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InterviewSolution
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.
| 151. |
Assing a reason for each of the following observation : (i) The transition metals (with the exception of Zn, Cd and Hg) are hard and have high melting and boiling points (ii) The ionisation enthalpies (first and second) in the first series of the transition elements are found to vary irregularly |
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Answer» (i) Transition metals have high melting and boiling point and are hared to strong metallic bonds between the atoms of these elements .Greater the unpaired electrons stronger the metallic bond (ii) The irregular trend is due to the fact that removel of e -s alters the relative energy of 4s and 3d there is reorgansiation energy accompanying ionisation .Thuis result in relaease of has low I.E because a loss of one e gives stables configuation .Zn has very high I.E because electronn has to to be removed from 4s stable configuration |
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| 152. |
(a) A colloidal sol is prepared by the given method in figure. What is the charge of AgI colloidal particles in the test tube ? How is the sol formed, represented? (b) Explain how the phenomenon of adsorption finds application in Heterogeneous catalysis. (c) Which of the following electroytes is the most effective for the coagulation of `Fe(OH)_(3)` sol which is a positively charged sol? `NaCI, Na_(2)SO_(4), Na_(3)PO_(4)` |
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Answer» (a) Positive charge in developed on the sol. Sol is represented as `Agl/Ag^(+)` (b) Adsorption of reactants on the solid surface of the catalysts increases the rate of reaction. (c) `Na_(3)PO_(4)` Hardy-Schulze rule. |
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| 153. |
Write the formula of an oxo-anion of Manganese (Mn) in which it shows the oxidation state equal to its group number |
| Answer» `Mn_(2)O_(7)` is an oxo-anion of Manganese in which it shown the oxidation state equal to its group number | |
| 154. |
Write the structure of 1-Bromo-4-chlorobut-2-ene |
| Answer» `underset("1-Bromo-4-chlorobut-2-ene")(H_(2)underset(Cl)underset(|)(C) - HC = CH - C_(2) - Br)` | |
| 155. |
Write IUPAC name of the following compound `(CH_(3)CH_(2))_(2) NCH_(3)` |
| Answer» N-Ethyl-Nmethylethanmine | |
| 156. |
For a reaction `R rarr P`, half -life `(t_(1//2))` is observed to be independent of the initial concentration of reactants. What is the order of reaction ? |
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Answer» It is a first order reaction `|t_(1//2) = (0.693)/(k)|`[independent of the initial concentration of reactions] |
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| 157. |
Explain out breeding out crossing and cross breeding practices in animal husbandry. |
| Answer» (a) Chemical methods often kills both useful and harmful life forms indiscriminately.Eradication of the creatures that are often described as pests is not only possible but also undesirable , for without them the beneficial predatory and par4sitic insects which depends upon them as food or hosts would not be able to survive. | |
| 158. |
Arrange the following compounds in an increasing order of their solubility in water: `C-6H_5NH_2(C_2H_5)_2NH,C_2H_5NH_2` |
| Answer» `C_6H_5NH_2lt(C_2H_5)_2NH_2lt(C_2H_5)NH` | |
| 159. |
Calculate the energy released in MeV in the following nuclear reaction : `._(92)^(238)Urarr._(90)^(234)Th+._(2)^(4)He+Q ["Mass of "._(92)^(238)U=238.05079 u` Mass of `._(90)^(238)Th=234.043630 u` Massof `._(2)^(4)He=4.002600 u` `1u = 931.5 MeV//c^(2)]` |
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Answer» Given nuclear reaction is `._(92)^(238)U rarr ._(90)^(234)Th + ._(2)^(4)He+Q` Mass defect `=M_(U)-M_(Th)-M_(He)` `=238.05079-234.043636-4.002600=0.00456 u`. Energy released `=(0.00456 u) xx (431.5 MeV//C^(2))=4.25 MeV`. |
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| 160. |
(i) Write the scientific names of the two species of filarial worms causing filariasis. (ii) How do they affect the body of infected persons(s)? (iii) How does the disease spread? |
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Answer» (i) Filarial worms species. 1. Wuchereria bancrofti, 2. Wucheria Malayai. (i) They normally affect the lymph vessels of the lower limbs causing them to swell like that of an elephant , hence called elephantiasis. (iii) The diasease spreads through Female Culex mosquito which is the vector . The pathogens are treansmitted to a healthy person through the bite by the female mosquito vectors. |
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| 161. |
What is coagulation process ? |
| Answer» Coagulation : The process of setting of colloidal particles is called coagulation or precipitation of the sol. | |
| 162. |
