This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 3151. |
the common impurities present in bauxite areA. CuOB. ZnOC. CaOD. `SiO_(2)` |
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Answer» Correct Answer - D Impurities likes `Fe_(2)O_(3)` and `SiO_(2)` are present in bauxite . |
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| 3152. |
Name the principle ore of aluminium. How is the metal extracted from this ore ? |
| Answer» Bauxite. By electrochemical reduction. | |
| 3153. |
Common impurities present in bauxite areA. CuOB. ZnOC. `Fe_(2)O_(3)`D. `SiO_(2)` |
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Answer» Correct Answer - C::D Both `Fe_(2)O_(3) " and " SiO_(2)` are the common impurities in Bauxite. |
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| 3154. |
Which of the following ores are concentrated by froth floatation ?A. HaematiteB. GalenaC. Copper pyritesD. Magnetite |
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Answer» Correct Answer - B::C Both Galena (PbS) and Copper pyrites `(CuFeS_(2))` are the sulphide ores. |
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| 3155. |
The process of removing lighter gangue particles by washing in a current if water is called:A. levigationB. liquationC. leachingD. cupellation |
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Answer» Correct Answer - A Lighter gangue particles are washed in a current of water by a process called levigation. In levigation the powdered ore is agitated with water or washed with a upward stream of running water, the lighter particles of sand, clay etc. are washed away leaving behind heavier ore particles. |
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| 3156. |
What is flux ? Give one example each of acidic and basic flux. |
| Answer» (i) ` SiO_(2) (ii) MgO`. | |
| 3157. |
Name a method used for removing gangue from sulphide ore. |
| Answer» The method is known as Froth Floatation process. | |
| 3158. |
Assertion: Tungsten has very high melting point. Reason: Tungsten is a covalent compound.A. If both Assertion `&` Reason are True `&` the Reason is a correct explanation of the Assertion.B. If both Assertion `&` Reason are True but Reason is not a correct explanation of the Assertion.C. If Assertion is True but the Reason is False.D. If both Assertion `&` Reason are False. |
| Answer» Correct Answer - C | |
| 3159. |
Why is it not possible to measure the single electrode potential ? |
| Answer» Oxidation and reduction do not occur alon. Moreover, it is relative tendency and can be measured with respect to a reference electrode only. | |
| 3160. |
Wrtie down the IUPAC name for each of the following complexes and indicate the oxidation state, electronic configuration and coordination number. Also give stereochemistry and magnetic moment of the complex. (i). `K[Cr(H_2O)_2(C_2O_4)_2].3H_2O` (ii). `[Co(NH_3)_5Cl]Cl_2` (iii). `[CrCl_3(py)_3]` (iv). `Cs[FeCl_4I` (v). `K_4[Mn(CN)_6]` |
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Answer» (a). Potassium diaquadioxalatochromate(III) hydrate. Coordination no. `=6`. Shape`=`octahedral. Oxidation state of `Cr=+3` `Cr(Z=24)=3d^(5)4s^(1),Cr^(3+)=3d^(3),n=3(t_(2g)^(3)e_(g)^(0))` `=mu=sqrt(3xx5)=sqrt(15)BM=3.87BM` (b). Trichloridotripyridinechrornium(III) Coordination no. Of `Cr=6`. Shape `=` Octahedral Oxidation state of `Cr=+3`. `Cr(Z=34)=3d^(5)4s^(1),Cr^(3+)=3d^(3)=t_(2g)^(3)e_(g)^(0),n=3,mu=sqrt(1(1+2))sqrt(3)=1.73BM` (d). Pentaaniminechloridocobalt(III) chloride. C.N. of `Co=6`, shape`=`octahedral O.S. of `Co:x+0-1=+2orx=+3` `Co(Z=27)=3d^(7)4s^(2)`, `Co^(3+)=3d^(6)=t_(2g)^(6)e_(g)^(0),n=0,mu=0` (e). Caesium tetrachloroferrate(III) C.N. of `Fe=4`, shape `=` tetrahedral O.N. Of `Fe:x-4=-1orx=3` `Fe(Z=26)=3d^(6)4s^(2)` `Fe^(3+)=3d^(5)=e^(2)t_(2)^(3),n=5` `mu=sqrt(5+(5+2))=sqrt(35)=5.92BM`. |
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| 3161. |
While numbering , which is given first preference?A. functional groupB. double bondC. triple bondD. substituents |
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Answer» Correct Answer - A |
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| 3162. |
`CH_(4)overset(1000^(@)C)(rarr) A+B_((g))`, `CH_(3)CH_(2)COONaoverset(Sodalime)underset(Delta)(rarr) C+Na_(2)CO_(3)` `C overset(Delta)underset(450^(@)C)(rarr) B+E,E+B overset(Ni)rarrX` If `E` is a hydrocarbon, then `X` is identical withA. (a)`A`B. (b)`B`C. (c )`C`D. (d)`D` |
