InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 2301. |
2 Out of CHs Coo and OH which is stronger base and why? |
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Answer» OH- is the stronger base because its conjugate Acid which is H + is a strong acid. |
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| 2302. |
What is acid rain? How is it caused?Mention its harmful effects. |
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Answer» acid rain case by the harmful effects of camicle Acid rainis arainor any other form ofprecipitationthat is unusuallyacidic, meaning that it has elevated levels of hydrogen ions (lowpH). It can have harmful effects on plants, aquatic animals and infrastructure. Acid rain is caused by emissions ofsulphur dioxideandnitrogen oxide, which react with thewater moleculesin theatmosphereto produce acids. Some governments have made efforts since the 1970s to reduce the release of sulphur dioxide and nitrogen oxide into the atmosphere with positive results. Acid rain has been shown to have adverse impacts on forests, freshwaters and soils, killing insect and aquatic life-forms as well as causing damage to buildings and having impacts on human health. |
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| 2303. |
4. Why does the pointer of spring balance move up when the object suspended from it is immersed in watÄr? |
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Answer» Aa we know spring balance measures weight of body .hence when we put the substance in waterthe force of buoyancy reduces the net weight of that body and hence the spring balance shows a decrease. |
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| 2304. |
रण R का W so 5 7<SWOX S §i0 lw— I |
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Answer» 1)ethoxyaceticacid2)cyclohex-1,3-diene hit like if you find it useful |
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| 2305. |
WS4- SynthesizeI. Answer the following questions in one word.1. Name some natural fibres.2. Name a polymer that occurs naturally3. Name a synthetic fibre having properties similar to that of silk4. Name the synthetic fibre that resembles wool5. Name the plastic that is used for making crockery |
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Answer» 1. Cotton, Silk2. Silk, wool, DNA, cellulose and proteins3. Rayon4.Acrylic5.Melamine formaldehyde resinis thepolymer 1.cotton, silk2.silk,dna,cellulose3.rayon is also known as artificial silk4.Acrylic5.melamine formaldehyde resin is the used polymer |
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| 2306. |
he solution contains HCI (wlIl I0 IlWHAT DO ALL ACIDS HAVE IN COMMONin acids is such that wh |
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Answer» When you dissolve an acid or base in water, it makesions. This makes the water conduct electricity better. The stronger the acid (or base), the more ions are produced, so the conductivity of the solution increases. They can both conduct electricity. When an acid reacts with a base, a salt is formed they both can carry and share ions |
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| 2307. |
. Why are we able to sip hot tea or milk faster from a saucer than a cup ? |
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Answer» It is easier tosip hot teaor milkfaster from a saucer rather than from a cup because of evaporation . ... And evaporation causes cooling. So if the surface area is more rate of evaporation isfasterand so is the cooling rate. Henceweareable to sip hot teaor milkfasterfrom a saucer rather than from a cup |
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| 2308. |
4. Why are we able to sip hot tea ormilk faster from a saucer ratherthan a cup? |
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Answer» because of its surface area evaporation takes place according to surface area |
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| 2309. |
3. Why are we able to sip hot tea or milk faster from a saucer rather than from a cup ?11 |
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Answer» We are able to sip hot tea or milk faster from a saucer rather than from a cup because saucer has a larger surface area due to which evaporation occurs faster. The vapours take up the latent heat from hot milk or tea kept in the saucer faster, than in a cup which has a smaller surface area than the saucer and make the contents of the saucer cool down faster |
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| 2310. |
NCLUSION:Onecrystalofpotassiumpermanganatewhichsetally made up of large number of tiny particles.CHARACTERISTICS OF PARTICLES OF MATTEREvery matter is made up of particles. |
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Answer» 1.They are of small size.2) They are in continuous motion because they possess high kinetic energy.3) They occupy space4) They attract each other due to high intermolecular forces |
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| 2311. |