In the figure a long uniform potentiometer wire AB is having a constant potential gradient along its length. The null points for the two primary cells of emfs `epsilon_(1) " and " epsilon_(2)` connected in the manner shown are obtained at a distance of 120 cm and 300 cm from the end A. Find (i) `epsilon_(1)//epsilon_(2)` and (ii) position of null point for the cell `epsilon_(1)`. How is the sensitivity of a potentiometer increased ? |
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Answer» In first combination `epsilon_(1) " and " epsilon_(2)` are opposing each other while in second combination `epsilon_(1) " and " epsilon_(2)` are adding each other, so, `epsilon_(1)-epsilon_(2)=Kl_(1)` `epsilon_(1)+epsilon_(2)=Kl_(2)` `implies (epsilon_(1)-epsilon_(2))/(epsilon_(1)+epsilon_(2))=(l_(1))/(l_(2)) implies (epsilon_(1)-epsilon_(2))/(epsilon_(1)+epsilon_(2))=(120)/(300)` `5 epsilon_(1)-5epsilon_(2)=2 epsilon_(1)+2epsilon_(2)` `5 epsilon_(1)-2epsilon_(1)=2epsilon_(2)+5epsilon_(2)` `implies 3 epsilon_(1)=7epsilon_(2) implies (epsilon_(1))/(epsilon_(2))=(7)/(3) implies epsilon_(1) : epsilon_(2)=7 : 3` The sensitivity of a potentiometer can be increased by reducing the potential gradient. |
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| 163. |
A rectangular loop of wire of size `4 cm xx 10 cm` carries a steady current of 2A. A straight long wire carrying 5 A current is kept near the loop as shown. If the loop and the wire are coplanar, find (i) the torque acting on the loop and (ii) the magnitude and direction of the force on the loop due to the current carrying wire. |
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Answer» `(i) |vec(tau)|=|vec(M)xx vec(B)|=MB sin theta` `implies tau =MB sin 0^(@)` `implies tau =0` (ii) Force between two currents Carrying wires, `F=(mu_(0))/(2 pi).(I_(1)I_(2))/(r ).l` `therefore` Force on arm `AB, F_(AB)` `=(mu_(0))/(2 pi)xx(5xx2)/(1xx10^(-2))xx(0.10)` `F_(AB)=(mu_(0))/(2pi)xx100N` [Attraction towards the wire] Force on arm `CD, F_(CD) =(mu_(0))/(2pi)xx(5xx2)/(5xx10^(-2))xx(0.10)` `F_(CD)=(mu_(0))/(2pi)xx20N` [Repulsion away from the wire] Force on arm BC and AD, `F_(BC)=F_(AD)=0` `therefore` Resultant force on loop, `F=F_(AB)-F_(CD)=(mu_(0))/(2pi)(100-20)` `implies F=2xx10^(-7)xx80==16xx10^(-6)N`. [Towards the straight wire i.e., force of attraction]. |
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| 164. |
A rectangular loop of wire of size `2 cm xx 5cm ` carries a steady current of 1A. A straight long wire carrying 4 A current is kept near the loop as shown. If the loop and the wire are coplanar, find (i) the torque acting on the end (ii) the magnitude and direction of the force on the loop due to the current carrying wire. |
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Answer» (i) Torque `tau=0`, as net acting on the coil is zero. (ii) Forces on the loop `=(mu_(o)I_(1)I_(2))/(2pi r)` length (Newton) `=2xx10^(-3)((4xx1)/(1xx10^(-2))-(4xx1)/(3xx10^(-2)))xx5xx10^(-2)` towards the straight wire, `=2xx10^(-5)xx(4-(4)/(3))xx5xx10^(-2)` `=(80)/(3)xx10^(-7)` N. towards the straight wire. |
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| 165. |
A rectangular loop of wire of size `2.5 cm xx 4 cm` carries a steady current of 1A. A straight long wire carrying 2 A current is kept near the loop as shown. If the loop and the wire are coplanar, find (i) the torque acting on the loop and (ii) the magnitude and direction of the force on the loop due to the current carrying wire. |
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Answer» (i) Torque `tau=0`, as there will be forces on the plane of the rectangular coil and wire. (ii) Forces on the loop `=(mu_(o)I_(1)I_(2))/(2 pi r)` length (Newton) `= 2xx10^(-7) ((2xx1)/(2xx10^(-2)))xx4xx10^(-2)` Towards the straight wire, `=2xx10^(-5)xx(1-(2)/(5))xx4xx10^(-2)=4.8xx10^(-7)` N. towards the straight wire. |
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| 166. |
(i) What is population density ? Why are ecologists interested in measuring it ? (ii) Write the different ways of measuring population density.Explain any two with the help of specific examples. |
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Answer» Ecologists are interested in measuring population density for the following reasons : 1. The size of the population tells us a lot about its status in the habitat. 2. Ecological processes such as outcome of compettion with another species, the impact of a predator or the effect of pesticide application can be easily can be easily evaluated in terms of change in the population size. (ii) Population density means number of individuals present per unit area, population density can be measured by determining the population size.The different methods to study population size are as follows , (1) Per cent cover or biomass : In an area with 200 parthernium plants and only one banyan tree with large conopy, the density of banyan tree is small but does not reflect its important role in the community. Here percent cover or biomass is a more meaningful method of assessing population density. (2). Total Number : It involves the counting of organisms in the given area. (3) Relative Density : In this method, there is no need to count the organisms individully. Example the number of fishes caught per trap gives the measure of their total density in a given water body. (4) Indirect Assessment : The tiger census is based on pug marks and faecal pellets. |
