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Tue September 30, 2014  

Ans: C o n s i d e r space t h e space f o l l o w i n g space g r a p h : C o n s i d e r space t h e space g i v e n space e q u a t i o n s : 2 x minus 3 y equals 4 space a n d space 6 y equals 4 x plus 3 rightwards double arrow 2 x minus 4 equals 3 y space a n d space 6 y equals 4 x plus 3 rightwards double arrow y equals fraction numerator 2 x over denominator 3 end fraction minus 4 over 3 space a n d space y equals 4 over 6 x plus 3 over 6 C o m p a r e space t h e space a b o v e space e q u a t i o n s space w i t h space t h e space g e n e r a l space e q u a t i o n space y equals m x plus c comma space w h e r e space m space i s space t h e space s l o p e r space o f space t h e space w q u a t i o n a a n d space c space i s space t h e space y space i n t e r c e p t. L e t space m subscript 1 space a n d space m subscript 2 space b e space t h e space s l o p e s space o f space t h e space l i n e s space 2 x minus 3 y equals 4 space a n d space 6 y equals 4 x plus 3 rightwards double arrow m subscript 1 equals 2 over 3 space a n d space m subscript 2 equals 4 over 6 equals 2 over 3 rightwards double arrow m subscript 1 equals 2 over 3 equals m subscript 2 S i n c e space t h e space s l o p e s space a r e space e q u a l comma space t h e space l i n e s space a r e space p a r a l l e l.

Tue September 30, 2014  

Ans: Copper vessels get tarnished by reaction with oxygen in the air (black copper oxide is formed) or by reaction with water (green copper hydroxide is formed).
These substances are difficult to remove and they aren't very soluble in water. Lemon juice contains citric acid. This reacts with the copper oxide and copper hydroxide to produce copper citrate. This is readily soluble in water and washes off easily, leaving behind shiny copper surface.

Tue September 30, 2014  

Ans: begin mathsize 12px style L a w s space o f space i n d i c e s space a to the power of m cross times a to the power of n equals a to the power of m plus n end exponent  fraction numerator x over denominator square root of 2 cross times 3 root of 2 end fraction equals fraction numerator x over denominator open parentheses 2 close parentheses to the power of begin display style 1 half end style end exponent cross times open parentheses 2 close parentheses to the power of 1 third end exponent end fraction equals x over open parentheses 2 close parentheses to the power of begin display style 1 half plus 1 third end style end exponent equals x over open parentheses 2 close parentheses to the power of begin display style fraction numerator 3 plus 2 over denominator 6 end fraction end style end exponent equals x over open parentheses 2 close parentheses to the power of begin display style 5 over 6 end style end exponent end style

Tue September 30, 2014  

Ans: The van der Waals equation contains a pair of constants ( a and b ) that change from gas to gas. The value of ‘a’ accounts for the intermolecular attractive forces between gas molecules. The magnitude of ‘a’ is indicative of the strength of the intermolecular attractive force. Molecules experiencing the weakest attractive forces will have the smallest ‘a’ constant while those with the strongest attractive forces will have the largest values. Eg: Consider N 2 gas and NH 3 gas. N 2 gas has non-polar bonds between their components atoms which results in weaker attractions between molecules. In case of NH 3 gas hydrogen bonds between molecules which are stronger than non-polar bonds. So NH 3 gas has higher ‘a’ constant value (4.170) than N 2 gas (1.390). The factor ‘b’ accounts for the volume occupied by the gas molecules. Small molecular volume results in small b values and a large molecular volume corresponds to a large 'b' constant. Eg: CH 4 gas has higher molecular volume as compare to H 2 gas so CH 4 gas will show higher ‘b’ constant value (0.04278) while H 2 gas will show lower ‘b’ constant value (0.02661).

Tue September 30, 2014  

Ans: Dear ankit.jatiya715@gmail.com

Thanks for asking us a question in Ask the Expert section of TopperLearning.com.
The van der Waals equation becomes the ideal gas law as the constant 'a' or constant 'b' value approach to zero.

Regards

Topperlearning Team.

Tue September 30, 2014  

Ans: A vapour is a substance which experiences a phase change while gas does not. A substance is called as in vapour state when it is in solid or liquid state at room temperature. That means its natural state of matter is solid or liquid. The natural state of matter for a gas is gas. For example, steam is vapour because it is in liquid state at room temperature. N 2 , O 2 are gases as they are in gaseous state at room tempearture.

