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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your Class 11 knowledge and support exam preparation. Choose a topic below to get started.
| 2551. |
Which of the following has highest boiling point NO2 PH3 H2s H2o |
| Answer» Hay<br>No2 | |
| 2552. |
An electron cannot exist in the nucleus whose radius is 10^-15 meter . Why ? |
| Answer» | |
| 2553. |
Isotope of hydrogen? |
| Answer» 1) Protium (H with A = 1 and Z = 1.) 2) Deuterium (H with A = 2 and Z = 1.) 3) Tritium ( H with A = 3 and Z = 1) | |
| 2554. |
All masses of atom |
| Answer» Same | |
| 2555. |
Need of atom |
| Answer» Every atom combines with other atoms to lower it\'s inbuilt ionisation energy and thereby, attain stability. This is the BASIC need for all atoms (Noble gases from Kr)<br>Need of atom | |
| 2556. |
Which is the basic theme of organisations in the periodic table |
| Answer» Atomic number<br>1) Similar physical and Chemical properties of elements down the group. 2) Increasing order for atomic number by 1 unit across every period. 3) Similarity in electronic configuration of atoms down a group. | |
| 2557. |
Change of protons Charge of neutronsCharge of electrons |
| Answer» Charge of protons = + 1.6 × 10 ^-19 C. Charge of electron = - 1.6 × 10^ -19 C. Charge of neutron = 0<br>Fast | |
| 2558. |
Which of the following hydrides are generally nonstochiometric in nature? |
| Answer» Saline hydrides | |
| 2559. |
How to define electronic configuration |
| Answer» The electron configuration is the distribution of electron of an atom or molecule in atomic or molecular orbitals.<br>The arrangement of electrons in energy levels around an atomic nucleus...is defined as electronic configuration... | |
| 2560. |
Name the alkines |
| Answer» NaOH, KOH | |
| 2561. |
Find the conjugate of the complex numbers (i) -i√5 |
| Answer» i√5 ok.na<br>change of iorta sign<br>Root 5 plus iorta | |
| 2562. |
Calculate all the four quantum no. Of last electron of Cr |
| Answer» Cr = 1s2 2s2 2p6 3s2 3p6 3d5 4s1. Therefore, last electron is in 3d orbital. Quantum numbers are :- n = 3.l = 2. ml = 2.ms = 1/2. | |
| 2563. |
How to learn chemistry?? Especially GOC.. |
| Answer» 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.<br>Also be confident<br>By revising again and again and by doing lots of question practice | |
| 2564. |
The length of h bonding in solid state of water is |
| Answer» | |
| 2565. |
How many Sigma and Pi bonds are present in Pentene |
| Answer» Pentene structure is CH2=CHCH2CH2CH3. Therefore, number of σ bonds = 8. And total number of π bonds = 1. | |
| 2566. |
FeCl3 + H2 S \uf0ae FeCl2 + 5 + HCl |
| Answer» | |
| 2567. |
unit of length. |
| Answer» Here you have three units.1st is meter.2nd is foot.3rd is centimetres.<br>Meter<br>Meter<br>metere | |
| 2568. |
SIMILARITIES AND DISIMILARITIES BETWEEN HYDROGEN,HALOGEN AND ALKALI METALS |
| Answer» Similarities between hydrogen and alkali metals :Electronic configurationElectropositive characterValency and oxidation stateReducing agentsDissimilarities between alkali metals:Ionisation enthlapy Non metallic in natureAtomicitySimilarities between hydrogen and halogenElectronic configurationIonisation enthlapyElectronegative characterOxidation stateDiatomic natureNature of compoundsDissimilarities between hydrogen and halogenLess tendency for hydride formationAbsence of unshared pair of electronOxides of halogens are acidic | |
| 2569. |
Calculate the volume of 0.015M HCL solution required to prepare 250ml of a 5.25×1/1000M HCL solution |
| Answer» M1*V1 = M2*V2(0.015×250)÷(5.25÷1,000) = V2V2 = 714.285 ml<br>M1*V1 = M2*V2 | |
| 2570. |
How many moles of methane required to produce 88g of CO2 after combustion |
| Answer» 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. // | |
| 2571. |
Electronic configuration of florine |
| Answer» 1s*2 2s*2p*5<br>??<br>1s² 2s² 2p⁵ | |
| 2572. |
Which out of the following contain least number of atoms?A. 1gm of H2OB. 1gm of CH4C. 1gm of N2 |
| Answer» 1×4×6•023×1023 | |
| 2573. |
Which out of the following contain least number of atoms? |
| Answer» | |
| 2574. |
Short note on Development of Chemistry |
| Answer» What is this<br>Hdhdjsjvkvvieie | |
| 2575. |
What are the limitations of octect rule? |
| Answer» 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\u200b , BeCl 2\u200b , BH 3\u200b , BF 3\u200b , 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\u200b , SF 6\u200b , IF 7\u200b , H 2\u200b SO 4\u200b , the central atom has more than 8 valence electrons, yet the molecule is stable. | |
| 2576. |
Calculate the number of moles in 7.9mg of calcium |
| Answer» 2.0×power of 10 is minus3 | |
| 2577. |
Notes for structure of atom |