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| 167. |
(b) Mention any two situations when the technique is useful. |
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Answer» The technique is useful in the following situations : 1. Paternity testing in case of disputes. 2. In determining population and genetic diversities. |
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| 168. |
(a) How is DNA fingerprinting done ? Name any two types of human samples which can be used for DNA fingerprinting. Explain the process sequentially. |
| Answer» (a) The human samples which can be used for DNA fingerprinting are blood and hair follicle. | |
| 169. |
Explain the significance of satellite DNA in DNA fingerprinting technique. |
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Answer» Satellite DNA consists of highly repetive and generally non-coding DNA.It is very significant in DNA finger printing for the following reason. It is speciffic for every species. It is a highly polymorphic DNA. An example which is variable number of Tandom Repeats (VNT Rs). In different individuals the different chromosomes will vary in the number of repeats. This can be used for identification. VNTRs sequences are inheritable and an individual inherits it from both parents and this can be used for indentifiaction easier. |
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| 170. |
Mention any two conditions that enhance the chances of syngamy in organisms exhibiting external fertilization. |
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Answer» (i) Organisms exhibiting external fertilization show great synchrony between the sexes and release a large no. of gametes into the surrounding medium waters in order to enhance the chances of syngamy. (ii) This happens in the bony fishes, algae and frogs where a large no. of offsprings are produced. |
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| 171. |
When Sunita, a class XII student, came to know that her parents are planning to rent out the top floor of their house to a mobile company she protested. She tried hard to convince her parents that this move would be a health hazard. (i) Ultimately her parents agreed. (1) In what way can the setting up to transmission tower by a mobile company in a residential colony prove to be injurious to health? (2) By objective to this move the parents, what value did Sunita display? (3) Estimate the range of e.m. waves which can be transmitted by an antenna of height 20m. (Given radius of the earth = 6400 km). |
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Answer» (1) A transmitting tower makes use of electromagnetic waves such as microwaves, exposure to which can cause sever health hazards like tumour and cancer. Also, the transmitting antenna operates on a very high power, so the risk of someone getting severely burnt in a residential area increases. (2) By objecting to this move to her parents, Sunita has displayed awarness towards the health and environment of her society. (3) Range of the transmitting antenna, `d=sqrt(2hR)` Here, h is the height of the transmitting antenna and R is the radius of the Earth. `R=6400km=64xx105m` `d=sqrt(2xx20xx64xx10^(5))=16,000m` `16km` |
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| 172. |
How does the process of natural selection affect Hardy-Weinberg equilibrium? Explain List the other four factors that disturb the equilibrium. |
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Answer» The principal says that allete frequencies in a population are stable and constant from generation to generation. The gene pool (total genes and their alleles in a population) remains constant, this is called genetic equilibrium. The sum of all the all the frequencies is 1. For exmple- In a diploid p and q denote the frequency of A and a. The frequency of AA is `p^(2)` and aais `q^(2)` and of Aa is 2pq. This `p^(2)+q^(2)2pq=1` which means `(p+q)^(2)=1`. When frequency measured differs from expected frequencies the difference in frequencies indicates the extent of evolutionary change. The equilibrium is distributed by 1-gene migration 2-mutation 3- genetic drift 4- natural selection. Distribution in genetic equilibrium, or Hardy-Weinberg equilibrium, i.e., change of frequency of alleles in a population would then be interpreted as resulting in evolution. |
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| 173. |
differentiable between somaclones and somatic hybrids give one example of each . |