Tue September 30, 2014  

Ans: Dear ankit.jatiya715@gmail.com

Thanks for asking us a question in Ask the Expert section of TopperLearning.com.
This is observed because the gases like hydrogen and helium shows intermolecular repulsive forces which cause the actual volumes to be greater than the ideal values.

Regards

Topperlearning Team.

Tue September 30, 2014  

Ans: x equals 2...... left parenthesis i right parenthesis y equals 5...... left parenthesis i i right parenthesis x plus y equals 6... left parenthesis i i i right parenthesis S o l v i n g space left parenthesis i right parenthesis space a n d space left parenthesis i i right parenthesis comma space w e space g e t x equals 2 comma space y equals 5 A left parenthesis 2 comma space 5 right parenthesis S o l v i n g space left parenthesis i right parenthesis space a n d space left parenthesis i i i right parenthesis comma space w e space g e t x equals 2 comma space y equals 4 B left parenthesis 2 comma space 4 right parenthesis S o l v i n g space left parenthesis i i right parenthesis space a n d space left parenthesis i i i right parenthesis comma space w e space g e t x equals 1 comma space y equals 5 C left parenthesis 1 comma space 5 right parenthesis O r t h o c e n t e r space o f space a space r i g h t space a n g l e d space t r i a n g l e space i s space a t space t h e space a n g l e space c o n ta i n i n g space r i g h t space a n g l e. F o r space increment A B C comma space o r t h o c e n t r e space i s space a t space A left parenthesis 2 comma 5 right parenthesis

Tue September 30, 2014  

Ans: fraction numerator x squared plus 5 x plus 6 over denominator x squared plus 4 x plus 4 end fraction equals fraction numerator 3 x plus 1 over denominator 5 x minus 2 end fraction rightwards double arrow fraction numerator x squared plus 3 x plus 2 x plus 6 over denominator left parenthesis x plus 2 right parenthesis squared end fraction equals fraction numerator 3 x plus 1 over denominator 5 x minus 2 end fraction rightwards double arrow fraction numerator x left parenthesis x plus 3 right parenthesis plus 2 left parenthesis x plus 3 right parenthesis over denominator left parenthesis x plus 2 right parenthesis squared end fraction equals fraction numerator 3 x plus 1 over denominator 5 x minus 2 end fraction rightwards double arrow fraction numerator left parenthesis x plus 2 right parenthesis left parenthesis x plus 3 right parenthesis over denominator left parenthesis x plus 2 right parenthesis squared end fraction equals fraction numerator 3 x plus 1 over denominator 5 x minus 2 end fraction rightwards double arrow fraction numerator x plus 3 over denominator x plus 2 end fraction equals fraction numerator 3 x plus 1 over denominator 5 x minus 2 end fraction left square bracket because x not equal to minus 2 right square bracket rightwards double arrow left parenthesis x plus 3 right parenthesis left parenthesis 5 x minus 2 right parenthesis equals left parenthesis x plus 2 right parenthesis left parenthesis 3 x plus 1 right parenthesis rightwards double arrow 5 x squared minus 2 x plus 15 x minus 6 equals 3 x squared plus x plus 6 x plus 2 rightwards double arrow 2 x squared plus 6 x minus 8 equals 0 rightwards double arrow x squared plus 3 x minus 4 equals 0 rightwards double arrow x squared plus 4 x minus x minus 4 equals 0 rightwards double arrow x left parenthesis x plus 4 right parenthesis minus 1 left parenthesis x plus 4 right parenthesis equals 0 rightwards double arrow left parenthesis x minus 1 right parenthesis left parenthesis x plus 4 right parenthesis equals 0 therefore x equals 1 comma minus 4

Mon September 29, 2014  

1) What is coordination no of atoms :
 
A) IN A CUBIC CLOSE PACKED STRUCTURE
 
b) In abcc
 
2) A cubic solid is made of two elements P & Q . Atoms of Q are at the corners of the cube and P at the body centre.What is formula of the compound? What are the coordination no P & Q ?
 
3) If the radius of octahedral void is are r and radius of the atoms in close packing is R derive relation between r & R ?
 
4 ) Ferric oxide crystallise in a hexagonal close packed array of oxides ions with two out of every three octahedral holes occupied by ferric ions. derive the formula of ferric oxide.
 
