Sigma bonds are a result of the head-to-head overlapping of atomic orbitals whereas pi bonds are formed by the lateral overlap of two atomic orbital.

Yes, with disublimation

Yes of course with the help of deposition or desublimation process.

Grignard reagents are used synthetically to form new carbon–carbon bonds. A Grignard reagent has a very polar carbon–magnesium bond in which the carbon atom has a partial negative charge and the metal a partial positive charge.

According to Arrhenius,acids dissociate H+ ions in water and bases dissociateOH- ion in water.

Ohh god ???

Element

The s-block andp-block elements are called representative elements.

The Representative Elementsare those elements within the first two families (Groups I and II on the far left) and the last six families or groups (on the right) of the Periodic Table. TheTransition Metals are the elements in those Groups within the middle of the Table.

Elements in which all the inner shells are complete but outer shell is incomplete is known as representative elements i.e Those elements which have less than 8

Ineed answer

Add1 carbon also in methane so it 2 hydrogen is coming so...

Methane is treated with chlorine gas to form chloro  methane. Chloro methane is treated with sodium and dry ether to form ethane

Just put m before ethane?

Hybridisation of B is SP2

A dynamic equilibrium exists once a reversible reaction occurs. Substances transition between the reactants and products at equal rates, meaning there is no net change. Reactants and products are formed at such a rate that the concentration of neither changes.

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I also...

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The catalyst is used for the hydrogenation of alkynes to alkenes (i.e. without further reduction into alkanes). The lead serves to deactivate the palladium sites, further deactivation of the catalyst with quinoline or 3,6-dithia-1,8-octanediol enhances its selectivity, preventing formation of alkanes.

Yes

0.0125 moles then 0.925/74=0.0125 moles

No

Yes!!!

no it is not in s block because it shows some properties of metals and some of non metals

Yes hydrogen exist in s block element because it is 1spower2 last configuration s is coming so it is exist in s block

Yes hydrogen exist in s block element

First is ka molar mass nikalo phir iska molecule or phir molecule/ NA -6.022×10 (23)

The smallest unit into which a substance can be divided without changing it's chemical nature

A group of atoms bonded together, representing the smallest fundamental unit of a chemical compound that can take part in a chemical reaction.

Atoms together joined my a bond is called Molecule

Molecule is made up of atoms.

