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    • Posted by Vivek Rawat 5 years ago

    CBSE > Class 11 > Chemistry
    • 1 answers

    Akriti Singh Rathore 5 years ago

    The van der Waals constant 'a' represents the magnitude of intermolecular forces of attraction and the Van der Waals constant 'b' represents the effective size of the molecules.

    0Thank You

    ANSWER

    • Posted by Prithviraj Jangu Bishnoi 5 years ago

    CBSE > Class 11 > Chemistry
    • 1 answers

    Divyansh Rai 5 years ago

    Along the period ionisation enthalpy increases while down the group decrease Along the period election affinity increases while down the group decrease

    0Thank You

    ANSWER

    • Posted by Royal Thakur ? 5 years, 1 month ago

    CBSE > Class 11 > Chemistry
    • 3 answers

    Vishist Bollu 5 years ago

    288(v1) * 301 (t2) / 288 (t1) = 301 (v2)

    2Thank You

    Royal Thakur ? 5 years, 1 month ago

    Thank you ?

    3Thank You

    Ranger King 5 years, 1 month ago

    Charles formula =v1/t1=v2/t2 T=15+273=288k V=288ml T=28+273=301k V1t2/t1=v2 288×288/301= 275.56

    2Thank You

    ANSWER

    • Posted by Royal Thakur ? 5 years, 1 month ago

    CBSE > Class 11 > Chemistry
    • 2 answers

    Royal Thakur ? 5 years, 1 month ago

    Thanks a lot ?

    3Thank You

    Ranger King 5 years, 1 month ago

    Pv=nrt T=pv/nr T=2.46×10/1×0.0821 =29.96

    2Thank You

    ANSWER

    • Posted by Killers Gaming 5 years, 1 month ago

    CBSE > Class 11 > Chemistry
    • 1 answers

    Pankaj Ponia 5 years ago

    H-o-h/h+

    0Thank You

    ANSWER

    • Posted by Mohit Yadav 5 years, 1 month ago

    CBSE > Class 11 > Chemistry
    • 1 answers

    Aryan Singh 5 years, 1 month ago

    Oxygen doesn't burn it is an oxidizer

    1Thank You

    ANSWER

    • Posted by Laishram Archana 5 years, 1 month ago

    CBSE > Class 11 > Chemistry
    • 1 answers

    Ranger King 5 years, 1 month ago

    1. 2s 2. 3d

    1Thank You

    ANSWER

    • Posted by Titiksha Kakran 5 years, 1 month ago

    CBSE > Class 11 > Chemistry
    • 1 answers

    Ranger King 5 years, 1 month ago

    Mass of 1 electron = 9.1×10power - - 31 1gm=1/9.1×10power - - 31=1.09×10power30

    1Thank You

    ANSWER

    • Posted by Mayank Pushkar 5 years, 1 month ago

    CBSE > Class 11 > Chemistry
    • 1 answers

    Bhawna Dahiya 5 years ago

    CH4N2O

    0Thank You

    ANSWER

    • Posted by Ishü Mîrzâ 5 years, 1 month ago

    CBSE > Class 11 > Chemistry
    • 4 answers

    Pankaj Ponia 5 years ago

    Si unit of mass is kilogram it can be written in short form as (kg).

    0Thank You

    Bhawna Dahiya 5 years ago

    Mass has dimension less and according to SI system of units mass has unit Kilograms (Kg).

    0Thank You

    Ranger King 5 years, 1 month ago

    S I unit of mass is kg

    1Thank You

    Amisha Sharma 5 years, 1 month ago

    Kg

    2Thank You

    ANSWER

    • Posted by Royal Thakur ? 4 years, 9 months ago

    CBSE > Class 11 > Chemistry
    • 0 answers
    ANSWER

    • Posted by Subhranshu Behera 5 years, 1 month ago

    CBSE > Class 11 > Chemistry
    • 1 answers

    Shaik Habeeb Ahmed 5 years, 1 month ago

    h÷v=c

    3Thank You

    ANSWER

    • Posted by Suma Suma 4 years, 10 months ago

    CBSE > Class 11 > Chemistry
    • 1 answers

    Sia ? 4 years, 10 months ago

    The photoelectric effect is the emission of electrons when electromagnetic radiation, such as light, hits a material. ... The experimental results instead show that electrons are dislodged only when the light exceeds a certain frequency—regardless of the light's intensity or duration of exposure.

