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    • Posted by Lovish Kumar 2 years ago

    CBSE > Class 11 > Chemistry
    • 1 answers

    Anjan Karthi 2 years ago

    Wavenumber = 1/Wavelength = Frequency/Speed of light.

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    • Posted by Vikas Mishra Mishra 2 years ago

    CBSE > Class 11 > Chemistry
    • 1 answers

    Anjan Karthi 2 years ago

    In Cr (Z = 24), expected configuration is [Ar] 3d4 4s2, but due to extra stability offered by half filled orbital configuration, obtained is [Ar] 3d5 4s1.

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    • Posted by Susma Rabidas 2 years ago

    CBSE > Class 11 > Chemistry
    • 0 answers
    ANSWER

    • Posted by Gayatri Pathade 2 years ago

    CBSE > Class 11 > Chemistry
    • 1 answers

    Anjan Karthi 2 years ago

    E = hneu

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    • Posted by Dhaneswar Chalan 2 years ago

    CBSE > Class 11 > Chemistry
    • 4 answers

    Sariful Miah 2 years ago

    What is the atomic number of carbon

    3Thank You

    Tejas T. 2 years ago

    This content has been hidden. One or more users have flagged this content as inappropriate. Once content is flagged, it is hidden from users and is reviewed by myCBSEguide team against our Community Guidelines. If content is found in violation, the user posting this content will be banned for 30 days from using Homework help section. Suspended users will receive error while adding question or answer. Question comments have also been disabled. Read community guidelines at https://mycbseguide.com/community-guidelines.html

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    3Thank You

    Tejas T. 2 years ago

    This content has been hidden. One or more users have flagged this content as inappropriate. Once content is flagged, it is hidden from users and is reviewed by myCBSEguide team against our Community Guidelines. If content is found in violation, the user posting this content will be banned for 30 days from using Homework help section. Suspended users will receive error while adding question or answer. Question comments have also been disabled. Read community guidelines at https://mycbseguide.com/community-guidelines.html

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    Tejas T. 2 years ago

    This content has been hidden. One or more users have flagged this content as inappropriate. Once content is flagged, it is hidden from users and is reviewed by myCBSEguide team against our Community Guidelines. If content is found in violation, the user posting this content will be banned for 30 days from using Homework help section. Suspended users will receive error while adding question or answer. Question comments have also been disabled. Read community guidelines at https://mycbseguide.com/community-guidelines.html

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    • Posted by Dhaneswar Chalan 2 years ago

    CBSE > Class 11 > Chemistry
    • 2 answers

    Account Deleted 2 years ago

    It can be explain through an example Na==no of p+=no of e– In Na+= no of p+ > no of e–, so more attraction takes place and hence size is reduced In Na-= no of p+ < no of e–, so less attraction takes place and hence size is increased Therefore, Na+ < Na < Na-

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    Tejas T. 2 years ago

    It's easy

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    • Posted by Rewati Raman 🤓🤓🤓 2 years ago

    CBSE > Class 11 > Chemistry
    • 2 answers

    Jatt Attri 1 year, 11 months ago

    U

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    Tejas T. 2 years ago

    This content has been hidden. One or more users have flagged this content as inappropriate. Once content is flagged, it is hidden from users and is reviewed by myCBSEguide team against our Community Guidelines. If content is found in violation, the user posting this content will be banned for 30 days from using Homework help section. Suspended users will receive error while adding question or answer. Question comments have also been disabled. Read community guidelines at https://mycbseguide.com/community-guidelines.html

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    • Posted by Mohit Arora 2 years ago

    CBSE > Class 11 > Chemistry
    • 2 answers

    Tejas T. 2 years ago

    This content has been hidden. One or more users have flagged this content as inappropriate. Once content is flagged, it is hidden from users and is reviewed by myCBSEguide team against our Community Guidelines. If content is found in violation, the user posting this content will be banned for 30 days from using Homework help section. Suspended users will receive error while adding question or answer. Question comments have also been disabled. Read community guidelines at https://mycbseguide.com/community-guidelines.html

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    Ruchi . 2 years ago

    According to orbitals overlap concept the formation of a covalent bond between two atoms result by pairing of elements present in the valence shell having opposite spins.

