Correct option is (D) Concave, convex Myopia means short-sightedness, in this, a person has a clear vision when looking at objects close to them, but distant objects will appear blurred, to correct...
Correct option is (A) active mass The rate at which a substance reacts depends upon its active mass as the rate of reaction is directly proportional to concentration of each reactant and product.
Correct option is (A) $\mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{7}$ Oleum, or fuming sulfuric acid, is a solution of various compositions of sulfur trioxide in sulfuric acid, or sometimes more...
and 
, internal resistance 
and 
, connected in parallel. The equivalent emf of the combination is- (A) 
(B) 
(C) 
(D) 
Answer: Option (B)$\frac{\varepsilon_{1} r_{2}+\varepsilon_{2} r_{1}}{r_{1}+r_{2}}$
Solution: Option 3 is correct Sodium hydroxide (NaOH) reacts with acetic acid (CH3COOH) to form sodium acetate (CH3COONa) and water (H2O). The equation for this reaction can...
$ 1.\,\,7.22\,c{{m}^{3}} $ $ 2.\,\,1.38\times {{10}^{-1}}\,c{{m}^{3}} $ $ 3.\,\,2.38\times {{10}^{-2}}\,c{{m}^{3}} $ $ 4.\,\,3.023\times {{10}^{-2}}\,c{{m}^{3}} $ Solution: $ 1.\,\,7.22\,c{{m}^{3}}...
1. 1 → 6 β 2. 1 → 4 β 3. 1 → 6 α 4. 1 → 4 α Solution: 1 → 4 α Maltose is a disaccharide made up of alpha 1,4 glucose molecules linked together. Maltase is the enzyme that breaks it down. It acts on...
1M NaCl and 1M MgCl2 1.5M KCl and 2.5M Urea 1.5M AlCl3 and 2M Na2SO4 1M NaCl and 2M MgCl2 Solution: 1.5M AlCl3 and 2M Na2SO4 The osmotic pressure and molar concentration in isotonic solutions should...
1. 2 - chloropropanal 2. 2 - chloropentane 3. 2 - chloro,2-methylbutane 4. 2 - chlorobutane Solution: 2 - chloro,2-methylbutane 2-chloro-2-methylbutane is optically inactive as no chiral center is...
$ 1.\,\,{{\left[ Ni{{(CN)}_{4}} \right]}^{2-}} $ $ 2.\,\,[Ni{{(CO)}_{4}}] $ $ 3.\,\,{{\left[ CO{{(N{{H}_{3}})}_{6}} \right]}^{3+}} $ $ 4.\,\,{{[NiC{{l}_{4}}]}^{2-}} $ Solution: $...
1. -42.35 kJ 2. -169.4 kJ 3. -50.82 kJ 4. -236.7 kJ Solution: -169.4 kJ Given mass = 0.06 kg = 60 g Molar mass of ethane = 2(12) + 6(1) = 30 g number of moles of ethane in 60 grams = 60/30 = 2 We...
1. Temperature increases 2. Temperature decreases 3. Ea increases 4. Ea decreases Solution: Ea decreases A catalyst is a material that speeds up a reaction without being consumed in the process....
Solution: (i) The point O where the angle bisectors meet is called the incenter of the triangle. (ii) The perpendicular drawn from point O to AB and CA are equal. i.e., OR and OQ. (iii) ∠ACO = ∠BCO....
1. Buna-N 2. Neoprene 3. Butyl rubber 4. Buna-S Solution: Buna-S Styrene-butadiene rubber is another name for the rubber produced (SBR). Bu stands for butadiene, Na for sodium, and S for styrene in...
Solution: Steps to construct: Step 1: Draw a line segment BC = 5cm. Step 2: With Center as B and radius 5cm, with center as C and radius 5cm draw two arcs which intersect each other at point A. Step...
Solution: Steps to construct: Step 1: Mark a point O. Step 2: With center O and radius 4cm and 6cm, draw two concentric circles. Step 3: Join OA and mark its mid-point as M. Step 4: With center M...
2 dm3 3 dm3 4 dm3 8 dm3 Solution: 4 dm3 From ideal gas equation, we have: $ \frac{{{P}_{1}}{{V}_{1}}}{{{T}_{1}}}=\frac{{{P}_{2}}{{V}_{2}}}{{{T}_{2}}} $ $ We\,have\,: $ $...
1. NCl5 is unstable 2. larger size of nitrogen 3. inertness of nitrogen 4. non-availability of vacant d- atomic orbitals Solution: non-availability of vacant d- atomic orbitals By employing the...
1. Reaction with calcium oxide 2. Reaction with sodium 3. Reaction with ammonia 4. Photosynthesis reaction Solution: Photosynthesis reaction The overall balanced equation of photosynthesis reaction...
Solution : (a) Given: O is the center of the circle. To Prove : ∠AOC = 2 (∠ACB + ∠BAC). Proof: In ∆ABC, ∠ACB + ∠BAC + ∠ABC = 180° (Angles of a triangle) ∠ABC = 180o – (∠ACB + ∠BAC)….(i) In the...
1. Gabriel phthalamide synthesis 2. Nef carbonyl synthesis 3. Etard reaction 4. Hoffmann degradation Solution: Nef carbonyl synthesis Nef carbonyl synthesis Acid hydrolysis of a salt of a...
Solution: (a) triangle ABC is an equilateral triangle Each angle = 60o ∠A = 60o But ∠A = ∠D (Angles in the same segment) ∠D = 600 Now ABEC is a cyclic quadrilateral, ∠A = ∠E = 180o 60o + ∠E = 180o...
Solution: Three circles with centers A, B and C touch each other externally at P, Q and R respectively and the radii of these circles are 2 cm, 3 cm and 4 cm. By joining the centers of triangle ABC...
Solution: (a) Let the circle touch the sides BC, CA and AB of the right triangle ABC at points D, E and F respectively, where BC = a, CA = b and AB = c (as showing in the given figure). As the...
