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NEET & AIIMS 11th PCB-Chemistry -Mole Concept Demo Videos
Hello, students, let us continue our discussion on mole and equivalent concept. First we will take an example on percentage purity and then we will discuss, what is the percentage yield of a reaction.
Let us take an example. An impure 6g of NaCl is dissolved in water and then treated with excess of silver nitrate. The mass of AgCl precipitate is found to be 14.35g. What is the percentage purity of NaCl? Now, in this case first of all we will see the reaction what is happening. It is double displacementry. It is a double displacement action. NaCl plus silver nitrate it gives you AgCl plus NaNO3. And now we will first develop the strategy here. What should be our strategy here? That any AgCl will be precipitated here it will be precipitated by the actual AgCl only. So, using the mass of AgCl precipitate, we will first calculate the mass of NaCl that is present and then using this information we will find the percentage purity. Whatever is actual, 6g is theoretically given and we can calculate. So, we write down the reaction and then we first do the stoichiometric analysis. 1 mole NaCl gives you 1 mole AgCl. So, using the molar mass data, now they have given atomic rates, for silver it is 108, sodium 23 and chlorine 35.5. So, from here we can calculate the molar mass of NaCl as 58.5 and that of AgCl as 143.5. Now, from here converting the moles into mass, we can write down 58.5 g of NaCl is producing 143.5 g of AgCl. Now, we will do reverse analysis. 143.5 g of AgCl is going to be produced by 58.5 g of NaCl. So, 1g of AgCl will be produced by 58.5 divided by 143.5 g of NaCl. Now, how much we are getting, 14.35 g of AgCl so that is going to produced, rather that is going to be produced by 58.5 by 143.5 into 14.35 g of NaCl. Which comes out as 5.85 g of NaCl, this was the actual amount. Theoretically how much it is given? 6 g. so, percentage purity simply what, 5.85 divided by 6 times 100 that is equal to 97.5.
Now, let us discuss a very important term percentage yield. What is percentage yield? Sometimes the reaction may not get completed according to the initial amount taken. Means whatever amount you have taken, sometimes you predict that it will happen but that doesn’t happen only. What are the possible reasons? The first possible reason is establishment of equilibrium. Now, this chapter you will read later in the portion of chemistry, but let me tell you, the amount reactant taken by you after a certain time it stabilises, it doesn’t proceed further. So, that is called as equilibrium. And therefore the reaction that you expect for the corresponding product doesn’t occur. Now, for the second reason, some of the reactants are not able to react due to formation of a protective layer. Suppose there is one solid and if you pour liquid on top of it, that reacts with the surface and then covers it up and the remaining solid cannot react and therefore we will not get the actual amount that we are thinking. Now, the third possible case would be when we take gaseous reactants they may simply escape out of the reaction chamber and will not give you the products. And the fourth possible reason can be, other reactions happening in parallel for the same reactants. Like till now you have studied that, N2 and H2 can give you ammonia NH3 but they can also combine to give you N2H4 which is called as hydrazine. So, you may not get ammonia in total that you are thinking. Now, in such cases the efficiency of the process is measured using percentage yield. What is percentage yield? It is a product mass that you actually get divided by the product mass that you theoretically predict times 100.
Now, let us take an example for this. A on controlled oxidation gives X according to the reaction. We have got stoichiometric. They are saying some of the reactant A oxidises and only 70% of maximum yield is obtained. What mass of X is produced if 200 g of A is taken? Now, what should be our strategy? First of all we will calculate normally, we will think 100% yield is there and let us find out how much mass of A that we can produce. And then using the mass of X that we have calculated, we will calculate the actual mass by using the percentage yield 70%. So, let us see, how we do it. We write down stoichiometry of the reaction. 2A plus 9B gives you 2X plus 4Y plus 5Z but what is the interesting part for me, 2A and 2X. So, we write down 2 mole A is going to produce 2 mole X. So, 1 mole A is going to produce 1 mole X and we first convert the moles into mass. So, 128 g of A will give you 128 g of X based on a molecular mass data that was given to us. So, from here we can say 128 g of A produces 148 g of X and therefore 1 g of A is going to produce 148 by 128 g of X. Assuming 100% yield, 200 g of A is going to produce 148 by 128 times 200 g of X. Now, since percentage yield is equal to product mass actual divided by product mass theoretical times 100. We will rearrange this equation because we want the actual mass. So, we can write it down as mass actual equal to percentage yield by 100 times mass theoretical. Now, putting in the value, value of yield is 70%, so, 70 by 100 times this much mass and that is going to give us 161.875 g. So, first we calculate the actual amount or rather the theoretical amount and then we calculate the actual amount using yield.
