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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.
1051. |
Name the event during cell divison cycle that results in the gain or loss of chromosome. |
Answer» Aneuploidy- It is the everit during cell divising cycle that results in the gain or loss of chromosome and failure of segregation of chromosomes | |
1052. |
Name the type of cell division that takes place in the zygote on an organism exhibiting haplontic life cycle |
Answer» Mitosis takes place in the zygote of an organism exhibiting haplontic life cycle | |
1053. |
List the two methodologies which were involved in human genome project . Mention how they were used. Explain YAC and mention what was it used for. |
Answer» (a) The methods involved two major approaches . One approach focused on identifying all the genes that are expressed as RNA to as expressed sequence sequence tags . The other took the blind approach of simply sequencing the whole set of genome that contained all the coding and non - cording seqeuences and later assigning different regions in the sequence with functions. YAC (Yeast Articial Chromosomes) for sequencing , the total DNA from a cell is isolated and converted into random fragments of relatively smaller sizes and cloned in suitable host using specialised vectors . the cloning resulted into amplification of each pices of DNA fragment so that it subsequently could be sequenced with ease. |
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1054. |
How do copper and hormone releasing IUDs act as contraceptives? Explain. |
Answer» The copper releasing IUDs release Cu ions which suppress sperm motility and the fertilising capacity of sperms. The hormone releasing IUDs make the uterus unsuitable for implantation and the cervix hostile to the sperms. | |
1055. |
State the role of C peptide in human insulin. |
Answer» The C-peptide is an extra polypeptide stretch present in human pro-insulin. This C-peptide is removed on maturation of pro-insulin to insulin. C-peptide in efficient assembly folding processing of insulin. | |
1056. |
What is the major difference you observe in the offsprings produced by asexual reproduction and in the progeny produced by sexual reproduction ? |
Answer» The progeny formed from asexual reproduction is genetically similar to parents and it does not show variation. But in aexual reproduction, individuals produced as a result of meiosis and gametic fusion exhibit genetic variation and difference from either of the two parents as well as among themselves. |
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1057. |
Mention the role of ribosomes in peptide-bond formation. How does ATP facilitate it? |
Answer» There are two sites in the large subunit of the ribosome, for subsequent amino acids to bind to and thus, be close enough to each other for the formation of a peptide bond. The ribosome also acts as a catalyst for the formation of peptide bond. Amino acids become activated by binding with their aminoacyl-tRNA synthetase in the presence of ATP. | |
1058. |
Offsprings derived by asexual reproduction are called clones. Justify giving two reasons. |
Answer» In asexual reproduction an indiviual parent brings forth the offsprings. These offsprings are identical to each other and also exact copies of their parents. Such morphologically and genetically similar individual are called clones. | |
1059. |
Classify colloids where dispersion medium is water. State their characterstics and write one example of each of these classes. |
Answer» (i)Sol =When solid is dipersed is water eg-solid,starch sol. (ii) Emulsion : When liquid is dispersed in water ex=milk. (iii) From : When gas is dipresed in water eg.Soap lather , whipped cream . |
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1060. |
(i) Give reasons : (a) HCHO is more reactive than `CH_3-CHO` towards addition of HCN . (b) pKa of `O_2N-CH_2-COOH` is lower than that of `CH_3-COOH`. (c) Alpha hydrogen of aldehydes and ketones is acidic in nature. (ii) Give simple chemical tests to distinguish between the following pairs of compounds : (a) Ethanal and Propanal (b) Pentan-2 one and Pentan-3-one |