5) Gold (atomic radius = 0.144 nm) crystallise in a ff centred unit cell.what is the length of a side of the cell ?
 
6) ALUMINIUM CRYSTALLISE IN A CUBIC CLOSE PACKED STRUCTURE . Its metallic radius is 125 pm
 
How many unit cells are there in 1.00 cm3 of aluminium ?
 
7) If NaCl is dopped with 10-3 mol % of SrCl2.what is the concentration of cation vacancies ?
 
 
 
 
 
 
Ans: Dear pandeyknk@gmail.com

Thanks for asking us a question in Ask the Expert section of TopperLearning.com.
In case of multiple questions within a query, please post each question individually and let us know where you are getting stuck so that we would be able to explain things better.     Solution for your first query: The hexagonal closest packed (hcp) has a coordination number of   12 and contains 2 atoms per unit cell. The face-centered cubic (fcc) or cubic closed packed has a coordination number of   12   and contains   4   atoms per unit cell. The body-centered cubic (bcc) has a coordination number of   8   and contains 2 atoms per unit cell.     Regards

Topperlearning Team.

Mon September 29, 2014  

Ans: The seven fundamental or base quantities chosen in SI units are the seven dimensions. They are denoted within square brackets[ ] . Length is represented by [L], mass by [M] time by [T] ,electric current by [A] temperature by [K], luminous intensity by [Cd] and amount of substance by [mol]. The dimensions of a physical quantity are the powers or exponents to which the units of base quantities are raised to represent a derived unit of that quantity. All physical quantities can be represented in terms of the dimensions of length[L] , mass[M], and time[T]   Example: Area = length × breadth = L× L Dimensional formula = [L] 2 Thus to represent area we have to raise [L] to the power 2. Therefore Area is said to have two dimensions in length . As unit of mass and time are not required in representing area , we can write the dimensional formula of area as: [M 0 L 2 T 0 ] and hence we say that area has zero dimension in mass and time in addition to 2 dimensions in length. Similarly Velocity =displacement / Time                         =[L] / [T] Velocity =[LT -1 ] =[M 0 L 1 T -1 ]. This is the dimensional formula of velocity The dimensions of velocity are : zero in mass. +1 in length and -1 in time. To find the dimensions : First the relation of the physical quantity with respect to other quantities have to be known, like density = mass/ volume. And then its written in terms of the dimensions of the fundamental quantities . In such a way you can find the dimensions of any physical quantity. 

Mon September 29, 2014  

Ans: C o n s i d e r space t h e space e v a l u a t i o n space o f space t h e space i n t e g r a l comma space integral fraction numerator d x over denominator 1 plus x to the power of 4 end fraction : I equals integral fraction numerator d x over denominator 1 plus x to the power of 4 end fraction equals 1 half integral fraction numerator 1 plus x squared plus 1 minus x squared over denominator 1 plus x to the power of 4 end fraction d x equals 1 half open square brackets integral fraction numerator 1 plus x squared over denominator 1 plus x to the power of 4 end fraction d x plus integral fraction numerator 1 minus x squared over denominator 1 plus x to the power of 4 end fraction d x close square brackets equals 1 half open square brackets integral fraction numerator 1 plus begin display style 1 over x squared end style over denominator x squared plus 1 over x squared end fraction d x minus integral fraction numerator 1 minus begin display style 1 over x squared end style over denominator x squared plus 1 over x squared end fraction d x close square brackets equals 1 half open square brackets integral fraction numerator 1 plus begin display style 1 over x squared end style over denominator open parentheses x minus begin display style 1 over x end style close parentheses squared plus 2 end fraction d x minus integral fraction numerator 1 minus begin display style 1 over x squared end style over denominator open parentheses x plus 1 over x close parentheses squared minus 2 end fraction d x close square brackets N o w space s u b s t i t u t e space x minus 1 over x equals t space a n d space x plus 1 over x equals u ; rightwards double arrow open parentheses 1 plus 1 over x squared close parentheses d x equals d t space a n d space open parentheses 1 minus 1 over x squared close parentheses d x equals d u rightwards double arrow I equals 1 half open square brackets integral fraction numerator begin display style 1 end style over denominator t squared plus 2 end fraction d x minus integral fraction numerator begin display style 1 end style over denominator u squared minus 2 end fraction d u close square brackets rightwards double arrow I equals 1 half open square brackets integral fraction numerator begin display style 1 end style over denominator t squared plus 2 end fraction d x minus integral fraction numerator begin display style 1 end style over denominator u squared minus 2 end fraction d u close square brackets rightwards double arrow I equals 1 half open square brackets fraction numerator 1 over denominator square root of 2 end fraction tan to the power of minus 1 end exponent fraction numerator t over denominator square root of 2 end fraction minus fraction numerator 1 over denominator 2 square root of 2 end fraction log fraction numerator u minus square root of 2 over denominator u plus square root of 2 end fraction close square brackets rightwards double arrow I equals 1 half open square brackets fraction numerator 1 over denominator square root of 2 end fraction tan to the power of minus 1 end exponent fraction numerator open parentheses x minus 1 over x close parentheses over denominator square root of 2 end fraction minus fraction numerator 1 over denominator 2 square root of 2 end fraction log fraction numerator space x plus 1 over x minus square root of 2 over denominator space x plus 1 over x plus square root of 2 end fraction close square brackets