Thank you.
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NEET & AIIMS 11th PCB-Biology- Kingdom Plantae Demo Videos
Hello, dear students, welcome to the chapter Kingdom Plantae. Students, in this module we are just going to discuss about the introduction part.
Let’s start, students, the entire plant kingdom is broadly divided into various kind of components. Let’s see its components. The very first picture that I want to show you is of Volvox and they are included into a category which is known as algae, okay. Let’s see what is next, what you can see in the picture, some leaf like structure. Yes and they are included in a category which is known as Bryophytes, alright. So, students you must be aware of the term ‘ferns’. You must have seen in the jungles or forests, isn’t it? So, they are included in a category which is known as Pteridophytes, alright. Students, you must be visiting various gardens. Yes, have you seen this picture earlier, well this is the picture of Cycus and they are included into a category which is known as Gymnosperms, fine. And last but not the least look at the beautiful picture of the orchids, students, yes and they are included into a category which is very well known as Angiosperms. Fine, so, students, so far we have seen that the plant kingdom is divided into various kinds of components and they are Algae, Bryophytes, Pteridophytes, Gymnosperms and Angiosperms.
Well, so we have seen that the plant kingdom has various components, isn’t it. So, let’s see what is the position of the Algae, Bryophytes, Pteridophytes, Gymnosperms and Angiosperm?
Let’s see what is next. Students, the entire plant kingdom is broadly divided into two groups. And they are the Cryptogamae or Cryptogams which are non-flowering plants and Phanerogamae or Phanerogams which are flowering and seed bearing plants. Well, students, when I say there are seed bearing plants, there are two possibilities the seed may not be included or enclosed within the fruit. As you can in the Gymnosperms and the example is Cycus just now we have seen, isn’t it. And another example is Christmas tree. Well, so there is the second category which is known as Angiospermae. In this case the seed is enclosed within the fruit. Well, you can see the seed in mango, guava, isn’t it. So, they all are the Angiosperms.
So, students, the Cryptogamae which are further divided into Thallophyta, Bryophyta and Pteridophyta.
So, students in our coming module we are just going to discuss about the Thallophyta or Algae.
So, students, as you can see we have just completed the introduction and classification of plant kingdom.
See you in next module.
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NEET & AIIMS 11th PCB-Biology-Body Fluids & Circulation Demo Videos
Hello, students, in this module we are going to be drawing a so called very difficult diagram of biology. Which is the diagram I am talking about? I am talking about human heart and that too the internal structure of human heart. But you are going to observe within minutes this diagram is not going to be staying difficult for you. In fact you will enjoy drawing this diagram. But before we start with the diagram, students, you must know which side is which one. Now, in biology, if this is a sheet of paper in front you. Now, this is going to be the right side and this is going to be the left side of the diagram. So, let’s start drawing our diagram with first drawing the upper chambers and which are the chambers I am talking about the right atrium, students. So, first draw the right atrium. Similarly, a beautiful curve for the left atrium also. Now, this completes the upper two chambers of the heart. Now, we need to draw the lower chambers of the heart. So similarly, a nice curve for the right ventricle in here and then the left ventricle, students, on the other side. Now, this completes the outline of our heart.
