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]]>We have already studied about irrational numbers in previous class. Now, we will study the real numbers and also about the important properties of positive integers for Euclid’s division algorithm and the Fundamental Theorem of Arithmetic.
Euclid’s division algorithm says us about divisibility of integers. It states that any positive integer ‘a’ can be divided by any other positive integer ‘b’ in such a way that it has ‘r’ remainder which is smaller than ‘b’. In fact it is a long division method. We also use it to compute the HCF of two positive integers.
The Fundamental Theorem of Arithmetic says us about the expression of positive integers as the product of prime integers. It states that every positive integer is either prime or a product of powers of prime integers. We have already known how to find HCF and LCM of positive integers by using the Fundamental Theorem of Arithmetic in previous class. Now, we will learn about irrationality of many numbers like etc. by applying this theorem. We know that the decimal representation of a rational number is either terminating or if repeating if not terminating. We have to use the Fundamental Theorem of Arithmetic to determine the nature of the decimal expansion of rational numbers.
EUCLID’S DIVISION LEMMA:
Euclid was the first Greek Mathematician who gave a new way of thinking the study of geometry. He also made important contributions to the number theory. Euclid’s Lemma is one of them. It is a proven statement which is used to prove other statements.
Let ‘a’ and ‘b’ be any two positive integers. Then, there exist unique integers q and r such that
;
Now, we say ‘a’ as dividend, ‘b’ as divisor, ‘q’ as quotient and ‘r’ as remainder.
Dividend = (divisor quotient) + remainder
Example 1: Let 578 be divided by 16.
36 is as quotient and 2 as remainder.
In this case :
Dividend = 578
Divisor = 16 Quotient = 36 And remainder = 2 |
Dividend = (quotient divisor) + remainder
Hence, 578 = (36 16) + 2
Note:
To get the HCF of two positive integers, let ‘c’ and ‘d’, with c > d, follow as :
(i) Applying Euclid’s Lemma,
(ii) If r = 0, d is the HCF of c and d.
If then applying the division lemma to d and r.
(iii) We can continue the process till the remainder is zero. The divisor at this stage will be the required HCF.
This algorithm works because HCF (c,d) = HCF (d,r)
Where the symbol HCF (c,d) denotes the HCF of c and d etc.
Example 2: Use Euclid’s algorithm to find HCF of 425 and 40
Solution: Let a = 425 and b = 40
By Euclid’s division lemma; we have
…. (i) |
By using the above theorem, we observe that the common divisors of a = 425 and b = 40 are also the common divisors of b = 40 and r_{1} = 25 and vice-versa.
Applying Euclid’s division lemma on divisor b = 40 and remainder r_{1} = 25,
We get …. (ii) |
; where q_{2} = 1 and r_{2} = 15
Again using the above theorem, we find that, the common divisors of r_{1} = 25 and r_{2} = 15 are the common divisors of b = 40 and r_{1} = 25 and vice-versa. But the common divisors of b = 40 and r_{1} = 25 are the common divisors of a = 425 and b = 40 and vice-versa.
Applying Euclid’s division lemma on r_{1} = 25 and r_{2} = 15, we get
…. (iii) |
r_{1 }= r_{2}q_{3} + r_{3}, where q_{3} = 1 and r_{3} = 10
Again by using the above theorem, we find that common divisors of r_{2} = 15 and r_{3} = 10 are the common divisors of a = 425 and b = 40 and vice-versa.
Now, Using Euclid’s division lemma on r_{2} = 15 and r_{3} = 10, we get
…. (iv) |
; where q_{4} = 1 and r_{4} = 5
Again by using the above theorem, we find that, the common divisors of r_{2} = 15 and r_{3} = 10 are the common divisors of a = 425 and b = 40 and vice-versa.
Using Euclid’s division lemma on r_{3} = 10 and r_{4} = 5, we get
…. (v) |
Hence, r_{4} = 5 is a divisor of r_{3} = 10 and r_{4} = 5. Also, it is the greatest common divisor (or HCF) of r_{3} and r_{4}. So, r_{4} = 5 is the greatest common divisor (or HCF) of a = 425 and b = 40. We also observe that r_{4} = 5 is the last non-zero remainder in the above process of repeated application of Euclid’s division lemma on the divisor and the remainder in the next step.