    0Thank You

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    • Posted by Class 12 5 years, 1 month ago

    CBSE > Class 11 > Chemistry
    • 2 answers

    Om P̤A̤T̤E̤L̤ 5 years ago

    The elements that can easily lose electrons to form positive ions are called electropositive elements, for example: metals. ... The elements that can easily accept electrons to form negative ions are called electronegative elements, for example: non-metals.

    0Thank You

    Swastik . P 5 years, 1 month ago

    Electronegative atom pulls electron towards it and Electropositive atom is opposite of electronegative

    2Thank You

    ANSWER

    • Posted by Fear Fighter 5 years, 1 month ago

    CBSE > Class 11 > Chemistry
    • 0 answers
    ANSWER

    • Posted by Johal ? 5 years, 1 month ago

    CBSE > Class 11 > Chemistry
    • 2 answers

    Anuj Kumar 5 years, 1 month ago

    <cml><MDocument><MChemicalStruct><molecule molID="m1"><atomArray><atom id="a1" elementType="C" x2="-5.3125" y2="1.4322916666666667"/><atom id="a2" elementType="C" x2="-3.9788208781719643" y2="2.2022916666666665"/><atom id="a3" elementType="C" x2="-2.645141756343929" y2="1.4322916666666665"/><atom id="a4" elementType="C" x2="-1.311462634515<cml><MDocument><MChemicalStruct><molecule molID="m1"><atomArray><atom id="a1" elementType="C" x2="-5.3125" y2="1.4322916666666667"/><atom id="a2" elementType="C" x2="-3.9788208781719643" y2="2.2022916666666665"/><atom id="a3" elementType="C" x2="-2.645141756343929" y2="1.4322916666666665"/><atom id="a4" elementType="C" x2="-1.3114626345158946" y2="2.2022916666666665"/><atom id="a5" elementType="C" x2="0.022216487312141098" y2="1.4322916666666665"/><atom id="a6" elementType="C" x2="1.3558956091401768" y2="2.2022916666666665"/><atom id="a7" elementType="C" x2="-2.645141756343929" y2="-0.10770833333333352"/><atom id="a8" elementType="C" x2="-1.1504888010837082" y2="3.7338553855338077"/><atom id="a9" elementType="O" x2="-3.208820878171964" y2="3.5359707884947023" lonePair="2"/><atom id="a10" elementType="O" x2="-0.13875734612004503" y2="-0.09927205220047464" lonePair="2"/><atom id="a11" elementType="N" x2="2.895895609140177" y2="2.2022916666666665" lonePair="1"/><atom id="a12" elementType="C" x2="3.9263567429328186" y2="1.0578486354314793"/><atom id="a13" elementType="C" x2="3.371781780477596" y2="3.666918701761203"/><atom id="a14" elementType="C" x2="3.092652500649646" y2="-1.1867091514227464"/><atom id="a15" elementType="C" x2="3.997841789180055" y2="-2.4325953227601653"/></atomArray><bondArray><bond atomRefs2="a1 a2" order="1" id="b1"/><bond atomRefs2="a2 a3" order="1" id="b2"/><bond atomRefs2="a3 a4" order="1" id="b3"/><bond atomRefs2="a4 a5" order="1" id="b4"/><bond atomRefs2="a5 a6" order="1" id="b5"/><bond atomRefs2="a3 a7" order="1" id="b6"/><bond atomRefs2="a4 a8" order="1" id="b7"/><bond atomRefs2="a2 a9" order="1" id="b8"/><bond atomRefs2="a5 a10" order="1" id="b9"/><bond atomRefs2="a6 a11" order="1" id="b10"/><bond atomRefs2="a11 a12" order="1" id="b11"/><bond atomRefs2="a11 a13" order="1" id="b12"/><bond atomRefs2="a14 a15" order="1" id="b13"/></bondArray></molecule></MChemicalStruct><MElectronContainer occupation="0 0" radical="0" id="o1"><MElectron atomRefs="m1.a9" difLoc="0.0 0.0 0.0"/><MElectron atomRefs="m1.a9" difLoc="0.0 0.0 0.0"/></MElectronContainer><MElectronContainer occupation="0 0" radical="0" id="o2"><MElectron atomRefs="m1.a9" difLoc="0.0 0.0 0.0"/><MElectron atomRefs="m1.a9" difLoc="0.0 