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Mr. Singh 5 years ago (5002243)

Satellite DNA is DNA sequence that contain highly repetitive DNA which means that a single sequence is repeated many times over. The variations between individuals in the lengths of their DNA satellite forms the basis of DNA fingerprint..
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Mr. Singh 5 years ago (5002243)

A Probe is a single stranded sequence of DNA or RNA used to search for its complementary sequence in a sample genome. ...The Probe is labbled with a radioactive or chemical tag that allows its binding to be visualized. In a similar way, labbeled antibodies are used to probe a sample for the presence of a specific protein .
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Mr. Singh 5 years ago (5002243)

No it is not a n type of semiconductor
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Ananya Singh 5 years ago (10515241)

गीत का दूसरा meaning hai संगीत hai
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Riya Sharma 5 years ago (9203792)

The poet's mother laughed at the happy memories of the moments that had passed long back. She looked back to her childhood with nostalgia and remembered the innocent joys of her childhood days.
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Sea which are on the west coast of India is Arabian sea branch and bay which is on the east coast of India is Bay of Bengal branch.
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Reetika Mewade 3 years, 10 months ago (9367583)

Charge particles are known as ions. There are two types of ions- Anion- The positive charge ion is called anion. Cation- The negative charge ion is called cation.

Gauri Thite 4 years, 10 months ago (9800626)

First lesson notes
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Palak Kumari 4 years, 11 months ago (8478441)

Need

A S 5 years ago (9161785)

Should

Vdjssss 9471 5 years ago (10529847)

should / need/must , hope it helps

Gunn Bhatia 5 years ago (10477788)

'need' will be the appropriate word for the blank
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Sia ? 4 years, 7 months ago (6945213)

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Aditya Yadav 5 years ago (4282869)

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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Mr. Singh 5 years ago (5002243)

Means uss coupon ka code kya hai
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  • 4 answers

Winn Still 5 years ago (10169082)

9

Rekha Choudhary 5 years ago (10035948)

9

Mili Khamar 5 years ago (10494199)

Value of k=9

Riddhi Sharma 5 years ago (10230939)

Value of k is 9
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Harsh Garg 5 years ago (4325535)

Computer applications dr.vipan arora
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Khushbu Singh✍️ 4 years, 10 months ago (8696220)

Covalent solid

I A 4 years, 11 months ago (10276972)

Crystalline solid

Insha Khalid 5 years ago (10011493)

Covalent solid

Preeti Kannaujiya 5 years ago (3676868)

Crystalline- Covalent

Mr. Singh 5 years ago (5002243)

Covalent-network (also called atomic) solid- Made up of atoms connected by covalent bonds, the intermolecular forces are covalent bonds as well. Characterized as being very hard with very high melting points and being poor conductors. Examples of this type of solid are diamond and graphite, and the fullerences.
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Gunn Bhatia 5 years ago (10477788)

Modern forms of money include currency (paper notes) and coins. 1.It is accepted as a medium of exchange because the currency is authorised by the government of India. 2.No individual in India can legally refuse a payment made in rupee. 3.Any person holding money can easily exchange it with any commodity or service that he desires. 4.It acts as intermediate in the exchange process of different countries.
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Sukhjinder Jandu 4 years, 5 months ago (3408231)

Nicolas likes to play with her friends.
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A S 4 years, 11 months ago (9161785)

Define translocation ? Name 2 conductive tissues and their importance? Explain the process of translocation? Why transpiration is necessary? How plants transport water & minerals?elucidate the transportation system in plants ?
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It is suggested to undertake this project after completing the unit on preparation of financial statements. The student(s) will be allowed to select any business of their choice or develop the transaction of imaginary business. The project is to run through the chapters and make the project an interesting process. The amounts should emerge as more realistic and closer to reality. Specific Guidelines for Teachers Give a list of options to the students to select a business form. You can add to the given list: 1. A beauty parlour 10. Men's wear 19. A coffee shop 2. Men's saloon 11. Ladies wear 20. A music shop 3. A tailoring shop 12. Kids wear 21. A juice shop 4. A canteen 13. A Saree shop 22. A school canteen 5. A cake shop 14. Artificial jewellery shop 23. An ice cream parlour 6. A confectionery shop 15. A small restaurant 24. A departmental store 7. A tyre repair shop 16. A sweet shop 25. A flower shop 8. A dry cleaner 17. A grocery shop 26. A gift shop 9. A stationery shop 18. A shoe shop 27. A photostat shop After selection, advise the student(s) to visit a shop in the locality (this will help them to settle on a realistic amounts different items. The student(s) would be able to see the things as they need to invest in furniture, decor, lights, machines, computers etc. A suggested list of different item is given below. 1. Rent 19. Wages and Salary 2. Advance rent [approximately three months] 20. Newspaper and magazines 3. Electricity deposit 21. Petty expenses 4. Electricity bill 22. Tea expenses 5. Electricity fitting 23. Packaging expenses 6. Water bill 24. Transport 7. Water connection security deposit 25. Delivery cycle or a vehicle purchased 8. Water fittings 26. Registration 9. Telephone bill 27. Insurance 10. Telephone security deposit 28. Auditors fee 11. Telephone instrument 29. Repairs & Maintenance 12. Furniture 30. Depreciations 13. Computers 31. Air conditioners 14. Internet connection 32. Fans and lights 15. Stationery 33. Interior decorations 16. Advertisements 34. Refrigerators 17. Glow sign 35. Purchase and sales 18. Rates and Taxes At this stage, performas of bulk of originality and ledger may be provided to the students and they may be asked to complete the same with the help of computers. In the next step the students are expected to prepare the trial balance and the financial statements.
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Kajal Rani 4 years, 11 months ago (10521666)

Helo Dear, From which chapter you are asking this questions??, Are you are asking from chapter 10 "An alien hand"

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