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    • Posted by Shreya Rana 2 years ago

    CBSE > Class 11 > Chemistry
    • 1 answers

    Gayatri Pathade 2 years ago

    The process of emission of e- from the metal surface when a light of suitable frequency strikes on certain metals. This is called Photoelectriceffect.

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    • Posted by Koshal Chouhan 2 years ago

    CBSE > Class 11 > Chemistry
    • 1 answers

    Nidhi Deshwal 2 years ago

    A chemical bond is a formation of bond due to the electrostatic forces of attraction due to sharing of electrons or donating electrons. The attractive force which holds up the constituent particles (atoms, ions or molecules) together in a chemical species is known as chemical bond.

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    • Posted by Sonali Singh 2 years ago

    CBSE > Class 11 > Chemistry
    • 2 answers

    Anjan Karthi 2 years ago

    No,not sp3. Here hybridisation h = 1/2 (Valence e + no of monovalent atoms + no of -ve charge - no of +ve chahe) = 1/2 (7+5) = 12/2 = 6 ~~ sp3d2 hybridisation. Hope you got what you wanted.

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    Udit Kumar Dhruw 2 years ago

    1/2(3+5)=8/2=4 sp³hybridisation

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    • Posted by Satya Vati 2 years ago

    CBSE > Class 11 > Chemistry
    • 2 answers

    Diya Rana 1 year, 11 months ago

    According to the plank quantum theory any particle like electrons not continuosly but discontinuosly in the form of small discrete packet known as quanta or in the case of light this discrete packet remember that energy of this quantas are directly proportional the the frequency ....and plank quantum denoted by h and h= 6.626×10*34 jsec

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    Anjan Karthi 2 years ago

    1) Matter emit radiation which constitute small packets of energy called quanta (For light, photon is the quanta). 2) The total energy of the radiation emitted is an integral multiple of the individual energies of each quanta, which is expressed as hv (where, h = Planck's constant = 6.626 × 10^-34 Jsec and v = frequency)

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    • Posted by Chandni Chaudhary 2 years ago

    CBSE > Class 11 > Chemistry
    • 1 answers

    Aman Acharya 2 years ago

    Mole is a unit and it's concept is called mole concept and it's simply means a avogrado no as example 1 mole means 6.023×10²³ no

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    • Posted by Navrose Singh 2 years ago

    CBSE > Class 11 > Chemistry
    • 0 answers
    ANSWER

    • Posted by Zaarakhan Khan 2 years, 1 month ago

    CBSE > Class 11 > Chemistry
    • 1 answers

    Aman Acharya 2 years ago

    The Formula Of Average Atomic mass = Atomic Mass 1 × % Abudance + Atomic mass 2 × % abundance/ 100 10.8 = 11(x) + 10( 100-x ) / 100 1080 = 11x + 1000 -- 10x 1080 - 1000 = x x = 80 So boron having Atomic mass 11 have 80% abundance in nature and 10 has 20%

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    • Posted by Pranjal Mishra 2 years, 1 month ago

    CBSE > Class 11 > Chemistry
    • 1 answers

    Abhrangshu Chakraborty 2 years, 1 month ago

    22.7L (0°C, 1bar) 22.4L (0°C, 1 atm) 《From Avogadro's hypothesis》

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    • Posted by Redmi 10S 2 years, 1 month ago

    CBSE > Class 11 > Chemistry
    • 0 answers
    ANSWER

    • Posted by Satyam Kumar Mishra 2 years, 1 month ago

    CBSE > Class 11 > Chemistry
    • 1 answers

    Anjan Karthi 2 years ago

    Q = ±ne. ///// n = Q/e = 2.5 × 10^-16 /1.6 × 10^-19 = 25 × 10^3 / 16 = 1.5625 × 10^3 = 1563 electrons.

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    • Posted by Alina Patnaik 2 years, 1 month ago

    CBSE > Class 11 > Chemistry
    • 1 answers

    Anjan Karthi 2 years ago

    No of moles = 9.5 / 95 = 0.1 mol. No of ions = 0.1 × 6.022 × 10^23 = 6.022 × 10^22 ions. (NOW, 1 (PO4)3- HAS 15 + 32 + 3 = 50e) Therefore, no of el = 50 × 6.022 × 10^23 = 3.011 × 10^25 electrons //

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    • Posted by Deeksha Sohal 2 years, 1 month ago