Now observe, students, this is the base of the heart and now here this is the apex of the heart. Apex should be slightly tilted towards the left side. That’s a structurally correct diagram. Now further, students, draw the septum in between the ventricles. This is the inter-ventricular septum, students. After this a beautiful blood vessel arising from right ventricle, which is the pulmonary trunk, students, we are needing to draw. Now, pulmonary trunk, this trunk is dividing into right and the left branch. At this stage you need to draw only the left side of the pulmonary trunk, like this branching it beautifully in here, this is the left pulmonary artery. Now how about the right side, not drawing right now, students, because we need to draw another vessel right now which is the vessel arising from the left ventricle. I am talking about your systemic aorta arching beautifully in here. Three branches you draw in here and then curve it beautifully down here like this. This is your systemic aorta. After this, students, draw the superior vena cava in here and once we have drawn superior vena cava, draw down in here the inferior vena cava here. Now after drawing superior and inferior vena cava, students, now you need to complete the pulmonary trunks right side and observe in here, students, now herein right side of the pulmonary trunk. This is the right pulmonary artery branching beautifully just like the left side. So, this completed the pulmonary trunks both the sides. Now, observe the walls of all the four chambers. You need to draw the walls and how to draw the walls. Make thinner walls of atria as you can see in here. Here the atria have got thin walls, students, compared to the ventricle. Observe the walls of ventricle in here, students, the wall is uneven because of muscular ridges present and there are three papillary muscles also present in here and observe right ventricle wall is thicker than the wall of the right atrium. Now, left ventricles, students, when we need to draw, observe the wall of the left ventricle. It is uneven having two papillary muscles in here. Now, also note, students, when you are drawing the four walls of these four chambers, the thickest wall you will be making of the left ventricle. And I talked about the uneven line, don’t make a smooth line here. Why this uneven line because there are muscular ridges called columnae carnae present in here, also the muscles down here these are the papillary muscles in here, students.
Now, we need to see, we need to draw in each chamber what structure has to be present. Right now starting with the right atrium here, you need to draw the opening of the superior vena cava. Similarly you will be drawing the opening of the inferior vena cava also in here and also, students, in the right atrium you observe a depression down here. Now this depression is called as Fossa Ovalis. Now this was the right atrium structure.
How about the left atrium, in the left atrium, students, you need to draw four openings in here, two on one side and two on the other side. Now, these are the opening of the four pulmonary veins. Now, students, observe another important structure you need to draw. At this stage you will be drawing the valves which are the semi lunar valves. Here you can see this is the semi lunar valve of the pulmonary trunk and another semi lunar valve here of aorta, students. Now, these are the two pocket like valves we call as the semi lunar valves, the half moon shaped valves you need to draw.
After this an important structure we are drawing which is the valve present in between the right atrium and the right ventricle. Now these valves, students, this is known as the tricuspid valve because it is got three flaps present in it. On the other side that means in the left ventricle you can see, say in between the left atrium and the left ventricle, the valve which is present, this valve is the bicuspid valve because there are two flaps present in here. Observe in here, students, this is the bicuspid valve in here. Now, bicuspid valve is attached to the papillary muscles. These are the papillary muscles. By some fibres, now these are the fibres which are present, these fibres are known as chordae tendinae. So, three structures in here, this is the bicuspid valve also called as mitral valve then chordae tendinae, these fibrous structures, and down here you can see the papillary muscles in here.
Now, students, here in we are going to be colouring our heart. You don’t need to colour the heart. You need to just draw the pencil sketch. But let’s see how beautiful our heart looks in here. Now, look at that, pulmonary trunks dividing into pulmonary arteries, that systemic aorta then superior vena cava and the inferior vena cava also observe, students, the walls of all the four chambers, the thickest being the left ventricle wall, the openings of the atria, the bicuspid and the tricuspid valve and the pulmonary veins.
Now, this is the diagram of internal structure of heart, students, and the labelling which are required by you are these labelling only.
So, enjoy drawing the structure of human heart, take care.
