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Manivannan
04 Apr 2017
Assertion: Gases do not settle to the bottom of the container.
Reason: Gases have high kinetic energy.

Both assertion and reason are correct statements, and reason is the correct explanation of the assertion.

Both assertion and reason are correct statements, but reason is not the correct explanation of the assertion.
Assertion is correct, but reason is wrong statement.
Assertion is wrong, but reason is correct statement.
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Ionic Solid Defect Density Ionic Solid DecreasesFollowing Exhibit Frenkel Defect Tetragonal Crystal System Unit Cell Dimensions50 Mangoes Box Write Total Number Mangoes Terms Number Boxes Use B Number Boxes Metal Crystallizes Face Centred Cubic Lattice Edge Unit Cell 408 Pm Diameter Metal AtomNumber Octahedral Voids Per Atom Present Cubic Close Packed StructureCalcium Fluoride Structure Coordination Number Cation Anion RespectivelySpace Occupied Bcc Arrangement ApproximatelyFollowing Substances Considered Pseudo Solid Following Show Anisotropy Nacl Doped Mgcl2 Nature Defect Produced Coordination Number Cation Body Centred Cubic LatticeSilicon Doped Arsenic Example Type Semiconductor Metal Crystallises Simple Body Centred Face Centred Cubic Structure Whose Unit Cell Lengths A1 A2 A3Element Atomic Mass 100 G Mol Bcc Structure Unit Cell Edge 400 Pm Density ElementSemiconductor Ge Made P Type AddingJar Contains 24 Marbles Green Others Blue Marble Drawn Random Jar Probability Green Find Number BlueSolid Ab Nacl Structure Atoms Occupy Corners Cube Unit Cell Face Centred Atoms Along One Axes PassinSolid Structure W Atoms Located Corners Cubic Lattice Atoms Centre Edges Na Atom Centre Cube FormulaNumber Unit Cells 58 5 G Nacl NearlyCompoundhas Cubic Close Ccp Arrangement X Unit Cell Structure Shown Empirical Formula CompoundX Two Sets X 18 Elements X 8 Elements 15 Elements Many Elements X Explain Difference Properties Diamond Graphite Basis Structures Compound Elements B Crystallizing Cubic Close Packing Four Atoms Unit Cell Atoms Occupy Alternate CoSuggest Reasons B F Bond Lengths Bf3 130 Pm 143 Pm Differ Group 400 People 250 Speak Hindi 200 Speak English Many People Speak Hindi English Using Section Formula Show Points 2 3 4 B 1 2 1 Collinear Find Coordinates Points Trisect Line Segment Joining Points P 4 2 6 Q 10 16 6 Evaluate Given Limit Nancy Archeologist Found Piece Solid Iron Tomb Old Palace Determined Density Solid Iron 233 C 1 65 GEvaluate Given Limit Evaluate 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Impossible C Unforeseeable Ii Pick Two Lines Stanza 2 JusAverage Fixed Cost Average Variable Cost Average Cost Firm Related Supply Curve Firm Long Run Potassium Crystallizes Bcc Lattice Approximate Number Unit Cells 2 G Potassium Atomic Mass PotassiumKcl Forms Fcc Structure Density 1 984 G Cm 3 Edge Length Unit Cell 629pm Molar Mass KclSuspense Account Necessary Suspense Account Balance Rectification Errors Detected Accountant HappensKinds Errors Would Cause Difference Trial Balance Also List Examples Would Revealed Trial Balance Compound Xy2 O4 Oxide Ions Arranged Ccp Cations X Present Octahedral Voids Cations Equally DistributDensity Potassium Bromide 2 75 G Cm3 Length Unit Cell 654 Pm ExhibitsMetal Crystallizes Two Cubic Systems E Face Centred Cubic Fcc Body Centred Cubic Bcc Whose Unit CellUnit Cube Length Licl Nacl Structure 5 14 Assuming Anion Anion Contacts Radius Chloride Ion Solid Compound Oxide Ions Arranged Ccp Cations X Occupy One Sixth Tetrahedral Voids Cations Occupy ORupak Ltd Issued 10 000 Shares Rs 100 Payable Rs 20 Per Share Application Rs 30 Per Share Allotment Following Statements Correct Respect Defects Solids Element Atomic Mass 100 G Mol Bcc Structure Unit Cell Edge Length 400 Pm Density Element Comment Influence English Language Way Life Indian Life Reflected Story Narrator Attitude English Following Statements Regarding Cao Incorrect Crystalline Solids Anisotropic Nature Statement Mean Silver Crystallizes Face Centred Cubic Unit Cell Side Unit Cell Length 400 Pm Calculate Radius Silve Explain Alkyl Amine Basic Ammonia B Would Convert Aniline Nitrobenzene Ii Aniline Iodobenzene Stoichiometric Defect Crystals Increases Density Solid Density Lead 11 35 G Cm 3 Metal Crystallizes Fcc Unit Cell Estimate Radius Lead Atom Mass Lead 207 GRohit Identified X Ray Diffraction Studies Copper Crystallizes Face Centered Cubic Fcc Crystal LattiStatement 1 Geometrical Isomers Complex Optically Inactive Statement 2 Geometrical Isomers Complex PNumber Atoms Hcp Unit CellVolume Hcp Unit CellEmpty Space Hcp Unit CellCompoundhas Cubic Close Ccp Arrangement X Unit Cell Structure Shown Empirical Formula CompoundElement Crystallizes Fcc Lattice Edge Length 400 Pm Maximum Diameter Atom Placed Interstitial Site WCrystal Structure Formula Ionic Solids RepresentsUse Following Information Answer Next Question Precipitation Reaction Silver Chloride Represented 2aMetal Crystallizes Body Centered Cubic Lattice Density 1 51gcm 3 Radius Metal Atom 248 Pm Calculate Crystal Ab Rock Salt Structure Molecular Weight 6 023 U Arbitrary Number U Minimum Distance Cation AElement X Crystallizes Cubic Structure One Atom Corner Two Atoms One Diagonal Volume Unit Cell 48 X Unit Cell Calcium Sulphide Consists Sulphide Ions Corner Calcium Ion Body Center Cube Edge Length UnFollowing Unit Cells Show Atom Different Locations Edge Length Cells Pm Radius Sphere Present CornerFollowing Statements Regarding Crystalline Amorphous Solids Incorrect Use Following Information Answer Next Question Ice Molecular Solid So2 Molecular Solid Information AUse Following Information Answer Next Question Compound Formed Two Elements X Atoms Element X Form HEfficiency Packing Case Body Centred Cubic StructureUse Following Information Answer Next Question X Ray Diffraction Studies Show Silver Crystallizes CcNumber Particles Per Unit Cell Face Centred Cubic Fcc LatticeUse Following Information Answer Question Substances Strongly Attracted Whilesubstances Weakly AttraUse Following Information Answer Question Crystal Si Doped Leads Formation Semi Conductor InformatioUse Following Information Answer Next Question Given Figure Represents Gap Filled Valence Band Next Use Following Information Answer Next Question Empirical Formula Metal Oxide Crystal Respective PercUse Following Information Answer Next Question Copper Crystallizes Fcc Lattice Edge Length 3 61 10 8Following Defects Arise Solid Heated Use Following Information Answer Next Question X Ray Diffraction Shows Element Crystallizes Bcc UnitFollowing Figures Represents F Centre Type Solid Soft Waxy Nature Crystal Structure Formula Ionic Solids RepresentsFind Three Rational Numbers 23 34 Evaluate 31336Malleability Ductility PropertiesCommon Table Sugar E Sucrose ExampleEvaluate 7056Find Square Root Number Using Method Prime Factorisation 9216Metals Act Good Conductors Electricity Express 81 Sum 9 Odd Numbers Ii Express 100 Sum 10 Odd Numbers