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    • Posted by Poonam Dhakar 3 years, 10 months ago

    CBSE > Class 11 > Chemistry
    • 1 answers

    Om P̤A̤T̤E̤L̤ 3 years, 9 months ago

    A charge contain by the atom during reaction. Charge should be positive or negative or 0

    1Thank You

    ANSWER

    • Posted by Poonam Dhakar 3 years, 10 months ago

    CBSE > Class 11 > Chemistry
    • 1 answers

    Om P̤A̤T̤E̤L̤ 3 years, 9 months ago

    Favourable factors for ionic bond formation are as follows: Low ionization enthalpy of metal atom. High electron gain enthalpy (Δeg H) of a non-metal atom. High lattice energy of the compound formed.

    1Thank You

    ANSWER

    • Posted by Prithviraj Jangu Bishnoi 3 years, 10 months ago

    CBSE > Class 11 > Chemistry
    • 0 answers
    ANSWER

    • Posted by Shipra Tiwari 3 years, 10 months ago

    CBSE > Class 11 > Chemistry
    • 5 answers

    Vedha ..... 3 years, 9 months ago

    100 degree Celsius

    0Thank You

    Ranger King 3 years, 9 months ago

    100degree Celsius

    1Thank You

    Poonam Dhakar 3 years, 9 months ago

    373 degree Kelvin

    0Thank You

    Shubham Gurjar 3 years, 10 months ago

    100 degree celsius

    2Thank You

    Priya ? Raj 3 years, 10 months ago

    100°C

    5Thank You

    ANSWER

    • Posted by Achsah Ajay 3 years, 10 months ago

    CBSE > Class 11 > Chemistry
    • 1 answers

    Swastik . P 3 years, 10 months ago

    (H2)^4/(H2O) ^4

    0Thank You

    ANSWER

    • Posted by Deepak Singh 3 years, 10 months ago

    CBSE > Class 11 > Chemistry
    • 0 answers
    ANSWER

    • Posted by Nikku Ad 3 years, 10 months ago

    CBSE > Class 11 > Chemistry
    • 2 answers

    Manisha Choudhary 3 years, 10 months ago

    2.E=hc/wavelength E= 6.626×10^-34×3×10^8/0.50×10^-15s

    1Thank You

    Manisha Choudhary 3 years, 10 months ago

    1. the energy of photon E=hv E= 6.626×10^-34×3×10^5

    1Thank You

    ANSWER

    • Posted by Simran Jit 3 years, 10 months ago

    CBSE > Class 11 > Chemistry
    • 4 answers

    Tanisha Verma 3 years, 10 months ago

    Ideal gas Z=0

    2Thank You

    Ranger King 3 years, 10 months ago

    Ideal gas. - 1.pv=nrt . 2. No force of interaction. 3. Volume particles is very small volume container so it is ignore. Real gas- 1. Pv is not equal to nrt. 2. This is force present. 3. Volume of particles can not be ignore

    3Thank You

    Divyansh Rai 3 years, 10 months ago

    Gases that follows Boyle's law , Charles'law ,Lussac's law and avogadro's law are called ideal gases

    2Thank You

    Mayank Sharma 3 years, 10 months ago

    Pta ni

    3Thank You

    ANSWER

    • Posted by John David 3 years, 10 months ago

    CBSE > Class 11 > Chemistry
    • 1 answers

    Om P̤A̤T̤E̤L̤ 3 years, 9 months ago

    Electronic configuration of nitrogen (Z = 7) is 1s2 2s2 2p3. Since nitrogen atom has 7 electrons, the molecular orbitals of nitrogen molecule (N2) has 14 electrons which are distributed as below : Molecular orbital energy level diagram of N2 molecule • Bond order =  (8 2)/2 = 3 (N ≡ N) • Absence of unpaired electrons showed that N2 molecule is diamagnetic. MOED of 'O2' : Electronic configuration of Oxygen (Z = 8) is 1s2 2s2 2p4 . Since Oxygen atom has 8 electrons, the molecular orbitals of Oxygen molecule (O2) has 16 electrons, which are distributed as below : Molecular orbital energy level diagram of O2 • Bond order = (10 - 6)/2 = 2(O = O) • Presence of two unpaired 6 electrons (π* 2p1y π*2p1z,) showed that O2 molecule is paramagnetic. Read more on Sarthaks.com - https://www.sarthaks.com/535006/give-the-molecular-orbital-energy-diagram-of-n2-and-o2-write-the-bond-order-of-n2-and-o2

    0Thank You

    ANSWER

    • Posted by Vedha ..... 3 years, 10 months ago

    CBSE > Class 11 > Chemistry
    • 0 answers
    ANSWER

    • Posted by Shivam Kumar 3 years, 7 months ago

    CBSE > Class 11 > Chemistry
    • 1 answers

    Sia ? 3 years, 7 months ago

    In the thermodynamics of equilibrium, a state function, function of state, or point function is a function defined for a system relating several state variables or state quantities that depends only on the current equilibrium thermodynamic state of the system[1] (e.g. gas, liquid, solid, crystal, or emulsion), not the path which the system took to reach its present state. A state function describes the equilibrium state of a system, thus also describing the type of system.