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Answer» Correct Answer - C `(a) -Carbon, (B) - H_(2), (C ) - C_(2)H_(6), (D)=H_(2), (E)=C_(2)H_(4), (X) = C_(2)H_(6)` |
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| 3163. |
The correct sequence of thermal stability of the following carbonates is :(1) BaCO3 < SrCO3 < CaCO3 < MgCO3 (2) MgCO3 < CaCO3 < SrCO3 < BaCO3(3) BaCO3 <CaCO3 <SrCO3 < MgCO3 (4) MgCO3 < SrCO3 < CaCO3 < BaCO3 |
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Answer» Correct option (2) MgCO3 < CaCO3 < SrCO3 < BaCO3 Explanation: In 2nd group carbonates thermal stability increases down the group due to increase in lattice energy. |
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| 3164. |
Which of the options in the given table are correct? Option | Natural Source | Acid Present (i) Orange | Oxalic acid (ii) Sour milk | Lactic acid (iii) Ant sting | Methanoic acid (iv) Tamarind | Acetic acid (a) (i) and (ii) (b) (i) and (iv) (c) (ii) and (iii) (d) (ii) and (iv) |
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Answer» (c) (ii) and (iii) |
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| 3165. |
Which of the following salts do not have the water of crystalisation? (i) Bleaching Powder (ii) Plaster of Paris (iii) washing soda (iv) Baking soda (a) (ii) and (iv) (b) (i) and (iii) (c) (ii) and (iii) (d) (i) and (iv) |
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Answer» Correct option is (d) (i) and (iv) |
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| 3166. |
Negation of (p ∧ ~ q ) → (~ p ∨ q) is (1) ~(p → q)(2) (~p → q)(3) (p → q)(4) p → ~q |
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Answer» Correct option is (1) ~(p → q) (p ∧ ~ q ) ∧ (p ∨ ~ q) = (p ∧ ~ q) = ~(p → q) |
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| 3167. |
Which ion is not present in teeth enamel(1) Ca2+(2) F-(3) P3+(4) P5+ |
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Answer» Correct option is (3) P3+ Calcium and phosphate are the major component of hydroxyapatite crystal that form the inorganic portion of the teeth. |
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| 3168. |
Boyle’s law states that, when temperature is constant, the volume of a given mass of a perfect gas (a) varies directly as the absolute pressure (b) varies inversely as the absolute pressure (c) varies as square of the absolute pressure (d) does not vary with the absolute pressure. |
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Answer» (b) varies inversely as the absolute pressure |
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| 3169. |
What should be the temperature of 1.3 kg of CO2 gas in a container at a pressure of 200 bar to behave as an ideal ? |
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Answer» Pressure, p = 200 bar Temperature, T = ? For CO2 pc = 73.86 bar Tc = 304.2 K As the gas behaves like an ideal gas, Z = 1 pr = \(\cfrac{p}{p_c}\) = \(\cfrac{200}{73.86}\) = 2.7 From compressibility chart for Z = 1, pr = 2.7 Tr = 2.48 ∴ T = Tr × Tc = 2.48 × 304.2 = 754.4 K. |
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| 3170. |
Joule’s law states that the specific internal energy of a gas depends only on (a) the pressure of the gas (b) the volume of the gas (c) the temperature of the gas (d) none of the above. |
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Answer» (c) the temperature of the gas |
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| 3171. |
The equation of state of an ideal gas is a relationship between the variables : (a) pressure and volume (b) pressure and temperature (c) pressure, volume and temperature (d) none of the above. |
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Answer» (c) pressure, volume and temperature |
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| 3172. |
The equation of the state per kg of a perfect gas is given by (a) p2v = RT (b) pv = RT (c) pv2 = RT (d) p2v2 = RT. where p, v, R and T are the pressure, volume, characteristic gas constant and temperature of the gas respectively. |
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Answer» Correct option is (b) pv = RT |
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| 3173. |