Fill Blanks 360000 Written Standard Form Ii 0 0000123 Written Standard Form Iii 23 2 Iv 3 10 3 UsualTick Correct Answer 0 000367 104 Usual Form 3 67 B 36 7 C 0 367 0 0367Tick Correct Answer 0 0000463 Standard Form 463 10 7 B 4 63 10 5 C 4 63 10 9 46 3 10 6 Point Defects Crystals B Frenkel Defect Terms Band Theory Write Difference Conductor Insulator Type Packing Efficient Packing Efficiency Terms Percentage Understand Ferromagnetism Ferrimagnetism Give Example Types Lattice Imperfections Found Crystals Name Type Structure Possessed Unit Cell Cscl Average Distance Sun Earth 15 Crore Km B 1 5 Crore Km C 150 Crore Km 1 500 Crore Km Planet Solar System Associated Following 1 Messenger Roman Gods 2 Roman God War 3 Greek God Sky 4 RoSignificance Sun Solar System Moon Support Form Life Distinguish Following 1 Parallels Meridians 2 Local Time Standard Time 3 Latitude Longitude 4 TorridFollowing Years February 29 Days 2000 B 2010 C 2014 2006Speed Earth Rotation North Pole Days Nights Equal Duration Throughout World Equinoxes Bus Ride Comfortable Metro Ride Court Sessions Judge Highest Court Civil Cases District Write Names Three Public Conveniences Municipalities Municipal Corporation Provide Regular Collection Disposal Waste Within Municipal Area Looked Department Municipal Corporation B MuNyaya Panchayat Function Panchayati Raj System Important Large Country Like Period Institution Panchayats Decline Voter Elect Right Candidate Political Party Group Parties Combining Form Government Called Opposition B Authoritarian C Coalition PresidentialConstitution List Different Types Constitutions Name Categories People Special Steps Taken Government Weaker Backward Sections Society Protected Police B People C State DalitsReligious Music Folk Music Following Solids Amorphous One Point Defect Crystals Solid Decreases Density Solid Density Copper Metal 8 95 G Cm 3 Radius Copper Atom 127 8 Pm Copper Unit Cell Simple Cubic Body CentType Unit Cell Cinnabar Hgs Cadmium Sulphide Cds Correspond Type Molecular Solid Ccl4 Classified Formula Compound Cubic Structure Formed Elements B Atoms Present Corner Atoms B Present Face CentresCorundum Oxide Ions Arranged Hexagonal Close Packing Aluminium Ions Occupy Two Thirds Octahedral VoiUrea Exhibit Definite Heat Fusion Glass Exhibits Indefinite Heat Fusion Calculate Number Atoms Present Face Centred Cubic Lattice Fcc Density Kbr 2 75 Gcm 3 Length Edge Unit Cell 654 Pm Predict Type Cubic Lattice Unit Cell Kbr BelongsTotal Number Voids Present 1 Mole Compound Forming Hexagonal Close Packed Structure B Write CoordinaAtoms Element Cubic Unit Cell Radius R Calculate Length Face Diagonal Ii Body Diagonal B Schottky DeAgi Crystallizes Cubic Close Packed Zns Structure Fraction Tetrahedral Sites Occupied Ag Ions B CompAmorphous Solid Considered Super Cooled Liquid Metal Crystallizes Two Cubic Phases E Face Centred Cubic Fcc Body Centred Cubic Bcc Phase Whose UnitHelp Diagram Illustrate Hexagonal Close Packing Two Dimensions Diamond Rhombic Sulphur Covalent Solids Latter Low Melting Point Compared Former B Nacl Cscl DiffereAccount Following Statement Quartz Crystalline Nature Quartz Glass Amorphous Despite Molecular FormuMetallic Gold Crystallises Fcc Lattice Calculate Radius Gold Atom Atomic Mass Gold 197 Density 19 3 Element Atomic Mass 78 G Mol 1 Crystallises Face Centred Cubic Lattice Edge Length Unit Cell 400 Pm Number Atoms Present Cubic Closed Unit Cell One Atom Corner Two Atoms Body Diagonal Packing Efficiency Case Body Centred Cubic Structure Solid Ab Nacl Structure Radius Cation 200 Pm Radius Anion B Radius Ratio Octahedral Void Given 0 414Compound Formed Two Elements X Atoms Element X Form Hcp Lattice Element Occupy Tetrahedral Voids ForConcentration Cation Vacancies Nacl Doped 10 3 Mol Srcl2 Calculate Approximate Number Unit Cells Present 1 G Ideal Zns Crystal Atomic Mass Zn 65g Mol Derive Relation R R Radius Octahedral Void R Radius Atoms Close Packing R Gold Crystallises Face Centred Unit Cell Length Side Unit Cell Given Atomic Radius Gold 0 144 Nm Group 13 Elements Increase Electrical Conductivity Group 14 Elements Compound Pq Crystallises Bcc Lattice Unit Cell Edge Length 480 Pm Calculate Distance Oppositely CharAnswer Following Questions Using Given Illustration Defective Crystal Interstitial Defect Depicted GChromium Metal Atomic Mass 52 G Mol Crystallizes Body Centred Cubic Lattice Length Unit Edge Found 2Element E Crystallizes Atomic Mass 150 G Mol Fcc Lattice Edge Length 600 Pm Number Atoms Present 15 Compound Ab2c4 C Ions Arranged Ccp Pattern Cations Present Octahedral Voids Cations B Equally DistriAgf Crystallizes Lattice Similar Nacl Density Agf 5 85 G Cm3 Distance Cation Anion Agf Lattice Following Figures Represents F Centre Packing Efficiency Hcp StructureFollowing Statements Regarding Crystalline Amorphous Solids Incorrect Following Alternatives Correctly Classifies Materials Crystalline Amorphous Solids Use Following Information Answer Next Question Ice Molecular Solid So2 Molecular Solid Information AFollowing Solids Covalent Solid Use Following Information Answer Next Question Compound Formed Two Elements X Atoms Element X Form HEfficiency Packing Case Body Centred Cubic StructureUse Following Information Answer Next Question X Ray Diffraction Studies Show Silver Crystallizes CcNumber Particles Per Unit Cell Face Centred Cubic Fcc LatticeUse Following Information Answer Next Question Substances Strongly Attracted Whilesubstances Weakly Use Following Information Answer Next Question Crystal Si Doped Leads Formation Semi Conductor InforUse Following Information Answer Next Question Given Figure Represents Gap Filled Valence Band Next Use Following Information Answer Next Question Copper Crystallizes Fcc Lattice Edge Length 3 61 10 8Following Defects Arise Solid Heated Use Following Information Answer Next Question X Ray Diffraction Shows Element Crystallizes Bcc UnitUse Following Information Answer Next Question Empirical Formula Metal Oxide Crystal Respective PercExplain Following Suitable Example Paramagnetism Piezoelectric EffectX Ray Diffraction Shows Element Crystallises Bcc Unit Cell Cell Edge 2 55 10 8 Cm Density Element FoDescribe Two Main Types Semiconductors Formed Crystal Si Doped BCubic Solid Made Two Elements X Atoms Corners Cube Atoms X Body Centre Formula Compound Iron Body Centred Cubic Unit Cell Cell Edge 286 65 Pm Density Iron 7 87 G Cm 3 Use Information CalcuGive Brief Classification Molecular Solids Copper Crystallises Fcc Lattice Edge Length 3 61 10 8 Cm Calculate Atomic Radius Justify Given Statements Phosphorus Doped Silicon Semiconductor Schottky Defect Reduces Density CrysX Ray Diffraction Studies Show Silver Crystallises Ccp Unit Cell Edge Length 408 6 Pm Separate ExperAccount Following Statement Quartz Crystalline Nature Quartz Glass Amorphous Despite Molecular FormuNh3 Exist Liquid Room Temperature Values Axial Angles Unit Cell Graphite Number Atoms Present Cubic Closed Unit Cell One Atom Corner Two Atoms Body Diagonal Name Type