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Answer» Somaclones : these are the plants obtained thriugh tissure culture and are genetically identical to the parent plant example : tomato apple . Somatic hybrids : these are obtained by fusing the protoplasts from two different varieties of plants Example pomato obtained by protoplast of tomato and potato. |
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| 174. |
Explain the relevance of "Totipotency" and "Somaclones" in raising healthy banana plants from virus infected banana plants. |
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Answer» Totipotency is the ability of a living cell to express all of its genes to regenerate a whole new individual plant. Somaclones are the plants obtained through tissue culture and are genetically identical to the parent plant. From virus infected bana plant tissues are excised, disinfected and tissue cultured to raise virus free genetically identical banana plants. |
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| 175. |
When and why do some animals like frogs hibernate ? |
| Answer» Hibernation or winter sleep is a resting stage where animals such as frogs escape winters (cold) by hiding themselves in their shelters. They escape the winter season by entering a state of inactivity by slowing their metabolism. | |
| 176. |
How do neutrophils act as a cellular barrier to pathogens in humen ? |
| Answer» Nuetrophils dispose off microbes and dead cells by feeding on them . Then are called the soldiers of animals body. Neutrophils in blood can phagocytose and destroy the microbes. | |
| 177. |
What was the challenge for production of insulin using rDNA techniques ? How did Eli Lilly produce insulin using rDNA technology? |
| Answer» The main challenge for production of insulin assembled into a mature form. In 1983, Eli Lilly, an American company prepared two DNA sequences corresponding to A and B, chains of human insulin and introduced them in plasmids of E. coli to produce insulin chains. Chains A and B were produced separately, extracted and combined by creating disulfide bonds to form hormone insulin. | |
| 178. |
List the key tools used in recombinant DNA technology |
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Answer» The Key tools used in recombinant DNA technology are : (1) Restriction enzym (ii) Polymerase enzyme (iii) Ligase enzyme (iv) Vectors (v) host organism/cell |
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| 179. |
Name the enzyme produced by Streptococcus bacterium. Explain its importance In medical sciences. |
| Answer» Streptococcus bacterium producess treptokinase. It is used for removing clots from the blood vessels in a patient suffering from myoc rdial infarction/or in a heart patient. | |
| 180. |
Why should biological control of pests and pathogens be preferred to the conventional use of chemical pesticides? |
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Answer» (a) Bacillus thuringiensis-Bacillus thuringiensis is a bacterium. B-toxin gene is cloned from bacterium and expressed in plants to provide resistance from insect without using insecticides. Thus, by the application of biotechnology pest resistant plant are developed. Some other examples are Bt corn, Bt ricé, Bt tomato, Potato and Soyabean. When they are eaten by the insect larvae, the toxin is released in the gut and becomes active and kills the larvae. Specific Bt toxin genes obtained from Bacillus thuringiensis are used in several crop plants which make them resistant to the insect Pest. (b) Nucleopolyhedrovirus-These viruses are excellent candidates for species speciiic, narrow spectrum insecticidal applications. They have been shown to have no negative impacts on plants mammals, birds, fish or even I on non-target insects. They are especially desirable when beneficial insects are being conserved r to aid in an overall integrated pest management (IPM) programme, or when an ecologically sensitive area is being treated. |
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| 181. |
List any two characters of Pea plants used by Mendal in his experiments other than height of the plants and the colour of the seed. |
| Answer» Two characters of Pea plant and the colour of the seed shape. | |
| 182. |
(i) Which organ of the human body is initially affected when bitten by an infected female Anopheles ? Name the stage of the parasite that infects this organ. (ii)Explain the events that are responsible for chill and high fever in the patient. |
| Answer» See Q. 25, Set-I., Delhi Board-2008. | |
| 183. |
List the salient features of double helix structure of DNA. |
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Answer» The salient features of double helix structure of DNA are : (i) The DNA is made up of two polynucleotide chains, where the backbone is made up of sugar and phosphate groups and the nitrogenous base projecting to the centre. (ii) The diameter of the Strand is always equal due to pairing of purine and pyramidine i.e. , adenine is complementary to thymine While guanine is complementary to cytosine. (iii) The stability of the DNA structure is due to strong bond between negetively charged nucleic acid and positively charged basic histone protein, along with hydrogen bonds between the base pairs of two strands. |