Equation for specific heat at constant pressure of an ideal gas is given by (a) cp = a + KT + K1T2 + K2T3 (b) cp = a + KT2 + K1T3 + K2T4 (c) cp = a + KT2 + K1T4 + K2T (d) cp = a + KT2 + K1T3 + K2T2. where a, K, K1 and K2 are constants. |
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Answer» (a) cp = a + KT + K1T2 + K2T3 |
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| 3174. |
Van der Waals’ equation may be written as(a) \(\left(p+\cfrac{a}{v}\right)\) (v – b) = RT(b) \(\left(p+\cfrac{a}{v^2}\right)\)(v – b) = RT(c) \(\left(p+\cfrac{a}{v^2}\right)\) (v2 – b) = RT(d) \(\left(p+\cfrac{a}{v^2}\right)\) (v2 – b) = RT2 |
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Answer» (b) \(\left(p+\cfrac{a}{v^2}\right)\)(v – b) = RT |
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| 3175. |
The ionization constant of phenol is `1.0xx10^(-10)`. What is the concentration of phenolate ion in `0.05 M` solution of phenol? What will be its degree of ionization if the solution is also `0.01 M` in sodium phenolate? |
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Answer» i. `{:(C_(6)H_(5)OH,hArr,C_(6)H_(5)O^(Θ),+,H^(o+)),(C(1-alpha),,Calpha,,Calpha):}` `K_(a)` of phenol`=1.0xx10^(-10) (Phenol is a W_(A))` `:. alpha=sqrt(K_(a)/C)=sqrt((1.0xx10^(-10))/0.05)=sqrt((10^(-10)xx100)/5)` `sqrt(10^(-10)xx20)=4.47xx10^(-5)M` ii. Phenate ion, `[C_(6)H_(5)O^(Θ)]=Calpha` `=(0.05xx4.47xx10^(-5))` `=2.2xx19^(-6)M` iii. `pH=- log(2.2xx10^(-6))M=5.65` iv. When the concentration of phenate is `0.01 M` (salt) and concentration of phenol (acid) is `0.05 M`. It forms acidic buffer. `:. pH=pK_(a)+log [("Salt")/("Acid")]` `=10+log(0.01/0.05)` `=10-0.7=9.3` |
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| 3176. |
The first ionization constant of `H_(2)S` is `9.1xx10^(-8)`. Calculate the concentration of `HS^(Θ)` ion in its `0.1 M` solution. How will this concentration be affected if the solution is `0.1 M` in `HCl` also? If the second dissociation constant if `H_(2)S` is `1.2xx10^(-13)`, calculate the concentration of `S^(2-)` under both conditions. |
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Answer» i. To calculate `[HS^(Θ)]` `{:(,H_(2)S,hArr,H^(o+),+,HS^(Θ)),("Initial",0.01 M,,,,),("After disso",0.01-x,,x,,x),(,~~0.1,,,,):}` `{:(Initial,0.01 M,,,),(After disso,0.01-x,,x,x),(,~~0.1,,,):}` `K_(a)=(x xx x)/0.1=9.1xx10^(-8) or x^(2)=9.1xx10^(-9)` or `x=9.54xx10^(-5)=[HS^(Θ)]=[H^(o+)]` ii. In `0.1 M HCl:` `{:(HCl,rarr,H^(o+),+,Cl^(Θ)("completely ionised")),(0.1,,0,,0),(0,,0.1,,0.1):}` Due to common ion effect `(H^(o+))`, the disociation of `H_(2)S` is supresed. `:. K_(a_(1))=([H^(o+)][HS^(o+)])/([H_(2)S])=((0.1)[HS^(o+)])/0.1` `:. K_(a_(1))=[HS^(Θ)]=9.1xx10^(-8) M`. iii. `HS^(Θ)rarr H^(o+)+S^(2-)` `K_(a_(2))=([H^(o+)][S^(2-)])/([HS^(Θ)])` `1.2xx10^(-3)=((9.54xx10^(-5))[S^(2-)])/((9.54xx10^(-5)))` `:. [S^(2-)]=1.2xx10^(-13) M`. iv. In presence of `0.1 M HCl`. `K_(a_(1))xxK_(a_(2))=([H^(o+)]^(2)[S^(2-)])/([H_(2)S])` `9.1xx10^(-8)xx1.2xx10^(-13)=((0.1)^(2)[S^(2-)])/((0.1))` `:. [S^(2-)]=(9.1xx10^(-8)xx1.2xx10^(-13))/0.1` `=1.092xx10^(-19) M` |
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| 3177. |
Write the IUPAC name of (i) CH3 – CO – CH2 – CH2 – CH3 |
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Answer» (i) 2-Pentanone (ii) 3-Methyl-1-pentanal |
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| 3178. |
The IUPAC name of is (a) 2-pentyl-4-en-1-ol (b) 4-penten-2-yn-1-ol (c) 1-pentene-3-yn-5-ol (d) 5-Hydroxy-1-pentene-3-yne |
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Answer» (b) 4-penten-2-yn-1-ol |
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| 3179. |
1. What is metamerism? Give example for metamers. 2. What are free radicals? How are they formed? |
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Answer» 1. It is the isomerism which arises due to different alkyl chains on either side of the functional group in the molecule. e.g. Methoxy propane(CH3OC3H7) and ethoxyethane (C2H5OC2H5) are metamers. 2. Free radicals are highly reactive species containing unpaired electrons. They are formed by homolytic cleavage of covalent bond. e.g. H3 |
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| 3180. |