Stoichiometric Defect Decreases Density Crystal Heated Solid Ab Nacl Structure Radius Cation 200 Pm Radius Anion B Radius Ratio Octahedral Void Given 0 414Solid Made Two Elements X Atoms X Ccp Arrangement Atoms Yoccupy Tetrahedral Sites Formula Compound Zinc Oxide Turn Yellow Heated Solids Schottky Frenkel Defects Conduct Electricity Though Small Extent Stoichiometric Defects Also Called Intrinsic Defects Window Glass Old Buildings Look Milky Naf Mgo Number Electrons Also Internuclear Distances Approximately 235 Pm 215 Pm Reason Difference MCrystal Nacl Turn Yellow Heated Atmosphere Sodium Vapour Solid Oxide Ions Arranged Ccp Cations X Occupy One Sixth Tetrahedral Voids Cations Occupy One Third Many Lattice Points Unit Cell Face Centred Cubic Lattice Calculate Approximate Number Unit Cells Present 1 G Ideal Zns Crystal Atomic Mass Zn 65g Mol Fraction Ni2 Ni3 Ni0 98o1 00 Concentration Cation Vacancies Nacl Doped 10 3 Mol Srcl2 Gold Crystallises Face Centred Unit Cell Length Side Unit Cell Given Atomic Radius Gold 0 144 Nm Octahedral Void Close Packing Spheres Always Size Justify Answer Derive Relation R R Radius Octahedral Void R Radius Atoms Close Packing R Niobium Crystallises Body Centred Cubic Structure Calculate Edge Length Atomic Radius Niobium 143 PmSilicon Germanium Behave Semiconductors Group 13 Elements Increase Electrical Conductivity Group 14 Elements Calculate Distance Oppositely Charged Ions Lattice Calculate Radius P Radius Q 275 Pm Assume Cation Fits Exactly Cubic Void Metallic Gold Crystallises Fcc Lattice Calculate Radius Gold Atom Atomic Mass Gold 197 Density 19 3 Packing Efficiency Case Body Centred Cubic Structure Differentiate Amorphous Crystalline Solids Element Atomic Mass 78 G Mol 1 Crystallises Face Centred Cubic Lattice Edge Length Unit Cell 400 Pm Compound Formed Two Elements X Atoms Element X Form Hcp Lattice Element Occupy Tetrahedral Voids ForCoordination Number Atom Face Centred Cubic Lattice Calculate Density Silver Crystallises Face Centred Cubic Structure Distance Nearest Silver Atoms StrExplain Conductivity Semiconductors Increases Increase Temperature Metals Decreases Write Short Notes Following Interstitial Defect Frenkel DefectCalculate Length Side Unit Cell Many Unit Cells 1 00 M3 Aluminium Type Substances Would Make Good Permanent Magnets Explain Type Unit Cell Cinnabar Hgs Correspond Formula Compound Cubic Structure Formed Elements B Atoms Present Corner Atoms B Present Face CentresMeant Term Coordination Number Corundum Oxide Ions Arranged Hexagonal Close Packing Aluminium Ions Occupy Two Thirds Octahedral VoiPoint Defect Results Decrease Density Crystal Help Diagram Illustrate Hexagonal Close Packing Two Dimensions Amorphous Solid Considered Super Cooled Liquid Urea Exhibit Definite Heat Fusion Glass Exhibits Indefinite Heat Fusion Calculate Number Atoms Present Face Centred Cubic Lattice Fcc State Four Main Characteristics Metallic Solids Br Ions Form Close Packed Structure Radius Br Ion 195 Pm Radius Cation Fits Tetrahedral Hole Cation Interstitial Defect Depicted Given Ionic Crystal Cation Usually Dislocated Normal Site Interstitial Site Density Kbr 2 75 Gcm 3 Length Edge Unit Cell 654 Pm Predict Type Cubic Lattice Unit Cell Kbr BelongsTotal Number Voids Present 1 Mole Compound Forming Hexagonal Close Packed Structure Write Coordination Number Ion Caf2 Diamond Rhombic Sulphur Covalent Solids Latter Low Melting Point Compared Former Nacl Cscl Different Structures Yet Similar Formula Metal Crystallizes Two Cubic Phases E Face Centred Cubic Fcc Body Centred Cubic Bcc Phase Whose UnitFrenkel Defect Found Halides Alkali Metals Coordination Number 12 Found Ionic Crystals Using X Ray Diffraction Methods Unit Length Nacl Observed 0 5627 Nm Density Nacl Found 2 164 Gcm 3 TAtoms Element Cubic Unit Cell Radius R Calculate Length Face Diagonal Ii Body DiagonalSchottky Defect Observed Crystal Agi Crystallizes Cubic Close Packed Zns Structure Fraction Tetrahedral Sites Occupied Ag Ions Compound Ab2 Possesses Caf2 Type Crystal Structure Coordination Number A2 B Crystals Cscl Forms Body Centred Cubic Lattice Caesium Chloride Ions Contact Along Body Diagonal Cell Length Type Solid Soft Waxy Nature Class Solids High Melting Boiling PointsMalleability Ductility PropertiesCommon Table Sugar E Sucrose ExampleUse Following Information Answer Next Question Given Chemical Equation Represents Anionic Single RepUse Following Information Answer Next Question Decomposition Reaction Potassium Chlorate RepresentedUse Following Information Answer Next Question Decomposition Reaction Potassium Chlorate RepresentedUse Following Information Answer Next Question Decomposition Reaction Potassium Chlorate RepresentedFollowing Solutes Dissolve Form Molecular Solution Covalent Compounds SolubleMetals Act Good Conductors ElectricityCalculate Number Unit Cells Present 0 2 G Silver Given Silver Face Centered Cubic Lattice Atomic MasCertain Element X Present Gaseous State 65 0c 2 Atm Pressure Cooled 00c Forms Bcc Type Crystal LattiUse Following Information Answer Question Unit Cell Axial Distance X Orthorhombic B C Cubic Ii B C ZUse Following Information Answer Question Angles Edges Crystal X Equal Angles Equal 90 Example Type Use Following Information Answer Question Magnetite Fe3o4 Weakly Attracted Magnetic Field Heating BeUse Following Information Answer Question Crystal Class Configuration Axial Distances Axial Angles 1Following Unit Cells Primitive Triclinic Structure Following Schematic Alignments Magnetic Moment Correctly Represents Magnetite Molecule Use Following Information Answer Question Substance X Gets Magnetised Magnetic Field Direction LosesFollowing Elements Permanently Magnetised Following Compounds Antifluorite Structure Use Following Information Answer Question Angles Crystal X Equal 90 Edges Different Lenghts Example Following Oxides Metal Like Electrical Property Solid Bcc Structure Distance Nearest Approach Two Atoms 1 73 Edge Length CellFollowing Best Description Water Solid Phase Among Following Incorrect Statement Presence Excess Lithium Makes Licl Crystals Pink B Solid Cubic Crystal Made Two Elements P Q Atoms Aluminium Crystallizes Fcc Structure Atomic Radius Metal 125 Pm Length Side Unit Cell Metal Many Atoms Constitute One Unit Cell Face Centered Cubic Crystal Type Semiconductor Obtained Silicon Doped Boron B Type Magnetism Shown Following Alignment MagneticType Stoichiometric Defect Shown Agcl Type Semiconductor Obtained Silicon Doped Boron 3 B Type Magnetism Shown Following Alignment MagnetType Substances Would Make Better Permanent Magnets Ferromagnetic Ferrimagnetic Type Semiconductor Obtained Silicon Doped Boron 3 B Type Magnetism Shown Following Alignment MagnetMeant Doping Semiconductor Tungsten Crystallizes Body Centred Cubic Unit Cell Edge Unit Cell 316 5 Pm Radius Tungsten Atom OrirWrite Point Distinction Metallic Solid Ionic Solid Metallic Lustre Tungsten Crystallizes Body Centred Cubic Unit Cell Edge Unit Cell 316 5 Pm Radius Tungsten Atom OrirMeant