Answer» (i)(a) In ethanl , the `(+)` ve charge on the carbonyl carbon is lowerd by high `+I` effect of two alkyl groups in comparison to methanol in which 2 hydrogens are present . Further increase in the no. of alkyl groups also increases the steric hindrance making it less reactive. Thus, in general, HCHO is more reactive than `CH_3CHO` towards addition of nucleophile HCN. (b) `NO_2` is withdrawing group, which disperse the negative charge on carbonlate ion (decreases electron density) stabilize the carboxylate anion and thus, increases the acidity of carboxylic acid. Greater the acidic character, lower will be the pKa value. Hence, `O_2N-CH_2COOH` is having lower value of pKa than `CH_3COOH`. (c) `alpha`-hydrogen of aldehyde and ketone is attached to carbonyl carbon. After removal of `alpha`-Hydrogen, conjugated base so obtained is resonance Hence `alpha-H` in aldehyde and ketone is acidic in nature. (ii) `overset(|)underset(Theta)underset (..)(C)-overset(..)overset(O: )ChArr-overset(|)C=overset(-)overset( :O: )overset(|)(C)-` (ii) (a) Ethanal `(CH_2CHO)` and propanal `(CH_3CH_2CHO)` `CH_3CHO+l_(2)+NaOHoverset("Iodoform")underset("Reaction")to` no reaction (NO yellow ppt) (b) Penton -2-one pentan-3-one `underset("Penton-2-one")(C-C-C-overset(O)overset(||)C-C+l_2)+NaOHoverset("Iodoform")underset("Reaction")tounderset("Yellow ppt")(CHI_3)+CH_3CH_2CH_2-COONa+Nal+H_2O` `C-C-C-overset(O)overset(||)underset("Penton-2-one")(C-C+l_2+NaOH)overset("Iodoform")underset("Reaction")tounderset("No Yellow ppt")("No reaction")` |
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1061. |
What are the main difference of glass, made up `SiO_(4)` tetrahedral ? |
Answer» Glass is an amorphous solid. | |
1062. |
(a) Complete the following chemical equations : (i) `MnO_(4)^(-) (aq) + S_(2)O_(3)^(2-) (aq) + H_(2)O(l) to ` (ii) ` Cr_(2)O_(7)^(2-)(aq)+Fe^(2+) (aq) + H^(+) (aq) to ` (b) Explain the following observatons : (i) `La^(3+) (Z = 57) and Lu^(3+) (Z = 71)` do not show any colour in solutions. (ii) Among the divalent cations in the first series of transition elements, manganese exhibits the maximum paramagnetism. (iii) `Cu^(+)` ion is not known in aqueous solutions. |
Answer» (a) (i) `8 MnO_(4)^(-)(aq)+ 3 S_(2)O_(3)^(2-) (aq) + H_(2)O(l) to 8 MnO_(2) (s) + 6 SO _(4)^(2-) (aq) + 2OH^(-) (aq)` (ii) `Cr_(2)O_(7)^(2-)(aq)+6 Fe^(2+) (aq) + 14 H^(+) (aq) to 2 Cr^(3+) (aq) + 6 Fe^(3+) (aq) + 7 H_(2) O (l) ` (b) (i) It is due to obsence of unpaired electrons, `(e^(-)s)` they cannot undergo f - f transion. (ii) ` Mn^(2+)` has maximum number of unpaired electrons (iii) Since` Cu^(+)` in aqueous solution undergoes disproportionation i.e. ` 2 Cu^(+) (aq ) to Cu^(2+) (aq ) + Cu(s)` ` Cu^(2+)`(aqu) has more negative hydration energy which can over come ` 2nd` I. E energy. |
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1063. |
Describe the roles of heat , primers and the bacterium Thermus aquaticus in the process of PCR. |
Answer» (i) Heat : To denature or separation of DNA into two strands. (ii) Thermus aquaticus : Amplification is achieved by DNA polymerase isolated form bacteria thermus qauaticus, which remain active dursig high temperature induced desaturation of double stranded DNA. |
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1064. |
Describe the roles of heat , primers and the bacterium Thermus aquaticus in the process of PCR. |
Answer» (i) Heat : To denature or separation of DNA into two strands. (ii) Thermus aquaticus : Amplification is achieved by DNA polymerase isolated form bacteria thermus qauaticus, which remain active dursig high temperature induced desaturation of double stranded DNA. |
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1065. |
Justify that electrostatic potential is constant throughout the volume of charged conductor and has same value on its surface as inside it. |
Answer» E=0 in a conductor makes `Q_(cm)=0` and charges stay on the surface. So, the potential is constant and will be the same as on the surface.. | |
1066. |
Identify the picture and mention the vegetative part that helps it to propagate. |
Answer» Rhizome of ginger, the vegetative propagation occurs through the underground stem called Rhizome. A rhizome is an underground horizontal stem with stored fo9d and bud. | |