Mon September 29, 2014  

Ans: begin mathsize 14px style Given : Dielectric space strength space of space air space straight V equals 3 cross times 10 to the power of 6 space straight V divided by straight m Area space straight A equals 20 space cm squared Plate space separation space straight d equals 0.1 space mm straight E equals straight V over straight d straight V equals straight E. straight d equals 3 cross times 10 to the power of 6 cross times 10 to the power of minus 4 end exponent equals 300 space straight V straight V subscript rms equals fraction numerator straight V subscript straight p over denominator square root of 2 end fraction equals fraction numerator 300 over denominator square root of 2 end fraction equals 300 cross times 0.707 equals 210 space straight V Hence comma space maximum space rms space voltage space of space an space AC space source space which space can space be space connected space is space 210 space straight V end style

Mon September 29, 2014  

Ans: ¤® ¤¿ ¤¤ ¥ ¤° ¤µ ¤° ¤ ¤ª ¤• ¥‡ ¤ª ¥ ¤° ¤¶ ¥ ¤¨ ¤• ¥‡ ¤² ¤¿ ¤ ¤§ ¤¨ ¥ ¤¯ ¤µ ¤¾ ¤¦ ¥¤ ¤ ¤• ¤¸ ¤® ¤¯ ¤ª ¤° ¤ ¤ª ¤ ¤• ¤¹ ¥€ ¤ª ¥ ¤° ¤¶ ¥ ¤¨ ¤• ¥€ ¤¸ ¥ ¤µ ¤¿ ¤§ ¤¾ ¤• ¤¾ ¤² ¤¾ ¤­ ¤‰ ¤ ¤¾ ¤¸ ¤• ¤¤ ¥‡ ¤¹ ¥ˆ ¤‚ ¤¹ ¤® ¤ ¤ª ¤• ¥‹ ( ¤ª ¥ ¤° ¤¶ ¥ ¤¨ 1) ¤• ¤µ ¤¿ ¤— ¤¿ ¤° ¤¿ ¤§ ¤° ¤• ¥€ ¤• ¥ ¤£ ¥ ¤¡ ¤² ¤¿ ¤¯ ¤¾ ¤ ¤¯ ¤¹ ¤¾ ¤ ¤ª ¤° ¤‰ ¤ª ¤² ¤¬ ¥ ¤§ ¤• ¤° ¤µ ¤¾ ¤° ¤¹ ¥‡ ¤¹ ¥ˆ ¤‚ ¥¤ ( ¤ª ¥ ¤° ¤¶ ¥ ¤¨ 2 ) ¤µ ¤¾ ¤¦ - ¤µ ¤¿ ¤µ ¤¾ ¤¦ ¤¦ ¥ ¤° ¥ ¤­ ¤¾ ¤— ¥ ¤¯ ¤¸ ¥‡ ¤ ¤ª ¤• ¥‡ ¤¹ ¤® ¤¾ ¤° ¥€ ¤ª ¥ ¤° ¤¦ ¤¾ ¤¨ ¤¸ ¥‡ ¤µ ¤¾ ¤“ ¤‚ ¤• ¥‡ ¤¦ ¤¾ ¤¯ ¤° ¥‡ ¤¸ ¥‡ ¤¬ ¤¾ ¤¹ ¤° ¤¹ ¥‹ ¤œ ¤¾ ¤¤ ¤¾ ¤¹ ¥ˆ ¥¤ " ¤µ ¤¿ ¤¶ ¥‡ ¤· ¤œ ¥ ¤ž ¤¸ ¥‡ ¤ª ¥‚ ¤› ¥‹" ("Ask the Expert" ) ¤• ¥€ ¤¸ ¥‡ ¤µ ¤¾ ¤› ¤¾ ¤¤ ¥ ¤° ¥‹ ¤‚ ¤• ¥‡ ¤¸ ¤‚ ¤¦ ¥‡ ¤¹ ¥‹ ¤‚ ¤• ¥‹ ¤¸ ¥ ¤ª ¤· ¥ ¤Ÿ ¤• ¤° ¤¨ ¥‡ ¤• ¥‡ ¤² ¤¿ ¤ ¤¹ ¥ˆ ¤¨ ¤• ¤¿ ¤² ¤‚ ¤¬ ¥‡ ¤‰ ¤¤ ¥ ¤¤ ¤° ¤ª ¥ ¤° ¤¦ ¤¾ ¤¨ ¤• ¤° ¤¨ ¥‡ ¤• ¥‡ ¤² ¤¿ ¤ ¥¤ ¤¯ ¤¦ ¤¿ ¤ ¤ª ¤¨ ¥‡ ¤‡ ¤¸ ¤µ ¤¿ ¤· ¤¯ ¤ª ¤° ¤œ ¤µ ¤¾ ¤¬ ¤¤ ¥ˆ ¤¯ ¤¾ ¤° ¤• ¤¿ ¤¯ ¤¾ ¤¹ ¥ˆ ¤¤ ¥‹ ¤ ¤• ¤µ ¤¿ ¤¶ ¥‡ ¤· ¤® ¤¾ ¤® ¤² ¥‡ ¤• ¥‡ ¤° ¥‚ ¤ª ¤® ¥‡ ¤‚ ¤¹ ¤® ¤‰ ¤¸ ¤• ¥€ ¤¸ ¤® ¥€ ¤• ¥ ¤· ¤¾ ¤… ¤µ ¤¶ ¥ ¤¯ ¤• ¤° ¥‡ ¤‚ ¤— ¥‡ ¥¤ ¤‡ ¤¸ ¤ª ¥ ¤° ¤¶ ¥ ¤¨ ¤• ¥‡ ¤¸ ¥ ¤¥ ¤¾ ¤¨ ¤ª ¤° ¤ ¤ª ¤ ¤• ¤” ¤° ¤ª ¥ ¤° ¤¶ ¥ ¤¨ ¤ª ¥‚ ¤› ¤¸ ¤• ¤¤ ¥‡ ¤¹ ¥ˆ ¤‚ ¥¤ ¤— ¤¿ ¤° ¤¿ ¤§ ¤° ¤• ¥€ ¤• ¥ ¤£ ¥ ¤¡ ¤² ¤¿ ¤¯ ¤¾ ¤ - 1)      ¤¬ ¥€ ¤¤ ¥€ ¤¤ ¤¾ ¤¹ ¤¿ ¤¬ ¤¿ ¤¸ ¤¾ ¤° ¤¿ ¤¦ ¥‡, ¤ ¤— ¥‡ ¤• ¥€ ¤¸ ¥ ¤§ ¤¿ ¤² ¥‡ ¤‡ ¥¤
¤œ ¥‹ ¤¬ ¤¨ ¤¿ ¤ ¤µ ¥ˆ ¤¸ ¤¹ ¤œ ¤® ¥‡ ¤‚, ¤¤ ¤¾ ¤¹ ¥€ ¤® ¥‡ ¤‚ ¤š ¤¿ ¤¤ ¤¦ ¥‡ ¤‡ ¥¥

¤¤ ¤¾ ¤¹ ¥€ ¤® ¥‡ ¤‚ ¤š ¤¿ ¤¤ ¤¦ ¥‡ ¤‡, ¤¬ ¤¾ ¤¤ ¤œ ¥‹ ¤ˆ ¤¬ ¤¨ ¤¿ ¤ ¤µ ¥ˆ ¥¤
¤¦ ¥ ¤° ¥ ¤œ ¤¨ ¤¹ ¤‚ ¤¸ ¥‡ ¤¨ ¤• ¥‹ ¤‡, ¤š ¤¿ ¤¤ ¥ ¤¤ ¤® ¥ˆ ¤‚ ¤– ¤¤ ¤¾ ¤¨ ¤ª ¤¾ ¤µ ¥ˆ ¥¥