Vrite the conclusions drawn by Rutherford when he observed the folloMost of the alpha particles passing straight through the gold foilSome alpha particles getting deflected from their path.Very small fraction of alpha particles getting deflected by 1800 |
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Answer» Conclusion of Rutherford's scatteringexperiment: 1)Most of the space inside the atom is empty because most of the α-particles passed through the gold foil without getting deflected. 2)Very few particles were deflected from their path, indicating that the positive charge of the atom occupies very little space. 3)A very small fraction of α-particles were deflected by very large angles, indicating that all the positive charge and mass of the gold atom were concentrated in a very small volume within the atom Surprisingly one out of every 12,000 alpha particles appeared to rebound. |
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| 2312. |
| Why are we able to sih hot tea haster fromla saucer rather than รก cup? |
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| 2313. |
whybehaves like bose |
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Answer» It behaves like base because in acid medium, O atom can donate its Lone pair e to accept H+ ion. |
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| 2314. |
Very small fraction of alpha particles getting deflected by 180°. |
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Answer» small number ofalpha particlesweredeflectedby large angles (> 4°) as they passed through thefoil. There is a concentration of positive charge in the atom. Like charges repel, so the positivealpha particleswere being repelled by positive charges. |
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| 2315. |
alpha-particles can be detected by using |
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Answer» Zinc sulfidecan detect alphaparticles. In an experiment, Thealpha particlesweredetectedby a zinc sulfide screen, which emits a flash of light upon analpha particlecollision. alpha particles can be detected by using ratheford model |
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| 2316. |
Predict the products of the following reactions:O,N |
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| 2317. |
Predict the sign of ΔH and ΔS for reaction :- 2Cl(g)->Cl2(g) |
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| 2318. |
Predict the shape of XeFmolecule, according to VSEPRtheory. |
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Answer» thank you |
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| 2319. |
(a) Predict the product in the following reactions :NBS/hv |
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| 2320. |
Predict the shape of XeFmolecule, according to VSEPRtheory.4 |
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| 2321. |
CHARACTERISTICS OF MATTER |
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Answer» The two main characteristics features of matter are :-(i) It occupies space .(ii) It has mass and volume . The characteristics of matter are:- 1⃣Particles of matter have space between them. 2⃣Particles of matter are continuously moving. 3⃣Particles of matter attract each other. it occupied spaceit have massit is smallest particle it occupies space and has a mass.it's presence can be feel by one of our five senses. |
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| 2322. |
Q.2 Very short answer.1) State octet rule. |
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Answer» Theoctet ruleis a chemicalruleof thumb that reflects observation that elements tend to bond in such a way that each atom has eight electrons in its valence shell, giving it the same electronic configuration as a noble gas. |
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| 2323. |
what is covalent bond? |
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Answer» Ans :- A covalent bond, also called a molecular bond, is a chemical bond that involves the sharing of electron pairs between atoms. These electron pairs are known as shared pairs or bonding pairs, and the stable balance of attractive and repulsive forces between atoms, when they share electrons, is known as covalent bonding. PLEASE LIKE THE ANSWER |
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| 2324. |
a] Describe the estimation of Carbon and Hydrogen by Liebig's method. 4 |
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Answer» Liebig's method is used for the estimation of "Carbon and hydrogen".A known mass of organic compound is heated in the presence of pure oxygen. The carbon dioxide and water formed are collected and weighed. The percentages of carbon and hydrogen in the compound are calculated from the masses of carbon dioxide and water. Estimation of carbon and hydrogen in an organic compound is based on their conversion to CO and H2O respectively. |
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| 2325. |
(2) Covalent bond |
