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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.
| 49951. |
A list of species having the formula XZ_(4)is given below : XeF_(4) ,SF_(4), SiF_(4) ,BF_(4)^(-),BrF_(4)^(-),[Cu(NH_(3))_(4)]^(2+) ,[FeCl_(4)]^(2-) ,CoCl_(4)^(2-) and [PtCl_(4)]^(2-) Defining shape onthebasis of the lacation ofX and Z atoms , thetotal number of species having a square planar shape is |
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Answer» `SiF_(4)` = Tetrahedral , `BF_(4)^(-)` = Tetrahedral , `BrF_(4)^(-)` = Square planar . `[Cu(NH_(3))_(4)]^(2+)` = Squara planar, `[FeCl_(4)]^(2-)` = Tetrahedral , `[CoCl_(4)]^(2-) `= Tetrahedral , `[PtCl_(4)]^(2-)` = Squara planar. |
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| 49952. |
A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased. What happens when equilibrium is restored finally and what will be the final vapour pressure? |
| Answer» Solution :When equilibrium is RESTORED, RATE of evaporation become equal to the rate of CONDENSATION. The final VAPOUR pressure will be the same as the original vapour pressure because vapour pressure depends only on temperature. | |
| 49953. |
A liquit is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased. How do rate of evaporation and condensation change initially? |
| Answer» Solution :The rate of evaporation REMAINS constance TEMPERATURE. But the rate of condensation will decrease INITIALLY because there are fewer MOLECULES per unit volume in the vapour PHASE. | |
| 49954. |
A liquid is in equilibrium with its vapour in a sealed container at a fixed temoperature. The volume of the container is suddenly increased. What is the initial effect of the change on vapour pressyre? |
| Answer» SOLUTION :INITIALLY, the VAPOUR pressure will decrease because the same amount of vapour will be distributed in LARGER VOLUME. | |
| 49955. |
A liquid with high boiling point decomposes on simple distillation but it can be steam distilled for its purification. Explain how is it possible? |
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Answer» Solution :In vapour phase distillation vapour of organic compound and water is distilled out water vapour is passed in LIQUID which should be purified. So, vapour of liquid and vapour of water rises up and pass in condenser. So, liquid and water remain non-misible and form separate layer. When the total vapour PRESSURE of liquid and water becomes equal to ATMOSPHERIC pressure than distillation take place. so, `(p_(1) + p_(2))= p " " p_(1) lt p and p_(2) lt p` `:. p_(1) = (p- p_(2))` Vapour pressure of organic liquid (A) is less than total pressure (p). so, vapour distillation take place at lower temperature than BOILING POINT |
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| 49956. |
A liquid which decomposes ator below its boiling point can be purified by |
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Answer» STEAM distillation |
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| 49957. |
A liquid which decomposes at its boiling point can be purified by |
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Answer» A)distillation at ATMOSPHERIC pressure |
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| 49958. |
A liquid which decomposes at its boiling point can be purified by ........ |
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Answer» DISTILLATION at ATMOSPHERIC PRESSURE |
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| 49959. |
A liquid weighing 86.44 g occupies a volume of 76.3 mL. Calculate the density of the liquid to the proper number of significant figures. |
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Answer» Solution :Density = `("MASS")/("VOLUME")` `=(86.44 G)/(76.3 ML)` `=1.1328964 g//mL` =1.13 g/mL (after rounding off) In this calculation, the least precise number has only three SIGNIFICANT figures. Therefore, the result is also reported upto three significant figures. |
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| 49960. |
A liquid non0volatile impurities. How will you obtain pure liquid ? |
| Answer» SOLUTION :By DISTILLATION PROCESS. | |
| 49961. |
A liquid is tranferrred from a smaller vessel to a bigger at the same temperature . What will be the effect on the vapour pressure ? |
| Answer» SOLUTION :No EFFECT as it DEPENDS only on the nature of the LIQUID and temperature. | |
| 49962. |
A liquid is inequilibrium with its vapur in a sealed container at a fixed temperature. The volume of the container is suddentlyincreases. (a)What is the intial effect of the change on vapour pressure ? (b) How do rates of evaporation and consideration change intially ? (c) What happens when equilibrium is restored finally and what will be the final vapour pressure ? |
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Answer» Solution :(a) Intially, the vapour pressure will decrease because the same account of vapour are now distributed in larger space. (b) The RATE of EVAPORATION REMAINS constant at constant temperature in a closed vessel 9discussed in unit 5 under vapur pressure. ). However, the rate of condensation will be low INITIALLY because there are fever molecules per unit volume in the vapour phase and hence the number ofcollisions per unit time with the liquid surface DECREASES. (c ) When equilibriumis restored , rate of evaporation = rate of consensation . The final valour pressure will be same as it was originally because vapour pressure of a liquid depends only on temperature and not volume. |