    CBSE > Class 11 > Chemistry
    • 1 answers

    Prathameh Kanhere 2 years, 1 month ago

    BeCl2 ka Sp hybridization

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    • Posted by Account Deleted 2 years, 1 month ago

    CBSE > Class 11 > Chemistry
    • 0 answers
    ANSWER

    • Posted by Mayank Sharma 2 years, 1 month ago

    CBSE > Class 11 > Chemistry
    • 1 answers

    Tejas T. 2 years ago

    This content has been hidden. One or more users have flagged this content as inappropriate. Once content is flagged, it is hidden from users and is reviewed by myCBSEguide team against our Community Guidelines. If content is found in violation, the user posting this content will be banned for 30 days from using Homework help section. Suspended users will receive error while adding question or answer. Question comments have also been disabled. Read community guidelines at https://mycbseguide.com/community-guidelines.html

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    Remember the goal of this website is to share knowledge and learn from each other. Ask questions and help others by answering questions.

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    • Posted by Riya Chand 2 years, 1 month ago

    CBSE > Class 11 > Chemistry
    • 0 answers
    ANSWER

    • Posted by Aleena Ss 2 years, 1 month ago

    CBSE > Class 11 > Chemistry
    • 1 answers

    Anjan Karthi 2 years, 1 month ago

    When no of electrons in the valence shell is less than 4, then, these electrons have a tendency to be lost from the atom so that the atom can attain the nearest (penultimate shell) octet. When no of electrons is greater than 4, there is a tendency that the atoms gain electrons in their neighborhood to attain nearest (valence shell) octet. But, when no of valence electrons is equal to 4, gaining 4 electrons is a matter outside the control of the nucleus and losing 4 electrons causes huge energy loss to the molecule, disturbing it's thermodynamic stability. So, here electrons are just shared, and not completely transferred as in the previous cases.

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    • Posted by Sarveshaa Ss 2 years, 1 month ago

    CBSE > Class 11 > Chemistry
    • 1 answers

    Anjan Karthi 2 years, 1 month ago

    M = Wb/Mb V(L) = 5 ×1000/40×450 = 0.27 M. //

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    • Posted by Pankaj Kaushish 2 years, 1 month ago

    CBSE > Class 11 > Chemistry
    • 1 answers

    Anjan Karthi 2 years, 1 month ago

    n(Fe) = 69.9/56 = 1.24. n(O2) = 30.1/32 = 0.94. Therefore, n(Fe) / n(O2) = 1.3 : 1. Therefore, Empirical formula = FeO (nearly).

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    • Posted by Tamanna Tirkey 2 years, 1 month ago

    CBSE > Class 11 > Chemistry
    • 1 answers

    Paras Rawat 2 years, 1 month ago

    Frequency=6.91

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    • Posted by Pranshu Patel Patel 2 years, 1 month ago

    CBSE > Class 11 > Chemistry
    • 4 answers

    Anshika Gupta 2 years, 1 month ago

    Nitric oxide

    1Thank You

    Bhavya Garg 2 years, 1 month ago

    Hello

    0Thank You

    Rishita Saw 2 years, 1 month ago

    Nitric oxide

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    Pranshu Patel Patel 2 years, 1 month ago

    Sir

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    • Posted by Mohd Arsalan 2 years, 1 month ago

    CBSE > Class 11 > Chemistry
    • 0 answers
    ANSWER

    • Posted by Anushya G 2 years, 1 month ago

    CBSE > Class 11 > Chemistry
    • 1 answers

    Abhipsa Behera 2 years, 1 month ago

    Electrons:They are the negatively charged subatomic particles that were discovered by scientist J.J. Thomson in the cathode ray-discharge tube experiment. J.J. Thomson's Cathode Ray Experiment:In this experiment, he used a special hard, cylindrical glass tube known as a discharge tube that consists of two, metal plates.These metal plates were called electrodes, each sealed at the two ends of the tube.These metal plates are connected to the terminals of a high-voltage battery.The metal connected to the positive terminal is called the anode and the metal plate connected to the negative terminal is called the cathode.A vacuum pump is connected to one side of the tube to lower the pressure inside the tube.Very high voltage electricity through the discharge tube was passed at very low pressure. Observations:From the experiment, he observed that a continuous beam of rays was emitted.A green fluorescent glow was observed at the end of the tube.They were found moving from the cathode region to the anode region. This proved that these rays consisted of negatively charged particles and these particles were called electrons.

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