    0Thank You

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    • Posted by Vedha ..... 3 years, 10 months ago

    CBSE > Class 11 > Chemistry
    • 2 answers

    Umesh Kadam 3 years, 9 months ago

    G

    0Thank You

    Abhinav Upadhyay 3 years, 10 months ago

    Carbon-12,Carbon-13 and Carbon-14 are three isotopes of carbon.

    3Thank You

    ANSWER

    • Posted by Vedha ..... 3 years, 10 months ago

    CBSE > Class 11 > Chemistry
    • 1 answers

    Abhinav Upadhyay 3 years, 10 months ago

    Since KO2 is superoxide and contains odd number of electrons or unpaired electrons that is why it is paramagnetic.

    1Thank You

    ANSWER

    • Posted by Vivek Rawat 3 years, 10 months ago

    CBSE > Class 11 > Chemistry
    • 1 answers

    Akriti Singh Rathore 3 years, 10 months 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 3 years, 10 months ago

    CBSE > Class 11 > Chemistry
    • 1 answers

    Divyansh Rai 3 years, 10 months 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 ? 3 years, 10 months ago

    CBSE > Class 11 > Chemistry
    • 3 answers

    Vishist Bollu 3 years, 10 months ago

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

    2Thank You

    Royal Thakur ? 3 years, 10 months ago

    Thank you ?

    3Thank You

    Ranger King 3 years, 10 months 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 ? 3 years, 10 months ago

    CBSE > Class 11 > Chemistry
    • 2 answers

    Royal Thakur ? 3 years, 10 months ago

    Thanks a lot ?

    3Thank You

    Ranger King 3 years, 10 months ago

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

    2Thank You

    ANSWER

    • Posted by Killers Gaming 3 years, 10 months ago

    CBSE > Class 11 > Chemistry
    • 1 answers

    Pankaj Ponia 3 years, 10 months ago

    H-o-h/h+

    0Thank You

    ANSWER

    • Posted by Mohit Yadav 3 years, 10 months ago

    CBSE > Class 11 > Chemistry
    • 1 answers

    Aryan Singh 3 years, 10 months ago

    Oxygen doesn't burn it is an oxidizer

    1Thank You

    ANSWER

    • Posted by Laishram Archana 3 years, 10 months ago

    CBSE > Class 11 > Chemistry
    • 1 answers

    Ranger King 3 years, 10 months ago

    1. 2s 2. 3d

    1Thank You

    ANSWER

    • Posted by Titiksha Kakran 3 years, 10 months ago

    CBSE > Class 11 > Chemistry
    • 1 answers

    Ranger King 3 years, 10 months 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 3 years, 10 months ago

    CBSE > Class 11 > Chemistry
    • 1 answers

    Bhawna Dahiya 3 years, 10 months ago

    CH4N2O

    0Thank You

    ANSWER

    • Posted by Ishü Mîrzâ 3 years, 10 months ago

    CBSE > Class 11 > Chemistry
    • 4 answers

    Pankaj Ponia 3 years, 10 months ago

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

    0Thank You

    Bhawna Dahiya 3 years, 10 months ago

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

    0Thank You

    Ranger King 3 years, 10 months ago

    S I unit of mass is kg

    1Thank You

    Amisha Sharma 3 years, 10 months ago

    Kg

    2Thank You

    ANSWER

    • Posted by Royal Thakur ? 3 years, 7 months ago

    CBSE > Class 11 > Chemistry
    • 0 answers
    ANSWER

    • Posted by Subhranshu Behera 3 years, 10 months ago

    CBSE > Class 11 > Chemistry
    • 1 answers

    Shaik Habeeb Ahmed 3 years, 10 months ago

    h÷v=c

    3Thank You

    ANSWER

    • Posted by Suma Suma 3 years, 7 months ago

    CBSE > Class 11 > Chemistry
    • 1 answers

    Sia ? 3 years, 7 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

    ANSWER

    • Posted by Class 12 3 years, 10 months ago

    CBSE > Class 11 > Chemistry
    • 2 answers

    Om P̤A̤T̤E̤L̤ 3 years, 9 months 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 3 years, 10 months ago

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

    2Thank You

    ANSWER

    • Posted by Fear Fighter 3 years, 10 months ago

    CBSE > Class 11 > Chemistry
    • 0 answers
    ANSWER

    • Posted by Johal ? 3 years, 10 months ago

    CBSE > Class 11 > Chemistry
    • 2 answers

    Anuj Kumar 3 years, 10 months 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 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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 ? 3 years, 10 months 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 3 years, 10 months ago

    CBSE > Class 11 > Chemistry
    • 1 answers

    Mayukh Karmakar 3 years, 10 months ago

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

    0Thank You

    ANSWER

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