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NEET & AIIMS 11th PCB Biology-Control and Co-ordination-Nervous Demo Videos
Hello students, so in this module we will be dealing with Hindbrain or Rhombencephalon. And in Hindbrain we will be taking care of Cerebellum in this module.
Now, first let’s see where is your Hindbrain? That’s your Hindbrain. It comprises of three parts. The first part is called as Cerebellum. Right ahead is Pons, the second part of Hindbrain. Below Pons is the last part called as Medulla Bblongata.
In this module, students, we will be talking only about Cerebellum. Now Cerebellum is this structure, students. So where is it present in your brain? This is the side view of the brain. So, Cerebellum is present right below Occipital lobe or you can also say it is present behind the Medulla. Now, let’s see what is the structure of Cerebellum, how does it look like externally as well as internally? Externally it has got two lateral lobes which are known as cerebellar hemispheres and right in between is a worm like lobe which is known the Vermis, the median lobe is called as Vermis. Now, Cerebellum, students, it comprises of 11% of the total brain weight. This is the second largest part of the brain. And which is the largest part of the brain, that being Cerebrum. Cerebellum, students, outside has got folds which are known as Folia.
Cerebellum, let’s see how it looks inside when we cut this section of Cerebellum. It looks like this. These are two halves of the cerebellum. Now, what are these two halves? The outer peripheral region is known as Cerebellar Cortex and it is made up of grey matter, whereas inner called as Cerebellar Medulla, is made up of white matter. But white matter in this structure as you can see, students, is in the form of a branching tree like structure and that branching tree like structure is called as Arbor Vitae known as tree of life. Arbor Vitae an important structure. Let’s see, how it actually in real brain looks like? It looks like this, this is Arbor Vitae the internal structure, the white matter of Cerebellum.
Students, what is the function of Cerebellum? Important functions which are co-ordination of voluntary muscles, also maintaining your equilibrium. That means what? That means it helps develop motor skills, the movement like walking or even finer skills like knitting. Now, there is a very well trained Cerebellum in this picture. Let’s observe, now you can see the co-ordination of voluntary muscles and maintaining equilibrium through beautifully trained Cerebellum in here.
Cerebellum, students, is highly sensitive to alcohol. Alcohol makes a person to lose the two important functions of cerebellum, co-ordination of muscular movements and loss of equilibrium. So, when Cerebellum is affected by alcohol. This is the impact of alcohol on Cerebellum, loss of equilibrium and co-ordination of muscles. In law it’s a criminal offence to be driving under the influence of alcohol.
So, in this module, students, we took care of Cerebellum. Next module we will be studying Pons Varolii as well as Medulla Oblongata. Till then take care.
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NEET & AIIMS 11th PCB- Physics Circular Motion Demo Videos
Hello, students, welcome back to the chapter circular motion. In the previous module we discussed about centripetal force which is necessary for a particle to move in a circular motion, be it a uniform circular motion or non-uniform circular motion. The value of centripetal force is equal to MV square by R or M omega square R. We also discussed that the particles would be acted upon by a tangential force which is M alpha R for non-uniform circular motion. In this particular module, we will discuss about the examples of the same.