The set of equation (i) to (v) is called Euclid’s division algorithm for 425 and 40. The last divisor, or the last but non-zero remainder 4 is the HCF (or GCD) of 425 and 40.
The above process of finding HCF can also be carried out by successive divisions as follows:
Note:
(i) Euclid’s division lemma and algorithm are so closely interlinked that it is often called division algorithm.
(ii) Euclid’s Division Algorithm is stated for only positive integers except zero. i.e., .
Example 3: A sweet seller has 840 kaju barfis and 260 badam barfis. He wants to stack them in such a way that each stack has the same number, and they take up the least area of the tray. What is the number of barfis that can be placed in each stack for this purpose?
Solution. The area of the tray that is used up will be the least. For this, we find HCF (840, 260). Then this number will give the maximum number of barfis in each stack and the number of stacks will then be the least.
Now, applying Euclid’s algorithm to find their HCF, we have
840 = 260 3 + 60
260 = 60 4 + 20
60 = 20 3 + 0
Hence, HCF of 840 and 260 is 20.
So, the sweet seller can make stacks of 10 for both kinds of barfi.
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]]>Answer.
The statements (a) and (b) are incorrect because PbO is getting reduced, not Pb, similarly C is getting oxidized not CO_{2}.
Answer.
(d) The given reaction is an example of displacement reaction as in this reaction Al displaces, Fe from Fe_{2}O_{3}.
Answer.
(a) Fe_{(s)} _{+} 2HCl_{(aq)} FeCI_{2(aq)} + H_{2(g)}
Answer.
Balanced chemical equation: – A chemical equation is said to be balanced if the number of atoms of each element participating in the reaction are equal on both side of the equation.
The chemical reaction should be balanced because law of conservation of mass holds good which states that “in a chemical reaction total mass of the reactant must be equal to the total mass of the product”.
Answer.
(a) N_{2(g)} + 3H_{2(g)} 2NH_{3(g)}
(b) 2H_{2}S_{(g)} +3O_{2(g) }2H_{2}O_{(l)} +2SO_{2(g)}
(c) 3BaCl_{2(aq) }+ Al_{2} (SO_{4})_{3(s)} 3BaSO_{4(s)} _{+} 2AlCl_{3(aq)}
(d) 2K_{(s)} + 2H_{2}O_{(l)} 2KOH_{(aq)}+H_{2(g)}
Answer.
(a) 2HNO_{3} + Ca(OH)_{2 } Ca(NO_{3})_{2} + 2H_{2}O
(b) 2NaOH + H_{2}SO_{4} Na_{2}SO_{4} + 2H_{2}O
(c) AgNO_{3} + NaCl AgCl + NaNO_{3}
(d) BaCl_{2} + H_{2}SO_{4} BaSO_{4} + 2HCl
Answer.
(a) Ca(OH)_{2} + CO_{2} CaCO_{3} + H_{2}O
(b) Zn+2AgNO_{3} Zn(NO_{3})_{2} + 2Ag
(c) 2Al + 3CuCl_{2} 2AlCl_{3} + 3Cu
(d) BaCl_{2} + K_{2}SO_{4} BaSO_{4}+2KCl
Answer.
(a) 2KBr_{(aq)} + Bal_{2(aq)} 2KI_{(aq)} + BaBr_{2(aq)}
(b) ZnC0_{3(s)} ZnO(s) + CO_{2}(g)
(c) H_{2(g}) + Cl_{2(g) } 2HCl_{(g)}
(d) Mg_{(s)} + 2HCl_{(aq)} MgCl_{2(aq)} + H_{2(g)}
Answer.
Exothermic reaction: – Those reactions which occur with evolution of heat are exothermic reactions. e.g.
2NaOH_{(aq)} + H_{2}SO_{4(aq) }Na_{2}SO_{4(aq)} + 2H_{2}O_{(l)} + Q
Enothermic reaction: – Those reactions which require heat to occur are endothermic reactions, e.g.
2Pb(NO_{3})_{2(s)} + Q 2PbO_{(s)} + 4NO_{2(g)} + O_{2(g)}
Answer.
During respiration oxidation of glucose occur which produce heat energy.
Answer.
During decomposition reaction a single reactant decompose to form two or more products, whereas in combination two or more reactant react to form a single product, e.g.
N_{2} + 3H_{2 }2NH_{3} (Combination reaction.)
2H_{2}O 2H_{2} + O_{2} (Decomposition reaction.)
Answer.