Assertion (A) : Silver salts are used in black and white photography.Reason (R) : Silver salts do not decompose in the presence of light.(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).(C) Assertion (A) is true, but Reason (R) is false.(D) Assertion (A) is false, but Reason (R) is true. |
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Answer» Correct answer is (C) Assertion (A) is true, but Reason (R) is false. Silver salt (AgCl) are used in black and white photography. silver salt (AgCl) is photo sensitive compound, it decomposes into elemental chlorine (Cl2) and Ag(metal). AgBr is also used as black and white photography. 2AgCl \(\overset{h\lambda}{\longrightarrow}\) 2Ag + Cl2 |
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| 3181. |
Minerals enter a plant mainly byA) DiffusionB) Pressure flowC) TranslocationD) Active Transport |
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Answer» Answer: (D) ∙ Option A Diffusion is a process in which what is random movement of the molecules from origin of their concentration to a region of lower concentration. ∙ Option B Pressure flow is a hypothesis based on which the translocation mechanism of the phloem is explained. This is also known as mass flow hypothesis. According to this theory the water containing food molecules flows under pressure through the phloem. ∙ Option C Translocation is the movement of sucrose and other food substances around the plant. It helps in delivering nutrients to all the parts of the plants. ∙ Option D Active transport is the process in which the molecules can move against the concentration gradient as well. This transport process helps the mineral to enter into the plant. The plants take up the mineral ions you are and move them afterwards where the concentration is higher as compared to the concentration in the soil. |
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| 3182. |
Explain the process of producing alumina from bauxite. |
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Answer» Bauxite is treated with hot concentrated NaOH. Aluminum oxide reacts with NaOH solution forming sodium aluminate solution (NaAlO2). The unreacted impurities are removed from the solution. To the remaining solution of sodium aluminate, add Al(OH)3 in small quantity and dilute with water. Then the whole aluminum in the solution gets precipitated as Al(OH)3 . The Al(OH)3 precipitated is separated from the solution. Then it is washed and heated strongly. Then Al(OH)3 decomposes to give pure AL2O3 or alumina. 2Al(OH)3(s) → Al2O3(s) + 3H2O(1). |
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| 3183. |
Manufacturing of iron. a. Name the furnace used for producing iron. b. Name the materials using for producing iron. c. Write down the reaction occurring on coke when hot is blasted on it? d. Why CaCO3 is dropping inside the furnace? e. Write down the nature of gangue with iron ore. f. Gangue + flux → ……….. Write down the uses of the product formed in blast furnace. g. Reducing agent in blast furnace h. Write down the reactions taking place inside the blast furnace. i. Iron formed from the blast furnace is called …………… j. How can we change iron into steel? k. What are the different types of steel? |
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Answer» a. Blast Furnace b. Mixture of roasted haematite, coke and limestone. c. At the bottom of the blast furnace, coke combines with oxygen in the hot current of air. The CO2 which rises up along with the hot air current is reduced by coke. d. CaCO3 in the furnace decomposes to form CaO and CO2 . CaO which is basic combines with SiO2 (acidic) to form slag. Molten slag which is lighter floats over the heavier molten iron. CaCO3 is added to the furnace for the production of slag. e. Acidic f. Slag. (CaSiO3) used for the production of cement and in the Construction of Road. g. Carbon monoxide h. CaCO3 (s) → CaO + CO2 CaO + SiO2 → CaSiO3 C + O2 CO2 + Heat CO2 (g) + C(s) + Heat → 2CO (g) Fe2O3 +3CO → 2Fe + 3CO2 i. Pig iron j. Different types of steel can be prepared by varying the amount of carbon from 0.1 to 1.5 k. Mild steel, Medium steel, High carbon steel Mild steel:- If the carbon content in steel is from 0.05% to 0.2 % then it in called mild steel. I Medium steel :- Medium steel contains carbon from 0.2 % to 0.6 % medium steel High carbon steel: – If the content of carbon is from 0.61 % to 15 % then it is known as high carbon steel. |