Solution: (i) Join OB ∠OBA = 90° (Radius through the point of contact is perpendicular to the tangent) OB2 = OA2 – AB2 r2 = (r + 7.5)2 – 152 r2 = r2 + 56.25 + 15r – 225 15r = 168.75 r = 11.25 Hence,...
Solution: (a) From A, AP and AS are the tangents to the circle ∴AS = AP = 6 From B, BP and BQ are the tangents ∴BQ = BP = 5 From C, CQ and CR are the tangents CR = CQ From D, DS and DR are the...
1. Violet 2. Yellow 3. Black 4. Prussian blue Solution: Prussian blue The reaction equation is: $ FeS{{O}_{4}}+NaOH\,\,\to \,\,Fe{{(OH)}_{2}}+N{{a}_{2}}S{{O}_{4}} $ $ 6NaCN+Fe{{(OH)}_{2}}\,\,\to...
Solution: Radii of the circles are 5 cm and 2.8 cm. i.e. OP = 5 cm and CP = 2.8 cm. (i) When the circles touch externally, then the distance between their centers = OC = 5 + 2.8 = 7.8 cm. (ii) When...
Solution: CT is the radius CP = 13 cm and tangent PT = 12 cm CT is the radius and TP is the tangent CT is perpendicular TP Now in right angled triangle CPT, CP2 = CT2 + PT2 [using Pythagoras axiom]...
Nd La Dy Yb Solution: La We have Z = 57 for La. In +3 oxidation state Lanthanum(La) achieves the noble gas configuration. So, Lanthanum exhibits only +3 oxidation state.
Solution: (a) Construction: Join BC, and AC then ABCD is a cyclic quadrilateral. Now in ∆DCF Ext. ∠2 = x + z and in ∆CBE Ext. ∠1 = x + y Adding (i) and (ii) x + y + x + z = ∠1 + ∠2 2 x + y + z =...
(b) In the figure given below, if ∠ACE = 43° and ∠CAF = 62°, find the values of a, b and c (2007) Solution: (a) In ∆PQR, ∠PRQ = 90° (Angle in a semi-circle) and ∠PQR = 58° ∠RPQ = 90° – ∠PQR = 90° –...
A. Eo increase three times B. Eo is unchanged C. Eo decreases three times D. Go remains unchanged Solution: Eo is unchanged $ Nernst's\text{ }equation\text{ }states~: $ $...
A. sp3d2 B. sp3d3 C. sp3d D. dsp2 Solution: sp3d2 PCl5 is expected to have a hybridization of sp3d and a shape of trigonal bipyramidal. But in solid-state PCl5 exists in ionic form...
Solution: (a) ADFE is a cyclic quadrilateral Ext. ∠FEB = ∠ADF ⇒ ∠ADF = 80° ABCD is a parallelogram ∠B = ∠D = ∠ADF = 80° or ∠ABC = 80° (b)In trapezium ABCD, AD || BC (i) ∠B + ∠A = 180° ⇒ 70° + ∠A =...
B2O3 CO ZnO CaO Solution: ZnO Zinc oxide (ZnO) is known as amphoteric oxide because it has both acidic and basic properties.
Solution (a) ∠NYB = 50°, ∠YNB = 20°. In ∆YNB, ∠NYB + ∠YNB + ∠YBN = 180o 50o + 20o + ∠YBN = 180o ∠YBN + 70o = 180o ∠YBN = 180o – 70o = 110o But ∠MAN = ∠YBN (Angles in the same segment) ∠MAN = 110o...
Solution: (a) Construction: Join AB ∠A = ∠C = 350 (Alt Angles) ∠ABC = 35o (b) ∠AOC + reflex ∠AOC = 360o 130o + Reflex ∠AOC = 360o Reflex ∠AOC = 360o – 130o = 230o Now arc BC Subtends reflex ∠AOC at...
Solution: (i) ∠ADB and ∠ACB are in the same segment. ∠ADB = ∠ACB = 50° Now in ∆ADB, ∠DAB + X + ∠ADB = 180° = 42o + x + 50o = 180o = 92o + x = 180o x = 180o – 92o x = 88o (ii) In the given figure we...
1. Cyclo - sulphur 2. α - Sulphur 3. β - Sulphur 4. plastic sulphur Solution: Cyclo - sulphur
decreasing the concentration of KCl solution Increasing the quality of Hg2Cl2 decreasing the quantity of calomel Increasing the concentration of KCl solution Solution: decreasing the concentration...
1. 2412.5 C 2. 965 C 3. 4825 C 4. 9650 C Solution: 4825 C We have: Mass of one electron = 9.10×10−31 kg Charge of one electron = 1.602×10−19 coulomb $...
1. +2 2. +3 3. +4 4. +5 Solution: +5 Ammonium metavanadate, NH4VO3, is the most common source of vanadium in the +5 oxidation state. This isn't particularly water soluble, thus it's normally...
1. nucleophilic addition 2. nucleophilic substitution 3. electrophilic substitution 4. electrophilic addition Solution: electrophilic substitution Bromination is the reaction that is taking place...
. Identify the correct relation from the following relations:Solution: the correct option is A As the reaction progresses, the reactant concentration decreases and the product concentration rises. As a result, the rate of...
Solution: The correct option is 2. Hydride stability reduces as one moves along the group from NH3 to SbH3. This is because their bond dissociation enthalpy has decreased. The...
is1. Nature of solute only 2. Nature of the solute and the concentration of solution 3. Nature of the solvent and the concentration of solution 4. Nature of solvent only Solution: 3. Nature of the...
1. Ampere 2. 0C 3. Kelvin 4. Candela Solution: Candela The SI unit of luminous intensity is candela.
- 4.5 J 2×10-2 J - 506.5 J 225 J Solution: - 506.5 J The following equation can be used to describe the work done by a system against an external pressure: We know that: w = -Pext ∆V 1atm...
