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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.
| 2251. |
Power point presentation on chemical bonding |
| Answer» | |
| 2252. |
What are the three laws of thermodynamics? |
| Answer» there ....is only two laws not 3<br>?final velocity=initial velocity+acceleration*time.................1?displacement=initial velocity*time+1/2acceleration*(time)..,...............,...2?(final velocity)square=(initial velocity)square+2acceleration*displacement................,3here initial velocity(u) , final velocity(v),acceleration(a),time(t),displacement(x) | |
| 2253. |
What is enthalpy or heat content |
| Answer» Enthalpy is a property of a thermodynamic system, defined as the sum of the system\'s internal energy and the product of its pressure and volume. | |
| 2254. |
Model related to hydrogen chapter |
| Answer» | |
| 2255. |
Impotant chapters of chemistry |
| Answer» Every chapter are compulsory<br>Every chapter is Important but you can read redox, organic chemistry, hydrocarbon, unit 5 , 4 , 7and 2<br>Because this chapter related to class12<br>Goc (organic chemistry), redox and thermodynamics<br>all chapters are very very important | |
| 2256. |
Explain : wurtz reaction is carried out in dry ether |
| Answer» Alkyl halide react with Na in presence of dry ether to give alkane.CH3-Cl + 2Na + Cl-CH3==Ch3- Ch3 +2Nacl | |
| 2257. |
What is (n+l) rule |
| Answer» In cases where (n+l) is the same for two orbitals (eg, 2p and 3s) the (n+l) rule says that the orbital with lower n has lower energy. In other words, the size of the orbital has a larger effect on orbital energy than the number of planar nodes. | |
| 2258. |
calculate the volume occupied by 0.5moles of co2 at -25°c and 760mm pressure |
| Answer» | |
| 2259. |
Stability due to hyperconjugation |
| Answer» | |
| 2260. |
Draw Lewis dot structure of H2S2O7 and H2O2 |
| Answer» | |
| 2261. |
Aromatic compound , non aromatic, antiaromatic ??????? |
| Answer» | |
| 2262. |
Can we change gas into solid state ? |
| Answer» Yes, with disublimation<br>Yes of course with the help of deposition or desublimation process. | |
| 2263. |
3,3-Dimethylpentane open structure |
| Answer» | |
| 2264. |
What is grignard\'s reagents and it\'s use |
| Answer» Grignard reagents\xa0are\xa0used\xa0synthetically to form new carbon–carbon bonds. A\xa0Grignard reagent\xa0has a very polar carbon–magnesium bond in which the carbon atom has a partial negative charge and the metal a partial positive charge. | |
| 2265. |
Physical and chemical properties of acids and bases |
| Answer» According to Arrhenius,acids dissociate H+ ions in water and bases dissociateOH- ion in water.<br>Ohh god ??? | |
| 2266. |
Dual nature of matter and light |
| Answer» Element | |
| 2267. |
What is dynamic equilibrium ? ? |
| Answer» A\xa0dynamic equilibrium\xa0exists once a reversible reaction occurs. Substances transition between the reactants and products at equal rates, meaning there is no net change. Reactants and products are formed at such a rate that the concentration of neither changes. | |
| 2268. |
Explain the last point of set of rules for calculation of oxidation number. |
| Answer» ?<br>I also...<br>I don\'t know | |
| 2269. |
What is lindlar\'s catalyst and it\'s use |
| Answer» The\xa0catalyst\xa0is\xa0used\xa0for the hydrogenation of alkynes to alkenes (i.e. without further reduction into alkanes). The lead serves to deactivate the palladium sites, further deactivation of the\xa0catalyst\xa0with quinoline or 3,6-dithia-1,8-octanediol enhances\xa0its\xa0selectivity, preventing formation of alkanes. | |
| 2270. |
3.011×10²⁰ molecules are of Al2O3 = 2×3.011×10²⁰ Al³+ ions |
| Answer» | |
| 2271. |
How many silver atoms are present in a piece of jewellery weighing 10.78?(Ag=107.8a.m.u) |
| Answer» First is ka molar mass nikalo phir iska molecule or phir molecule/ NA -6.022×10 (23) | |
| 2272. |
manufacture of benzene from phenol |
| Answer» Add -OH functional group | |
| 2273. |
Prepare 100ml of M/20 ferrous Ammonium Sulphate solution find the molarity and strength of KMnO4?? |
| Answer» | |
| 2274. |
A gas expands isothermally and reversibly. Work done by the gsa |
