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.
| 3051. |
30. Calculate the mass of l molecule of N2 [N-BII ) 4.05 × 10.22 gm |
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Answer» I mole has 6.02 × 10²³ molecules having total mass = 28g so, 1 molecules will have mass = 28/(6.02) × 10_²³ = 4.65×10-²³ gm thank you |
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| 3052. |
17. How many grams of sugar would contain 4.4 gram atoms of oxygen? |
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Answer» thanku |
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| 3053. |
Calculate the number of gram atoms and gram moleculesof sulphur (S3) in 8 g of sulphur. Atomic mass of sulphur- 32 amu. |
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Answer» we know that ,m=8g ,molar mass of s8 =32×8 =256g so,we got m=8g ,M=256g. now,N=?,n=?,No=6.022×1 sorry went mistakely 1Gram atom means 1 mole No. Of moles = Weight in gram/Molar mass No. Of moles =8/32 =1/4moles .25 gram atom (Moles)of sulphur is there in 8 gm of sulphur |
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| 3054. |
Calculate(a) mass of 1.5 gram atoms of calcium (at. mass 40) |
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Answer» 1 gram atoms = mass of 1 moles so, mass of 1.5 moles of Ca is = 1.5*40 = 60 gm. |
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| 3055. |
kinetic1.1.T.)6. A gas occupied a volume of 84 mL at -73째C and 380 mm. What would its volume beat 27째C and 570 mm pressure ? |
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Answer» Use P1V1/ T1= P2V2/ T2Very important : Before using this formula, make sure Temperature is in Kelvin.To convert °C to Kelvin, add 273 to it. So, T1= -73°C = 200 KAlso, T2 = 27°C = 300 KSo, we have,P1 = 380 mm. P2 = 570 mmV1 = 84 ml. V2 = ?T1 = 200 K. T2 = 300 K As, P1V1/ T1= P2V2/ T2=> V2 = P1V1T2 / ( P2T1) |
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| 3056. |
Density of a gas at 276K and 380 mm pressure is 3.0 gram/litre. Weightof 1.25 mole of the gas in grams is |
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Answer» PV=nRTPV=(m/M)RTPM=dRT P:Pressurem:mass of the gasM:Molecular Massd:DensityR:Universal Gas ConstantT:Temperature. P=380 mm= 0.5 atm=50662.5T=276 Kd=3g/l=3 kg/m3 M=dRT/PM=0.06793 kg1.23 M=.08492 kg=84.92 gm |
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| 3057. |
The Molecular weight of a gas is 40. At 400Kif 120 g of this gas has a volume of 20 litres,the pressure of the gas in atm is |
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| 3058. |
Density of a gas at 276K and 380 mm pressureis 1.5 gram/litre. Weight of 1.25 mole of thegas in gramS IS |
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| 3059. |
If the mass of Ais+ must be the mass of BSb. The raulu.radius of planet B. If the mass ofM., what must be the massthat the value of g on B is half that ofits value on A?Ans: 2Mc. The mass and weight of an object onearth are 5 kg and 49 N respectively.What will be their values on themoon? Assume that the accelerationdue to gravity on the moon is 1/6thof that on the earth.Ans: 5 kg and 8.17 N |
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Answer» Answer is 5KG AND 8.17 Newton mass will remain same and weight =1/6 *49 = 8.166(approx) |
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| 3060. |
PVFor one gram molecular weight ofa gas- |
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Answer» we know the relation as PV = nRT where n = no. of moles = (total mass)/(molecular weight ) = m/M but here M = 1 gm , so n = m so PV = mRT therefore PV/T = mR. value in calories Pls |
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| 3061. |
estionsIf one mole of carbon atomsweighs 12 grams, what is themass (in grams) of 1 atom ofcarbon? |
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Answer» If one mole of carbon atoms weighs 12 grams, what is the mass(in grams)of 1 atom of carbon?the molecular mass of carbon atom= 12gan atom of the carbon mass= 12g1 atom of the carbon mass= 12/6.022×10²³g= 1.99×10-²³g |
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| 3062. |
An element crystallizes in a b.e.c lattice with cell edge of 500pm. The densityof the element is 7.5g em-3. How many atoms are present in 300 g of theelement ? |
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Answer» thanks |
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| 3063. |
50. Ionic mass of X3-is 31 g mol-, lt has 16 neutrons. Thus, number of electrons in X3-isA) 15B) 16C) 17D) Is |
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Answer» The Atomic mass = 31so, no..of proton = 31-16 = 15 , now in neutral atom no. of P = no. of e- but since it has -3 charge so, no. of e- = 15+3 = 18 |
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| 3064. |
