11th Standard Syllabus & Materials
11th Standard
TN 11th Tamil இயற்கை வேளாண்மை,சுற்றுச்சூழல் -செய்யுள் - மனோன்மணீயம் Important Questions And Answers Study Material - QB365 Set A
NEW11th Standard
TN 11th Tamil என்னுயிர் என்பேன் -துணைப்பாடம் - இசைத்தமிழர் இருவர் Important Questions And Answers Study Material - QB365 Set A
NEW11th Standard
TN 11th Tamil மொழி கலை -செய்யுள் - ஒவ்வொரு புல்லையும் Important Questions And Answers Study Material - QB365 Set A
NEW11th Standard
TN 11th Tamil பீடு பெற நில் - இலக்கணம் - பகுபத உறுப்புகள் Important Questions And Answers Study Material - QB365 Set A
NEW11th Standard
TN 11th Tamil பீடு பெற நில் - துணைப்பாடம் - வாடிவாசல் Important Questions And Answers Study Material - QB365 Set A
NEW11th Standard
TN 11th Tamil பீடு பெற நில் - செய்யுள் - குறுந்தொகை Important Questions And Answers Study Material - QB365 Set A

Published on: 25/06/2021
QB365 provides detailed and simple solution for every
Creative Questions in class 11 Chemistry Subject. It will helps to get more idea about question pattern in every Creative questions with solution.
Download Tamil Nadu 11th Standard Chemistry question papers, model tests, one-mark questions, important questions, and public exam papers in PDF format. Free study materials and answer keys for TN State Board students.
Questions + Answers key
Take MCQ Chemistry Test1.
CO2 gas cannot be liquefied at room temperature. Give reason.
2.
What is compression factor?
3.
Explain about the applications of Graham's law of diffusion.
4.
Helium diffuses more than air. Give reason.
5.
Define Graham's law of diffusion.
1.
Only below the critical temperature, by the application of pressure, a gas can be liquefied. CO2 has critical temperature as 303.98 K. Room temperature means (30 + 273 K) 303 K. At room temperature, (critical temperature) even by applying large amount of pressure CO2 cannot be liquefied. Only below the critical temperature, it can be liquefied. At room temperature, CO2 remains as gas.
2.
The deviation of real gases from ideal behaviour is measured in terms of a ratio of PV to nRT. This is termed as compression factor.
Compression factor = Z = \(PV\over nRT\)
For ideal gases Z = 1 at all temperature and pressures.
3.
(i) Graham's law of diffusion is useful to determine the molecular mass of the gas if the rate of diffusion is known.
(ii) Graham's law forms the basis of the process of enriching the isotopes of U235 from other isotopes and also useful in isotopic separation of deuterium and protium.
4.
Take two balloons, one is filled with air and another with helium. After one day, the helium balloon was shrunk, because helium being lighter diffuses out faster than the air.
5.
Graham's law of diffusion: The rate of diffusion or effusion is inversely proportional to the square root of molecular mass of a gas through an orifice.
\(\frac{r_A}{r_B}=\sqrt{\frac{M_B}{M_A}}\)
rA, rB = rate of diffusion of gases A, B
MA, MB = Molecular mass of gases A, B.
11th Standard Syllabus & Materials
11th Standard
TN 11th Tamil பீடு பெற நில் - செய்யுள் - காவடிச்சிந்து Important Questions And Answers Study Material - QB365 Set A
NEW11th Standard
TN 11th Tamil பீடு பெற நில் - உரைநடை - மலை இடப்பெயர்கள் : ஓர் ஆய்வு Important Questions And Answers Study Material - QB365 Set A
NEW11th Standard
TN 11th Tamil மாமழை போற்றுதும் - துணைப்பாடம் - யானை டாக்டர் Important Questions And Answers Study Material - QB365 Set A
NEW11th Standard
TN 11th Tamil மாமழை போற்றுதும் - செய்யுள் - ஐங்குறுநூறு Important Questions And Answers Study Material - QB365 Set A
Tamilnadu Stateboard 11th Standard Subjects

Maths

Commerce

Economics

Biology

Business Maths and Statistics

Accountancy

Computer Science

Physics

Chemistry

Maths

Biology

Economics

Physics

Chemistry

History

Business Maths and Statistics

Computer Science

Accountancy

Computer Applications

History

Computer Technology

Commerce

Computer Applications

Computer Technology

Tamil

English

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Tamilnadu Stateboard Standards