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Answer» Ans :- A covalent bond, also called a molecular bond, is a chemical bond that involves the sharing of electron pairs between atoms. These electron pairs are known as shared pairs or bonding pairs, and the stable balance of attractive and repulsive forces between atoms, when they share electrons, is known as covalent bonding. PLEASE LIKE THE ANSWER |
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| 2326. |
2. In haemoglobin 0.3357 Mr Pod prosut, Atomic mass offe . S. 3 we Moon Как илаха о На вилоя осі ісHOTY 10%.. Find the no. of iron atoms inhaemoglobin. |
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Answer» Number of iron atoms present in haemoglobin is 4. |
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| 2327. |
68) In the oxeacids of chlorine bond containsa) dt-dπ bondingb) prebondingc) pt-pπ bondingd) none of the aboveer dr honds are equal to those present in CIO+ |
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Answer» pπ-pπ bond is found in oxacids of chlorine. |
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| 2328. |
19. Which of the following molecule contains a bondhaving highest polarity?a) CH3-SHc) CH,-NH2b) CH,-OHd) CH,-CI. |
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Answer» Ch3 Clcoz it will cause high+I effect option D should be correct because Cl is most electronegative out of all these |
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| 2329. |
7 Which molecule contains both polar and nonpolarcovalent bond?(1) NH4(2) HCI(3) CH4H202 |
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Answer» Because of the difference in electronegativity between Hydrogen and Oxygen, the single bonds between them are polar, with a partial negative charge on oxygen and partial positive on Hydrogen. Since Oxygen has the same electronegativity as itself, there is no difference in polarity across the bond, so it isnonpolar.option 4 thank you thank you |
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| 2330. |
terogeneous equilIBHum (heoniculThe reaction A(g) + B(g)C(g)+ D(g) is studiedin a one litre vessel at 250°C. The initial concentrationof A was 3n and that of B was n. When equilibrium wasattained, equilibrium concentration of C was found tothe equal to the equilibrium concentration of B. Whatis the concentration of D at equilibrium?(A)n/2(C) (n -n/3)(B) (3n-1/2)D)n |
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| 2331. |
à Caicvlale the essure exerted by 8 50 of ammonia (NHs) contained in 0.5 litre vessel of |
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| 2332. |
Two moles of NH, when put into a previously evacuatedvessel (one litre), partially dissociate into N2 and H2. If atequilibrium one mole of NH3 is present, the equilibriumconstant is-2(b) 27/64 mol2 litre2(d) 27/16 mo litre2(a) 3/4 mol litre(c) 27/32 mol litre |
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Answer» Rx 2NH3 = N2 + 3H2 Initial mole= 2 0 0 Final Mole = 1 1/2 3/2 Equilibrium constant = 3/2*3/2*3/2*1/2= 27/16 ans
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| 2333. |
At constant temperature in one litre vessel, when the reaction, 2SO(g) 2SO2 (g)+ O2 (g) is at equilibrium, theso2 concentration is 0.6 M, initial concentration of SO3 is 1 M. The equilibrium constant is(A) 2.7(B) 1.36(C) 0.34(D) 0.675 |
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Answer» thank you so much snighda snigdha |
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| 2334. |
7.To what temperature must a neon gas sample beheated to double its pressure if the initial volume ofgas at 75°C is decreased by 150%? |
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| 2335. |
ate10. For gaseous reaction, rate k [A] [B]. If volumeofof container is reduced toof initial, then therate of the reaction will be...... times of initial:-(3) 16hene(2) 8(4) 16 |
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| 2336. |
(पद्खथ S८ ०« ०्ु्स अं USSR ता. _you_ con vt §gotuon into Pop? |
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Answer» An olfactory indicator is a substance whose smell varies depending on wether it's mixed with acidic solution or basic solution. when gypsum(CaSO4.2H2O) is heated to about 100 degree celcius it loses some of water of crystallisation and is converted to platter of Paris. |
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| 2337. |
Ethyl bromide, Differentiate between soaps and synthetic deterpentsL.P. 2007, 2008, 2010, 2011 Set-II, 2015 s |
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Answer» soap is for bathing and contains no harmful chemicals and detergent has harmful chemical and for wasting clothes |
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| 2338. |
Dote4+6)5 + 3 (3- |
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Answer» [(4+6)5+3(3÷2)+(3*2)][10*5+9/2+6][50+9+12/2][50+21/2]100+21/2121/2 |
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| 2339. |
A 7.0 M solution ofKOH in water contains 28% by mass of KOH, what is density of solution ingm/ml? |
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Answer» here, is the solution for a similar ques.just substitute the numeeical values of your doubt and proceed accordingly A 6.9 M solution of KOH in Water contain 30% by mass of KOH what is the density of the solution |