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| 3184. |
a. Write down the names of Anode, Cathode, Electrolyte used in the Electrolyte cell for the manufacturing of copper, b. Write down the equations for the reaction in anode and cathode. |
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Answer» Anode – Copper to be refined Cathode – Pure Copper Electrolyte – Aqueous Copper Sulphate Solution mixed with H2SO4 b. Anode – Cu → Cu2 + 2e- Cathode – Cu2+ +2e- → Cu |
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| 3185. |
Analyze the IUPAC names given to the following organic compounds and correct them, if incorrect.iii. CH3 – CH2 – CH = CH – CH3 : Pent – 3 -ene |
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Answer» i. 3 – Methylhexane ii. Pent – 2 – ene |
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| 3186. |
Given below are the equations for the reaction taking place inside the blast furnace.C + O2 → CO2CO2 + C → 2CO CaCO3 + SiO2 → CaSiO3 Fe2O3 +3CO → 2Fe + 3CO2 a. Name the ore of iron. b. Which is the gangue in iron ore? c. Name the flux used in blast furnace, d. Gangue + flux → ………… Which product is formed in blast furnace? e. Reducing agent used in blast furnace. f. Subjects dropped in blast furnace are …., ……… |
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Answer» a. Haematite b. SiO2 c. CaO d. slag(CaSiO3) e. Carbon monoxide f. A mixture of roasted haematite, coke and limestone. |
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| 3187. |
An organic compound is given below. CH3 – CH2 – CH2 – CH2 = CH2 – CH3 Pick out suitable statements for the given compound from below. a. It’s a saturated compound b. The general formula is CnH2n c. It’s an alkene d. IUPAC name is hex – 4 – ene e. Similar with the molecular formula of cyclohexane. f. IUPAC name is hex – 2 – ene. |
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Answer» b. The general formula is CnH2n c. It’s an alkene e. Similar with the molecular formula of cyclohexane. f. IUPAC name hex – 2 ene. |
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| 3188. |
1. What are the functional groups in CH3 – CH2 – OH and CH3 – O – CH3 ?2. Try to write down their molecular formula3. Are they isomers? |
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Answer» 1. -OH,CH3-O- 2. C2H6O 3. Yes. They show functional isomerism |
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| 3189. |
CH3 – CH2 – C = C – CH2 – CH3 a. Write the IUPAC name of this organic compound. b. Write the structure of any two isomers of this compound. |
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Answer» a. Hex-3-yne b. CH3 – C = C – CH2 – CH2 – CH3 CH3 – CH2 – C = C – CH2 – CH3 |
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| 3190. |
The structural formulae of two organic compounds are given. i. CH3 – O – CH2 – CH3 ii. CH3 – CH2 – CH2 – OHa. What are the IUPAC name of these compounds? b. Write one similarity and one difference between these two compounds, c. What is this phenomenon known as? |
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Answer» a. i. Methoxy ethane ii. Propan – l – ol b. Similarity: Molecular formula is same Difference: Functional group is different c. Functional group isomerism. |
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| 3191. |
a. Complete the table. Method of preparation & Content Pig ironCast ironWrought ironSteel ...i........ii.......iii.......iv....b. Stainless steel and Nichrome are having same content (Fe, Ni, Cr, C). But nature of both alloys are different Why? c. Bauxite and clay are minerals of aluminum. But bauxite is the only ore of Aluminium. Why?. |
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Answer» a. i. Obtained from Blast Furnace Contains 4% Carbon and other impurities like manganese silicon, phosphorus, etc. ii. Pig iron mixed with scrap iron and coke melted in a special furnace contains 3% carbon. iii. Made by purifying cast Iron. iv. Prepared by varying the amount of carbon from 0.1 to 1.5%. b. Stainless steel and Nichrome are having the same content but the nature of both alloys are different because ratio of constituent elements are different. c. A mineral from which a metal is economically, easily and quickly extracted is called the core of the metal. Among the minerals of aluminum, bauxite possesses these properties. Hence bauxite is the ore of Aluminium. Since clay does not possess these properties it is not an ore of Aluminium. |