5. What is the formula of a compound in which the element Y forms ccp lattice andoccupy 2/3rd of tetrahedral voids? |
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Answer» thank you |
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| 3065. |
U Q.6. What is the formula of a compound in whichthe element P forms ccp lattice and atoms of Qoccupy 2/3rd of tetrahedral voids? |
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| 3066. |
Silver forms ccp lattice. X-ray studies of its crystals show that the edge length of its unit cellis 408.6 pm. Calculate the density of silver (Atomic mass 107.9 u) |
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| 3067. |
l nu3 |
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| 3068. |
Calculate the eve hucletCharge on delectron |
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Answer» effective charge is Z - e , where Z = 21 ignore the 4s electron as it is ahead of 3d and e is. 0*0.35 ( for 0 d electrons beside the 1 that we are calculating ) + 1*18 ( other 18 electron at the back) so, e = 18 and Z-e = 21-18 =3 |
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| 3069. |
ए याद कोई भी बुलबुला काला न किया हो।||JIT पुणपुल का काला किया जाए।A balanced chemical equation is in accordance with -(1) Avogadro's law13/ law of conservation of mass(2) law of constant proportionsलत रासायनिक समीकरण निम्न नियम पर आधारित होती है :(4) law of gaseous volumes(1) आवोगादो नियम(2) स्थिर समानुपात का नियम |
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Answer» A balanced chemical equation always follow the law of mass of conservationas it cant be created nor be destroyed. |
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| 3070. |
1. They eve wall of a hurrleane is very calm |
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| 3071. |
CERT Numerical'scompound is formed by two elements M and N. The element N forms ecp and atoms of Mсиру ///3rd of tetrahedral voids. What is the formula of the compound ?Ans- M2N1 |
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| 3072. |
What is the formula of a compound in which the element Y forms cep lattice and atoms of X occupy 2/3rd oftetrahedral voids? |
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| 3073. |
instrument which is use to know theWhich is thetemperature ?az volt meter> thermometerb) Galvanometerd) beam balance |
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Answer» c is the right answer bhaiyya jii option c is the right answer c is the right answer option c is correct answer |
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| 3074. |
25. Indicate the principle behind the method for refining of zine. (CBSE-201 |
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Answer» thanks |
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| 3075. |
QUESTIONSWhy should a magnesium ribbon be cleaned before burning in an |
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Answer» when magnesium ribbon is expose to air it forms a thick coating of magnesium oxide which prevents the burning. So to get pure magnesium it is cleaned with sandpaper before burning. |
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| 3076. |
| 46 T Servaton Consiokvees omy axb/fijmic कैकी P fxamp/oin. |
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Answer» Respirationisconsideredasexothermic reactionbecause energy is released in thisprocess. Glucose combines with oxygen present in our cells to form carbon dioxide and water along with energy. Respirationis consideredasexothermic reactionbecause energy is produced during thisprocess. |
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| 3077. |
iunoIn. How will you find the valencyof chlorine, sulphur' anmagnesium? |
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Answer» Form the electronic configuration and the electrons are left in the last is your valency chlorine = at. no.=17.then electronic configuration is 2,8,7. so valency is -1.sulphur=16 .then electronic configuration is 2,8,6. so valency is -2.magnesium=12. then el.conf. is 2,8,2. so valency is +2. |
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| 3078. |
2.II. Fill in the blanksnamed carbon.1. |
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Answer» A.L. Lavoisier proposedcarbonin 1789 from the Latin carbo meaning "charcoal |
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| 3079. |
Kerosene has a lower ignition temperaturethan diesel. What does this statementmean? |
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Answer» A fuel withlowest ignition temperature meansit catches fire more quickly andiscalled highly inflammable.Kerosene has lower ignition temperaturewhen compared todiesel. Thismeans kerosene willstart to catch fire faster than diesel |
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| 3080. |