1. Dolomite 2. Siderite 3. China clay 4. Calamine Solution: China Clay $ Dolomite-[CaMg{{(C{{O}_{3}})}_{2}}] $ $ Siderite-[FeC{{O}_{3}}] $ $ China\text{ }Clay-[Al2{{O}_{3}}.2Si{{O}_{2}}.2{{H}_{2}}O]...
1. Tertiary alcohol 2. Secondary alcohol 3. Primary alcohol 4. Dihydric alcohol Solution: Primary alcohol The primary alcohols are those that have only one alkyl group linked to the carbon atom of...
1. Be 2. Mg 3. Sr 4. Ba Solution: Ba Because of its high reactivity, barium (Ba) is only found in nature in conjunction with other elements. Sulfate and carbonate are the most common compounds...
Solution: Option 1. is the correct answer Consider methyl bromide's alkaline hydrolysis to yield methanol. $ C{{H}_{3}}-Br+NaOH\,\,\xrightarrow{\Delta }\,\,C{{H}_{3}}-OH+NaBr $...
H2 SO2 N2 O2 Solution: SO2 SO2 is an easily liquefiable gas and easily liquefiable gases are adsorbed to a greater extent than elemental gases like N2, O2, and H2.
$ 1.\,A{{l}_{2}}{{O}_{3}} $ $ 2.\,{{N}_{2}}O $ $ 3.\,N{{a}_{2}}O $ $ 4.\,S{{O}_{2}} $ Solution: $ 2.\,{{N}_{2}}O $ Oxides that are neither acidic nor basic are known as neutral oxides. In other...
1. Nylon-6 2. Terylene 3. Bakelite 4. Nylon-6,10 Solution: terylene Ethylene glycol (1,2 ethanediol) and terephthalic acid are the monomers of terylene (1,4 benzene dicarboxylic acid). Terylene...
Cu2+ Fe3+ Na+ Cr3+ Solution: Na+ Cu2+ = 1s22s22p63s23p63d9 Fe3+ = 1s22s22p63s23p63d5 Cr3+ = 1s22s22p63s23p63d3 Na+ = 1s22s22p6 All the given ions except Na+ have unpaired electrons in their...
$ C{{l}_{2}}+2B{{r}^{-}}\,\to \,2C{{l}^{-}}+B{{r}_{2}} $ $ 1.\,C{{l}^{-}} $ $ 2.\,B{{r}^{-}} $ $ 3.\,B{{r}_{2}} $ $ 4.\,C{{l}_{2}} $ Solution: $ 4.\,C{{l}_{2}} $ Br is the reducing agent because it...
1. Sucralose 2. Alitame 3. Aspartame 4. Saccharine Solution: Sucralose Alitame, Aspartame and Saccharine all three contain the CO-NH linkage whereas Sucrolose contains the C-O linkage....
2-methyl-2-nitropropane 2-nitropropane Nitroethane 1-nitropropane Solution: 2-methyl-2-nitropropane The structure of 2-methyl-2-nitropropane is as follows: We can see that...
Solid Sol Emulsion Aerosol Solid foam Solution: Solid Foam Pumice stone is an example of solid foam. In this type of colloid, the dispersion medium is solid and the dispersion phase is...
molecules of urea are present in 100 mL of its solution. The concentration of solution is0.1 M 0.02 M 0.01 M 0.001 M Solution: 0.01 M Given: 6.02×1020 molecules of urea The volume of solution is 100/1000 = 0.1 L 1 mole of urea have 6.02×1023 molecules Number of moles present = 6.02×1020...
$Chlorobenzene+{{H}_{2}}O\xrightarrow[\Pr essure]{Cu,\,673K}A\xrightarrow[373K]{conc.\,{{H}_{2}}S{{O}_{4}}}B$ 4-hydroxybenzene Sulphonic acid Benzene Sulphonic acid 2-hydroxybenzene Sulphonic acid...
4.5 g 4.8 g 6.07 g 20.7 g Solution: 6.07 g Excess reagents are reactants that are not used up when a chemical reaction is completed. Because its quantity limits the amount of product generated, the...
$ N{{i}_{(s)}}|(1M)\,Ni_{(aq)}^{2+}||(1M)\,Au_{(aq)}^{3+}|Au\left( s \right) $ $ if\,E_{Ni}^{\circ }=-0.25V,\,E_{Au}^{\circ }=1.50V $ -1.25 V 1.75 V 1.25 V -1.75 V Solution: 1.75 V $...
5600 sec 360.0 sec 560.0 sec 3364 sec Solution: 560 sec $ Given: $ $ {{[R]}_{0}}=0.0210\,M~ $ $ [R]=0.0150\,M~ $ $ k=6\times {{10}^{-4}}\,{{\sec }^{-1}} $ $ For\text{ }a\text{ }first\text{...
Phenyl Methanol 2-Methyl Propan-2-ol Propan-2-ol Vinyl Alcohol Solution: Vinyl Alcohol Vinyl Alcohol is represented as follows: The simplest enol is vinyl alcohol, commonly...
Zero Three Two One Solution: One Lactic acid (2-hydroxy propionic acid) is a bifunctional molecule that has both a carboxylic acid and a hydroxyl group, making it useful in a...
$ 1.\,\,{{[Ir{{({{C}_{2}}{{O}_{4}})}_{2}}C{{l}_{2}}]}^{3-}} $ $ 2.\,\,{{[CoC{{l}_{2}}{{(en)}_{2}}]}^{+}} $ $ 3.\,\,{{\left[ Co{{(en)}_{2}}{{(N{{O}_{3}})}_{2}} \right]}^{+}} $ $ 4.\,\,\left[...
+3 +4 +6 +2 Solution: +6 Potassium ferrate has the chemical formula K2FeO4. This purple salt is paramagnetic, and is a rare example of an iron(VI) compound. $...