| Answer» In isothermal process-- T=const.So, ∆U=0∆U=q+wq+w=0q=-w | |
| 2275. |
Structural representation of organic compounds |
| Answer» | |
| 2276. |
No of fore brain |
| Answer» | |
| 2277. |
Identify the oxidation and reduction.Ch4 + o2 ➡ co2 + h2o |
| Answer» | |
| 2278. |
Explain formal charge |
| Answer» A charge contain by the atom during reaction. Charge should be positive or negative or 0 | |
| 2279. |
Explain formal change |
| Answer» A charge contain by the atom during reaction. Charge should be positive or negative or 0 | |
| 2280. |
Write the expression for equilibrium constant Kp of the reaction 3Fe(s) + 4H2O(g) ⇌Fe3O4(s) + 4H2(g) |
| Answer» (H2)^4/(H2O) ^4 | |
| 2281. |
Differentiate between ideal gas and real gas |
| Answer» Ideal gas Z=0<br>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<br>Gases that follows Boyle\'s law , Charles\'law ,Lussac\'s law and avogadro\'s law are called ideal gases<br>Pta ni | |
| 2282. |
Molecular orbital theory for o2, n2 ,f2 |
| Answer» 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\xa0= 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*\xa02p1y π*2p1z,) showed that O2 molecule is paramagnetic.\xa0Read 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 | |
| 2283. |
Why does the solubility of alkaline earth metal hydroxides in water increase down the group? |
| Answer» | |
| 2284. |
What is state functions |
| Answer» 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. | |
| 2285. |
Write the significance of Vander wall constant |
| Answer» 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. | |
| 2286. |
Explain about the periodicity in ionisation enthalpy and electron afficincy in a group and period |
| Answer» Along the period ionisation enthalpy increases while down the group decreaseAlong the period election affinity increases while down the group decrease | |
| 2287. |
At 2.46 ATM pressure the volume of 28g N2 gas is 10L then calculate the temperature....?? |
| Answer» Thanks a lot ?<br>Pv=nrt T=pv/nrT=2.46×10/1×0.0821=29.96 | |
| 2288. |
Structure of H2O2 |
| Answer» H-o-h/h+ | |
| 2289. |
why does water (H2O) does not burn though it has oxygen which is required for oxidation? |
| Answer» Oxygen doesn\'t burn it is an oxidizer | |
| 2290. |
Calculate the number of electrons which together weight 1 gm |
| Answer» Mass of 1 electron = 9.1×10power - - 311gm=1/9.1×10power - - 31=1.09×10power30 | |
| 2291. |
Who is the reporter...? |
| Answer» | |
| 2292. |
Draw the total possible isomers of C6H14. |
| Answer» <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><br>#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 ??? | |
| 2293. |
Can anyone tell me the syllabus of organic chemistry in class 11 ? |
| Answer» 1. Chp. 12 -Organic chemistry. 2. Chp.13 -Hydrocarbons. | |
| 2294. |
How many chapters are deleted in chemistry, please tell me ? |
| Answer» Ch 2 delete<br><a href="http://cbseacademic.nic.in/web_material/CurriculumMain21/revisedsyllabi/Deduction/DELETEDChemistry_2020-21.pdf">Tap here</a> to get them<br>I think Chapters delete nhi huye hai ...topics delete huye h... | |
| 2295. |
Chemistry deleted syllabus / topics. |
| Answer» | |
| 2296. |
Datlon\'s atomic theory |
| Answer» Matter consists of individual particals namely atom which cannot devided further.and atoms of a particular element are identical<br>Postulate of atomic theory - matter consist of indivisible and indescrible particle-atom of same elements are identical to each other.<br>Is called dalton partial pressure<br>In a mixture of gases the pressure applied by pressure is the sum of pressure of individual gases | |
| 2297. |
Justify that work done in a reversible manner represents the maximum work. |
| Answer» | |
| 2298. |
Hume project Banna h Kya koi Meri help ke sakta h |
| Answer» Ya<br>i will try but please tell me on which topic | |
| 2299. |
Thermodynamics physics chemistry dono m h ?? |
| Answer» Yes as well as in bio also<br>Thermodynamics in physics and chemistry has a little difference which you must know. Else you will be in troublw | |
| 2300. |
the production of lime not satisfactory when the top of the chimney was closed |
| Answer» 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. | |