0.0 0.0"/></MElectronContainer><MElectronContainer occupation="0 0" radical="0" id="o3"><MElectron atomRefs="m1.a10" difLoc="0.0 0.0 0.0"/><MElectron atomRefs="m1.a10" difLoc="0.0 0.0 0.0"/></MElectronContainer><MElectronContainer occupation="0 0" radical="0" id="o4"><MElectron atomRefs="m1.a10" difLoc="0.0 0.0 0.0"/><MElectron atomRefs="m1.a10" difLoc="0.0 0.0 0.0"/></MElectronContainer><MElectronContainer occupation="0 0" radical="0" id="o5"><MElectron atomRefs="m1.a11" difLoc="0.0 0.0 0.0"/><MElectron atomRefs="m1.a11" difLoc="0.0 0.0 0.0"/></MElectronContainer></MDocument></cml>8946" y2="2.2022916666666665"/><atom id="a5" elementType="C" x2="0.022216487312141098" y2="1.4322916666666665"/><atom id="a6" elementType="C" x2="1.3558956091401768" y2="2.2022916666666665"/><atom id="a7" elementType="C" x2="-2.645141756343929" y2="-0.10770833333333352"/><atom id="a8" elementType="C" x2="-1.1504888010837082" y2="3.7338553855338077"/><atom id="a9" elementType="O" x2="-3.208820878171964" y2="3.5359707884947023" lonePair="2"/><atom id="a10" elementType="O" x2="-0.13875734612004503" y2="-0.09927205220047464" lonePair="2"/><atom id="a11" elementType="N" x2="2.895895609140177" y2="2.2022916666666665" lonePair="1"/><atom id="a12" elementType="C" x2="3.9263567429328186" y2="1.0578486354314793"/><atom id="a13" elementType="C" x2="3.371781780477596" y2="3.666918701761203"/><atom id="a14" elementType="C" x2="3.092652500649646" y2="-1.1867091514227464"/><atom id="a15" elementType="C" x2="3.997841789180055" y2="-2.4325953227601653"/></atomArray><bondArray><bond atomRefs2="a1 a2" order="1" id="b1"/><bond atomRefs2="a2 a3" order="1" id="b2"/><bond atomRefs2="a3 a4" order="1" id="b3"/><bond atomRefs2="a4 a5" order="1" id="b4"/><bond atomRefs2="a5 a6" order="1" id="b5"/><bond atomRefs2="a3 a7" order="1" id="b6"/><bond atomRefs2="a4 a8" order="1" id="b7"/><bond atomRefs2="a2 a9" order="1" id="b8"/><bond atomRefs2="a5 a10" order="1" id="b9"/><bond atomRefs2="a6 a11" order="1" id="b10"/><bond atomRefs2="a11 a12" order="1" id="b11"/><bond atomRefs2="a11 a13" order="1" id="b12"/><bond atomRefs2="a14 a15" order="1" id="b13"/></bondArray></molecule></MChemicalStruct><MElectronContainer occupation="0 0" radical="0" id="o1"><MElectron atomRefs="m1.a9" difLoc="0.0 0.0 0.0"/><MElectron atomRefs="m1.a9" difLoc="0.0 0.0 0.0"/></MElectronContainer><MElectronContainer occupation="0 0" radical="0" id="o2"><MElectron atomRefs="m1.a9" difLoc="0.0 0.0 0.0"/><MElectron atomRefs="m1.a9" difLoc="0.0 0.0 0.0"/></MElectronContainer><MElectronContainer occupation="0 0" radical="0" id="o3"><MElectron atomRefs="m1.a10" difLoc="0.0 0.0 0.0"/><MElectron atomRefs="m1.a10" difLoc="0.0 0.0 0.0"/></MElectronContainer><MElectronContainer occupation="0 0" radical="0" id="o4"><MElectron atomRefs="m1.a10" difLoc="0.0 0.0 0.0"/><MElectron atomRefs="m1.a10" difLoc="0.0 0.0 0.0"/></MElectronContainer><MElectronContainer occupation="0 0" radical="0" id="o5"><MElectron atomRefs="m1.a11" difLoc="0.0 0.0 0.0"/><MElectron atomRefs="m1.a11" difLoc="0.0 0.0 0.0"/></MElectronContainer></MDocument></cml>

    2Thank You

    Royal Thakur ? 5 years, 1 month ago

    #Anubhav misra get out from here... Kyunki Ye app Tum jaise logo ke liye nhi h.....jha se Ye likhna sikhe Ho vha se Thodi respect Krna bhi sekh lete to Accha rehta.....dusro ke maa ke nhi to km se km apne maa ke respect kro Aur unke sanskaro ke bhi..... Same on you and also on people like you ??...Discusting you are ???

