No products in the cart.

Ask questions which are clear, concise and easy to understand.

Ask Question

    • Posted by Ashman Preet Kaur Narwal 3 years, 3 months ago

    CBSE > Class 11 > Chemistry
    • 0 answers
    ANSWER

    • Posted by Aagney S Madhavan 3 years, 3 months ago

    CBSE > Class 11 > Chemistry
    • 1 answers

    Anjan Karthi 3 years, 3 months ago

    Though both fluorine and chlorine have same number of valence electrons, due to smaller valence orbital size of fluorine, the electron density in valence orbital is greater in fluorine than in chlorine. So, more effort would be required to insert a new electron in fluorine than in chlorine. So, chlorine has more negative electron gain enthalpy than in fluorine.

    0Thank You

    ANSWER

    • Posted by Amit Kumar 3 years, 3 months ago

    CBSE > Class 11 > Chemistry
    • 0 answers
    ANSWER

    • Posted by Nidhi Kharwal 3 years, 3 months ago

    CBSE > Class 11 > Chemistry
    • 1 answers

    Aashay Tembhurne 3 years, 3 months ago

    1 mole CH4 weighs 16g according to reaction 1mole of methane is required to produce 1 mole CO2 hence amount of CO2 produced is 12+32=44g

    3Thank You

    ANSWER

    • Posted by Jin Woo Sung 3 years, 3 months ago

    CBSE > Class 11 > Chemistry
    • 1 answers

    Anjan Karthi 3 years, 3 months ago

    Wavelength = 6000 Å = 6 × 10^-7 m. V = 3 × 10^8 m/s. Therefore, frequency = v/wavelength = 3 × 10^8 / 6 × 10^-7 = 0.5 × 10^15 Hz. // Energy = hv/wavelength = 6.626 × 10^-34 × 0.5 × 10^15 = 3.313 × 10^-19 J. //

    0Thank You

    ANSWER

    • Posted by Priyanka Khatri 3 years, 3 months ago

    CBSE > Class 11 > Chemistry
    • 0 answers
    ANSWER

    • Posted by Khushi Yadav 3 years, 3 months ago

    CBSE > Class 11 > Chemistry
    • 0 answers
    ANSWER

    • Posted by Himanshi Tyagi 3 years, 3 months ago

    CBSE > Class 11 > Chemistry
    • 1 answers

    Anjan Karthi 3 years, 3 months ago

    Molar mass of C12H22O11 = 342 g. Given mass = 3.42 g. No. of atoms of C = 12 × 6.022 × 10^23 × 3.42/342 = 12 × 6.022 × 10^23 × 10^-2 = 72.264 × 10^21 atoms. // Similarly, find H and O atoms also.

    0Thank You

    ANSWER

    • Posted by Rosa Deepti 3 years, 3 months ago

    CBSE > Class 11 > Chemistry
    • 0 answers
    ANSWER

    • Posted by Rebu Do 3 years, 3 months ago

    CBSE > Class 11 > Chemistry
    • 1 answers

    Ansuman Rath 3 years, 3 months ago

    Pagal

    0Thank You

    ANSWER

    • Posted by Nima Wongmu Bhutia 3 years, 3 months ago

    CBSE > Class 11 > Chemistry
    • 1 answers

    Sia ? 3 years, 3 months ago

    (a) s-block elements:

    All of the s-block elements are unified by the fact that their valence electrons (outermost electrons) are in an s orbital. The s orbital is spherical and can be occupied by a maximum of two electrons. All of the s-block elements are metals. In general, they are shiny, silvery, good conductors of heat and electricity and lose their valence electrons easily. In fact, they lose their trademark s orbital valence electrons so easily that the s-block elements are the most reactive elements.

    (b) p- block elements:

    Following are the properties of p-block elements.