Answer.
Displacement reaction: In this reaction an element which is more active displaces the other element which is less active from its salt solution.
e.g. Fe_{(s)} + CuSO_{4(aq)} FeSO_{4(aq)} + Cu_{(s)}
Double displacement reaction: In this reaction exchange of ions between two reactants take place, e.g.
BaCl_{2(aq)} + Na_{2}SO_{4(aq) } BaSO_{4(s)} + 2NaCl_{(aq)}
Answer.
Cu_{(s)} + 2AgNO_{3(aq)} Cu(NO_{3})_{2(aq)} + 2Ag_{(s)}
Answer.
Those reactions which produce an isoluble product or precipitate are known as precipitation reaction. e.g.AgNO_{3(aq)} + NaCl_{(aq)} AgCl_{(s)} +NaNO_{3(aq)}
Answer.
Oxidation: Oxidation involve addition of oxygen e.g.
C + O_{2} CO_{2}, C is oxidised to CO_{2}.
N_{2} + O_{2} 2NO, N_{2} is oxidised to NO.
Reduction: Reduction involve removal of oxygen, e.g.
C + H_{2}O CO + H_{2}, H_{2}O is reduced to H_{2}.
CuO + H_{2} Cu + H_{2}O, CuO is reduced to Cu.
Answer.
Answer.
To prevent iron articles from rusting, we apply paint on it, by which iron does not come in contact with air and moisture which cause rusting.
Answer.
When the oil and fat containing food is surrounded by unreactive gas nitrogen, there is no oxygen to cause its oxidation and make it rancid.
Answer.
Corrosion: It is a process in which metals are eaten up gradually by the action of air, moisture or a chemical on their surface.
Rancidity: When fats and oil containing food are oxidised with air or oxygen, their smells and tastes changes. This process is called rancidity.
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]]>Answer.
The magnesium ribbon which we use has a coating of “basic magnesium carbonate” on its surface, which is formed by a slow action of moist air on it. Therefore before burning it in air, it is cleaned by a sand paper which remove the protective layer of basic magnesium carbonate from the surface of magnesium ribbon, so that it can easily burn.
Answer.
(i) H_{2(g)} + Cl_{2(g) } 2HCl_{(g)}
(ii) 3BaCl_{2(s)} +Al _{2}(S0_{4})_{3(aq) } 3BaSO_{4{s)} _{+} 2AlCl_{3(aq)}
(iii) 2Na_{(s)} + 2H_{2}O_{(l)} 2NaOH_{(aq)} + H_{2(g)}
Answer.
(i) BaCl_{2(aq)} + Na_{2}SO_{4(aq)} BaSO_{4(s)} + 2NaCl_{(aq)}
(ii) NaOH_{(aq)} _{+} HCl_{(aq)} NaCl_{(aq)} + H_{2}O_{(aq)}
Answer.
(i) Quick lime-CaO
(ii) CaO_{(s)} + H_{2 }O_{(l)} Ca(OH)_{2(aq)}
Answer.
The gases formed at two electrodes are produced due to decomposition of water on passing electricity. Since the electrolysis of water produces 2 volume of hydrogen gas and 1 volume of oxygen gas, thus we can say that the ratio of hydrogen and oxygen in water is 2:1 by volume and hence the amount of gas collected in one test tube will be double of the amount collected in other.
Answer.
As iron is more reactive than copper thus it displaces copper from copper sulphate solution.
Fe(s) +CuSO_{4(aq)} FeSO_{4(aq)} + Cu_{(s)}
(Blue) (Green)
In this way concentration of copper sulphate decreases while concentration of ferrous sulphate increases. As a result colour changes from blue to green.
Answer.
Ans. (i)
Answer (ii)
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]]>It is a simple way to represent any chemical reaction in the form of word equation. A chemical reaction involve reactants and the products separated by an arrow e.g. in the given reaction
Zn + H_{2} SO_{4} ZnSO_{4} + H_{2}
Zn and H_{2}SO_{4} are reactants and ZnSO_{4} and H_{2} are products.
A chemical reaction is said to be balanced if number of atoms of each element on the RHS and LHS of the arrow are same.
To balance a chemical equation we should, follow the following steps.