    8Thank You

    ANSWER

    • Posted by Fear Fighter 5 years, 1 month ago

    CBSE > Class 11 > Chemistry
    • 1 answers

    Mayukh Karmakar 5 years, 1 month ago

    1. Chp. 12 -Organic chemistry. 2. Chp.13 -Hydrocarbons.

    0Thank You

    ANSWER

    • Posted by Pratyasha Das 5 years, 1 month ago

    CBSE > Class 11 > Chemistry
    • 0 answers
    ANSWER

    • Posted by Anjali ✍️ 5 years, 1 month ago

    CBSE > Class 11 > Chemistry
    • 3 answers

    Pankaj Ponia 5 years ago

    Ch 2 delete

    0Thank You

    First Name 5 years, 1 month ago

    <a href="http://cbseacademic.nic.in/web_material/CurriculumMain21/revisedsyllabi/Deduction/DELETEDChemistry_2020-21.pdf">Tap here</a> to get them

    4Thank You

    Royal Thakur ? 5 years, 1 month ago

    I think Chapters delete nhi huye hai ...topics delete huye h...

    5Thank You

    ANSWER

    • Posted by Har Har Mahadev 🙏 5 years, 1 month ago

    CBSE > Class 11 > Chemistry
    • 0 answers
    ANSWER

    • Posted by Class 12 5 years, 1 month ago

    CBSE > Class 11 > Chemistry
    • 0 answers
    ANSWER

    • Posted by Varsha Baboria 5 years, 1 month ago

    CBSE > Class 11 > Chemistry
    • 4 answers

    Ankur Yadav 5 years ago

    Matter consists of individual particals namely atom which cannot devided further. and atoms of a particular element are identical

    0Thank You

    Ranger King 5 years, 1 month ago

    Postulate of atomic theory - matter consist of indivisible and indescrible particle -atom of same elements are identical to each other.

    3Thank You

    Ranger King 5 years, 1 month ago

    Is called dalton partial pressure

    3Thank You

    Ranger King 5 years, 1 month ago

    In a mixture of gases the pressure applied by pressure is the sum of pressure of individual gases

    2Thank You

    ANSWER

    • Posted by Harish Tode 5 years, 1 month ago

    CBSE > Class 11 > Chemistry
    • 1 answers

    Ranger King 5 years, 1 month ago

    n=?v=500ml=0.5 L,760 mm of Hg=1 atm.  Now apply the gas equation, n=RTPV​ n=0.0821×3001×0.5​ ⟹n=2.03×10−2mol

    0Thank You

    ANSWER

    • Posted by Laishram Archana 5 years, 1 month ago

    CBSE > Class 11 > Chemistry
    • 0 answers
    ANSWER

    • Posted by Amrita Kumari 5 years, 1 month ago

    CBSE > Class 11 > Chemistry
    • 0 answers
    ANSWER

    • Posted by Archana Meena 5 years, 1 month ago

    CBSE > Class 11 > Chemistry
    • 2 answers

    Pankaj Ponia 5 years ago

    Ya

    0Thank You

    Tec Om 5 years, 1 month ago

    i will try but please tell me on which topic

    5Thank You

    ANSWER

    • Posted by सत्य सनातन? 5 years, 1 month ago

    CBSE > Class 11 > Chemistry
    • 2 answers

    Pankaj Ponia 5 years ago

    Yes as well as in bio also

    0Thank You

    First Name 5 years, 1 month ago

    Thermodynamics in physics and chemistry has a little difference which you must know. Else you will be in troublw

    3Thank You

    ANSWER

    • Posted by Vishal Saigal 5 years, 1 month ago

    CBSE > Class 11 > Chemistry
    • 3 answers

    Ranger King 5 years, 1 month ago

    Extensive properties, such as mass and volume, depend on the amount of matter being measured

    1Thank You

    Vishal Saigal 5 years, 1 month ago

    It is too short . I don't understand.

    1Thank You

    Aadya Singh 5 years, 1 month ago

    which depends upon amount of matter...

    3Thank You

    ANSWER

    • Posted by Niranjan Sindhur 5 years, 1 month ago

    CBSE > Class 11 > Chemistry
    • 0 answers
    ANSWER

    • Posted by Pooja Singh 4 years, 8 months ago

    CBSE > Class 11 > Chemistry
    • 1 answers

    Sia ? 4 years, 8 months ago

    Quicklime is an important compound. It is prepared by thermal decomposition of limestone in tall furnaces called kiln. During such an operation the top of the Chimney of the kiln was closed. This time also there was no improvement in the production of lime.

    0Thank You

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