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| 49963. |
A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased. (a) What is the initial effect of the change on vapour pressure ? (b) How do rates of evaporation and condensation change initially? (c) What happens when equilibrium is restored finally and what will be the final vapour pressure ? |
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Answer» Solution :(a) Initially the vapour pressure will increase. (B) The rate of evaporation will remain constant at constant temperature in closed vessel. HOWEVER, rate of CONDENSATION will be low initially, because there are lesser number of molecules per UNIT volume in the vapour PHASE. Therefore, the number of collisions per unit volume with the liquid surface decreases. (c) When equilibrium is restored, rate of evaporation becomes equal to rate of condensation. The final vapour pressure will be same as it was originally. |
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| 49964. |
A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased. How do rates evaporation and condensation change initially ? |
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Answer» Solution :`underset("Low pressure")(A_((l)))tounderset("High pressure")(A_((g))` If volume is increased at CONSTANT temperature, pressure DECREASES, since, `p prop 1//V` at constant temperature. Rate of EVAPORATION INCREASES and rate of condensation decreases. |
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| 49965. |
A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased. What happens when equilibrium is restored finally and what will be the final vapour pressure ? |
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Answer» Solution :`underset("Low pressure")(A_((l)))tounderset("HIGH pressure")(A_((g))` If volume is increased at constant temperature, pressure decreases, since, `p prop 1//V` at constant temperature. When EQUILIBRIUM is RESTORED finally the rate of EVAPORATION again becomes equal to the rate of condensation and the final vapour pressure becomes equal to the vapour pressure that was before the sudden INCREASE in the volume of the container. |
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| 49966. |
A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased. What is the initial effect of change on vapour pressure ? |
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Answer» Solution :`underset("Low pressure")(A_((l)))tounderset("High pressure")(A_((g))` If volume is INCREASED at constant temperature, pressure DECREASES, since, `p prop 1//V` at constant temperature. Decrease in pressure SHIFT the equilibrium in the direction of high pressure i.e. more vapours are formed HENCE vapour pressure INCREASES. |
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| 49967. |
A liquid is in equilibrium with its vapours at its boiling point. On the average, the molecules in the two phases have euqal : |
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Answer» POTENTIAL energy |
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| 49968. |
A liquid is in equilibrium with its vapour at its boiling point. On the average, the molecules in the two phases have equal: 1)potential energy 2)total energy 3)kinetic energy 4)intermolecular forces. |
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Answer» POTENTIAL ENERGY |
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| 49969. |
A liquid is found to scatter a beam of light but leaves no residue when passed through the filter paper. The liquid can be described as |
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Answer» A suspension |
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| 49970. |
A liquid has non-volatile impurities. Which method will you use for its purification ? |
| Answer» SOLUTION :DISTILLATION | |
| 49971. |
A liquid compound starts decomposing well before its boiling point under normal pressure. How will you purify it ? |
| Answer» SOLUTION :By DISTILLATION under REDUCED PRESSURE. | |
| 49972. |
A liquid can exist only, |
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Answer» Between triple POINT and critical point. |
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| 49973. |
A limiting reagent is the reactant which gets completely ..........during the reaction. |
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Answer» |
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| 49974. |
A lighter gas diffuses more rapidly than a heavier gas At a given temperature the rate of diffusion of a gas is inversely proportional to the square root of its density . |
| Answer» Solution :RATE of DIFFUSION `PROP sqrt((1)/(M))` . | |
| 49975. |
A light source of wavelength lambdailluminates a metal and ejects photo-electrons with (K.E.)_("max") = 4 eV . Find the value of work function ? |
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Answer» 1eV `phi ` for each metal is constant . if incident ENERGY is KNOWN then `phi` can be calculated. |
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| 49976. |
The work function for a metal is 4 eV. To eject the photo electrons with zero velocity the wavelength of the incident light should be |
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Answer» On solving `phi = 0.5` |
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| 49977. |