1067. |
State the type of interaction that exists between ticks and dogs. |
Answer» Correct Answer - Parasitism. | |
1068. |
A first order reaction has a rate constant of `0.0051 "min"^(-1)`. If we begin with `0.10M` concentration of the reaction , what concentration of reactant will remain in solution after 3 hours ? |
Answer» Here , `[R]_(o) = 0.10M` , [R] = ? t = 3 hours = `3 xx 60 = 180` min `k = 0.0051 "min"^(-1)`. Using the formula k = `(2.303)/(t) log ([R]_(o))/([R]) = 0.0051 = (2.303)/(180) log.(0.10)/([R])` `log.(0.10)/([R]) = (0.0051 xx 180)/(2.303)` `log.(0.10)/([R]) = (0.918)/(2.303) = 0.3986` `therefore " " (0.10)/([R])` = antilog `(0.3986)` `(0.10)/([R]) = 2.503 implies [R] = (0.10)/(2.503) = 0.039` M |
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1069. |
(a) What are fuel cells ? Give an example of a fuel cell. (b) Calculate the equilibrium constant `(logK_(c))and Delta_(r)G^(@)` for the following reaction at 298 K. `Cu_((s))+2Ag^(+)""_((aq))hArrCu^(2+)""_((aq))+2Ag(s)` Given `E_(cell)^(@)=0.46V and IF=96500 C mol^(-1).` |
Answer» `Cu_((s))+2Ag^(+)""_((aq))hArrCu^(2+)""_((aq))+2Ag""_((s))` `E_(cell)^(@)=0.46V, " "1F=96500C mol^(-1)` `DeltaG^(@)=-2nFE_(cell)^(@)impliesn=2` `DeltaG^(@)=-2xx96500xx0.46 implies" "DeltaG^(@)=-88780J//mol` `-DeltaG^(@)=2.303RTlogkimplies" "88780=2.303xx8.314xx298logk_(c)` `logk_(c)=(88780)/(2.303xx8.314xx298)implies" "logk_(c)=15.56` |
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1070. |
(a) What are the two classifications of batteries ? What is the difference between them ? (b) The resistanc of `0.01` M NaCl solution at `25^(@)C is 200Omega.` The cell constant of the conductivity cell is unity. Calculate the molar conductivity of the solution. |
Answer» (a) Batteries are classified in : Primary batteries and secondary batteries Primary batteries are those in which the redox reaction occurs only once and the cell becomes dead after sometime and cannot be used againd. Example: Lead storage battery and nickel-cadmium storage battery. (b) `C=0.01MR=200Omega` `G^(**)("cell constant")[(1)/(a)]=1cm ^(-1)" "K_(c)=((1)/(R))G^(**)" "implies(1)/(200)xx1` `=(1)/(200)Scm^(-1),^^_(m)=(K_(c)xx1000)/(M)` `=(1xx1000)/(200xx0.01)" "implies" "^^_(m)=500Scm^(2)mol^(-1)` |
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1071. |
Write two uses of each of the following polymers. `"(i) Polypropylene (ii) PVC (iii) Nylon - 66"` |
Answer» (i) Polypropylene : (a) In manufacture of carpet fibres, ropes. (b) In manufacture of toys and bottles. (ii) PVC (Poly vinyl chloride) : (a) It is used for making rain coats, hand bags. (b) It is used for making water pipes. (iii) Nulon-6,6 : (a) It is used for making carpets, textile fibers. (b) It is used for making bristles for brushes. |
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1072. |
Write two uses of each of the following polymers. `(i)"Polypropylene."" "(ii)"PVC."" "(iii)"Nylon-6 6."` |
Answer» Same as Q. 21 of Delhi Board (Comptt.) 2016. | |
1073. |
Which mode of propagation is used by short wave broadcast services ?. |
Answer» Sky wave proagation is used by short wave boradcast services having frequnecy range from a few MHz to 30 MHz. | |
1074. |
Four nuclei of an elements undergo fusion to form a heavier nucles, with release of energy. Which of the two - the parent or the daughter nucleus - would have higher binding energy per nucleon ? |
Answer» Parent nuclei because when daughter nuclei is fromed, some energy gets released . So, there will be loss energy in daughter nuclei. So, the daughter nuclie has to be less binding energy than the parent nuclei. | |
1075. |
(a) Explain one application of each one of the following : (A) Amniocentesis. (B) Lactational amenorrhea. (C) ZIFT. (b) Prepare a poster for the school programme depicting the objectives of : "Reproductive and Child Healthe C are Programme." OR (a) Explain any two ways by which apomictic seed can develop. (b) List one advantage and one disadvantage of a apomictic crop. (c) Why do farmers find production of hybrid seeds costly ? |