| Answer» | |
| 2578. |
Molecular mass of na2co3 |
| Answer» 106<br>Na2CO3 = 2(23) + 12 + 3(16) = 46 + 12 + 48 = 106 amu. | |
| 2579. |
Properties of resonance |
| Answer» One object can\'t defined other object | |
| 2580. |
5 atomic number |
| Answer» Boron<br>.Boron.Z M ⬇ ⬇5 10.118u<br>Boron | |
| 2581. |
Draw ideal gas equation PV=nRT from different gas laws |
| Answer» Boyle law - V~1/P - (I) equationCharle\'s Law - V~T. -(ii) equationAvagardo law- V~n. -(iii) equation By usin1,2,3 equation: V~nT/P PV~nT PV=nRT | |
| 2582. |
What volume of co2 at STP can be obtained on the decomposition of 100 g of caco3 |
| Answer» Volume of co2 at STP is 22.4 LBecause molar mass of caco3 = 40 + 12 + 3 × 16 = 100 g. So, co2 = 12 + 2 × 16 = 44 g or 22.4 L | |
| 2583. |
What is Multiplication and Division of Significant Figures |
| Answer» No<br>When\xa0multiplying two numbers, the important value is the number of significant figures. ... If the numbers being multiplied have three significant figures, then the product will have three significant figures. | |
| 2584. |
Nitro\'s secondary suffix |
| Answer» -amine. | |
| 2585. |
difference between amylose and amylopectin |
| Answer» \tAmyloseAmylopectinIt is a straight-chain polymer of D-glucose units\xa0It is a branched-chain polymer of D-glucose unitsConstitutes 20% of starchConstitutes 80% of starchSolubility in water is lessMore soluble in waterStraight chain structureBranched structureAmylose stains blue with iodineAmylopectin stains reddish-brown with iodineSoluble in hot water without swellingSoluble in hot water with swellingDoes not form a gel when hot water is added to itForms a gel when hot water is added to it\t | |
| 2586. |
Neil Bohrs modle of atom |
| Answer» | |
| 2587. |
How many grams of sodium present in 4.6 mol of sodium? |
| Answer» Thank you<br>Molar mass of sodium = 23 g. Moles = 4.6 mol. Therefore, required mass = Molar mass × no of moles = 23 × 4.6 = 105.8 g. // | |
| 2588. |
Calculate the percentage of N in NH3 Molecule |
| Answer» Molar mass of NH3 = 14 + 3(1) = 17 g. // Therefore, percentage composition of N in NH3 = (molar mass of N / Molar mass of NH3) × 100 = 1400/17 = 82.35%.// | |
| 2589. |
Calculate the molar mass of the following:1. H2O2. CO23. CH4 |
| Answer» Co2=12+2(16)=44u and ch4 =12+4(1)=16u and H2O=2(1)+16=18<br>2) CO2 = 12 + 2(16) = 44 g. //3) CH4 = 12 + 4(1) = 16 g. //<br>1 . h ( 2 ) + O2 + 16 = 18 amu | |
| 2590. |
in group-15 trihalides bond angle increases from fluoride to iodide? why |
| Answer» | |
| 2591. |
Find the number of (a) H atoms, (b) H2 molecules in 12 gm of H2gas (H2=2.0 amu). |
| Answer» 1 | |
| 2592. |
Q10. What will be the energy of photon moving with frequency 500A°. |
| Answer» | |
| 2593. |
Unit1 |
| Answer» | |
| 2594. |
calculate co2 in gm evolved by combustion of 16gm methane |
| Answer» 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 | |
| 2595. |
Calculate the frequency and energy of photon having wavelength of 6000°A |
| Answer» 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. // | |
| 2596. |
You are given a thread and a metre scale . How will you estimate the diameter of the thread |
| Answer» | |
| 2597. |
I want all formulas related to chemistry in pdf format with their name too |
| Answer» | |
| 2598. |
What is formed when CH3-CH2BR-CH2BR is treated with KOH ---------- |
| Answer» Pagal | |
| 2599. |
D block .S block .P blockF blockPhysical properties and chemical properties and it\'s applications |
| Answer» (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.\tThey are solids/liquids/gases at room temperature (Br\xa0is liquid)\xa0\tThey have variable oxidation states\xa0\tThey form acidic oxides\xa0\tGenerally, they form covalent compounds\xa0\tHalogens form salts with alkali metals\xa0\tThey have high ionization potentials\xa0\tThey have very large electron gain enthalpies\xa0\tThe aqueous solutions their oxides are acidic in nature.(c) d-block elements:\tForm stable complexes\tHave high melting and boiling points\tContain large charge/radius ratio\tForm compounds which are often paramagnetic\tAre hard and possess high densities\tForm compounds with profound catalytic activity\tShow variable oxidation states\tForm coloured ions and compounds.(d) f-block elements:\tThey are usually heavy metals\tThese elements have a generally higher melting and boiling point\tThey display a variety of oxidisation states\tThese elements tend to form coloured ions\tThey form complex compounds | |
| 2600. |
How many grams of NaOH should be dissolved to make 100 cm*3 of 0.15 M NAOH solution |
| Answer» Kay aap muja chemistry ka chapter 2 nd ka notes send kar da ga plz<br>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. // | |