formation of CH4(g) will be
(i) –74.8 kJ mol–1 
(ii) –52.27 kJ mol–1
(iii) +74.8 kJ mol–1 
(iv) +52.26 kJ mol–1.



Class 11 Chemistry Chapter 1 Some Basic Concept of Chemistry Exercise Solution Question 1.11

 

Class 11 Chemistry Chapter 1 Some Basic Concept of Chemistry Exercise Solution Question 1.11

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