Answer» (a) (A) Amniocentesis : It is also known as amniotic fluid test. This test detects chromosomal abnormalities in the fetus. After 16 weeks of pregnancy, sample of amniotic fluid is taken from amniotic sac and its DNA is examined for genetic abnormalities. (B) Lactational amenorrhea : Absence of menstruation during the intense lactation following delivery. This period is used as a contrceptive method upto six months following delivery as long as the mother breast-feeds the child. (C) ZIFT : Zygote intra fallopian transfer in an invitro fertillization (IVF) technique where the zygote or early embryo having upto 8 blastomeres is transferred into the fallopian tube to complete its further development. (a) Apomixis is a special mechanism of production seeds by flowering plants without fertilisation. Apomictic seed can develop. (i) Diploid egg cell is formed without reduction division which develops into the embryo without the fertilisation. Example - Grassesss. (ii) Nucellar cells surrounding hte embryo sac divide and protrude into the embryo sac and develop into embryos. Example - citrus and mango. (b) Apomixis makes use of hybrid seed more economical. Apomixis provides on opportunity to the farmers to keep using same hybrid seeds year after year and not buy hybrid seed every year. One disadvantage of apomictic seeds is that it reduces genetic diversity from parent to offsprings due to lack of variations in asexual reproduction. (c) Hybrid seeds production is costlier because it novolves cross pollination of plants in an artificial manner. If require lots of scientific research, scientists try thousands of combinations of parents to identify a desired hybrid. It also takes a lot of time (nearly 10 years) to develop hybrid seeds. Hybrid seed production also require complete isolation. |
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1076. |
Justify with the help of an example where a deliberate attempt by humans has led to the extinction of a particular species. |
Answer» When Nile Perch, a large predator fish, was introduced in lake Victoria, it started feeding on the native fish, chichlid fish, As a result, chichlid fish became instinct and Nile perch, not finding any food for itself died too. | |
1077. |
Two primary cells of emfs `E_(1)` and `E_(2)` are connected to the potentiometer wire AB as shown in the figure . If the balancing lengths for the two combinations of the cells are 250 cm and 400 cm , find the ratio of `E_(1)` and `E_(2)` . |
Answer» Here , `E_(1) - E_(2) = K xx 250 " " ... (i) ` `E_(1) + E_(2) = K xx 400 = 400 K " " ... (ii)` `therefore` K is potential gradient On adding the equations (i) and (ii) `2 E_(1) = 650 K` `E_(1) = 325 K` On subtracting equations (i) from (ii) we get `2 E_(2) = 150 K` `E_(2) = 75K` Hence , `" " (E_(1))/(E_(2)) = (325K)/(75K) = (13)/(3) = 4.33`. |
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1078. |
For the reaction `2N_(2)O_(5)(g) to 4NO_(2) (g) + O_(2)(g)` , the rate of formation of `NO_(2)(g) ` is `2.8 xx 10^(-3) Ms^(-1)` . Calculate the rate of disapperance of `N_(2)O_(5)(g)`. |
Answer» `2N_(2)O_(5)(g) to 4NO_(2)(g) + O_(2)(g)` Rate of formation `(R_(f)) = (Delta[NO_(2)])/(Deltat) = 2.8 xx 10^(-3) ` M/s Rate of disapperance of `N_(2)O_(5)(g) = (Delta[N_(2)O_(5)])/(Deltat) = ?` Rate of reaction = `(1)/(2) (Delta[N_(2)O_(5)])/(Deltat) = (1)/(4) (Delta[NO_(2)])/(Deltat)` `(Delta[N_(2)O_(5)])/(Deltat) = (2)/(4) xx (Delta[NO_(2)])/(Deltat) = (1)/(2) xx 2.8 xx 10^(-3)` M/s `(Delta[N_(2)O_(5)])/(Deltat) = 1.4 xx 10^(-3) ` M/s |
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1079. |
Calculate the freezing point of a solution containing 60 g glucose (Molar mass = 180 g `mol^(-1)`) in 250 g of water . (`K_(f)` of water = `1.86 K kg mol^(-1)`) |
Answer» `DeltaT_(f) = K_(f) xx m ` `T_(f)^(@) - T_(s) = (K_(f) xx W_(B))/(M_(B) xx W_(A) (kg))` `T_(f)^(@)` for water = `0^(@)C = 273 K` `M_(B) (C_(6)H_(12)O_(6)) = 12 xx 6 + 12 xx 1 + 6 xx 16 = 180 g//mol` `273 - T_(s) = (1.86 xx 60 xx 1000)/(180 xx 250)` `273-T_(s) = 2.48` `T_(s) = 273 - 2.48` `T_(s) = 270.52 K` `{:(T_(f)^(@) = "freezing point of solvent") , (T_(s)^(@) = "freezing point of solution." ) , (W_(B) = "given mass of solute"), (M_(B) = "molar mass of solute") , (W_(A) = "mass of solvent") , (K_(f)= "mole depression constant"):}` |
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1080. |