A light is passed through a sample of a single elecrton excited spice which causes further excitation to some higher energy level and it emit a photon of energy 12.09 eV in back trabsition to some lower energy higher energy level .Then (i)Fing=d out the atomic number of specie ifDebroglie wave lenght of electronin first excited stateis 221.4 pm. (ii) Find out the value of higher energy level and lower energy levelin back transition . |
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Answer» `lambda^(2)=(150)/(V)xx10^(-20)` `V=(150)/(221.54)xx(10^(-20))/(10^(-24))=30.6` `therefore`(T.E) First excited state=-`13.6((Z^(2))/2^(2))``=-30.6 EV` `Z=3 Ans` (ii) `E_(n2)-E_(N1)`= `DeltaE=12.09=13.6xx9((1)/(n_(1)^(2))-(1)/(n_(2)^(2)))` `n_(1)=3`, `n_(2)=9` |
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| 49978. |
A light green coloured salt soluble in water gives black percipitate on passing H_(2)S which dissolves readily in HCl. The metal ion present is : |
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Answer» `Co^(2+)` |
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| 49979. |
Assertion: Li generally forms covalent compounds Reason : Li^+ ion is small and has high polarizing power |
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Answer» Both (A) and (R) are TRUE and (R) is the CORRECT EXPLANATION of (A) |
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| 49980. |
A leaser emits monochromatic radiation of wavelength 663 nm. If it emits 10^(5) quanta per second per square metre, calculate the power output of the laser in joule per square metre per second. |
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Answer» Solution :ENERGY of one quantum `= hv = h (c)/(lamda) = ((6.63 xx 10^(-34)Js) (3 xx 10^(8) ms^(-1)))/((663 xx 10^(-9)m)) = 3 xx 10^(-19) J` Energy emitted per SEC per square metre = No. of quanta emitted per sq. metre per sec `xx` Energy of one quantum. `= 10^(15) xx 3 xx 10^(-19) J m^(-2) s^(-1)` `= 3xx 10^(-4) J m^(-2) s^(-1)` |
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| 49981. |
A layer of coke is spread over bauxite during extraction of aluminium. This acts as a/an |
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Answer» Flux |
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| 49982. |
(A) Lassaigne's test is not shown by diazonium compounds. (R) Diazonium compounds lose N_(2) on heating |
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Answer» If both ASSERTION and reason are CORRECT and reason is the correct explanation of the assertion |
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| 49983. |
(A) Lassaigne's test can be used to detect nitrogen in hydrazine. (R) During fusion with sodium metal, nitrogen and carbon of the organic compound combine to form sodium cyanide. |
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Answer» If both ASSERTION and reason are correct and reason is the correct EXPLANATION of the assertion |
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| 49984. |
A large number of polymers are quite resistant to the environment degradation process and are thus, accumulation of polymeric solid waste material. Certain new biodegrdable synthetic polyermer have been designed and development. Among these the most important ar aliphatic polyeters and polyamides. Nylon-2-nylon-6 |
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Answer»
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| 49985. |
A large number of polymers are quite resistant to the environment degradation process and are thus, accumulation of polymeric solid waste material. Certain new biodegrdable synthetic polyermer have been designed and development. Among these the most important ar aliphatic polyeters and polyamides. In the manufacture of LDPE, it is a: |
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Answer» polymer of ethylene |
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| 49986. |
A large number of polymers are quite resistant to the environment degradation process and are thus, accumulation of polymeric solid waste material. Certain new biodegrdable synthetic polyermer have been designed and development. Among these the most important ar aliphatic polyeters and polyamides. In the manufacture of homopolymer HDPE, it is a: |
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Answer» POLYMER of ethylene |
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| 49987. |
A large number of fishes are suddenly found floating dead on a lake. There is no evidence of toxic dumping but you find an abundance of phytoplankton. Suggest a reason for the fish killing |
| Answer» SOLUTION :Excessive phytoplankton organic pollutants such as leaves, grass trash etc.) growth which is PRESENT in WATER is biodegradable. Bacteria decomposes this organic matter in water. During this process when large number of bacteria decomposes the organic matter, they CONSUME the dissolved oxygen in water. When the level of dissolved oxygen falls below 6 PPM then thefishes cannot survive and they die. | |
| 49988. |
A large number of fish are suddenly found floating dead on a lake. There is no evidence of toxic dumping but you find an abundance of phytoplankton. Suggest a reason for the fish kill |
| Answer» Solution :Excessive phytoplankton (organic pollutants such as leaves, grass, trash, etc.) present in water is biodegradable. large population of bacteria decomposes this organic matter in water. During this process, they consume the oxygen DISSOLVED in water. Water has already limited dissolved oxygen (10 ppm). Thus, it is further depleted When the LEVEL of dissolved oxygen falls below 6 ppm the fish cannot survive HENCE, they die and lot DEAD on the lake | |