Doping Semiconductor Tungsten Crystallizes Body Centred Cubic Unit Cell Edge Unit Cell 316 5 Pm Radius Tungsten Atom IronElement Density 11 2 G Cm 3 Forms F C C Lattice Edge Length 4 10 8 Cm Calculate Atomic Mass Element Examine Given Defective Crystal Answer Following Questions Type Stoichiometric Defect Shown Crystal Element Density 11 2 G Cm 3 Forms F C C Lattice Edge Length 4 10 8 Cm Calculate Atomic Mass Element Examine Given Defective Crystal Answer Following Questions Type Stoichiometric Defect Shown Crystal Element Density 11 2 G Cm 3 Forms F C C Lattice Edge Length 4 10 8 Cm Calculate Atomic Mass Element Examine Given Defective Crystal Answer Following Questions Type Stoichiometric Defect Shown Crystal Element Density 2 8 Gcm 3 Forms F C C Unit Cell Edge Length 4 10 8 Cm Calculate Molar Mass Element G Write Type Magnetism Observed Magnetic Moments Aligned Parallel Anti Parallel Directions Unequal NuElement Density 2 8 G Cm 3 Forms F C C Unit Cell Edge Length 4 10 8 Cm Calculate Molar Mass Element Type Non Stoichiometric Point Defect Responsible Pink Colour Licl Ii Type Stoichiometric Defect ShoFormula Compound Element Forms Ccp Lattice Atoms X Occupy 1 3rd Tetrahedral Voids Element Molar Mass 27 G Mol 1 Forms Cubic Unit Cell Edge Length 4 05 10 8 Cm Density 2 7 G Cm 3 NatuFormula Compound Element Forms Ccp Lattice Atoms X Occupy 1 3rd Tetrahedral Voids Element Molar Mass 27 G Mol 1 Forms Cubic Unit Cell Edge Length 4 05 10 8 Cm Density 2 7 G Cm 3 NatuFormula Compound Element Forms Ccp Lattice Atoms X Occupy 1 3rd Tetrahedral Voids Element Molar Mass 27 G Mol 1 Forms Cubic Unit Cell Edge Length 4 05 10 8 Cm Density 2 7 G Cm 3 NatuDefine Following Terms Primitive Unit Cells Ii Schottky Defect Iii FerromagnetismFormula Compound Element Forms Ccp Lattice Atoms X Occupy 2 3rd Octahedral Voids Element X Molar Mass 60 G Mol 1 Density 6 23 G Cm 3 Identify Type Cubic Unit Cell Edge Length Unit CRadius Iron Sphere 0 21 Cm Density Iron 7 80 G Cc Calculate Mass B Edge Aluminium Cube 0 18 Cm DensiType Magnetism Shown Substance Magnetic Moments Domains Arranged Direction Element Crystallizes F C C Lattice Cell Edge 250 Pm Calculate Density 300 G Element Contain 2 1024 AGive Example Molecular Solid Ionic Solid Convert Chlorobenzene Biphenyl Ii Propene 1 Iodopropane Iii 2 Bromobutane 2 Ene Write Major ProductsElement Crystallizes B C C Lattice Cell Edge 500pm Density Element 7 5g Cm 3 Many Atoms Present 300 Following Crystalline Solids Also Known Giant Molecules Following Statements Incorrect Regarding Doping Aip Cds Typical Compounds GroupsFollowing Molecules Weakly Attracted Magnetic Field Give Reasons 1 3 3 Acetylation Aniline Reduces Activation Effect Ii Ch3nh2 Basic C6h5nh2 Iii Althoug Element Atomic Mass 93 G Mol 1 Density 11 5 G Cm 3 Edge Length Unit Cell 300 Pm Identify Type Unit Write One Difference Following 3 Lyophobic Sol Lyophilic Sol Ii Solution Colloid Iii Homogeneous CatCalculate Number Unit Cells 8 1 G Aluminium Crystallizes Face Centred Cubic F C C Structure Atomic M Based Nature Intermolecular Forces Classify Following Solids Sodium Sulphate Hydrogen B Happens CdcAnalysis Shows Feo Non Stoichiometric Composition Formula Fe0 95o Give Reason Element X Mass 40 G Mol 1 F C C Structure Unit Cell Edge Length 400 Pm Calculate Density X Number UnNac L Agcl One Shows Frenkel Defect Element Crystallizes Fcc Lattice Cell Edge 300 Pm Density Element 10 8 G Cm 3 Calculate Number AtomsType Stoichiometric Defect Shown Zns Conductivity Silicon Increases Doping Phosphorus Closest Packing AtomsFollowing Unit Cells Nearer Atoms Touching Whenever Two Dimensional Square Packing Layers Kept Way Centres Aligned Three Dimensions CoordinatioFollowing Frenkel DefectMany Unit Cells Present Cubic Shaped Ideal Crystal Nacl Mass 1 0 G Get N Type Semiconductor Impurity Added Silicon Following Number Valence ElectronsAssertion Value Van Der Waal Constant Larger Kr Reason Size Kr Assertion Gases Settle Bottom Container Reason Gases High Kinetic Energy Number Atoms Per Unit Cell Body Centered Cube 1 B 2 C 4 6Classify Following Solids Different Types Silver B P4 C Diamond NaclUnit Cell Metal Edge Length 288 Pm Density 7 86 G Cm 3 Determine Type Crystal Lattice Atomic Mass MeRatio Octahedral Holes Number Anions Hexagonal Closed Packed Structure Define Anisotropy Distinguish Crystalline Solids Amorphous Solids Read Passage Given Answer Following Questions Solids Formed Many Small Crystals However Process CrysIonic Crystal Lattice R R Radius Ratio 0 320 Co Ordination Number 3 B 4 C 6 8Ionic Solids Hard Brittle Explain Density Iron Crystal 8 54 Gram Cm 3 Edge Length Unit Cell 2 8 Atomic Mass 56 Gram Mol 1 Find Number Ionic Compound Crystallises Fcc Type Structure Ions Centre Face B Ions Occupying Corners Cube FormulCalculate Percentage Efficiency Packing Case Simple Cubic Cell Answer Three Following Silver Crystallises Fcc Structure Density Silver 10 51 G Cm 3 Calculate VolumIodine Exists Polar Molecular Solid B Ionic Solid C Non Polar Molecular Solid Hydrogen Bonded Molecu Element Atomic Weight 93 G Mol 1 Density 11 5 G Cm 3 Edge Length Unit Cell 300 Pm Identify Type Uni B Calculate Radius Copper Atom Atomic Weight Copper 63 55 G Mol 1 Crystallises Face Centered Cubic Alloy Gold Au Cadmium Cd Crystallises Cubic Structure Gold Atoms Occupy Corners Cadmium Atoms Fit FaFollowing Statements True Amorphous Solids Following Covalent Solid Crystal System B C 90oCompound Formed Elements X Crystallizes Cubic Lattice X Atoms Corners Atoms Body Center Two Face CenPercentage Free Space Cubic Close Packed Structure Metal Crystallizes Two Forms E Fcc Bcc Edge Lengths Two Forms 4 9 Ao 4 Ao Respectively Ratio DensitiFollowing Defect Density Crystal Reamins Nickel Oxide Formula Ni0 98o1 00 Fraction Nickel Exists Ni 2 Small Amount Indium Added Germanium BecomesFollowing Example Ferrimagnetism Comparing Ratios Find Whether Following Pair Linear Equations Consistent Inconsistent Given Figure Pr Pq Ps Bisects Qpr Prove Psr Psq Correct Representation Charles Law GivenMolar Volume Co2 Maximum Arya Abhimanyu Vivek Shared Lunch Arya Brought Two Sandwiches One Made Vegetable One Jam Two Boys FoCorrect Order Effusion Gases Vessel Oxygen Filled Ntp Weighed Evacuated Filled So2 Temperature Pressure Weighed Weight Oxygen WouDrop Liquid Acquires Spherical ShapeAbc Triangle Locate Point Interior Abc Equidistant Vertices Abc Value R Stp Conditions 0 C 1 Atm PressureGas Shows Ideal Behaviour Ethane Hydrogen Mixed Equal Proportions Mass Empty Container 27 0c Ratio Total Pressure Exerted HydrWorking Pressure Cooker BasedCritical Temperatures O2 H2 Nh3 Co2 154 3 K 33 2 K 405 5 K 304 10 K Respectively Gas Liquefy First STriangle Locate Point Interior Equidistant Sides Triangle Property Responsible Rise Water PlantsAmorphous SolidsGiven Linear Equation 2x 3y 8 0 Write Another Linear Equations Two Variables Geometrical RepresentatFind Perimeter Rectangle Whose Length 40 Cm Diagonal 41 Cm One Mole Van Der Waals Gas Pv Vs 1 V Plot Shown Value Van Der Waals