After the ban on plastic bags, students of one school decided to make the people aware of the harmful effects of plastic bags on environment ans Yamuna River. To make the awareness more implctful, they organized rally by joining hands with other school and distributed paper bags to vegetable vendors, shopkeepers and departmental stores. All students pledged not to use polythene bags in future to save Yamuna River. After reading the above passage, answer the following questions : (i) What values are shown by the students ? (ii) What are biodegradable polymers ? Given one example. (iii) Is polythene a condensation or an addition polymer ? |
Answer» (i) From the given passage, we can conclude that the students show awareness about the enviromnent. (ii) A polymer that can be decomposed by microorganisms within a definite period of time, so thet the polymer or its deraded product does not cause any harm to the environment, is called a bio-degradable polymer. For example, poly - `beta` - hydroxybutyrate - Co - `beta` - hydroxy valerate (PHBV) is a bio-degradable aliphatic polyester. `[{:(-CH_(2) -CH_(2)-C-O-CH-CH_(2)-CH-C-),(" "|" "||" "|" "||),(" "CH_3" "O" "CH_2-CH_3),(" "PHBV):}]_(n)` (iii) Polythene is an addtion polymer that is formed by addition of ethene molecules. `underset("Polythene Polymer")underset("Ethane")(nCH_2=CH_2)overset("polim erisation")tounderset("Re peating unit")(n[-CH_2-CH_2-])to[CH_2-CH_2]_n` |
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1081. |
(a) why are transgenic animals so called? (b) Explain the role of transgenic animals in (i) vaccine safety and (ii) biological products with the help of an example each |
Answer» Transgenic animals are called so as these animals posses the deliberate modification in their genome. The changes in the genome of the organisms are brought about by recombinat DNA technology. Role of transgenic animal in vaccine safety-Transgenic mice are being developed for use in testing the safety of vaccines before they are used on humans. Example-Transgenic mice are being used to test the safety of the polio vaccines. If successful and found to be reliable, they could replace the use of monkeys to test the safety of batches of the vaccine. Role of transgenic animal in production of biological products-Transgenic cow, Rosie is used for the production of human protein-enriched milk, which contained Alpha lactalbumin and was nutritionally more suitable for human babies. |
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1082. |
In a pond there were 20 Hydrilla plants. Through reproduction 10 new Hydrilla plants were added in a year. Calculate the birth rate of the population. |
Answer» Hydrilla Plants in Pond = 20 New Plants by reproduction= 10 Total Plants= 30 The Birth rate of Population = `("No. of individuals born")/("Total No. of individual")=10/20` = 0.5 Plants Per year |
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1083. |
How have transgenic animals proved to be beneficial in : (a) Production of biological products. (b) chemical safety testing. |
Answer» (a) Transgenic animals : These animals have had their DNA manipulated to incorporate and express an extra gene. These have been useful for production of bioligical products such as production of human protein (alpha-1 antitrypsin) for treatment of emphysema. Transgenic cow Rosie produced human alpha lactalbumin enriched milk (2.4 gms protein per liter) which was more balanced product for human babies than natural cow milk. (b) Chemical safety testing : Transgenic animals are made more sensitive to toxic substance than non-transgenic animals by introduction of certain genes. These transgenic animals are then exposed to toxic substances under study. Toxicity testing in such animals takes less time. |
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1084. |
Describe the mutual relationship between fig tree and wasp and comment on the phenomenon that operates in their relationship. |
Answer» Fig plants share a strict one-to-one relationship with the wasp species. Here, the fig species can be pollinated only by partner wasp species and no other species, the female wasp user the fruit, as a site oviposition and nourishing the larval. The wasp pollinates the fig inflorescance while seaching for suitable egg-laying rites. This is an example of mutualism. | |