The statement of the first example says that a particle is projected with a speed u at the angle theta with the horizontal and we are required to find out the radius of curvature at the highest point of the trajectory for the projectile. Now, we know that in this particular case the particle is going in a projectile motion. It is not a circular motion. But every point in the path can be treated as a point of a circle. So, we can find out the radius of curvature in this particular case as well. So, what we know is, that the particle is initially thrown at an angle theta with the horizontal with speed u. If we try to resolve this speed we get the value of horizontal speed to be u cos theta and the vertical speed here is u sin theta. The trajectory that the particle would follow would look something like this and at the highest point we know that the particle would have a horizontal speed, the vertical speed vanishes. And since the horizontal speed is u cos theta it does not change if air resistance is neglected, at the highest point the particle would have a horizontal speed of u cos theta. Now, we know that for a particle moving in circle the velocity is always tangential. So, therefore at the highest point what we can say is, the horizontal direction is our tangential direction that is e theta cap and the vertical direction is our radial direction that is er cap. Let’s say that the centre of the circle here is C and the radius is R. We are required to find out this value of R. Now, if we recall this steps that we have to follow to solve this kind of question is, first we have to identify the forces and we do that by drawing the FBD. In this particular case the particle is thrown as projectile and therefore it is going under the influence of gravity. Thus the only force that acts on the particle is mg and that is also vertically down. That means it is towards the centre and the net force which is required for a particle in circular motion towards the centre is equal to mv square by R. Thus this mg is the only force and we can conclude that mg here is providing the necessary centripetal force. So, equating this we get mg is equal to mv square by R, which on solving gives us value of R to be equal to v square by g and we are putting down the value of instantaneous speed at the highest point which is u cos theta. So, we get finally that the radius of curvature at the highest point as required in the question is u square cos square theta divided by g. This is the answer to this particular example.
Let us move ahead with another example which says that there is a bead which is placed over a rod which is hinged at one end. Now, this bead is at a distance of r from the hinge and the rod is rotating with the constant angular acceleration of alpha. We are also given that there is a coefficient of friction between the rod and the bead and the value is Mu. We have to find out the value of time and the angle after which the bead starts slipping and we have to neglect the gravity in this particular question.
Let us try to visualise what is happening in this particular case. The rod is actually rotating horizontally in this fashion and we can conclude that the bead would be moving in a circle of radius r before it starts slipping. So, if we try to view this particular motion from the top end, we will get the motion of the rod looks something like, so we can conclude that the bead is moving in the circle of radius r. And for a particle to move in a circle, what is required is a radial force and this radial force should be equal to m omega square r that means if it has a angular acceleration then we can conclude that the bead would also experience tangential force which is equal to m into alpha r. Now, what provides this bead the radial force? There is no gravity and the normal reaction would provide the tangential force. So, the friction in this particular case provides the necessary centripetal force, the radial force. Therefore we can write down m omega square r is equal to f which is equal to Mu N, let us mark this equation as one. Also we will discuss that the tangential force is provided by the normal reaction. And thus we can say that m alpha r is equal to N, this is our second equation. Now, if we divide the equation one and two, what we will get is the value of Mu the coefficient of friction is equal to omega square r divided by alpha r. Now, what we can conclude from here is the value of omega is equal to under root Mu into alpha. So, this is the value of omega where the bead would actually start slipping, why, because after this omega the friction would not be able to provide the necessary centripetal force. The value of friction would be less than the required value. So, what we will do is, now we have to find out the value of time and angle. We will use the fact that we can use the three equations of motion for the constant angular acceleration, where the first equation is omega is equal to omega 0 plus alpha t which gives us the value of t to be equal to under root of Mu by alpha. Now, using the third equation we get omega square is equal omega 0 square plus 2 alpha theta which gives us the value of angle to be equal to under root of Mu by 2. These two values are the final answer to this particular example.
I hope you have understood the module. We will continue our discussion on circular motion, thank you.
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NEET & AIIMS 11th PCB-Chemistry-General Organic Chemistry Demo Videos
Welcome back, students. So, since we have learned quite a lot about the phenomena of resonance. Now, this would be the last module where we will learn the application of resonance and proceed to the next electronic effect that is the introduction of a Hyperconjugation, alright.