fat Why Li2CO3 decomposed at a lower temperature, where as Na2CO3 at higher temperature?Draw the structure of BeCl, (vapour) |
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Answer» 1.As we move down the alkali metal group, the electropositive character increases. This causes an increase in the stability of alkali carbonates. However, lithium carbonate is not so stable to heat. This is because lithium carbonate is covalent. Lithium ion, being very small in size, polarizes a large carbonate ion, leading to the formation of more stable lithium oxide. ∆Li2CO3--------->LiO2+ CO2 Therefore, lithium carbonate decomposes at a low temperature while a stable sodium carbonate decomposes at a high temperature. 2. |
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| 3081. |
DIIl DC CUilsumed7k Assuming fully decomposed, the volume of CO2 released atSTP on heating 9.85 g of BaCO, (Atomic mass, Ba-137) willbe |
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| 3082. |
16. Assuming fully decomposed, the volume ofCO2 released at STP on heating 9.85 g ofBaCO3 (atomic mass, Ba 137) will be(A) 1.12 L[AIPMT 2000](B) 0.84 L(D) 4.96 L(C) 2.24 L |
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| 3083. |
ILMILAL KINETICSDAILY PRACTICE PROBLEM-3The reaction SO,CI, - →SO, + CI, obeys firstorder kinetics with rate constant 3.2 x 10-5 sec-at 320 °C. What % of SO CI will be decomposedon heating gas for 90 minute?[Antilog (0.051.188)](A) 20.85%0%(B) 15.86%(C)5.5%(D) 10%Thermal decomnosition of |
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Answer» 20.85℅ is right answer this correct answer (A) B) is correct answer |
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| 3084. |
b) NaOHag) Haq) NaCl(a)H,0(0). A solution of the substance X' is used for whitewashing.(a) Name the substanceX and write its formulab) Write the reaction of the substance X' with water.ns. (a) lhe substance·X, is calcium oxide (also called lime or quicklime). Its fo(b) CaO(s) + H:0() → Ca(OH)2(s) |
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Answer» (i)The substance ‘X’ which is used in white washing is calcium oxide or quick lime and its formula is CaO.(i)The reaction involved is |
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| 3085. |
LZSg of CUso4 disolved in water for making 100g of solution.Find its concentration. |
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Answer» concentration of solution =mass of solute/mass of solution(water) multiply by 100= (25/100)*100= 25 % Concentration of solution = 25% |
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| 3086. |
I . S W ला*_PM&PQMM_—%—*A—« 1100 plociian L("’"M—:,L e1) - |
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Answer» In DNA and RNA, the phosphodiester bond is the linkage between the 3' carbon atom of one sugar molecule and the 5' carbon atom of another, deoxyribose in DNA and ribose in RNA. Strong covalent bonds form between the phosphate group and two 5-carbon ring carbohydrates (pentoses) over two ester bonds. |
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| 3087. |
the wavelength of a cricket ball weighing 100g and travelling with a velocity will be nearly of 50 m |
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Answer» accourding to Broglie relation wavelength=h/mvso, wavelength of cricket ball=(6.626×10^-34)/.1×50I hope you understand wavelength=h/mv (100)/50=2 |
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| 3088. |
Temperature dependence of the rate of a reaction. |
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| 3089. |
The enhanced force of cohesion in metals is due to(1) The covalent linkages between atoms(2) The electrovalent linkages between atoms(3) The lack of exchange of valency electrons(4) The exchange energy of mobile electrons |
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Answer» Option (4) is correct. |
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| 3090. |
HalogenExchange |
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Answer» Naturally producedhalogenatedcompounds are ubiquitous across all domains of life where they perform a multitude of biological functions and adopt a diversity of chemical structures. Accordingly, a diverse collection of enzyme catalysts to install and removehalogensfrom organic scaffolds has evolved in nature. Accounting for the different chemical properties of the fourhalogenatoms (fluorine, chlorine, bromine, and iodine) and the diversity and chemical reactivity of their organic substrates, enzymes performing biosynthetic and degradativehalogenationchemistry utilize numerous mechanistic strategies involving oxidation, reduction, and substitution. Biosynthetichalogenationreactionsrange from simple aromatic substitutions to stereoselective C-H functionalizations on remote carbon centers and can initiate the formation of simple to complex ring structures. Dehalogenating enzymes, on the other hand, are best known for removinghalogenatoms from man-made organohalogens, yet also function naturally, albeit rarely, in metabolic pathways. This review details the scope and mechanism of nature'shalogenationand dehalogenation enzymatic strategies, highlights gaps in our understanding, and posits where new advances in the field might arise in the near future. |