1085. |
Name the blank spaces a, b, c and d from the table given below : |
Answer» a = Lactobacillus b = Trichoderma Polysporum c = Fungus (Yeast) d = Penicillin |
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1086. |
Explain accelerated europhication. Mention any two consequences of this phenomenon |
Answer» Accelerated europhication is nutrient enrichment of water bodies due to human activities like passage of sewage. Its consequences are : Large amount of nutrients in waters causes excessive growth of planktonic (free-floating) algae, called algal bloom which impart characteristic colour to water bodies. Depletion of oxygen content of water leading to the death of the aquatic organisms. |
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1087. |
(a) Name the causative organisms for the following diseases : (i) Elephantiasis (ii) Ringworn (iii) Amoebiasis (b) How can public hygiene help control such diseases ? |
Answer» (a) (i) Elephantiasis : 1. It is caused by Wucheraria Bancrofti and Wucheraria Malayi. 2. They cause inflammation of the organs in which they live for many years. 3. They normally affect the lymph vessels of the lower limb causing them to swell like that of an elephant hence called elephantiasis. 4. Genital organs may also be affected leading to gross deformation. 5. Female culex mosquito is the vector. (ii) Ringworm : 1. They are caused by fungi like Microsporum, Epidermophyton and Trichophyton. 2. The symptoms include (I) dry scaly lesions on skins, nail and scalp. (II) Lesions are accompanied by itching. 3. Ringworms are generally acquired from soil or by direct contact with the contaminated articles used by the infected persons. (iii) Amoebiasis : 1. It is caused by Entamoeba histolytica. 2. Infection is through contaminated food and water. 3. the pathogen resides in the large intestine. Its symptoms include : 1. Abdominal pain and cramps. 2. Stool with excess mucuc and blood clots. 3. Constipation. Housefly acts as mechanical carrier and transmits the parasite from faces of infected person to the food articles. (b) Measures to prevent infectious diseases `**" "Maintenance of personal hygiene `**" "MAintenance of public hygiene `**" "` Prevention of close contact with infected persons `**" "`The elimination of vectors and their breeding places. `**" "`Use of mosquito nets, introducing fishes like Gambusia in ponds `**" "`Spraying insecticides in ditches, drainages, etc. `**" "`Using vaccines, antibiotics and drugs. |
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1088. |
a) Explain Darwinian theory of evolution with the help of one suitable example. State the two key concepts of the theory. b) Mention any three characteristics of Neanderthal man that lived in near east and central Asia. |
Answer» a) Darwinian theory of Evolution i) According to Darwin, evolution took place by natural selection. ii) the number of life forms depends upon their life span and their ability to muliply. iii) Another aspect of natural selection is the survival of the fittest where nauter selects the antibiotic resistance in bacteria. The two key concepts of the theory are: 1) Branching descent: According to this concept, various species have come into existene from a common ancestor. 2) Natural selection: According to this concept, nature selects the individuals, which are most to fit to adatp to their environment. b) Characteristics of Neanderthal man i) They posses a brain capacity of 1400cc. ii) They were short but very strong with outward curved thigh bones. iii) They used hides to protect their body and to bury the dead. |
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1089. |
According the Darwinian theory, the rate of appearance of new forms is linked to their life cycles. Explain. |
Answer» The essence of the Drawinian theory about evolution is natural selection. | |
1090. |
Write the IUPAC name of the following compound: |
Answer» 1,4- dichloro -2- methyl benzene | |
1091. |
What is the basicity of `H_(3)PO_(3)` ? |
Answer» Dibasic due to the presentce of 2-OH molecules. | |