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| 3192. |
Which of the following is not the characteristic of interhalogen compounds ?(a) They are more reactive than halogens(b) They are quite unstable but none of them is explosive(c) They are covalent in nature(d) They have low boiling points and are highly volatile. |
| Answer» (d) Interhalogen compounds are not highly volatile | |
| 3193. |
Interhalogen compounds are more reactive than the individual halogen because(a) two halogens are present in place of one(b) they are more ionic(c) their bond energy is less than the bond energy of the halogen molecule(d) they carry more energy |
| Answer» (c) The bond energy of interhalogen compounds is less than the bond energy of halogens. | |
| 3194. |
Which one of the following halogens forms only one oxo acid ?A. `Br_2`B. `Cl_2`C. `F_2`D. `l_2` |
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Answer» Correct Answer - C `F_2` is forms only one oxo acid , HOF (or HFO), which is prepared by passing `F_2` over ice at `-40^(@)C`. `F_2+H_2Ooverset(-40^@C)toHOF+HF` |
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| 3195. |
Statement-l: Fluorine form only one oxoacid.Statement-II: Due to high electronegativity and small size. (1) Both Statement-l & Statement-ll are true (2) Both Statement-l & Statement-ll are false(3) Statement-I is true & Statement-ll is false (4) Statement-I is false & Statement-ll is true |
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Answer» (1) Both Statement-l & Statement-ll are true Due to small size & high electronegativity F can not act as central atom in higher oxidation state so F form only one oxy acid HOF. |
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| 3196. |
Assertion: Fluorine forms only one oxo acidReason: It is small and electronegative(a) Both assertion and reason are true, reason is correct explanation of assertion.(b) Both assertion and reason are true, but reason is not a correct explanation of assertion.(c) Assertion is true, but reason is false(d) Assertion is false, but reason is true |
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Answer» (a) Both assertion and reason are true, reason is correct explanation of assertion. Due to high electronegativity and small size, fluorine forms only one oxoacid, HOF known as fluoric (I) acid or hypofluorous acid. |
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| 3197. |
The number of Bravais lattices in a cubic crystal is – (A) 1 (B) 3 (C) 4 (D) 14 |
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Answer» Correct answer is (B) 3 |
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| 3198. |
Two blocks of mass 2 kg are connected by a massless ideal spring of spring constant capital K=10 N/m. The upper block is suspended from roof by a light string A .The system shown is in equilibrium. The string is now cut the acceleration of upper block just after the string is cut will be?Please answer in detail!! |
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Answer» Let m1 be the mass of first block and m2 of second block.. When the system is in equilibrium .. m2g=kx 2×10=kx Kx=20 m1g +kx = T 20+20=T T=40 Now when the string A is cut , we have removed the tension that balances m1g + kx.. This leads to acceleration of the block. Acceleration= 40/2 =20m/s Which is the intial acc just after the string is cut. But when the block comes in contact with m2 and the spring contracts both will fall with the same acc g. Hope u find it helpful. |
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| 3199. |
In the fig. The block of mass M is at rest on the floor. The acceleration with which a monkey of mass m should climb along the rope of negligible mass so as to lift the block from floor is? |
| Answer» The answer will be option (B) | |
| 3200. |
The atomic radius of strontium `(Sr)` is `215 p m` and it crystallizes with a cubic. Closest packing . Edge length of the cube is `:`A. 430 pmB. 608.2 pmC. 496. pmD. None of these |
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Answer» Correct Answer - B For ccp Lattice a `=(4r)/sqrt(2)=(4xx215)/sqrt(2)=608.2` pm |
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