ConstantMany Elements P Pollution Check Car Exhaust Found Emits 56 G 44 G 16 G N2 Co2 Ch4 Respectively Vapour Pressure ExhauWrite Following Intervals Set Builder Form 3 0 Ii 6 12 Iii 6 12 Iv 23 5 Discuss General Characteristics Gradation Properties Alkaline Earth Metals Experiment Rahul Discovered Gas Behaved Ideally Collected 5 L Gas One Container 10 L Another ContainLet Us Assume Galaxy Consists 2 5 1011 Stars One Solar Mass Long Star Distance 50 000 Ly Galactic CeMolar Volume Volume Occupied 1 Mol Ideal Gas Standard Temperature Pressure Stp 1 Atmospheric PressurFigure 13 8 Shows Plot Pv Versus Pfor 1 00 10 3 Kg Oxygen Gas Two Different Temperatures Dotted Plot1 G 1 4 G Hydrogen Nitrogen Respectively Mixed Evacuated Vessel Volume 5 Dm3 Maintained 400 K ResultAir Bubble Volume 1 0 Cm3 Rises Bottom Lake 40 Deep Temperature 12 C Volume Grow Reaches Surface TemEstimate Total Number Air Molecules Inclusive Oxygen Nitrogen Water Vapour Constituents Room CapacitEstimate Average Thermal Energy Helium Atom Room Temperature 27 C Ii Temperature Surface Sun 6000 K Three Vessels Equal Capacity Gases Temperature Pressure First Vessel Contains Neon Monatomic Second Temperature Root Mean Square Speed Atom Argon Gas Cylinder Equal Rms Speed Helium Gas Atom 20 C AtomEstimate Mean Free Path Collision Frequency Nitrogen Molecule Cylinder Containing Nitrogen 2 0 Atm TMetre Long Narrow Bore Held Horizontally Closed One End Contains 76 Cm Long Mercury Thread Traps 15 Certain Apparatus Diffusion Rate Hydrogen Average Value 28 7 Cm3 1 Diffusion Another Gas Conditions Gas Equilibrium Uniform Density Pressure Throughout Volume Strictly True External Influences Gas ColGiven Densities Solids Liquids Give Rough Estimates Size Atoms Substance Atomic Mass U Density 103 KRectangular Container Movable Piston Capacity V Litres Filled 12 G Gas Degree Temperature Container Following Examples Represent Nearly Simple Harmonic Motion Represent Periodic Simple Harmonic MotionFigure 14 27 Depicts Four X Plots Linear Motion Particle Plots Represent Periodic Motion Period MotiFollowing Functions Time Represent Simple Harmonic B Periodic Simple Harmonic C Non Periodic Motion Particle Linear Simple Harmonic Motion Two Points B 10 Cm Apart Take Direction B Positive Direction Following Relationships Acceleration Displacement X Particle Involve Simple Harmonic Motion 0 7x B 2Flask Capacity 2 5 Litres Contains One Mole Gas 300 K Using Van Der Waal Equation Evaluate Pressure Spring Balance Scale Reads 0 50 Kg Length Scale 20 Cm Body Suspended Balance Displaced Released OsciSpring Spring Constant 1200 N 1 Mounted Horizontal Table Shown Fig Mass 3 Kg Attached Free End SprinIsolated Vessel Weighs 138 2410 G Moist Conditions 40 1305 G Dry Conditions Maintained 298 K 0 9970gBottle Long Neck Filled Oxygen Gas Conditions 1 Atmosphere Pressure 298 K Temperature Diameter OxygePlot Corresponding Reference Circle Following Simple Harmonic Motions Indicate Initial 0 Position PaFigure 14 30 Shows Spring Force Constant K Clamped Rigidly One End Mass Attached Free End Force F ApPiston Cylinder Head Locomotive Stroke Twice Amplitude 1 0 Piston Moves Simple Harmonic Motion AngulAcceleration Due Gravity Surface Moon 1 7 Ms 2 Time Period Simple Pendulum Surface Moon Time Period Vessel Containing High Pressure Gas Like Hydrogen Tends Rise Gas Allowed Escape Orifice Bottom VessMatch Following Column Column Ii High Temperature P B Extremely Low Pressure Q C High Pressure R Z 1Cylindrical Piece Cork Density Base Area Height H Floats Liquid Density Cork Depressed Slightly ReleOne End U Tube Containing Mercury Connected Suction Pump End Atmosphere Small Pressure Difference MaMolar Volume Substance Compressibility Factor Z Vapours Show Particle Linear Shm Average Kinetic Energy Period Oscillation Equals Average Potential Energy PTemperature Gas Molecule Root Mean Square Speed Gas Molecule Ratio Probable Speed Root Mean Square Speed 0 82 Calculate Probable Speed Molecules Temperature String Mass 2 50 Kg Tension 200 N Length Stretched String 20 0 Transverse Jerk Struck One End StringStone Dropped Top Tower Height 300 High Splashes Water Pond Near Base Tower Splash Heard Top Given SSteel Wire Length 12 0 Mass 2 10 Kg Tension Wire Speed Transverse Wave Wire Equals Speed Sound Dry AUse Formulato Explain Speed Sound Air Independent Pressure B Increases Temperature C Increases HumidLearnt Travelling Wave One Dimension Represented Function F X X Must Appear Combination X V X V E F Evacuated Vessel Movable Piston External Pressure 1atm 0 1mol 1 0 Mol Unknown Compound Vapour PressuDefine Following Related Proteins Peptide Linkage Ii Primary Structure Iii Denaturation Discuss Effect Pressure Temperature Adsorption Gases Solids Value Van Der Waal Constant Gases O2 N2 Nh3 Ch4 1 360 1 390 4 170 2 253 L2 Atm Mol 2 Respectively GaAerosols Deplete Ozone Write Minors Cofactors Elements Following Determinants Ii Various Constituents Domestic Sewage Discuss Effects Sewage Discharge River X Connected Parametrically Equation Without Eliminating Parameter Find Social Divisions Based Peculiar India Organisations Undertake Activities Promote Interests Specific Social Sections Workers Employees TeacAbsolute Temperature Gas Doubled Pressure Halved Volume Change Temperature K Given Volume Gas 0 C Become One Third Volume Constant Pressure Term Represents Attractive Forces Present Real Gas Van Der Waals Equation State Two Moles Ideal Gas Put Series Changes Shown Following Graph Pressures Stages X Z 25 30 Speaker Poem Environmental Management Complex Huge Task Society Suppose Consumer Preferences Monotonic Say Preference Ranking Bundles 10 10 10 9 9 9 P Pressure V Volume Following Statement Incorrect Regarding Gas Laws Think Advice Council Ministers Binding President Give Answer 100 Words Initial Temperature 25oc Desired Increase Volume 90 Cm3 Gas 25 Without Changing Pressure TemperatureSaid Much Political Interference Working Administrative Machinery Suggested Autonomous Agencies AnswWrite Essay Two Hundred Words Proposal Elected Administration Instead Appointed Administration Alok Thinks Country Needs Efficient Government Looks Welfare People Simply Elected Prime Minister MiClass Debating Merits Bicameral System Following Points Made Discussion Read Arguments Say Agree DisEnumerate Steps Ascertain Correct Cash Book Balance Bank Reconciliation Statement Prepared Explain Reasons Balance Shown Bank Passbook Agree Balance Shown Bank Column Cash Book Distinguish Generalisation Discrimination Following Statements Correct Dalton Law Partial Pressures Value Van Der Waals Constant B Nitric Oxide 0 02788 L Mol Molecule Considered Spherical Diameter Absolute Temperature Gas Doubled Pressure Halved Volume Change Following Conditions Ideal Gas Law Give Best Results Temperature K Given Volume Gas 0 C Become One Third Volume Constant Pressure Term Represents Attractive Forces Present Real Gas Van Der Waals Equation State Two Moles Ideal Gas Put Series Changes Shown Following Graph Pressures Stages X ZCalculate Debt