Add -OH functional group

In isothermal process-- T=const. So, ∆U=0 ∆U=q+w q+w=0 q=-w

JEE Main Mole Concept Previous Year Questions With Solutions 1. Number of atoms in the following samples of substances is the largest in : (1) 127.0g of iodine (2) 48.0g of magnesium (3) 71.0g of chlorine (4) 4.0g of hydrogen Solution: 1 mole represents 6.023×1023 particles. 1 mole of iodine atom= 6.023×1023 Given 127.0g of iodine. no. of iodine atom = 1 mole of iodine 1mole of magnesium = 24g of Mg = 6.023×1023no.of Mg Given 48g of Mg = 2×6.023×1023 no. of Mg = 2 moles of Mg 1 mole of chlorine atom= 6.023× 1023 no. of chlorine atom = 35.5g of chlorine atom Given 71g of chlorine atom=2× 6.023× 1023 no. of chlorine atom = 6.023×1023 2 moles of chlorine atom. Given that 4g of hydrogen atom. will be equal to 4 × 6.023 × 1023 no. of atoms of hydrogen= 4 moles of hydrogen atom. Hence option(4) is the answer. 2. The ratio of mass percent of C and H of an organic compound (CxHyOz) is 6 : 1. If one molecule of the above compound ( CxHyOZ ) contains half as much oxygen as required to burn one molecule of compound CXHY completely to CO2 and H2O. The empirical formula of compound CxHyOz is : (1) C2H4O (2) C3H4O2 (3) C2H4O3 (4) C3H6O3 Solution: Given the ratio of mass percent of C and H of an organic compound (CxHyOz) is 6 : 1. Atomic mass of carbon = 12 Atomic mass of Hydrogen = 1 If we have x atoms of Carbon and y atoms of Hydrogen, 12*x = 6(1*y) 12x = 6y So y = 2x Given one molecule of compound ( CxHyOZ ) contains half as much oxygen as required to burn one molecule of compound CXHY completely to CO2 and H2O CXHy + O2 → xCO2+ (y/2)H2O Put y = 2x in above equation CXH2x + O2 → xCO2+ xH2O Oxygen needed = 2x+x = 3x z is half of oxygen required to burn. So z = 3x/2 = 1.5 x Check the given options which satisfies z = 1.5x. So the empirical formula is C2H4O3. Hence option (3) is the answer. 3. The concentrated sulphuric acid that is peddled commercially is 95% H2SO4 by weight. If the density of this commercial acid is 1.834 g cm-3, the molarity of this solution is :- (1) 17.8 M (2) 15.7 M (3) 10.5 M (4) 12.0 M Solution: Given Density = 1.834 1 ml solution contains 1.834 g 1000 ml solution will contain 1834 g 95% H2SO4 means 100 gm contain 95 gm H2SO4 Mass of solute = (95/100)×1834 Molecular weight of H2SO4 = 98 Molarity = No. of moles/ volume = mass of solute/98 = (95/100)×(1834/98) = 17.8 M Hence option (1) is the answer. 4. The ratio of masses of oxygen and nitrogen in a particular gaseous mixture is 1 : 4. The ratio of number of their molecule is : (1) 1 : 8 (2) 3 : 16 (3) 1 : 4 (4) 7 : 32 Solution: Given ratio of masses of oxygen and nitrogen = 1:4 Let mass of O2 = w Mass of N2 = 4w Molecules of O2 = w/(32×NA) Molecules of N2 = 4w/(28×NA) Ratio of number of molecules = w/(32×NA)÷4w/(28×NA) = w/(32×NA)×(28×NA)/4w = 7/32 So the ratio is 7:32. Hence option (4) is the answer. 5. 3g of activated charcoal was added to 50 mL of acetic acid solution (0.06N) in a flask. After an hour it was filtered and the strength of the filtrate was found to be 0.042 N. The amount of acetic acid adsorbed (per gram of charcoal) is : (1) 42 mg (2) 54 mg (3) 18 mg (4) 36 mg Solution: Molarity of CH3COOH solution = mass of acetic acid/molar mass)/volume of solution in litre Acetic acid is monobasic. 0.042 = W/(60×0.05) W = 0.042×60×0.05 = 0.126 g Amount of acetic acid actually adsorbed = 0.180-0.126 = 54mg Amount of charcoal available = 3 g So amount of acetic acid adsorbed per gram of charcoal = 54mg×1g/3.0g = 18 mg Hence option (3) is the answer. 6. The density of a solution prepared by dissolving 120 g of urea (mol. mass = 60 u) in 1000 g of water is 1.15 g/mL. The molarity of this solution is (1) 2.05 M (2) 0.50 M (3) 1.78 M (4) 1.02 M Solution: Given density of solution = 1.15g/mL mass of solution = 1000+120 = 1120 gm Molar mass = 60 Volume = mass /density of solution = 1120/1.15 No. of moles = 120/60 = 2 Molarity = No. of moles/ volume = 2÷ (1120×10-3/1.15) = 2×1.15×1000 /1120) = 2.05 M Hence option (1) is the answer. 7. The ratio of number of oxygen atoms (O) in 16.0g oxygen (O), 28.0 g carbon monoxide (CO) and 32.0g oxygen (O2) is : (Atomic mass :C =12, O =16 and Avogadro’s constant NA = 6.0 * 1023 mol-1) (1) 3 : 1 : 1 (2) 1 : 1 : 2 (3) 3 : 1 : 2 (4) 1 : 1 : 1 Solution: Molar mass of O3 = 48 Given 16 g O3 . So no. of moles of O3 = 16/48 = ⅓ 1 mole = 3 ×NA oxygen atoms So 1/3 mole = NA×3×1/3 no of atoms = NA oxygen atoms Molar mass of CO = 28 Given 28 g CO. So no of moles = 28/28 = 1 No. of atoms = 1×NA = NA Molar mass of O2 = 32 Given 32g O2 No. of moles = 32/32 = 1 No.of atoms = 1×NA = NA So the ratio is 1:1:1 Hence option (4) is the answer. 