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| 2340. |
Calculate the molarity of a solution containing 5 g of NaOH in 450 ml.solution. |
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Answer» Molar mass of NaOH= 40g per mole No. of moles in 5g of NaOH = 5/40 = 0.125 Volume of solution= 450mL=0.45L Molarity of solution= No.of moles of solute/ volume of solution= 0.125/0.45= 0.28 |
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| 2341. |
(A)"co,-CO10 ml of a compound containing N'and 'O'is mixed with 30 ml of H2 to produce H2ofN2 (g). Molecular formula of compound if both reactants reacts completely, 15(A) N,OH2O () and 10 m(B) NO2(C) N,03(D) N2O5lor fommula |
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| 2342. |
A piece of copper of mass 106 g is dipped ina measuring cylinder containing water at themark 22 ml. The water rises to the 34 mlmark. What is the density of copper ? |
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| 2343. |
0. hhich 1's SmallestNat or No |
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Answer» Na+ is smaller than Na tq Na+ as it has 1e- less than Na |
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| 2344. |
Q. 1. What are different states of matter?examples. |
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Answer» There are five known phases, or states, of matter: solids(ice), liquids, )water)gases(air),plasmaand Bose-Einstein condensates. The main difference in the structures of each state is in the densities of the particles. |
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| 2345. |
Page NoDate:anic compound contain 4o5csin-muazos er--an.das 1.de --unitrda argmolecula |
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Answer» thnku so much |
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| 2346. |
2Mg + O2 â 2Mgo, calculate how much oxygen is required tocombust 10 moles of Magnesium from the given equation. |
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Answer» As per the equation 2 moles of Mg require 1 mole of O2hence 10 moles would require 5 moles of O2 |
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| 2347. |
What is the weight of oxygen required for thecomplete combustion of 2.8 kg of ethylene?3.[AIPMT 1989](A) 2.8 kg(C) 9.6 kg(B) 6.4 kg(D) 96 kg |
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Answer» The chemical equation for the said reaction is, C2H4+12×2+4=283O296→2CO2+2H2O the gram molar weight of ethylene is= 12*2+1*4 g =28gthe gram molar weight of oxygen = 3*32 g =96 g∵The weight of oxygen required for complete combustion of 28 g ethylene =96 gaccording to question, ∴ Weight of oxygen required for combustion of 2.8 kg ethylene =96×2.8×100028×1000kg =9.6kgThat is the answer. |
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| 2348. |
calculate the no.of moles in the 46 gm of Na. |
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Answer» Molecular weight of Na=23hence moles=amount given/molecular mass=46/23=2 moles. |
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| 2349. |
12ml of a mixture of Methane and Ethane on complete combustion give 19 ml of carbon dioxide and 31 ml of water vapour under a given conditions of temperature and pressure the mixture contains? |
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Answer» let the volume of methane be a and Ethane be b. CH4 + 2O2 ---> CO2 + 2H2O a 2a a 2a C2H4 + 3O2 ------> 2CO2 + 2H2O b 3b 2b 2b now a+b = 12 and a+2b = 19 , so, a+2b-(a+b) = 19-12 = 7=> b = 7ml , and a = 5 ml volume of methane = 5 ml and ethane = 7 ml |
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| 2350. |
A gas mixture of 3.67 lit of ethylene and methane on complete combustion at 25°C and at 1 atmpressure produce 6.11 lit of carbondioxide. Find out the amount of heat evolved in k], during thiscombustion. (ΔΗ-(CH4)--890 kl mol-1 and (AHc(CH)--1423 kJ mol-11. |
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Answer» follows : C2H4+ 3O2→ 2CO2+ 2H2O 1 vol 2 vol. CH4+ 2O2→ CO2+ 2H2O 1 vol. 1 vol Let the vol. of CH4in mixture = xl ∴ Vol. of C2H4in the mixture = (3.67 - x)l Vol. of CO2produced by xlof CH4= xland Vol. of CO2produced by (3.67 - x)lof C2H4= 2(3.67 - x)l ∴ Total vol. of CO2produced = x + 2 (3.67 - x) Or 6.11 = x + 2(3.67 - x) or x = 1.23l ∴ Vol. of CH4in the mixture = 1.23l and Vol. of C2H4in the mixture = 3.67 – 1.23 = 2.44l Vol. of CH4per litre of the mixture = 1.23/3.67 = 0.335l Vol. of C2H4per litre of the mixture = 2.44/3.67 = 0.665 l Now we know that volume of 1 mol. Of any gas at 25°C (298 K) = 22.4 * 298/273 = 24.45l [∵ Volume at NTP = 22.4L] Heat evolved due to combustion of 0.335lof CH4= - 0.335 * 891/24.45 = - 12.20 kJ [given, heat evolved by combustion of 1l= 891 kJ] Similarly, heat evolved due to combustion due to combustion of 0.665lof C2H4 = - 0.665 * 1423/24.45 = - 38.70 kJ ∴ Total heat evolved = 12.20 + 338.70 =50.90 kJ |
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