1092. |
(a) A steady current of 2 amperes was passed through two electrolytic cells X and Y connected in series containing electrolytes `FeSO_(4)` and `ZnSO_(4)` until 2.8 g of Fe deposited at the cathode of cell X. How long did the current flow ? Calculate the mass of Zn deposited at the cathode of cell Y (Molar mass : Fe= 56 g `mol^(-1)` Zn-65.3g `mol^(-1)`, 1F -96500 C `mol^(-1)`) (b) In the plot of molar conductivity `(wedge_(m))` vs Square root of concentration `(c^(1//2))`, following curves are obtained for two electrolytes. A and B: Answer the following : (i) Predict the nature of electrolytes A and B. (ii) What happens on extrapolation of `wedge_(m)` to concentration approaching zero for electrolytes A and B ? |
Answer» (a) For cell X `w=M/(nxx965000)xxIt` `2.8=56/(2xx96500)xx2xxt` `t=(2.8xx2xx96500)/(56xx2)=4825s` Now, for cell Y `w=M/(nxx96500)xxIxxt` `w=(65.3)/(2xx96500)xx4825xx2=3.625` g (b) (i) A is the strong electrolyte. B is the weak electrolyte (ii) When the concentration approaching to zero strong electrolyte given `wedge_(m)^(0)` (limiting molar conductivity). But when concentration approaching to zero the curve of weak electrolyte does not cut the axis it becomes parallel to axis. |
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1093. |
What is the basicity of `H_(3)PO_(2)` acid and why ? |
Answer» In `H_(3)PO_(2)P-OH` group is present , hence, its basicity is one . | |
1094. |
`E_(cell)^(@)` for the given redox reaction is 2.71 V `Mg_((s))+Cu_((0.01 M))^(2+)toMg_((0.001 M))^(2+)+Cu_((s))` Calculate `E_(cell)` for the reaction. Write the direction of flow of current when an external opposite potential applied is (i) less than 2.71 V and (ii) greater than 2.71 V. |
Answer» Cell reaction `Mg(s)+Cu^(2+)toMg^(2+)+Cu` `E_(cell)=E_(cell)^(0)-(0.0591)/nlog""([Mg^(2+)])/([Cu^(2+)])=2.71-(0.0591)/2log""((0.001)/(0.01))=2.71-(0.0591)/2log""(1/10)` `=2.71-(0.0591)/2xx(-1)=2.71+(0.0591)/2=2.739 V`. (i) The electrons flow from Mg to Cu i.e., the direction of flow of current is from Cu to Mg. (ii) The electrons flow from Cu to Mg i.e., the direction of current is from Mg to Cu. |
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1095. |
Write the IUPAC name of the following : |
Answer» 3-Bromo-2-methylpropene. | |
1096. |
Differentiate between a productivity and decomposer giving an examples. |
Answer» Productivity-The rate of biomass production per unit area over a time period by plants during photosynthesis is called Productivity. Decomposers-These organisms secrete enzymes to break down complex organic matter into inorganic substances. e.g. Some bacteria and fungi. |
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1097. |
"Stability of a community depends on its species richness." Write how did David Tilman show this experimentally. |
Answer» Ecologists believe that communities with more species tend to be more stable than those with less species. This was confirmed by David Tilman. David Tilman showed this experimented by: i) Productivity should not vary to much from year to year. ii) It should be resistent to occasional natural and man-made disturbances. iii) It should be resistant to invasions by alien species. |
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1098. |
India has more than 50,000 strains of rice. Mention the level of biodiversity it represents. |
Answer» The quantity 50,000 strains of nee represent specific biodiversity. | |
1099. |
Write the full form of VNTR. How is VNTR different from Probe |
Answer» VNTR stands for Variable Number of Tandem Repeats.Probe is labelled or radioactive single stranded polynucleotide that hybridises with DNA fragments. Whereas, VNTR is a class of sateltite that shows high degree of polymorphism. | |
1100. |
Complete the following chemical equations : (i) `Cr_(2)O_(7)^(-) + H^(+) + I^(-) rarr` (ii) `MnO_(4)^(-) + NO_(2)^(-) + H^(+) rarr` |
Answer» (i) `Cr_(2)O_(7)^(+) + H^(+) + 6I^(-) rarr 2Cr^(3+) + 7H_(2)O 2I^(-) rarr I_(2) + 2e^(-) xx 3` `overline(underline(Cr_(2)O_(7)^(2-)+6I^(-)+14H^(+)rarr2Cr^(3+)+3I_(2)+7H_(2)O` (ii) `2MnO_(4)^(-) + 5NO_(2)^(+) + 6H^(+) rarr 2 Mn^(2+) + 5NO_(2)^(-) + 3H_(2)O` |
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