8. When CO2 (g) is passed over red hot coke it partially gets reduced to CO(g). Upon passing 0.5 litre of CO2 (g) over red hot coke, the total volume of the gases increased to 700 mL. The composition of the gaseous mixture at STP is : (1) CO2 = 200 mL: CO = 500mL (2) CO2 = 350 mL: CO = 350mL (3) CO2 = 0.0 mL: CO = 700mL (4) CO2 = 300 mL: CO = 400mL Solution: CO2 (g) + C (s) → 2CO (g) Total volume = 700 ml = 0.7 L 0.5+x = 0.7 x = .2L = 200 mL CO2 (g) = 0.5-0.2 = 300ml CO (g) = 2x = 400 mL Hence option (4) is the answer. 9. An open vessel at 300 K is heated till ⅖ th of the air in it is expelled. Assuming that the volume of the vessel remains constant, the temperature to which the vessel is heated is : (1) 750 K (2) 400 K (3) 500 K (4) 1500K Solution: At constant V and P, n1T1 = n2T2 n1 = n n2 = n-2n/5 = 3n/5 T1 = 300 K 300 n = (3n/5) T2 T2 = 300×5/3 = 500 K Hence option (3) is the answer. 10. The density of 3M solution of sodium chloride is 1.252 g mL-1. The molality of the solution will be (molar mass, NaCl = 58.5 g mol-1) (1) 2.18 m (2) 3.00 m (3) 2.60 m (4) 2.79 m Solution: Given Molar mass of NaCl = 58.5 g M = 3 mol L-1 Mass of weight W2 of NaCl in 1L solution W2 = 3×58.5 = 175.5g Mass of L solution = V × d = 1000 ×1.25 = 1250g Mass of H2O in solution (W1) = 1250-175.5 = 1074g m = W2×1000/Mw2 ×W1 = (175.5×1000)/58.5×1074.5 = 2.79m Hence option (4) is the answer. 11. 0.6 g of urea on strong heating with NaOH evolves NH3. Liberated NH3 will combine with which of the following HCl solution? (1) 100 mL of 0.2 N HCl (2) 400 mL of 0.2 N HCl (3) 100 mL of 0.1 N HCl (4) 200 mL of 0.2 N HCl Solution: NH2CONH2 + 2NaOH → Na2CO3 + 2NH3 2 mole of urea ≡ one mole of NH3 one mole of NH3 = one mole of HCl So one mole of HCl = 2 mole of urea = 2×0.6/60 = 0.02 mol. Hence option (1) is the answer. 12. Calculate the mass of FeSO4.7H2O which must be added in 100 kg of wheat to get 10 PPM of Fe. Solution: Ppm = (Mass of Fe/total mass)×106 Total mass = 100 kg = 100 × 1000 g Mass of Fe = (ppm × total mass )/106 = 10× 100 × 1000/106 = 1 g Molecular mass of FeSO4.7H2O = 278 Mass of one Fe = 56 g 56 g of Fe → 278 g of FeSO4.7H2O So 1 g of Fe → 278/56 = 4.96 g Hence 4.96 g is the answer. 13. Given a solution of HNO3 of density 1.4 g/mL and 63% w/w. Determine the molarity of HNO3 solution. Solution: Density = mass/volume of solution Volume = mass / density = 100g/1.4 g/ml = (100/1.4)ml Molarity = no. of moles of solute /Volume of solution( l) = 1.4×1000/100 = 14 M Hence 14 M is the answer. 14. A transition metal M forms a volatile chloride which has a vapour density of 94.8. If it contains 74.75% of chlorine the formula of the metal chloride will be (1) MCl2 (2) MCl4 (3) MCl5 (4) MCl3 Solution: Given vapour density = 94.8 Vapour density = molecular mass/2 Molecular mass = 94.8×2 = 189.6 Given 74.75% chlorine. So 74.75/100 * 189.6 = 141.72 g of chloride is there. Then the number of atoms of chloride will be 141.72/35.5 =3.97 which is approximately 4. So the formula of metal chloride will be MCl4. Hence option (2) is the answer. 15. 10 mL of 2(M) NaOH solution is added to 200 mL of 0.5 (M) of NaOH solution. What is the final concentration? (1) 0.57 M (2) 5.7 M (3) 11.4 M (4) 1.14 M Solution: No. of moles of NaOH in 10 mL of 2 M solution = (10/1000)×2 = 0.02 mol Number of moles of NaOH in 200 mL of 0.5M solution = (200/1000)×0.5 = 0.1 mol Total number of moles of NaOH = 0.02+0.1 = 0.12 mol Total volume = 10+200 = 210 mL = 0.210 L Final concentration = 0.12/0.210 = 0.57 M Hence option (1) is the answer. 16. A 5.2 molal aqueous solution of methyl alcohol, CH3OH, is supplied. What is the mole fraction of methyl alcohol in the solution? (1) 0.086 (2) 0.050 (3) 0.100 (4) 0.190 Solution: We know mole fraction = moles of solute/(moles of solute + moles of solvent) Let mass of water is 1 kg . Moles of CH3OH is 5.2 Xsolute = 5.2/(5.2+1000/18) = 5.2/(5.2+55.556) = 5.2/60.756 = 0.086 Hence option (1) is the answer

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Molarity Molality Mole fraction Mole cocept Avogadro law

A charge contain by the atom during reaction. Charge should be positive or negative or 0