![[Co{{(N{{H}_{3}})}_{4}}C{{l}_{2}}]Cl](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-7f8a1797a451d4bad014fffc46745e19_l3.png "Rendered by QuickLaTeX.com")
1 mole 2 mole 3 mole 4 mole Solution: 1 mole Only one mole of AgCI is precipitated with an excess of AgNO3 because only one Cl- is in the ionization sphere.
Formation of sigma bonds Mixing and recasting of atomic orbitals Excitation of electrons Loss and gain of electrons Solution: Loss and gain of electrons The following ideas are important in...
It is a saturated aliphatic compound, It is a homocyclic compound. It is a heterocyclic compound with oxygen atom in the ring The molecular formula of Pyran is C5H5S Solution: It is a heterocyclic...
Fe (Z = 26) Cr (Z = 24) Cu (Z = 29) Mn (Z = 25) Solution: Cr (Z = 24) The following is the electrical configuration of the given elements: $...
Na Mg Li Ca Solution: Li Lithium has the highest ionization potential of all alkali metals, implying that its tendency to ionize to give Li+ ions should be the least, implying that Li should be the...
Conductance Conductivity Resistance Resistivity Solution: Conductivity The ability of an electrolyte solution to conduct electricity is measured by its conductivity (or specific conductance)....
3-Bromoaniline 2-Bromoaniline 4-Bromoaniline 2,4,6-Tribromoaniline Solution: 2,4,6-Tribromoaniline This happens due to the highly activating nature of -...
It is a sugar molecule It is optically active It contains two asymmetric carbon atoms It has carbonyl and hydroxyl group Solution: It contains two asymmetric carbon atoms ...
108.6 pm 127.65 pm 181.6 pm 157.6 pm Solution: 127.65 pm For F.C.C, a√2 = 4r 361×√2 = 4r r = 127.65 pm
$ 1.\,\,{{C}_{2}}{{H}_{4}}Br $ $ 2.\,\,{{C}_{2}}{{H}_{3}}Br $ $ 3.\,\,{{C}_{3}}{{H}_{5}}Br $ $ 4.\,\,{{C}_{3}}{{H}_{6}}Br $ Solution: $ 3.\,\,{{C}_{3}}{{H}_{5}}Br $ $ The\text{ }Allyl\text{...
and 
acting along the sides of an equilateral triangle. If the total torque acting at point 
‘ (centre of the triangle) is zero then the magnitude of is
Correct option is C ($\mathrm{F}_{1}+\mathrm{F}_{2}$) Explanation: On taking clockwise $ \begin{array}{l} \mathrm{T}=\mathrm{T}_{\mathrm{F} 1}+\mathrm{T}_{\mathrm{F} 2}+\mathrm{T}_{\mathrm{F} 3} \\...
154 pm 120 pm 133pm 112pm Solution: 112 pm
M Glucose M Sucrose M Urea M KCl Solution: M KCl ΔTf = i Kf m ΔTf is the freezing point depression, i is the van’t Hoff factor, Kf is the molal freezing point depression constant for the solvent,...
Terylene PHBV Nylon - 6,6 Dextron Solution: Nylon-6,6 Polyamides have the amide linkage. Example- Nylon 6-6, Nylon 6 Nylon 6, 6 is used in the manufacturing of sheets, brush bristles, and...
800 mm of Hg 1600 mm of Hg 200 mm of Hg 600 mm of Hg Solution: 800 mm of Hg This is an example of Boyle's law, which states that the volume of a gas is inversely proportional to the pressure at...
for HCl is 90 KJ/mol. Then HCl bond energy is360 KJ/mol 213 KJ/mol 180 KJ/mol 425 KJ/mol Solution: 425 KJ/mol $ \frac{1}{2}{{H}_{2}}(g)+\frac{1}{2}C{{l}_{2}}(g)\,\,\to \,\,HCl $ $ \Delta {{H}_{f}}=-90KJ $ $ \Delta {{H}_{f}}=~B.E\text{...
![{{\text{K}}_{\text{3}}}\text{ }\!\![\!\!\text{ Fe(CN}{{\text{)}}_{\text{6}}}\text{ }\!\!]\!\!\text{ }](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-1e79ee193daba828d4bfe9ad8e4b1525_l3.png "Rendered by QuickLaTeX.com")
is 3.333. What is it’s percentage dissociation?80% 70% 33.33% 77.7% Solution: 77.7% $ {{K}_{3}}[Fe{{(CN)}_{6}}]\to 3{{K}^{+}}+{{[Fe{{(CN)}_{6}}]}^{3-}} $ $ We\,get: $ $ i=\alpha n+(1-\alpha )=1+\alpha (n-1)=1+\alpha (4-1)~ $ $ Van't\text{...
473 K 1193 K 423 K 692 K Solution: 692 K The metal Zinc has a melting point of 692K.
$ 1.\,\,Ba{{O}_{2}}+{{H}_{2}}S{{O}_{4}} $ $ 2.\,\,N{{a}_{2}}{{O}_{2}}+{{H}_{2}}S{{O}_{4}} $ $ 3.\,\,Ba{{O}_{2}}+{{H}_{3}}P{{O}_{4}} $ $ 4.\,\,Ba{{O}_{2}}+{{H}_{2}}O+C{{O}_{2}} $ Solution: $...
0.1 NA / 2 2(0.1) NA 0.1 NA 2 NA Solution: 0.1 NA / 2 Given: Mass of element = 40 g and Z = 2 (bcc structure) The number of moles in 1 gram of element = 1/40 Similarly, number of moles in 4 gram of...
There should be very little difference in energy of involving orbitals The shape of hybrid orbitals is the same as that of atomic orbitals The number of hybrid orbitals formed is equal to the...