So, when you talk about the application of resonance is basically the phenomena which helps us to understand the stability in the compound. So, here we are going to learn the acidic strength of carboxylic acid. As we know the fact any compound is acid if its conjugate base is stable. So, let’s look at the structure of carboxylic acid and if it donates a proton it forms a conjugate base which is called as a carboxylate ion, as we have seen earlier. Whereas now we are having two structures one is carboxylic acid and other is carboxylate ion. We will try and draw resonating structure of both of these because both of them are having lone alternate pie condition present in them. Which is one of the condition, which make the compound eligible to show phenomena of resonance. So, look at the carboxylic acid structure, one which is to the left. If we show delocalisation here, we will get a resonating structure which can be simply drawn like this. So, since the lone pair from the oxy atom has delocalised it is experiencing loss of lone pair, so it will carry a positive charge and the double bonded action gaining one addition electron pair so carry a negative charge. Which means now the newly obtained resonating structure carries a positive and negative charge, which means this structure is a charged separated structure. Now, try to work on the phenomena of resonance in this structure right here, which is nothing but the carboxylate ion. So, we again have a lone alternate pie condition present over here. So, if we show delocalisation will get one more resonating structure, which is equivalent of this carboxylate ion and you can see that there is only negative charge which is getting dispersed, delocalised. So, I will say that in this structure, one and two, which are the equivalent structures, you will find that these equivalent structures are charged delocalised structure. And whenever you get a charged separated structure, it is relatively less stable structure, on the other hand charged delocalised which means if only charge is getting dispersed, delocalised over several atoms then that delocalised structure is always a more stable structure than the other. And now you think about why are we calling this as an application? We say that any compound is acid if its conjugate base is stable. So, in case of carboxylic acid we saw that the conjugate base was a carboxylate ion which is giving us charged delocalised structure which is more stable. The more stable the conjugate base is, more stable is the acid from which that conjugate base is obtained. So, likewise, students, if you try and work on the sulphonic acid or if you try and work on the phenol, you will find that the charged delocalised structures are going to be more stable due to phenomena of a resonance. So, phenomena of a resonance can simply justify the acidic strength of the compounds. Well, having known this, now we are going to learn or see a simple sequence of acidic strength but the most important one. So, remember if we talk about this several acids the sulphonic acid is the strongest acid among all because its resonating structures are more stable. And after sulphonic acid the next acidic is carboxylic acid followed by we have a carbonic acid and then we have phenol, after phenol the structure or the compound which comes most acidic is methanol. Check out here we have methanol then relatively weaker acid is water and then we again have any other alcohol. Students, please look at this, methanol is only exceptional alcohol which is more acidic than water, rest all other alcohols are weaker acid than water and after this we have alkyne then we have ammonia and then we get alkene followed by alkane. So, this is a simple order of acidic strength of several compounds and you got to remember this, alright, students.
Now, having said that we are going to learn application of resonance and this is what, was the application of resonance. Now, we will see introduction to phenomena of hyperconjugation and there are several more names coming up right up there. We will justify each one of thes names but for the time being let’s look at this reaction.
So, we have tertiary butyl halide, so when this tertiary butyl halide is brought in presence of a polar solvent like water, again we will learn what is meant by a polar, then the C-X bond heterolytically breaks and which means the bond pair between C and X have been completely taken by X, so, the carbon has lost bond pair and therefore produces a positively charged species like this. This positively charged species where positive charged is carried by carbon is called carbocation. Now, let’s take that carbocation over here and remove all these methyl group. We will get the structure like this. So, this is a general representation of carbocation. Let’s take a replicate of that, so we are taking four such general representations of a carbocation. And now, in structure number one, right there we are fulfilling all the valences let’s say by methyl groups. So, now that carbocationic carbon which is carrying a positive charge is attached to three more carbons. So, the nature of the carbon carrying positive charge is tertiary. So, we call it as tertiary carbocation. Same way if we take two methyl groups attached to such carbon you will call it as the secondary carbocation. Since it is attached to two other carbons. So, if we move further, if I am fulfilling two valences H’s and one with methyl the nature of carbocationic carbon is primary and I fulfil all the valences with H’s, I will call it as the methyl carbocation, since it looks like methyl. So, the nature of carbon carrying positive charge decides the nature of carbocation and it is said that among all this the order of stability is found to be tertiary the most, then secondary, then primary and methyl is least stable carbocation and why is the order so, can be proven by the phenomena of hyperconjugation.
So, in the very next module, we will try and prove this order and with respect to this we will learn the phenomena of hyperconjugation but we will stop here in this module, thank you.
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