Equity Ratio Following Information Rs Total Assets 15 00 000 Current Liabilities 6 00Volume Occupied Co2 Negligible Pressure Exerted 1 Mol Co2 Gas 273 K Given Van Der Waals Constant 3 5Meaning Line Wry Laboured Ease Loss Traditions Rituals Funerary Practices Must Respected Two Balloons P Q Volume Balloon P Filled 5 G N2 Gas 423 K Balloon Q Filled 8 G Co2 Gas 323 K Based IN2 Gas Leaked Vessel 9 L Tiny Hole 10 Minutes 30 C Unknown Gas Also Leaked Vessel Conditions Gases DSystem Set Connecting Two Balloons B Stop Cock 30oc Volume Balloon Contained 20 G N2 Gas 0 6 Atm StoGaseous Mixture Ch4 N2 H2o Vapour Enclosed Vessel Volume 0 821 Ml Pressure Exerted Mixture 1 95 Mm HVan Der Waals Equation Fails Determine Correct Behavior Real Gases Conditions Vessel 1 L Capacity Contains 5 Moles Co2 25 C Gas Follows Real Gas Behavior Temperature Pressure Gas1 Mole Gas 120 Atm Pressure 2oc Compressibility Factor 0 07 Value Van Der Waal Constant Assuming VolTwo Different Real Gases Present Containers B Gas Container Z 1 1 Atm Whereas Gas Container B Z 1 1 Liquefaction Real Gas Depends Upon Critical Temperature Critical Pressure Gas Critical Compression FValue Van Der Waal Constant Terms Critical Parameters 1 Mole Co2 Gas Given Tc 31 25oc Pc 72 9 Atm VcMixture Co2 0 44 G Ethane 0 3 G Nitrogen 0 28 G Gases Confined Vessel 5 L Capacity 32oc Assuming GasMixture Subjected Separation Collecting Lighter Gas Water Pressure Exerted Collected Gas Passing WatMatch Column Column Ii Column Column Ii High Pressure P Volume Decreases B Low Pressure Q Volume IncMatch List List Ii List List Ii P Low Molecular Weight Gas 1 Z 1 Q Low Pressure Gas 2 V 22 4 L R Z PMatch Column Column Ii Column Column Ii Root Mean Square Speed P 8a27rb B Critical Temperature Q ArbMixture Containing Co C2h6 Ne Gases Exploded Electric Discharge 310 K Presence Excess Oxygen Volume 10 G Sample Mixture Na2co3 K2co3 Treated Excess Hcl Solution Evolves 2 29 L Co2 30oc 0 987 Bar Press5 L Vessel Contains N2 Gas Water 32oc 0 987 Atm Pressure Container Subjected Electrolysis Water Spli2 L Vessel Contains 3 1022 Molecules H2 8000 Nm 2 Pressure Ratio Root Mean Square Speed Probable SpeAnalysis Dry Air Showed Contains 70 N2 21 O2 Volume Density Dry Air 2968 K 1 Atm 0 905 G L Density MHelium Filled Balloon Weighs 150 Kg Diameter 30 Pressure Helium Gas Balloon 1 5 Atm 298 K Pay Load BAmong H2 Ch4 O2 Gas Easily LiquefiedUse Following Information Answer Next Question Stp 1 Gram Co H2o Ch4 Gases Taken Gas Occupies GreateCorrect Order Increasing Surface Tension Among Water Alcohol HexaneDensity Ammonia Nh3 20 C 7 Bar PressureUse Following Information Answer Next Question Spherical Balloon Diameter 19 Cm Filled P Hydrogen CyCase Ideal Gas Plot Linear Slope Equal ZeroFinal Temperature Flask Containing 350 Mg Air 27 C Heated Till 35 Air Mass ExpelledUse Following Information Answer Next Question Certain Amount Gas Pressure 0 82 Bar Occupies Volume Initial Temperature 25 C Desired Increase Volume 90 Cm3 Gas 25 Without Changing Pressure TemperatureOne Mole Van Der Waals Gas Pv Vs 1 V Plot Shown Value Van Der Waals ConstantFollowing Statements Justify Use Boyle Law Arrangement Three States Matter Increasing Order Void Size Two Moles Ideal Gas Put Series Changes Shown Following Graph Pressures Stages X Z1 7 G Nh3 Gas Kept Vessel Capacity 1 Litre 0 C Pressure Exerted Gas Following Graphs Correctly Represents Charle Law Critical Volume Critical Pressure Co Gas 0 09 M3 0 1 Atm Respectively Calculate Value Critical TempeRatio Average Speed Probable Speed Two Gases 1 13 Ratio Molar Mass Rt 2 Calculate Probable Speed Calculate Van Der Waals Constant Low Pressure Condition Gas Pressure 15 Bar Volume 550 Ml Spherical Balloon Diameter 1 2 Filled Gas Ntp Sea Level Diameter Raised Altitude Pressure 8 10 3 AtmUse Following Information Answer Next Question Using Boyle Law Pressure Volume Given Mass Gas Known Use Following Information Answer Next Question Density Gas 27 C 760 Mm Pressure 2 4 Kg M3 Constant PExperimental Set Two Bulbs P Q Equal Size Containing 1 8 Moles Nitrogen Gas 0 8 Atm 30oc Connected SDensity Gas 27 C 760 Mm Pressure 2 4 Kg M3 Constant Pressure Density Gas Becomes 1 8 Kg M3 Certain TFollowing Statements Correct Dalton Law Partial Pressures Certain Amount N2 Gas Vessel Withdrawn Maintain Initial Pressure 1 Atm Vessel Heated 472 C InitiallyBulb Volume 2 Liters Contains 3 1 X 1022 Molecules Oxygen Gas Pressure 8 7 X 102 Nm2 Root Mean SquarGas X2 Allowed Escape Small Hole Another Gas O2 Also Allowed Pass Different Hole 5 Minutes Total 20 Following Statements Correct Regarding Maxwell Distribution Curve Molecular Velocity Dipole Dipole Forces Present Pair Molecules Use Following Information Answer Next Question Sodium Reacts Water Form Sodium Hydroxide Hydrogen GaPressure Exerted 2 Moles Real Gas Contained 1 L Vessel 77 C Gas 3 0 Atm L2 Mol 2 B 0 05 L Mol 1 Van Der Waals Equation Reduces Ideal Gas EquationUse Following Information Answer Next Question 2 5 L Gas 9 105 Nm 2 Pressure Allowed Contract Till PVolume Gas 273 15 C 2 5 Bar 150 Dm3 Pressure Volume Gas 0 C Many Grams Nitrogen Present 16 42 L Sample Gas 3 5 Atm 23 C Density Gaseous Sample 581 7 Nm 2 47 C 7 Gm 3 Gas LikelyVolume Temperature Graph Represents Behaviour 1 Mol Ideal Gas Constant Pressure Following Graphs Correct Graphical Representation Boyle Law Use Following Information Answer Next Question Flask Volume 640 Ml Contains Air Heated 47 C Sealed FGay Lussac Law ApplicableGives Rise Surface Tension Liquids Understand Viscosity Liquid Effect Increase Temperature Surface TKinds Intermolecular Forces Attraction Act Polar Non Polar Molecules Explain Diazonium Salts Show PoAgriculture Veld Important Prairies Local Warm Wind Prairies Bora B Chinook C Mistral SiroccoAnswer Following Questions Calculate Number Molecules 3 65 G Hcl Gas B Chemical Formula Compound XclNitrogen Gas Oxygen Gas React Illustrated Given Equation N2 G O2 G 2no G Calculate Volume Reacting G State Gay Lussac Law Ii State Avogadro Law Iii Numerical Value Avogadro Number Iv Define Atomicity Gas Occupies Volume 300 Cm3 Temperature 27 C Pressure 760 Mm Volume Gas 4 C 760 Mm Pressure Gas Occupies Volume 612 Cm3 P Find Volume Temperature 27 C Pressure 720 Mm Mercury State Dalton Law Partial Pressure B Mixture Gases Molecular Mass 23 B Molecular Mass 30 Contains 90Define Ideal Gas Derive Equation Define Boyle Law B Gas Occupies Volume 900 Cm3 Pressure P Pressure Altered 3 5 Atms Volume Gas FounVacuum Tube 1 4l Capacity Containing Argon Gas Evacuated Till Pressure Inside 10 4atm Temperature MaReduced Pressure Temperature Gas Found 0 125 2 3 Respectively Reduced Volume Gas Critical Volume Gas0 07 Mol Unknown Gas Allowed Compress Vessel Capacity 450 Ml Calculate Critical Pressure Initially PFollowing Statements True Regarding Ideal Gases Rational Number Multiplicative Inverse Explain Find Multiplicative Inverse Additive Inverse Main Function Abscisic Acid Plants Increase Length Cells B Promote Cell Division C Inhibit Growth PrFollowing Statements Solid Liquid