Nitrogen Sulphur Oxygen Carbon Solution: Nitrogen In analytical chemistry, the Dumas method is a method for quantifying nitrogen in chemical substances based on a method proposed by Jean-Baptiste...
is treated with excess 
in presence of pyridine?$ 1.\,\,{{C}_{2}}{{H}_{5}}N{{(COC{{H}_{3}})}_{2}} $ $ 2.\,\,{{\left( {{C}_{2}}{{H}_{5}} \right)}_{2}}NH $ $ 3.\,\,{{C}_{2}}{{H}_{5}}COOH $ $ 4.\,\,{{C}_{2}}{{H}_{5}}NHCOC{{H}_{3}} $ Solution: $...
Dacron Unrealformaldehyde Polymer Nylon-6 Polythene Solution: Polythene The addition polymers are formed by the repeated addition of monomer molecules possessing double or triple bonds, e.g., the...
Butanal Butanoic Acid Propanoic acid and carbon di-oxide Butan-2-one Solution: Butan-2-one Ketone is the first oxidation product of secondary alcohol. For example, butan-2-ol is oxidised to...
under high pressure?Toulene Benzoic Acid Benzaldehyde Acetophenone Solution: Benzaldehyde When the vapors of CO and HCI are passed into benzene in the presence of anhydrous AICI3/CuCI, benzaldehyde...
$ A)\,{{N}_{2}}{{O}_{4}} $ $ B)\,N{{O}_{2}} $ $ C)\,{{N}_{2}}{{O}_{5}} $ $ D)\,{{N}_{2}}{{O}_{3}} $ Solution: $ B)\,N{{O}_{2}} $ Among the oxides of nitrogen, only NO2 contains odd number of...
. if [A] = 0.3 M what is the arye$ A)\,\,2.1\times {{10}^{-5}}\,{{s}^{-1}} $ $ A)\,\,1.2\times {{10}^{-5}}\,{{s}^{-1}} $ $ A)\,\,1.3\times {{10}^{-5}}\,{{s}^{-1}} $ $ A)\,\,1.6\times {{10}^{-5}}\,{{s}^{-1}} $ Solution: $...
Solution:- From the question it is given that, In a ∆ABC, D and E are points on the sides AB and AC respectively. AD = 5.7cm, BD = 9.5cm, AE = 3.3cm and AC = 8.8cm Consider the ∆ABC, EC = AC – AE =...
Solution:- From the given figure, AB = AC. If PM ⊥ AB and PN ⊥ AC We have to show that, PM x PC = PN x PB Consider the ∆ABC, AB = AC … [given] ∠B = ∠C Then, consider ∆CPN and ∆BPM ∠N = ∠M … [both...
Solution:- From the question it is given that, (i) The area of ∆ABC to the area of ∆PQR if their corresponding sides are in the ratio 1 : 3 Then, ∆ABC ~ ∆PQR area of ∆ABC/area of ∆PQR = BC2/QR2 So,...
Solution:- From the figure, PQ || AC, AP = 4 cm, PB = 6 cm and BC = 8 cm ∠BQP = ∠BCA … [because alternate angles are equal] Also, ∠B = ∠B … [common for both the triangles] Therefore, ∆ABC ~ ∆BPQ...
Solution:- From the given figure it is given that, CM and RN are respectively the medians of ∆ABC and ∆PQR. (i) We have to prove that, ∆AMC ~ ∆PQR Consider the ∆ABC and ∆PQR As ∆ABC ~ ∆PQR ∠A = ∠P,...
Solution:- From the figure, DB ⊥ BC, DE ⊥ AB and AC ⊥ BC We have to prove that, BE/DE = AC/BC Consider the ∆ABC and ∆DEB, ∠C = 90o ∠A + ∠ABC = 90o [from the figure equation (i)] Now in ∆DEB ∠DBE +...
Solution:- From the question it is give that, AP = 2PB, CP = 2PD (i) We have to prove that, ∆ACP is similar to ∆BDP and AC || BD AP = 2PB AP/PB = 2/1 Then, CP = 2PD CP/PD = 2/1 ∠APC = ∠BPD [from...
(v) Let E be the event of getting a king or a queen. There will be 4 cards of king and 4 cards of queen. Number of favourable outcomes = 4+4 = 8 P(E) = 8/52 = 2/13 Hence the probability of getting a...
Solution: Edge of the cube, a = 44 cm Volume of cube = a3 = 443 = 85184 cm3 Diameter of shot = 4 cm So radius of shot, r = 4/2 = 2 cm Volume of a shot = (4/3)r3 = (4/3)×(22/7)×23 = 704/21 cm3 Number...
Solution: For same material, density will be same. Density = mass/Volume Mass of the smaller sphere, m1 = 1 kg Mass of the bigger sphere, m2 = 7 kg The spheres are melted to form a new sphere. So...
Solution: Given radius of the hemisphere, r = 8 cm Volume of the hemisphere, V = (2/3)r3 = (2/3)×83 = (1024/3) cm3 Radius of cone, R = 6 cm Since hemisphere is melted and recasted into a cone, the...
solution; Given height of the tent above the ground = 85 m Height of the cylindrical part, H = 50 m height of the cone, h = 85-50 h = 35 m Diameter of the base, d = 168 m Radius of the base of...
Given edge of the cube, a = 14 cm Radius of the cone, r = 14/2 = 7 cm Height of the cone, h = 14 cm Volume of the cube = a3 = 143 = 14×14×14 = 2744 cm3 Volume of the cone = (1/3)r2h =...
Solution: (i) Given radius of the sphere, r = 14 cm Surface area of the sphere = 4r2 = 4×(22/7)×142 = 4×22×14×2 = 2464 cm3 Hence the surface area of the sphere is 2464 cm2. (ii) Given radius of the...
Solution: Given base area of the cone = 616 cm2 r2 = 616 (22/7)×r2 = 616 r2 = 616×7/22 r2 = 196 r = 14 Given volume of the cone = 9856 cm3 (1/3)r2h = 9856 (1/3)×(22/7)×142 ×h = 9856 h =...
Solution: (i)Given height of the tent, h = 10 m Radius, r = 24 m We know that l2 = h2+r2 l2 = 102+242 l2 = 100+576 l2 = 676 l = √676 l = 26 (ii) Curved surface area = rl = (22/7)×24×26 = 13728/7 m2...