| 49989. |
A large number of fish are suddenly found floating dead on a lake. There is no evidence of toxic dumping but you find an abundance of phytoplankton. Suggest a reason for the fish kill. |
| Answer» Solution :Phytoplankton CONSISTS of organic contaminants such as grass leaves ETC. In water these materials are DEGRADED by BACTERIA and a large amount of dissolved oxygen will be used up for this process. As the amount of DO drops below 6 ppm, fish cannot SURVIVE and they float dead. | |
| 49990. |
A large flask fitted with a stopcock is evacuated and weighted. Its mass is found to be 134.567 g. It is then filled at a pressure of 735 mm and 131^(@)C with a gas of unknown molecular mass and then reweighed. Its mass is 137.456 g. The flask is then rewighed. Its mass weighted again, its mass is now 1067.9 g. Assuming that the gas is ideal, calculate the molar mass of the gas. |
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Answer» Solution :Mass of WATER filling the flask is `(10.67.9-134.567)g=933.333g` `therefore` volume of flask=volume of water filling the flask LTBR `=933.3cm^(3)` `(therefore` density of `H_(2)O=1gcm^(-3)`) Now `P=735mm`, `T=31+273K=304K` `V=933.3cm^(3)` applying `PV=nRT` `((735)/(760)atm)((933.3)/(1000)L)=nxx0.0821xx304` gives `n=0.036mo` Mass of 0.036 mol of the gas `=(137.456-134.567)g` `=2.889g` therefore mass of 1 mol of the gas `=(2.889)/(0.036)=80.25g` `mol^(-1)` |
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| 49991. |
A large flask fitted with a stop-cock is evacuatedand weighed, its massis found to be 134.567 g . It is the filled to a pressure of 735 mm at 31^(@)C with a gas of unknown molecular mass and then reweighed , its mass is 137.456g. The flask is then filled with water and weighed again , its mass is now 1067.9g. Assuming that the gas is ideal, calculate the molar mass of the gas. |
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Answer» `:.` Volume of flask =Volume of water filling the flask`=933.3 cm^(3)"" (`:'` "density of "H_(2)O=1 g cm^(-3))` Now,P=735 mm, T=31+273 K=304K, V`=933.3 cm^(3)` Applying PV=nRT, i.e., `((735)/(760)atm)((933.3)/(1000)L)=nxx0.0821 L atm K^(-1)mol^(-1)xx304 K` This gives n=0.036 mol Mass of 0.036 mol of the gas =(137.456-134.567) g=2.889 g `:.`MASSOF 1 mol of the gas `=(2.889)/(0.036)=80.25 g` |
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| 49992. |
A laboratory analysis of an organic compound gives the following mass percentage composition: C = 60%, H = 4.48% and remaining oxygen. Find out the Empirical Formula of the compound. |
Answer» Solution : `:.` The empirical FORMULA is `C_(9)H_(8)O_(4)` |
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| 49993. |
a KMnO_(4) + bH_(2) SO_(4) + c FeSO_(4) to K_(2) SO_(4) + MnSO_(4) + Fe_(2) (SO_(4))_(3) + H_(2) O . In this unbalanced stoichiometric equation the values of a , b and c respectively are |
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Answer» 2,8,10 |
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| 49994. |
A jug contains 2L of milk. Calculate the volume of the milk in m^(3). |
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Answer» SOLUTION :`"1 L "="1000 cm"^(3) and 1m=100 cm` `(1m)/(100CM)=1=(100cm)/(1m)` To GET `m^(3)` from the above UNIT factors, the first unit factor is taken and it is cubed. `((1m)/(100m))^(3)rArr(1m^(3))/(10^(6)cm^(3))=(1)^(3)=1` Now `2L=2xx1000cm^(3)` The above is mltiplied by the unit factor `2xx1000cn^(3)xx(1m^(3))/(10^(6)cm^(3))=(2m^(3))/(10^(3))=2xx10^(-3)m^(3)` |
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| 49995. |
A J-shaped tube with smaller end closed and longer end open was taken. Mercury was added into it till the level of mercury in both the limbs was same. Volume of air enclosed in the closed end was found to be 2.4 mL. Now more mercury was added and the air enclosed in the closed ed reduced to 1.9 mL. Now, the difference in the level of the two limbs will be |
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Answer» 43 cm  INITIALLY, `P_(1)=760" mm", V_(1)=2.4" mL"` Finally, `P_(2)=?, V_(2)=1.9" mL"` `P_(1)V_(1)=P_(2)V_(2)" or "P_(2)=(P_(1)V_(1))/(V_(2))` `=(760xx2.4)/(1.9)=960" mm Hg"` `:.` Difference in the two LIMBS =960-760 mm=200 mm=20 cm |
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| 49996. |
A IUPAC name for the group CH_3undersetoverset(|)(CH_2CH_3)(C )HCH_2 - is |
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Answer» ISOPENTYL |
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| 49997. |
A IUPAC name for the group CH_(3)underset(CH_(2)CH_(3)" ")underset("|")(CHCH_(2)) - is : |
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Answer» ISOPENTYL |
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| 49998. |
A isoelectric point |
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Answer» COLLOIDAL PARTICLES BECOME uncharged |
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| 49999. |
'A' is widely used as slow acting nitrogeneous fertilizer. |
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Answer» `CaNCN` |
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| 50000. |
A is a higher phenol and B is an aromatic carboxylic acid. Separation of a mixture of A andBcan be carried out easily by having a solution |
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Answer» NaOH |
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