Gaseous Molecules False Use Following Information Answer Next Question Gases Molecules Force Attraction Molecules Form RegulDipole Dipole Forces Present Pair Molecules Use Following Information Answer Next Question Flask Volume 640 Ml Contains Air Heated 47 C Sealed FVolume Temperature Graph Represents Behaviour 1 Mol Ideal Gas Constant Pressure Pressure Exerted 2 Moles Real Gas Contained 1 L Vessel 77 C Gas 3 0 Atm L2 Mol 2 B 0 05 L Mol 1 Van Der Waals Equation Reduces Ideal Gas EquationGiven Pair Molecules Dipole Dipole Force Present Hcl Hcl Cl2 Cl2Critical Temperatures Tc Four Gases E F G H 329 5 K 412 7 K 315 6 K 435 8 K Respectively Gas LiquefyCritical Volumes Four Gases P Q R 25 Ml 75 Ml 125 Ml 175 Ml Respectively Gas Least Value Van Der WaaWrite Expression Charles Law Volumes Gas 0 C C V0 Vtrespectively Pressure V Ml Dry Gas Increased 1 Atm 4 Atm Constant Temperature Final Volume Gas Four Flasks Ii Iii Iv Volumes 1000 Ml 500 Ml 250 Ml 125 Ml Respectively Contain Equal Number MoleculType Curve Obtained Pv Gas Plotted P Constant Temperature Given Z Values Three Gases B C Gases Compressible Expected Ideal Behaviour Gas Z 0 6 B 1 2 C 1 8Compressibility Factor H2 Gas Always Greater 1 Liquid Heated Three Temperatures Viz 310 K 640 K 230 K Case Liquid Highest Vapour Pressure Diethyl Ether Ethyl Alcohol Compound Lower Vapour Pressure Van Der Waals Constant Three Gases B C Given Table Gas Cannot Liquefied Gas 1 2 B 0 C 0 6Balloon Contains 22 4 L Gas 173 K 1 Atm Pressure Would Volume Pressure Gas Doubled Glycerine Water Viscous Temperature Volume Gas Get Tripled Temperature 273 K Pressure Constant 5 0 L Gas 9 5 105 Nm 2 Pressure Allowed Contract Constant Temperature Till Pressure Becomes 2 0 106 Give Explanation Given Observation Plastic Water Bottle Containing 25 Water Kept Refrigerator Lid ClSample Hydrogen Gas Collected Water 20 C Vapour Pressure Water Vapour 20 C 2 3 Kpa Pressure ResultanPhenomenon Explains Rise Ink Blotting Paper Magnitudes Coefficient Viscosity Liquid 20 C 35 C 19 C 63 C P Q R Respectively Correct Order IncreasChamber Pressure 500 Atm 1000 K Temperature Chamber Pressure Raised 800 Atm London Forces Induced Two Atoms Oxygen State Avogadro Law Derive Relationship Molar Mass Density Gas Dispersion Forces Act Short Distances Volume Gas 273 15 C 2 5 Bar 300 Dm3 Pressure Volume Gas 0 C Change Size Bubbles Observed Reach Bottom Beaker Surface Explain Plot Graph Variation Pressure Gas Temperatures 320 K 630 K 210 K Pressure Exerted 4 6 G N2 Gas Stp Gas Kept Litre Flask 0 C Two Containers B Gas Pressure Volume Temperature Container Twice Container B Ratio Number Molecules Pressure Exerted 4 Moles Real Gas Contained 2 L Vessel 27 C Gas 4 0 Atm L2 Mol 2 B 0 06 L Mol 1 Value Gas Constant Stp One Mole Gas Flask Volume 840 Ml Contains Air Heated 50 C Sealed Flask Cooled 30 C Immersed Water Opened AssumingShow Graphically Variation Pressure Gas Temperature Constant Volume Giving Reasons Arrange Co2 25 C 1 Atm Co2 0 C 2 Atm Co2 50 C 1 Atm Order Increasing Densities Density Ch4 50 C 105 Pa Pressure 1 6 Kg 3 Compressibility Factor Methane Density Gaseous Sample X 855 1 Nm 2 47 C 9 Gm 3 Gas X Mass Obtained Liquefaction 112 Ml O2 P Gas Occupies Volume 5 0 L Must Additional Pressure Applied Decrease Volume Gas 3 5 L Keeping TemperaPressure Volume Conditions Real Gas Show Ideal Behaviour Gases Compressible Large Extent Collision Particles Results Pressure Gas Change Speed Particles Gas Explain Effect Surface Tension Liquid Increase Temperature Describe Type Forces Exist Given Molecules Cl2 Molecule H2 Cl2 MoleculesP Gaseous Sample Occupies Volume 78 Dm3 Volume Pressure 0 5 Atm Temperature 47 C Ideal Gases Give Best Results Low Pressure Temperature Relatively Higher Compared Boiling Point Gas Give Relation Calculate Amount Dry Gas Collected Water Volume Moist Gas Many Grams Nitrogen Present 16 42 L Sample Gas Pressure 2 5 Atm Temperature 23 C Nitrogen Gas Filled Air Bag Produced According Given Reaction 6nan3 Fe2o3 3na2o 2fe 9n2 G Amount SodVolume Co2 Present Inside Vessel 3 5 L Partial Pressure 2 10 3 Atm Total Pressure Inside Vessel 1 3 Closed Container Contains Mixture Three Non Reacting Gases Hydrogen Oxygen Methane Total Pressure MiDerive Ideal Gas Equation Combining Gas Laws Given Figure Represents System Kept Constant Temperature Total Pressure Inside System Stopcock OpeneBoyle Point Gas 62 C Effect Gas Temperature Lowered 62 C Increased 62 C Postulates Kinetic Theory Gases Hold Good Real Gases Based Given Graph Answer Following Questions Gas Es Compressible Low Pressure Less Compressible HighConditions Van Der Waals Equation Reduces Ideal Gas Equation Consider Graph Given H2 O2 Gases Behaviour Gases H2 O2 Ascertained Graph Three Flasks Containing Equal Volumes Equal Number Gaseous Molecules Kept Temperatures 415 K 273 K 3Effect Volume Gas Obeying Boyle Law Pressure Increased Four Times Initial Pressure Write Expression Charles Law Volumes Gas 0 C C V0 Vtrespectively Type Graph Obtained Pressure Fixed Amount Gas Plotted Temperature Constant Volume Effect Gas Compressed Cooled Critical Temperature Liquid Temperature 30 C Would Effect Surface Tension Liquid Temperature Increased 80 C Explain Factors Strength Dispersion Forces Depend Using Boyle Law Derive Relation Density Pressure Gas Derive Ideal Gas Equation Using Gas Laws Write Expression Boyle Law Plot Isotherm Gas Kept Temperatures 400 K 600 K Graph Temperature K Given Volume Gas 0 C Become One Third Volume Constant Pressure Explain One Experiences Headache Uneasiness High Altitudes Gas Easily Liquefied Gas Larger Effective Size Alcohol Water Higher Vapour Pressure Particular Temperature Give Reasons Effect Temperature Vapour Pressure Liquids Listed Part Clothes Washed Efficiently Hot Water Cold Water Falling Drops Spherical Shape Calculate Mass 240 Ml N2 300 C 105 Nm 2 Pressure Value Gas Constant R Volume Measured Dm3 Pressure Bar Two Postulates Kinetic Theory Gases Hold True Real Gases Postulates Corrected Van Der Waals Equation Meant Polarizability Atom Explain Effect Polarizability London Force Meant Term Aqueous Tension Pressure Exerted Dry Gas Mixture Oxygen Gas 5 54 Bar Pressure Nitrogen Gas 2 27 Bar Pressure 291 K GDefine Viscosity Coefficient Given Melting Points Two Substances B Substance P K 1272 B 719 Substance Higher Viscosity Ii Would EState Dalton Law Partial Pressures Derive Relation Partial Pressure Mole Fraction Gas Total Pressure Exerted Mixture 3 2 G Methane 4 4 G Co2 Contained 9 Dm3 Flask 27 C Alternative Atmospheric Pressure Correctly Converted Pascal 1 2l Ne 0 47 Bar 0 7l Ar 0 96 Bar 1 5l 0 55bar Introduced 4 5l Vessel Calculate Total Pressure GasesFollowing Assumptions Kinetic Theory Gases Incorrect Following Statements True Regarding Compressibility Factor Z Following Statements Correct Dalton Law Partial Pressures Certain Amount N2 Gas Vessel Withdrawn Maintain Initial Pressure 1 Atm Vessel Heated 472 C InitiallyBulb Volume 2 Liters Contains 3 1 X 1022 Molecules Oxygen Gas Pressure 8 7 X 102 Nm2 Root Mean SquarNormal