Solution: Given height of a cone, h = 9 cm Volume of the cone = 48 (1/3)r2h = 48 (1/3)r2×9 = 48 3r2 = 48 r2 = 48/3 = 16 r = 4 So diameter = 2×radius = 2×4 = 8 cm Hence the diameter of the cone is 8...
Solution: Given slant height of the cone, l = 10 cm Base radius, r = 7 cm Curved surface area of the cone = rl = (22/7)×7×10 = 220 cm2 Hence the curved surface area of the cone is 220...
Solution: Given line: 3x – 2y = -4 Slope (m1) is given by 2y = 3x + 4 y = (3/2) x + 2 So, m1 = 3/2 Now, the slope of the line parallel to the given line will have the same slope as 3/2 = m And the...
Solution: Let the co-ordinates of P(x, y) divides AB in the ratio m:n. A(-3,3) and B(12,-7) are the given points. Given m:n = 2:3 x1 = -3 , y1 = 3 , x2 = 12 , y2 = -7 , m = 2 and n = 3 By Section...
and the number of photons emitted during the pulse source is 
, calculate the energy of the source.Frequency of radiation $(\nu)$, $\nu=\frac{1}{2.0 \times 10^{-9} s}$ $\nu=5.0 \times 10^{8} s^{-1}$ Energy $(E)$ of source $=$ Nhv Where, $N$ is the no. photons emitted $\mathrm{h}$ is Planck's...
If the threshold wavelength is $\lambda_{0} n m\left(=\lambda_{0} \times 10^{-9} \mathrm{~m}\right)$, the K.E. of the radiation would be: $ h\left(\nu-\nu_{0}\right)=\frac{1}{2} m \nu^{2} $ Three...
Given, the work function of caesium $\left(W_{0}\right)=1.9 \mathrm{eV}$ (a)From the $W_{0}=\frac{h c}{\lambda_{0}}$ expression, we get: $\lambda_{0}=\frac{h c}{W_{0}}$ Where, $\lambda_{0}$ is the...
from the radiations of 
, calculate the number of photons received by the detector.From the expression of energy of one photon (E), $ E=\frac{h c}{\lambda} $ Where, $\lambda$ denotes the wavelength of the radiation $\mathrm{h}$ is Planck's constant c denotes the velocity of the...
No. quanta in $2 \mathrm{~J}$ of energy $\frac{2 J}{32.27 \times 10^{-20} J}$ $ =6.19 \times 10^{18} $ $ =6.2 \times 10^{18} $
Energy of one quantum $(E)=h v$ $ =\left(6.626 \times 10^{-34} \mathrm{Js}\right)\left(4.87 \times 10^{14} \mathrm{~s}^{-1}\right) $ Energy of one quantum $(\mathrm{E})=32.27 \times 10^{-20}...
Speed of the radiation, $\mathrm{c}=3 \times 10^{8} \mathrm{~ms}^{-1}$ Distance travelled by the radiation in a timespan of $30 \mathrm{~s}$ $ \begin{array}{l} =\left(3 \times 10^{8}...
Wavelength of the emitted radiation $=616 \mathrm{~nm}=616 \times 10^{-9} \mathrm{~m}$ (Given) (a)Frequency of the emission $(\nu)$ $ \nu=\frac{c}{\lambda} $ Where, $c=$ speed of the radiation...
. If the number of photons emitted is 
, calculate the power of this laser.Energy with which it emits photons=Power of laser Power $=E=\frac{N h c}{\lambda}$ Where, $N=$ number of photons emitted $\mathrm{h}=$ Planck's constant $\mathrm{c}=$ velocity of radiation...
The following is the frequency order in ascending order: Radiation from FM radio < amber light < radiation from microwave oven < X- rays < cosmic rays The following is the increasing...
more neutrons than electrons. Assign the symbol to this ion.Let us consider the total no. electrons present in $A^{3+}$ be $x$. Now, total no. neutrons in it $=x+30.4 \%$ of $x=1.304 x$ Since the ion has a charge of $+3, \Rightarrow$ no. electrons in neutral...
The number of electrons in a negatively charged ion be $x$. Then, no. neutrons present $=x+11.1 \%$ of $x=x+0.111 x=1.111 x$ No. electrons present in the neutral atom $=(x-1)$ (When an ion carries...
Let us consider that the No.of protons in the element be $x$. No. of neutrons $=x+31.7 \%$ of $x$ $=x+0.317 x$ $=1.317 x$ According to the question, Mass number of the element $=81$, which...
and 
can be written, whereas symbols 
and 
are not acceptable. Answer briefly.The general convention followed while representing elements along with their atomic masses (A), and their atomic numbers $(Z)$ is ${ }_{Z}^{A} \mathrm{X}$. Therefore, ${ }_{35}^{79} \mathrm{Br}$ is...
The findings obtained with a foil made up of heavy atoms will differ from those obtained with a foil made up of comparatively light atoms. The magnitude of positive charge in the nucleus of a...
rays. If the static electric charge on the oil drop is 
, calculate the number of electrons present on it.Charge held by the oil drop $=1.282 \times 10^{-18} C$ Charge held by one electron $=1.6022 \times 10^{-19} C$ Therefore, No. electrons present in the drop of oil $\frac{1.282 \times 10^{-18}...
of static electric charge. Calculate the number of electrons present in it.Charge held by one electron $=1.6022 \times 10^{-19} C \Rightarrow 1.6022 \times 10^{-19} C$ charge is held by one electron. Therefore, electrons carrying charge of $2.5 \times 10^{-16} C \quad...
