CBSE > Class 11 > Chemistry | CBSE Board | Homework Help

First you go through concept of organic chemistry in NCERT Book. Then make question by itself and answer if fell confident ,then make exercise question and internal question also ,finally you fell something I know about what actually organic chemistry after that you go on goc side book.

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Hemang Dubey 2 days, 9 hours ago

Also be confident

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Harsh Singh 3 days, 2 hours ago

By revising again and again and by doing lots of question practice

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How many Sigma and Pi bonds are present in Pentene

Posted by Hemang Dubey 3 days, 5 hours ago

CBSE > Class 11 > Chemistry

1 answers

Adithya Dev A 3 days, 2 hours ago

Pentene structure is CH2=CHCH2CH2CH3. Therefore, number of σ bonds = 8. And total number of π bonds = 1.

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The length of h bonding in solid state of water is

Posted by Hemang Dubey 3 days, 5 hours ago

CBSE > Class 11 > Chemistry

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FeCl3 + H2 S  FeCl2 + 5 + HCl

Posted by Hemang Dubey 3 days, 6 hours ago

CBSE > Class 11 > Chemistry

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unit of length.

Posted by Abhash Tiwari 3 days, 16 hours ago

CBSE > Class 11 > Chemistry

4 answers

Mukul Kumar 3 days, 3 hours ago

Here you have three units. 1st is meter. 2nd is foot. 3rd is centimetres.

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Kiran Sambharwal 3 days, 8 hours ago

Meter

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Dhruvi Chauhan 3 days, 10 hours ago

Meter

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Abhash Tiwari 3 days, 16 hours ago

metere

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The first ionization enthalpy of O is less than N. Why ?

Posted by Shivani Kumari 4 days, 21 hours ago

CBSE > Class 11 > Chemistry

1 answers

Adithya Dev A 4 days, 10 hours ago

Electronic configuration of O = 1s2 2s2 2p4 and that for N = 1s2 2s2 2p5. Here, in N, the 2p subshell already has a HALF FILLED electronic arrangement unlike in O. This ensures greater stability for N atom than for O atom and thus, ionisation enthalpy (energy required to remove outermost electron) for O becomes lesser in magnitude than that for N.

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The first ionization enthalpy of B is less than first ionization enthalpy of Be . Why?

Posted by Shivani Kumari 4 days, 21 hours ago

CBSE > Class 11 > Chemistry

1 answers

Adithya Dev A 4 days, 9 hours ago

Be has an Electronic configuration of 1s2 2s2 and B has an electronic configuration of 1s2 2s2 2p1. The p 2p orbital, causing lesser shielding effect than 2s orbital, would be able to loose an electron much more easily than 2s orbital. Hence, Be has greater ionization enthalpy than B.

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Types of chemical reaction in chapter 12

Posted by Dharni Rathor 1 week, 1 day ago

CBSE > Class 11 > Chemistry

1 answers

Shravani Tambe 1 week ago

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SIMILARITIES AND DISIMILARITIES BETWEEN HYDROGEN,HALOGEN AND ALKALI METALS

Posted by Apurva J Patil 1 week, 1 day ago

CBSE > Class 11 > Chemistry

1 answers

Prabhjot Kaur 5 days, 17 hours ago

Similarities between hydrogen and alkali metals : Electronic configuration Electropositive character Valency and oxidation state Reducing agents Dissimilarities between alkali metals: Ionisation enthlapy Non metallic in nature Atomicity Similarities between hydrogen and halogen Electronic configuration Ionisation enthlapy Electronegative character Oxidation state Diatomic nature Nature of compounds Dissimilarities between hydrogen and halogen Less tendency for hydride formation Absence of unshared pair of electron Oxides of halogens are acidic

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For hydrogen atom the energy is expressed as En=-13.6/n ev/atom. The value of 2nd Ionisation energy of helium atom will be

Posted by Kiran Mandal 1 week, 1 day ago

CBSE > Class 11 > Chemistry

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Calculate the volume of 0.015M HCL solution required to prepare 250ml of a 5.25×1/1000M HCL solution

Posted by Pinki Borah 1 week, 1 day ago

CBSE > Class 11 > Chemistry

2 answers

Kaushik Ka 6 days, 19 hours ago

M1*V1 = M2*V2 (0.015×250)÷(5.25÷1,000) = V2 V2 = 714.285 ml

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Kaushik Ka 6 days, 19 hours ago

M1*V1 = M2*V2

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Calculate the number of molecules in 20 g aluminum sulphate

Posted by Baazigha Sultana Siddiqui 1 week, 2 days ago

CBSE > Class 11 > Chemistry

1 answers

Adithya Dev A 3 days, 11 hours ago

Molar mass of Al2(SO4) 3 = 2(27) + (32 + 64) 3 = 54 + 288 = 342 g. // No of molecules = (20/342) × 6.02 × 10^23 = 0.352 × 10^23 molecules//