Boiling Point Liquid Temperature Vapour Pressure Liquid Becomes Equal One Atmospheric TemperaUse Following Information Answer Question Temperature Vapour Pressure Liquid Equal External PressureFollowing Statements Correct Use Following Information Answer Next Question Open Container Contains 100 Mg Air 30 C Mass Air RemaP Gaseous Sample Occupies Volume 81 9 Dm3 Volume 0 5 Atm 27 C Use Following Information Answer Next Question Gaseous Mixture Oxygen Nitrogen Contains 44 8 G Oxyge2 L Vessel Contains 6 G Hydrogen 28 G Nitrogen Gas 300 K Total Pressure Gaseous Mixture Vessel Use Following Information Answer Question 2 Moles Gaseous Sample Occupy Volume 2 L 1296 36 Nm 2 27 CCritical Temperatures Tc Four Gases P Q R 329 5 K 412 7 K 315 6 K 435 8 K Respectively Gas Liquefy FUse Following Information Answer Question Critical Volumes Four Gases E F G H 75 Ml 25 Ml 125 Ml 175Increase Temperature Surface Tension LiquidUse Following Information Answer Next Question Magnitudes Coefficient Viscosity Liquid 25 C 42 C 19 Use Following Information Answer Question Given Representation Shows Density Order Three States MattUse Following Information Answer Question Height Himachal Pradesh Assam Kashmir Delhi Sea Level 6 81Use Following Information Answer Question Normal Boiling Point Boiling Point Pressure Standard BoiliUse Following Information Answer Question Due Liquids Tend Minimize Ii Also Causes Iii Following RowUse Following Information Answer Question Quantity Unit X Force Nsm 2 Surface Energy Ii Jm 2 Z ViscoClosed Container Unknown Volume V Contains Gas 101 325 Bar Pressure Container Connected Another EvacNormal Boiling Point Liquid Temperature Vapour Pressure Liquid Becomes Equal One Atmospheric TemperaValue Temperature Kelvin 100 G So2 Contained 4890 Ml Container 0 58 Bar Pressure So2 Van Der Waals CUse Following Information Answer Question Performing Experiment Student Obtained Given Graph FollowiUse Following Information Answer Question Rakesh Two Cylinders Volume First Cylinder 150 00 Ml ContaUse Following Information Answer Question Volume 400 00 Ml Gas Needs Increased 40 00 Keeping PressurAir Pressure Inside Cylindrical Vessel Increased 2 46 Bar 2 72 Bar 0 C Would Final Temperature InsidUse Following Information Answer Question Three Cylinders X Z Contain Co2 O2 H2 Gases Respectively TFollowing Statements Incorrect Use Following Information Answer Question 5 Moles Ideal Gas Occupies 40 L Certain Conditions 2 MolesMixture Contains 9 2 G Methane 5 4 G So2 Mixed 15 Dm3 Flask 37 C Pressure Mixture Following Laws Indicated Given Graph Use Following Information Answer Question Student Two Cylinders Volume First Cylinder 100 00 Ml ContUse Following Information Answer Question Volume 200 00 Ml Gas Needs Increased 20 00 Keeping PressurAir Pressure Inside Cylindrical Vessel Increased 2 40 Bar 2 70 Bar 0 C Would Final Temperature InsidUse Following Information Answer Question Three Cylinders X Z Contain O2 N2 H2 Gases Respectively ThFollowing Statements Correct Use Following Information Answer Question 5 Mole Ideal Gas Occupies 50 L Certain Conditions 3 Moles Gaseous Mixture Nitrogen Hydrogen Contains 11 2 G Nitrogen 56 0 G Hydrogen Pressure Mixture 500 Nm 2Kinds Forces Exist Iodine Molecule Write Expression Van Der Waals Equation State Significance Unit Vander Waals Constants Boyle Law Write Units Van Der Waals Constant B Mean Vapour Pressure Liquid Drops Spherical Shape Understand Compressibility Factor Z Br2 Icl Molecular Mass Number Atoms Br2 Liquid State Icl Solid State Show Graphical Representation Charles Law Gas Deviate Ideal Gas Behaviour Temporary Dipole Developed Atom Would New Volume Gas Pressure Gas Doubled Define Combined Gas Law Balloon Filled 4 L Helium 47 C Balloon Placed Oven Temperature Reaches 60 C New Volume Balloon Explain Dalton Law Partial Pressure Write Six Postulates Kinetic Theory Gases Define Critical Temperature Ii Critical Pressure Iii Critical Volume Reason Behind Existence Surface Tension Liquids Ii State Gay Lussac Law Complete Table Gas Laws Relation Charles Law Gay Lussac Law Boyle Law Avogadro Law Ii Iodine Molecu Food Cook Faster Speed Plain Area Hilly Area Explain Reason Ii Volume Hair Reduced Getting Drenched1 Mole Ideal Gas Constant Temperature Plot Log P Log V P Pressure V Volume V U Represent Probable Velocity Average Velocity Root Mean Square Velocity Respectively Gas ParticulRelationship Among Probable Velocity Average Velocity Root Mean Square Velocity RespectivelyIntermolecular Interaction Dependent Inverse Cube Distance Molecules Intermolecular Force Attraction Methane MoleculesLondon Forces Present SpeciesFollowing Species Form Hydrogen Bond WaterFollowing Species Form Hydrogen Bond Among Following Statements Wrong Temperature Real Gas Behaves Ideal Gas Appreciable Range Pressure Called Following Statements True Gay Lussac Law Combining Volumes Select Statement Accordance Boyle Law Following Statements Incorrect Graph Given Temperature Called Absolute Zero Centigrade Temperature Volume Gas 10oc Doubles Keeping Pressure Constant Volume Stp Occupied 512 Ml Gas Maintained 293k 74 Cm Hg Pressure1 Atm Pressure Equals Mountaineers Carry Oxygen Cylinders Explained Two Containers Filled Helium Gas Volumes 2 Dm3 0 5 Dm3 Respectively Pressure First Container 280 Mm Among Following Formulae Best Describes Aqueous Tension Two Litres Gas 20oc Heated Till Pressure Volume Doubled New Temperature Effect Root Mean Square Speed Gas Temperature Gas Raised 27oc 627oc Spherical Shape Rain Drops Due Phenomenon Following Car Parking Charges Near Railway Station 4 Hours Rs 60 8 Hours Rs 100 12 Hours Rs 140 24 HMixture Paint Prepared Mixing 1 Part Red Pigments 8 Parts Base Following Table Find Parts Base Need Question 2 1 Part Red Pigment Requires 75 Ml Base Much Red Pigment Mix 1800 Ml Base Machine Soft Drink Factory Fills 840 Bottles Six Hours Many Bottles Fill Five Hours Photograph Bacteria Enlarged 50 000 Times Attains Length 5 Cm Actual Length Bacteria Photograph EnlaModel Ship Mast 9 Cm High Mast Actual Ship 12 High Length Ship 28 Long Model Ship Suppose 2 Kg Sugar Contains 9 106 Crystals Many Sugar Crystals 5 Kg Sugar Ii 1 2 Kg Sugar Rashmi Road Map Scale 1 Cm Representing 18 Km Drives Road 72 Km Would Distance Covered Map 5 60 Cm High Vertical Pole Casts Shadow 3 20 Cm Long Find Time Length Shadow Cast Another Pole 10 50Loaded Truck Travels 14 Km 25 Minutes Speed Remains Far Travel 5 Hours Following Inverse Proportion Number Workers Job Time Complete Job Ii Time Taken Journey Distance TraTelevision Game Show Prize Money Rs 1 00 000 Divided Equally Amongst Winners Complete Following TablFactory Required 42 Machines Produce Given Number Articles 63 Days Many Machines Would Required ProdCar Takes 2 Hours Reach Destination Travelling Speed 60 Km H Long Take Car Travels Speed 80 Km H Two Persons Could Fit New Windows House 3 Days One Persons Fell Ill Work Started Long Would Job TakeSchool 8 Periods Day 45 Minutes Duration Long Would Period School 9 Periods Day Assuming Number Scho