    • They are solids/liquids/gases at room temperature (Br is liquid) 
    • They have variable oxidation states 
    • They form acidic oxides 
    • Generally, they form covalent compounds 
    • Halogens form salts with alkali metals 
    • They have high ionization potentials 
    • They have very large electron gain enthalpies 
    • The aqueous solutions their oxides are acidic in nature.

    (c) d-block elements:

    • Form stable complexes
    • Have high melting and boiling points
    • Contain large charge/radius ratio
    • Form compounds which are often paramagnetic
    • Are hard and possess high densities
    • Form compounds with profound catalytic activity
    • Show variable oxidation states
    • Form coloured ions and compounds.

    (d) f-block elements:

    • They are usually heavy metals
    • These elements have a generally higher melting and boiling point
    • They display a variety of oxidisation states
    • These elements tend to form coloured ions
    • They form complex compounds

    1Thank You

    ANSWER

    • Posted by Priyanka Khatri 3 years, 3 months ago

    CBSE > Class 11 > Chemistry
    • 2 answers

    Priyanka Khatri 3 years, 3 months ago

    Kay aap muja chemistry ka chapter 2 nd ka notes send kar da ga plz

    1Thank You

    Anjan Karthi 3 years, 3 months ago

    MOLARITY = 0.15 M. Volume = 100 cubic cm = 0.1 L. Therefore, number of moles of NaOH = M × V = 0.15 × 0.1 = 0.015 moles. // Amount of NaOH = number of moles × Molar mass = 0.015 × 40 = 0.15 × 4 = 0.60 g. //

    1Thank You

    ANSWER

    • Posted by Daisy Singh 3 years, 3 months ago

    CBSE > Class 11 > Chemistry
    • 0 answers
    ANSWER

    • Posted by Vishal Saini 3 years, 3 months ago

    CBSE > Class 11 > Chemistry
    • 1 answers

    Vishal Saini 3 years, 3 months ago

    The chemical reaction from the situation of the displacement reaction from glucose defence as kinetic energy of the chemical reaction and the the reaction of glucose breakdown in pyruvate from the situation of the the for 3mg of the the range range of the chemical reactions

    0Thank You

    ANSWER

    • Posted by Abhishek Meena 3 years, 3 months ago

    CBSE > Class 11 > Chemistry
    • 0 answers
    ANSWER

    • Posted by Arpita Priyadarshini 3 years, 3 months ago

    CBSE > Class 11 > Chemistry
    • 0 answers
    ANSWER

    • Posted by Manish Dabi 3 years, 3 months ago

    CBSE > Class 11 > Chemistry
    • 1 answers

    Anjan Karthi 3 years, 3 months ago

    Out of 100 L, 21 L is O2. Similarly, Out of 1 L of air, amount of oxygen is 0.21 L. Therefore, number of moles of O2 = Given volume/Molar volume = 0.21/22.4 = 0.0093 mol. //

    0Thank You

    ANSWER

    • Posted by Sakshi Sharma 3 years, 3 months ago

    CBSE > Class 11 > Chemistry
    • 1 answers

    Anjan Karthi 3 years, 3 months ago

    1) It rules down the limitations of the concept of definite paths or trajectories around the nucleus put forward by Neils Bohr. 2) It is applicable only to microscopic objects and is negligible in the case of macroscopic ones. 3) It can be applied only in case of objects of size milli to micro grams. Useless in case of gram and kilograms weight.

    3Thank You

    ANSWER

    • Posted by Abhishek Su 3 years, 3 months ago

    CBSE > Class 11 > Chemistry
    • 1 answers

    Sia ? 3 years, 3 months ago

    Meson, any member of a family of subatomic particles composed of a quark and an antiquark. Mesons are sensitive to the strong force, the fundamental interaction that binds the components of the nucleus by governing the behaviour of their constituent quarks.