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How many moles of methane required to produce 88g of CO2 after combustion

Posted by Akash Malsa 1 week, 5 days ago

CBSE > Class 11 > Chemistry

1 answers

Adithya Dev A 1 week, 4 days ago

According to balanced equation, CH4 + 2O2 → CO2 + 2H2O. No of moles of CO2 produced = W2/M2 = 88/44 = 2 mol// Therefore, no. of moles of Methane required = 2 mol. //

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Electronic configuration of florine

Posted by Kanak Sahu 1 week, 5 days ago

CBSE > Class 11 > Chemistry

3 answers

Khushi Rani 1 week, 4 days ago

1s*2 2s*2p*5

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Jatin Ola 1 week, 5 days ago

👍👍

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Jatin Ola 1 week, 5 days ago

1s² 2s² 2p⁵

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Which out of the following contain least number of atoms? A. 1gm of H2O B. 1gm of CH4 C. 1gm of N2

Posted by Anoubam Victor 1 week, 5 days ago

CBSE > Class 11 > Chemistry

1 answers

Ningmi Chon 1 week, 5 days ago

1×4×6•023×1023

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Which out of the following contain least number of atoms?

Posted by Anoubam Victor 1 week, 5 days ago

CBSE > Class 11 > Chemistry

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Calculate the percent ionic character of HCl. Given that the observed dipole moment is 1.03D and length of HCl is 1.275 Å.

Posted by Priyanka Khatri 1 week, 6 days ago

CBSE > Class 11 > Chemistry

2 answers

Priyanka Khatri 1 week, 2 days ago

Thanks

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Roopam Almadi 1 week, 5 days ago

Fraction of charge = 1.03 × 10^-18 esu. cm / 4.8 × 10 ^-10 esu × 1.275 × 10 ^-8 cm Fraction of charge = 0.1683 % Ionic character = Fraction of charge × 100 So% ionic character = 16.83 % So % covalent character = 100 - 16.83 = 83.17 % ....

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If 2.36grm of sodium sulphate is added to berium chloride the residue is formed to be 3.88grm of barium sulphate and 1.98grm of sodium chloride.What is the mass of barium chloride?

Posted by Deiba Pariong 2 weeks ago

CBSE > Class 11 > Chemistry

1 answers

Adithya Dev A 2 weeks ago

Law of conservation of mass. LHS = RHS. Therefore, 2.36g Na2SO4 + x g BeCl2 = 3.88g BaSO4 + 1.98g NaCl. Therefore, required mass of BeCl2 = 3.5 g. //

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A mixture of mgco3 and caco3 I was heated for a long time the wait is decreased by 50% the percentage composition of mgco3 and that of caco3 in the mixture will be respectively

Posted by Pushpa K 2 weeks, 2 days ago

CBSE > Class 11 > Chemistry

1 answers

Himanshu Mali 1 week, 6 days ago

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Short note on Development of Chemistry

Posted by Baisa Hkm 👸👸 2 weeks, 3 days ago

CBSE > Class 11 > Chemistry

2 answers

Baisa Hkm 👸👸 1 week, 5 days ago

What is this

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Amitesh Tiwari 2 weeks ago

Hdhdjsjvkvvieie

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What is shelding effect?

Posted by Aman Kushwaha 2 weeks, 3 days ago

CBSE > Class 11 > Chemistry

1 answers

Adithya Dev A 2 weeks, 2 days ago

In case of big atoms like Sulphur and Potassium, the electrons in outer shells is not given a significant part of the effective nuclear charge of the atom due to the inner electrons. This is called SHIELDING EFFECT.

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What are the limitations of octect rule?

Posted by Jissmol Jacob Varghese 2 weeks, 3 days ago

CBSE > Class 11 > Chemistry

1 answers

Diksha Thakur 2 weeks, 2 days ago

Limitations : (1) Hydrogen with 1 electron attains stability by sharing, gaining or losing 1 valence electron. It does not need to complete octet to attain stability. Also, He has only 2 electrons and is stable. (2) Incomplete octet: In certain molecules such as BeH 2 , BeCl 2 , BH 3 , BF 3 , the central atom has less than 8 electrons in its valence shell, yet the molecule is stable. (3) Expanded octet: In certain molecules such as PF 5 , SF 6 , IF 7 , H 2 SO 4 , the central atom has more than 8 valence electrons, yet the molecule is stable.

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Calculate the number of moles in 7.9mg of calcium

Posted by Devi Devi 2 weeks, 4 days ago

CBSE > Class 11 > Chemistry