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| 184. |
Explain the function of each of the following : (a) Coleorihiza , ( b) Umbilical cord ,( c) Germ pores |
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Answer» Coleorhiza-w Protects the radicle of monocot embryo. Umbilical cord : It develops from two separate fetal origins. The yolk sack and the allantosis both comprise the umbilical cord. It helps in transport of substances to and from the embryo. Germ Pores : The outer wall of pollen grain through which the germ tube or embryo tube makes its exist on Germination. |
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| 185. |
Inheritance pattern of ABO blood group in human shows dominance, co-dominance and multiple allelism. Explain each concept with the help of blood group genotypes |
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Answer» Dominance : It is the phenomenon where one allele expresses itself exclusive of the presence of other allele, The alleles JA and JB both are dominant over allele i as JA and JB forms antigen A and antigen B respectively but i does not form any antigen. . . Co-Dominance : It is the phenomenon when the `F_1` generation resembles both the parents, and both the parental characters are expressed simultaneously. Both the alleles JA and JB are co- dominant as both ofthem are able to express themselves in the presence of each other in blood group AB (JAJB) by forming antigen A and B. Multiple allelism : It is the phenomenon of occurrence of a gene in more than two allelic form on the same locus. The ABO blood groups in humans are determined by three different allelic forms `I^a`,`I^b`and i. |
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| 186. |
Trace the life-cycle of malarial parasite in the human body when bitten by an infected female Anopheles. |
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Answer» When an infected female Anopheles mosquito bites a person, the sporozoites are injected into the body. They reach the liver cells through blood. The paraite reproduces asexually in the liver cells and by the bursting of liver cells, new cells are released into the blood. They enter the red blood cells and reproduce asexually, by bursting the red blood cells, the cycles of fever, chill and shivering occur. The released parasites infect new red blood cells, some continue the asexual reproduction and cause the cycles of fever. The sexual stage of the parasite leads to formation of gametocytes in the RBCs, which are picked by the Anopheles mosquito along with the blood meal. |
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| 187. |
Explain the cause of global warming. Why is it a warning to mankind? |
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Answer» Cause of global warming: Deforestation Rise in the concentration of greenhouse gases ( `CO_2,CH_4,CFC_5,N_2O`) Burning of fossil fuels. Rise in industrial wastes and pollutants Global warming is a warning to mankind because : (i) Rise in temperature is leading to increased melting of polar ice-caps as well as of other places like the Himalayan snow caps. This will result in a rise in sea level that can submerge many coastal areas. Deleterious changes in the environment results in odd weather and climate changes, e.g., EI Nino effect. |
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| 188. |
Mention a product of human welfare obtained with the help of each one of the following microbes : (a) LAB (b) Sacchromyces cerevisiae (c) Propionibacterium shermanii (d) Aspergillus niger |
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Answer» LAB-lactic acid bacteria: lactobacillus is used in the production of curd from milk. ( ii) Saccharomyces cerevisiae It is used in the baking industry. (iii) Propionibacterium sharmanii : It is used in the production of cheese (iv) Aspergillus niger: It is used in the production of citric acid. |
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| 189. |
Where does triple fusion take place in a flowering plant? Why is it so called? Mention its significance. |
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Answer» Triple fusion : It is a fertilisation that involves fusion of one male gamete and two polar nuclei ( or secondary nucleus, if the two have already fused) in the central cell of embryo sac. It takes place in the central cell of an embryo sac. Three nuclei are involved in triple fusion, i.e., one male nucleus and two polar nuclei in the central cell therefore, the process is termed triple fusion. Significance: Endosperm is triploid and helps in providing nourishrment to the embryo |
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| 190. |