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| 3091. |
Write an electronic configuration of the following: Mn+, Fe+,Cr2+ |
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| 3092. |
35.Explain the structure of diborane. |
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Answer» Diborane structure has four terminal and two bridging hydrogen atoms. The simple Boron Hydride is Diborane B2H6. The structure can be explained as, electrons are required for the formation of conventional covalent bond structure, whereas in Diborane there are only 12 valence electrons, three from each Boron atoms. Thus B2H6 is an electron deficient component. The four terminal hydrogen atoms and two boron atoms lie on one plane. Above and below this plane, there are two bridging hydrogen atoms. Each Boron atom forms four bonds eventhough it has only three electrons. The terminal B-H bonds are regular bonds but the bridge B-H bonds are different. Each bridge hydrogen is bonded to the two boron atoms only by sharing of two electrons. Such covalent bond is called three centre electron pair bond or a multicentre bond or banana bond. |
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| 3093. |
UT 30 15 highest.]The second ionization enthalpy of which of the followingalkaline earth metals is the highest? (PET (Kerala) 2016](a) Ba(b) Mg(c) Ca(d) Sr(e) BeHint: On moving from top to bottom in a group, ionizationenthalpy decreases.] |
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Answer» c) is the right answer of the following ca is right answer of the following option c is the correct andwer (c). Ca :- CalciumCalcium is the right answer. Answer:- (c). Ca c calcium is the right answer calcium is the right answer mg is the correct answer option (e)is the correct answer option C is the correct answer. option C is the correct answer |
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| 3094. |
210 mL at 25째C is heated at constant pressure to 90째C. What volume wouldm mole?A gas at 710 mL air Dow occupy?cunies 83.72 litres at 850 mm of He and 2If the ancie com |
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Answer» I don't understand that question |
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| 3095. |
What charge does a cation carry? |
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Answer» An ion is an atom or group of atoms in which the number of electrons is not equal to the number of protons, giving it a net positive or negative electrical charge. Ananionis an ion that is negatively charged, and is attracted to the anode (positive electrode) in electrolysis. Acationhas a net positive charge, and is attracted to the cathode (negative electrode) during electrolysis. positivecharge is on caption. |
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| 3096. |
4. What charge does a cation carry? |
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Answer» positivecharge are calledcations. |
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| 3097. |
24. If enthalpies of formation for C2H4(g), CO2 (g) andH2 O(/)at 25 °C and I atm pressure be 52, -394 and -286kJ mol respectively, enthalpy of combustion of C2H4(g)will be:(a) +141.2 kJ mol(c) 141.2 kJ mol(CBSE 1995)(b) +1412 kJ mol-1(d) -1412 kJ mol-1 |
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| 3098. |
The equations representing the combustioncarbon and carbon monoxide are:11.ofco(s) + O2(g)-> CO2(g) ΔΗ -284.5 k/molthe heat of formation of 1 mol of CO(g) is:(1) -109.5 kJ/mol (2)+109.5 kJ/mol(3) +180.0 kJ/mol ((4) + 100 kJ/mol |
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Answer» Hence, option (A) is correct |
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| 3099. |
Heat of neutralisation for the reaction,NaOH + HCI — NaCl + H2Ois 57.1 kJ mol'. The heat released when0.25 mol of NaOH is titrated against0.25 mol of HCl will be:1) 22.5 kJ2) 57.1 kJ3) 28.6 kJ4) 14.3 kJ |
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| 3100. |
12. In the reactionOHO NaCHO)+ CHCI, + NaOH→The electrophile involved is(1) Dichloromethyl cation (CHCIZ)(2) Dichlorocarbene (:C012)(3) Dichloromethyl anion(4) Formy cation (HO) |
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Answer» Its reimer temer reaction so Option 2 is rightit will create free radical of ccl2 Option B 2 is right choice chcl3 + NaoH first react to carbene generate it is a electronphile |
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