Radius of carbon atom$ =\frac{2.6}{2} $ $ \begin{array}{l} =1.3 \times 10^{-10} \mathrm{~m} \\ =130 \times 10^{-12} \mathrm{~m}=130 \mathrm{pm} \end{array} $ (b) Length of the arrangement $=1.6...
atoms of carbon are arranged side by side. Calculate the radius of carbon atom if th length of this arrangement is 
.Length of the arrangement $=2.4 \mathrm{~cm}$ No. carbon atoms present $=2 \times 10^{8}$ The diameter of the carbon atom $=\frac{2.4 \times 10^{-2} m}{2 \times 10^{8} m}$ $=1.2 \times 10^{-10} m...
We know that$ 1 \mathrm{~cm}=10^{-2} \mathrm{~m} $ Length of the scale $=20 \mathrm{~cm}=20 \times 10^{-2} \mathrm{~m}$ Diameter of one carbon atom $=0.15 \mathrm{~nm}=0.15 \times 10^{-9}...
The ionization energy for the 
atom in the ground state is 
atom 
The energy associated with hydrogen-like species is: $ E_{n}=-2.18 \times 10^{-18}\left(\frac{Z^{2}}{n^{2}}\right) J $ For the ground state of the hydrogen atom, $ \begin{array}{l} \Delta...
The wave number associated with the Balmer transition for the He+ ion (n = 4 to n = 2 ) is given by: $ \bar{\nu}=\frac{1}{\lambda}=R Z^{2}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right) $...
Only one electron exists in hydrogen atoms. The angular momentum of this electron, according to Bohr's postulates, is: $m v r=n \frac{h}{2 \pi} \quad \ldots .(1)$ Where, $n=1,2,3, \ldots$ As per de...
b)n = 3, l = 0(a)The total number of electrons in the atom =$2n^2$ if n is the primary quantum number. Hence, For n = 4, Total no. electrons = 2 (16) = 32 An atom with 32 electrons has the following electron...
a) This is not possible. The number n cannot be zero. (b) Possible. (c) This is not possible. The value of l can't be the same as the value of n. (d) This is not possible. Because mt can't be 1 when...
(a)n = 1, l = 0 implies a 1s orbital. (b)n = 3 and l = 1 implies a 3p orbital. (c)n = 4 and l = 2 implies a 4d orbital. (d)n = 4 and l = 3 implies a 4f orbital.
(I) The range of potential values for 'l' is 0 to (n – 1). As a result, the possible values of l for n = 3 are 0, 1, and 2. The total number of potential ml = (2l + 1) values. It has a range of...
The number of electrons in H2 is 1 + 1 = 2. 2 – 1 = 1 number of electrons in H2+ H2: The number of electrons in H2 equals 1 + 1 = 2. Number of electrons in O2 = 8 + 8 = 16. In O2+, the number of...
(i)No.protons = no.electrons in a neutral atom. The number of protons in the atoms of the element is 29. (ii)The electronic configuration of this element (atomic number 29) is 1s...
For the 3d orbital: Principal quantum number ,possible values (n) = 3 Azimuthal quantum number, possible values(l) = 2 Magnetic quantum number ,possible values(ml) = – 2, – 1, 0, 1, 2
For g-orbitals, l = 4. The possible values of ‘l’ range from 0 to (n-1),. For any given value of ‘n’, Hence, least value of n = 5, l = 4 (g orbital),
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Ne] 
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$(I I I)(a)[H e] 2 s^{1}$ Complete electronic configuration: $1 \mathrm{~s}^{2} 2 \mathrm{~s}{ }^{1}$. $\therefore$ the element's atomic number is 3 . The element is lithium (Li) (b) $[\mathrm{Ne}]...
(b) 
and (c) 
(II) (a) $3 \mathrm{~s}^{1}$ Complete electronic configuration: $1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{6} 3 \mathrm{~s}^{1}$ Total no. electrons in the atom $=2+2+6+1=11 \quad...
(b) 
(c) 
(d) 
The electronic configuration of the Hydrogen atom (in its ground state) $1 \mathrm{~s}^{1}$. The single negative charge on this atom indicates that it has gained an electron. Hence, the electronic...
kg. If its K.E. is
J, calculate its wavelength.As per de Broglie’s equation, $ \lambda=\frac{h}{m v} $ Given, K.E of electron $=3.0 \times 10^{-25} \mathrm{~J}$ $ \begin{array}{l} \text { Since K.E.= } \frac{1}{2} m v^{2} \therefore...
As per de Broglie’s equation, $ \lambda=\frac{h}{m v} $ Where, $\lambda$ denotes thr wavelength of the moving particle $\mathrm{m}$ is the mass of the particle $v$ denotes the velocity of the...
. Calculate the energy required to remove an electron completely from the n = 2 orbit. What is the longest wavelength of light in cm that can be used to cause this transition?Required energy for the ionization from $\mathrm{n}=2$ is: $ \begin{array}{l} \Delta E=E_{\infty}-E_{2} \\ =\left[\left(\frac{-\left(2.18 \times...
ergs. The ground-state electron energy is ergs.$ E_{5}=\frac{-\left(2.18 \times 10^{-18}\right) Z^{2}}{(n)^{2}} $ Where, $Z$ denotes the atom's atomic number Ground state energy $=-2.18 \times 10^{-11}$ ergs $=-2.18 \times 10^{-11} \times...
The Balmer series of the hydrogen emission spectrum, ni = 2. Hence, wavenumber expression ν is: $ \bar{\nu}=\left[\frac{1}{(2)^{2}}-\frac{1}{n_{f}^{2}}\right]\left(1.097 \times 10^{7}...
. What is the energy associated with the fifth orbit? (ii) Calculate the radius of Bohr’s fifth orbit for the hydrogen atom.(i) Energy associated with the fifth orbit of hydrogen atom is calculated as: $E_{5}=\frac{-\left(2.18 \times 10^{-18}\right)}{(5)^{2}}=\frac{-\left(2.18 \times 10^{-18}\right)}{25}=-8.72 \times...
Option (ii) is the answer. Glycogen has a similar structure to amylopeptin. It's an a-D glucose unit branched chain polymer with C1-C4 glycosidic linkage for chain formation and C1-C6 glycosidic...