    1Thank You

    ANSWER

    • Posted by Khushi Gupta 3 years, 4 months ago

    CBSE > Class 11 > Chemistry
    • 0 answers
    ANSWER

    • Posted by Nikhil Kumar Singh 3 years, 4 months ago

    CBSE > Class 11 > Chemistry
    • 0 answers
    ANSWER

    • Posted by Anirudha Chakraborty 3 years, 4 months ago

    CBSE > Class 11 > Chemistry
    • 1 answers

    Anjan Karthi 3 years, 4 months ago

    As the frequency of light falling on the electrons increases, the intensity of light also increases and thereby, the number of ejected photoelectrons increases.

    2Thank You

    ANSWER

    • Posted by Ritesh Kumar 3 years, 4 months ago

    CBSE > Class 11 > Chemistry
    • 1 answers

    Sia ? 3 years, 4 months ago

    Significant figures of a number in positional notation are digits in the number that are reliable and absolutely necessary to indicate the quantity of something.

    1Thank You

    ANSWER

    • Posted by Sumit Tak 3 years, 4 months ago

    CBSE > Class 11 > Chemistry
    • 2 answers

    Priyanshi Sharma 3 years, 3 months ago

    Hydrogen is the limiting reagent.

    1Thank You

    Anjan Karthi 3 years, 4 months ago

    In H2O or H2O2 ?

    0Thank You

    ANSWER

    • Posted by Lk . 3 years, 4 months ago

    CBSE > Class 11 > Chemistry
    • 2 answers

    Anjan Karthi 3 years, 4 months ago

    Camphor not possible as it has no liquid state.

    0Thank You

    Arun Kushwaha 3 years, 4 months ago

    We take camphor as example

    0Thank You

    ANSWER

    • Posted by Rashi Jain 3 years, 4 months ago

    CBSE > Class 11 > Chemistry
    • 2 answers

    Anjan Karthi 3 years, 4 months ago

    Distance (in miles) = 18500 miles/ hour . Distance (in km) = 18500 × 1.609 = 29766.5 kms. // Therefore, 29766.5 km/h = 29766.5 × 5/18 = 8268.47 m/s.//

    2Thank You

    Mayank Raj 3 years, 4 months ago

    Hiii

    0Thank You

    ANSWER

    • Posted by Himanshu Rajput 3 years, 4 months ago

    CBSE > Class 11 > Chemistry
    • 2 answers

    Sahil Singh 3 years, 4 months ago

    Hydrogen spectrum

    0Thank You

    Anjan Karthi 3 years, 4 months ago

    An emission atomic spectrum with lights of various wavelengths and frequencies.

    0Thank You

    ANSWER

    • Posted by Ram Sohan 3 years, 4 months ago

    CBSE > Class 11 > Chemistry
    • 1 answers

    Ayush Sharma {Xl-A} 3 years, 4 months ago

    Charge of electron was given by RA Millikan by oil drop experiment in (1906-14)

    4Thank You

    ANSWER

    • Posted by Aasa Aasa 3 years, 4 months ago

    CBSE > Class 11 > Chemistry
    • 0 answers
    ANSWER

    • Posted by Sinchana Mali Patil 3 years, 4 months ago

    CBSE > Class 11 > Chemistry
    • 1 answers

    Anjan Karthi 3 years, 4 months ago

    Determined in terms of several factors including MOLARITY, MOLALITY, MASS% and VOLUME%, NORMALITY, PARTS PER MILLION, etc. Generally, we can define the CONCENTRATION of a solution as the amount of solute in a particular amountof the solution.

    2Thank You

    ANSWER

    myCBSEguide App

    myCBSEguide

    Trusted by 1 Crore+ Students

    Install Now

    Test Generator

    Test Generator

    Create papers online. It's FREE.

    Try Now

    CUET Mock Tests

    CUET Mock Tests

    75,000+ questions to practice only on myCBSEguide app

    Install Now
    Google Play

    Download myCBSEguide App

    All Courses

    Other Websites

    CBSE Courses

    Resources

    Corporate

    NCERT Solutions

    CBSE Sample Papers

    All Content and Intellectual Property is under Copyright Protection | myCBSEguide.com ©2007-2024