In a pond there were 200 frogs, 40 more were born in a year. Calculate the birth rate of the population. |
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Answer» Birth rate=`("No of individuals born")/("Total no. of individual")` `40/200`= 0.5 Frog per year |
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| 191. |
(a) Explain the mechanism of a nucleophilic attack on the carbonyl group of an aldehyde or a ketone. (b) An organic commpound (A) (molecular formula `C_(8)H_(16)O_(2)`) was hydrlysed with dilute sulphuric acid to give a carboxylic acid (B) and an alcohol (C). Oxidatin of (C) with chromic acid also produced (B). On dehydratin (C) gives but-1-ene. Write the equations for the reactions involved. |
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Answer» (a) Mechanism of nucleophilic addition reactions : attack from the top face A nucleaphile attacks the electrophilic carbon atom of the polar carbonyl group from a direction perpendicular to the plane of `sp^(2)` hybridised orbitals of carbonyl carbon. The hybridisation of carbon changes from `Sp^(2) to sp^(3)` in this process and a tetrahedral alkoxide intermediate is produced. This intermediate captures a proton from the reaction medium to give the electrically neutral product. The net result is additon of `Nu^(-)and H^(+)`across the carbon oxygen double bond. |
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| 192. |
Apant of Antirrhinum majus with red flowers was crossed with another plant of the same species with white flowers. The plants of the `F_(1)` generation bore pink flowers . Explain the patttern o inheritance with the help of a cross. |
| Answer» The progeny of first generation had pink flowers because the gane for the red flower colour was not completely dominant over the gene for the white flower. This pattern of inheritance s known as incomplate dominace. | |
| 193. |
Apant of Antirrhinum majus with red flowers was crossed with another plant of the same species with white flowers. The plants of the `F_(1)` generation bore pink flowers . Explain the patttern o inheritance with the help of a cross. |
| Answer» The progeny of first generation had pink flowers because the gane for the red flower colour was not completely dominant over the gene for the white flower. This pattern of inheritance s known as incomplate dominace. | |
| 194. |
Arrange the following in incresing order of basic strength : `C_(6)H_(5)NH_(2),C_(2)H_(5)NHCH_(3),C_(6)H_(5)N(CH_(3))_(2)`. |
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Answer» Increasing order of basic strength in gaseous state is as follows : `C_(6)H_(5)NH_(2)ltC_(2)H_(5)NHCH_(3)ltC_(6)H_(5)N(CH_(3))_(2)` |
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| 195. |
What type of intermolecular attractive interaction exits in the pair of methanol and acetone |
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Answer» Doplole-dipole intermolecular attractive interaction exists methanol and acetone, as both are polar molecular. |
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| 196. |
What are Associated Colloids ? Given an example. |
| Answer» Associated Colloids : The substances which when dissolved in a medium at low concentrations behave as normal, strong electrolyte but at higher concentration exhibit colloidal state prodperties due to the formation of aggereated particles are called associated colloids. For example: Micelles formation in soaps and detergents. | |
| 197. |
For a first order reaction, time taken for half of the reaction to complete is `t_(1)and 3/4` of the reaction to complete is `t_(2)`. How are `t_(1)and t_(2) related? |
| Answer» `t_(2)=2t_(1)` because for `3/4`th of the reaction to complete, time required is equal to two half lives. | |
| 198. |
Define osmotic pressure of a solution. How is the osmotic pressure orelated to the concentration of a solute in a solution ? |
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Answer» Osmotic pressure : Minimum excess pressure tht has to be applied on the solution to prevent the entry of the solvent into the solution through 5 P.M. Osmotic pressure is the collingative property. Hence it will increase if concentration of solute increases. |
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| 199. |
Define the following terms : (i) Half- life of a reaction `(t_(1//2))` (ii) Rate constant `(k)`. |
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Answer» (i) Half life period `(t_(1//2))` is the time in which half of the substance has reacted . `t_(1//2)=[A]_(0)/(2k)` (ii) Rate constant : It is the rate of reaction when the molar concentration of each reactant is taken as unity. |
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| 200. |
Calculate the energy in fusion reaction . `overset(2)underset(1).H +overset(2)underset(1).H` `to overset(4)underset(2).He + n` " where BE of " `overset(2)underset(1).H=2.23 MeV`" and of " `overset(4)underset(2).He =7.73 MeV` |
| Answer» Energy per fusion = 7.73 -2.23 - 2.23 =3.27 MeV | |