Option (iv) is the answer. Glycogen is a type of sugar that is stored in the liver of mammals.
Option (iii) is the answer. Cane sugar (sucrose) is a disaccharide. When sucrose is hydrolyzed, one molecule of glucose and one molecule of fructose are produced.
Option (C) is the answer. Anomers are isomers that differ only in the conformation of the hydroxyl group at C—1 and are known as - and -fonns.
Option ( iii) is the answer. Hydrogen bonding help to keep the -helix structure of proteins stable. By twisting into a right-handed helix and hydrogen bonding the -NH group of each amino acid...
Option (B) is the answer. This structure represents sucrose, in which the C1—C2 glycosidic bond connects -D glucose and -D-fructose. This is a non-reducing sugar since the reducing groups of glucose...
Option (ii) is the answer. Vitamin C is ascorbic acid. Amino acid aspartic acid is a kind of amino acid. Dicarboxylic acids include adipic acid and saccharic acid.
Option (i) is the answer. Between the pentose sugars of nucleotides, there are 5′ and 3′ connections.
Option (ii) is the answer. Nucleic acids are nucleotide polymers connected together by phosphodiester linkage.
Option (iii) is the answer. It's found in the pyranose structure.
Option (i) is the answer. The main structure of proteins is the sequence of amino acids in a polypeptide chain.
Option (iii) is the answer. Adenine, guanine, thymine, and cytosine are the four bases found in DNA. Adenine, uracil, guanine, and cytosine are the four bases found in RNA. As a result, thymine is...
Option (iv) is the answer. Because vitamin B12 is water insoluble, it can be stored in the body.
Option (iv) is the answer. In DNA, uracil is absent; instead, thymine is present.
Option (i) is the answer. Anomers are cyclic configurations of monosaccharides that differ in structure at carbon-1. I and II are anomers in this case because they differ solely in carbon-1.
Option (iii) is the answer. The absence of a free -CHO group is indicated by the fact that glucose pentaacetate does not react with hydroxylamine. Only the cyclic nature of glucose may explain this...
Option (i) is the answer. The -OH group is on the lowest asymmetric carbon on the right side of the I, II, and III structures, which is similar to (+) glyceraldehyde.
Option (iii) is the answer. Anomeric carbon is carbon that is next to an oxygen atom in the cyclic structure of glucose or fructose. 'a' and 'b' are next to the oxygen atom, as illustrated in the...
Option (iii) is the answer (A) and (C) are in the Cl-C4 range, while (B) is in the Cl-C6 range.
Option (ii) and (iv) are the answers. Sucrose is a non-reducing sugar and a disaccharide.
Option (i) and (iii) are the answers Globulular protein is the structure that develops when a chain of polypeptides coils around to form a spherical shape. Insulin and albumin, for example, are...
Option (i) and (iv) are the answers. Amylopectin and glycogen are both glucose branching polymers.
Option (ii) and (iv) are the answers. Acidic amino acids have more than one -COOH group one against the –NH2 group.
Option (i), (ii) and (iii) are the answers. (a)Lysine is a kind of amino acid with the structural formula . (b) Because the number of NH2 groups (2) is more than the number of COOH groups, it is a...
Option (i) and (iii) are the answers. The five-membered cyclic structure of ribose and fructose is shown (furanose structures). They have a five-membered ring, similar to the foran compound....
Option (ii) and (iv) are the answers. Disulphide linkage and hydrogen bonding hold polypeptide chains together in fibrous proteins.
Option (i) and (ii) are the answers. Purines are made up of a six-membered nitrogen-containing ring fused together with a five-membered nitrogen-containing ring. Purine bases guanine and adenine...
Option (i) and (iv) are the answers. Enzymes are protein molecules that act as biocatalysts in the body's chemical reactions.
Lactose is the sugar found in milk. Glucose and galactose are two monosaccharides found in lactose. Disaccharides are oligosaccharides that include two monosaccharide units.
When glucose is heated with HI for a long time, n-hexane develops, implying that all six carbon atoms are connected in a straight chain.
When a nitrogenous base is connected to the 1' position of a five-carbon sugar, a nucleoside is produced. The 5' carbon of the sugar in a nucleoside molecule is bonded to the 5' carbon of the sugar...
Glycosidic linkages connect the monosaccharide units of polysaccharides. When an oxide bond is created between two monosaccharide units with the loss of a water molecule, it is called a glycosidic...
When glucose is treated with a mild oxidising agent like Br2 water, it is transformed to gluconic acid, a six-carbon carboxylic acid. When glucose is treated with nitric acid, it is transformed to...
Carbonyl groups can be found in monosaccharides. As a result, they're categorised as either aldose or ketose. Aldose refers to monosaccharides that contain an aldehyde group. Ketose refers to...
On the left side of the C5 carbon atom, the –OH group is connected. As a result, the provided compound is in the 'L' configuration.
D-configuration is the configuration of both aldopentoses. -D-ribose is ribose, while -D-2-deoxyribose is 2-deoxyribose.
Invert sugar is another name for sucrose. It comes from sugarcane and sugarbeet and is a naturally occurring sugar. When sucrose is hydrolyzed, the sign of rotation changes from Dextro (+) to laevo...
The sort of amino acids that make up a polypeptide chain are -amino acids and alpha-amino acids, where the amino acid is linked to the -carbon in the molecule.
The –NH group of each amino acid residue hydrogen is bound to the –C=O of an adjacent turn of the helix, forming a right-handed screw helix shape.
Enzyme oxidoreductases is the name given to a group of enzymes that catalyse redox processes. Alcohol Dehydrogenase, for example, aids in the reduction of alcohol levels in the human body when...
The sugar found in milk, lactose, is transformed to lactic acid during curdling, which is produced by bacteria.
When glucose is acetylated using acetic anhydride (CH3CO)2O in the presence of ZnCl2, glucose pentaacetate is formed, confirming the presence of five –OH groups.