(i) First of all list the number of atoms of different elements present in the reactant and the product,
(ii) Now start from the element which contains the maximum number of atom whether it is a reactant or a product,
(iii) Now to equalise the number of atoms on both side multiply the compound with a suitable integer. Similarly balance the other elements which are not balanced,
(iv) To check wheather the reaction is balance or not count atoms of each element on both sides of the equation. If the number of atoms of elements on both sides are equal, the equation is balanced.
Example
FeSO_{4} Fe_{2}O_{3} + SO_{2} + SO_{3}
Total no. of Fe 1 (L.H.S) – 2 (R.H.S)
Total no. of S 1 ( L . H . S ) – 2 (R.H.S)
Total no. of O 4 (L.H.S) – 8 (R.H.S)
As FeSO_{4} contains maximum no. of atoms, thus start from this, in order to equalise no. of O atom on both side multiply FeS0_{4} by 2. We get
2FeSO_{4} Fe_{2}O_{3} + SO_{2} + SO_{3}
Now the total no. of atoms of all the elements on both sides are equal. Thus the reaction is balanced.
A chemical reaction involves the making and breaking of bonds between
atoms to produce new substances. Chemical reactions are of many types.
In these reactions two or more reactants combine to form a single product, e.g.
It has been observed that during the formation of Ca(OH)_{2}, a large amount of heat is evolved, such reactions in which heat is evolved along with the formation of products are known as exothermic reactions. Combination reactions are generally exothermic reactions.
These reactions are just opposite to the combination reactions, i.e. in such reactions a single reactant decomposes to give two or more products, e.g.
Decomposition reactions generally occur due to absorption of heat. i.e. endothermic. “Those reactions which proceed with absorption of heat are known as endothermic reactions”.
Those reactions in which one element displaces the other element from its aqueous solution are known as displacement Reactions, e.g.
Zn (s) + CuSO_{4} (aq) ZnSO_{4}(aq) + Cu (s)
Those reactions in which exchange of ions between the reactants occur are called double displacement reactions, e.g
Oxidation is a process which involve addition of oxygen or removal of hydrogen.
e.g.
C + O_{2}CO_{2} (Addition of oxygen)
H_{2}S + CI_{2} 2HCI + S (Removal of hydrogen)
Whereas reduction is a process which involve addition of hydrogen or removal of oxygen,
e.g.
H_{2} + Br_{2} 2HBr (Addition of hydrogen)
CuO + H_{2} Cu + H_{2}O (Removal of oxygen)
It is the deterioration of a metal as a result of its reaction with air or water surrounding it. Corrosion of iron is called rusting. Rust is Fe_{2}O_{3}. x H_{2}O.
Pure iron does not rust, also iron kept in vacuum does not rust because for rusting air is necessary. Rusting is faster in sea water than in marine water due to the presence of dissolved salt.
The rusting can be promoted by the presence of (i) impurities of metal (ii) moisture (iii) electrolyte.
When fats and oils are exposed to air they become rancid and their smell and taste
changes. To avoid this antioxidants are added to fats and oil. Keeping food in air tight jars also helps to prevent oxidation.
Here ZnO is reduced to Zn and C is oxidized to CO
Oxidizing and Reducing agent: The reactant which get oxidized itself but reduce others is called reducing agent, and the reactant which reduces itself and oxidizes others is called oxidizing agent.
e.g.
Oxidation is also known as a process of deelectronation – i.e. the process in which electrons are lost. e.g.
(loss of 2e^{–})
(loss of le^{–})
Reduction is the process of electronation i.e. the process in which electrons are gained. e.g.
(gain of 2e-)
(gain of 2e^{–})
In general almost all the reactions are redox reactions but there are some examples which neither show oxidation nor reduction e.g.
(i) BaCl_{2} + H_{2}SO_{4} BaSO_{4} + 2HCl
(ii) CuSO_{4} + 2H_{2}O Cu (OH)_{2} + H_{2}SO_{4}
(iii) NaOH + HCl NaCl + H_{2}O
(v) AgNO_{3} + HBr AgBr + HNO_{3}
(vi) Na_{2}O + H_{2}SO_{4} Na_{2}SO_{4} + H_{2}O
All these reaction shows neither any loss of electrons (oxidation) nor gain of electrons (Reduction)
The reaction which produce a precipitate is called precipitation reaction. Precipitation results insoluble substance known as precipitate, e.g
Na_{2}SO_{4}(aq) + BaCl_{2} (aq) BaSO_{4(s)} + 2NaCl(aq)
(ppt)
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