¤• ¤¹ ' ¤— ¤¿ ¤° ¤¿ ¤§ ¤° ¤• ¤µ ¤¿ ¤° ¤¾ ¤¯ ¤¯ ¤¹ ¥ˆ ¤• ¤° ¥ ¤® ¤¨ ¤ª ¤° ¤¤ ¥€ ¤¤ ¥€ ¥¤
¤ ¤— ¥‡ ¤• ¥‹ ¤¸ ¥ ¤– ¤¸ ¤® ¥ ¤ ¤¿, ¤¹ ¥‹ ¤‡ ¤¬ ¥€ ¤¤ ¥€ ¤¸ ¥‹ ¤¬ ¥€ ¤¤ ¥€ 2)      ¤¸ ¤¾ ¤ˆ ¤‚, ¤¬ ¥ˆ ¤° ¤¨ ¤• ¥€ ¤œ ¤¿ ¤, ¤— ¥ ¤° ¥, ¤ª ¤‚ ¤¡ ¤¿ ¤¤, ¤• ¤µ ¤¿, ¤¯ ¤¾ ¤° ¥¤      ¤¬ ¥‡ ¤Ÿ ¤¾, ¤¬ ¤¨ ¤¿ ¤¤ ¤¾, ¤ª ¤ ¤µ ¤° ¤¿ ¤¯ ¤¾, ¤¯ ¤œ ¥ ¤ž– ¤• ¤° ¤¾ ¤µ ¤¨ ¤¹ ¤¾ ¤° ¥¥     ¤¯ ¤œ ¥ ¤ž– ¤• ¤° ¤¾ ¤µ ¤¨ ¤¹ ¤¾ ¤°, ¤° ¤¾ ¤œ ¤® ¤‚ ¤¤ ¥ ¤° ¥€ ¤œ ¥‹ ¤¹ ¥‹ ¤ˆ ¥¤     ¤µ ¤¿ ¤ª ¥ ¤°, ¤ª ¤¡ ¤¼ ¥‹ ¤¸ ¥€, ¤µ ¥ˆ ¤¦ ¥ ¤¯, ¤ ¤ª ¤• ¥€ ¤¤ ¤ª ¥ˆ ¤° ¤¸ ¥‹ ¤ˆ ¥¥    ¤• ¤¹ ¤— ¤¿ ¤° ¤¿ ¤§ ¤° ¤• ¤µ ¤¿ ¤° ¤¾ ¤¯, ¤œ ¥ ¤— ¤¨ ¤¤ ¥‡ ¤¯ ¤¹ ¤š ¤² ¤¿ ¤ ¤ˆ,    ¤‡ ¤… ¤¨ ¤¤ ¥‡ ¤° ¤¹ ¤¸ ¥‹ ¤‚ ¤¤ ¤° ¤¹ ¤¦ ¤¿ ¤¯ ¥‡ ¤¬ ¤¨ ¤¿ ¤ ¤µ ¥‡ ¤¸ ¤¾ ¤ˆ ¤‚ ¥¥        ¤… ¤§ ¥ ¤¯ ¤¯ ¤¨ ¤• ¥‡ ¤² ¤¿ ¤ ¤¹ ¤® ¤¾ ¤° ¥€ ¤¹ ¤¾ ¤° ¥ ¤¦ ¤¿ ¤• ¤¶ ¥ ¤­ ¤• ¤¾ ¤® ¤¨ ¤¾ ¤ ¤ ¥¤

Mon September 29, 2014  

Ans: A left parenthesis 0 comma minus 3 right parenthesis comma space B left parenthesis 4 comma 0 right parenthesis space a n d space C left parenthesis minus 4 comma 0 right parenthesis U sin g space D i s tan c e space f o r m u l a comma space w e space g e t A B equals square root of left parenthesis 0 minus 4 right parenthesis squared plus left parenthesis minus 3 minus 0 right parenthesis squared end root equals 5 B C equals square root of left parenthesis 4 plus 4 right parenthesis squared plus left parenthesis 0 minus 0 right parenthesis squared end root equals 8 A C equals square root of left parenthesis 0 plus 4 right parenthesis squared plus left parenthesis minus 3 minus 0 right parenthesis squared end root equals 5 A s space A B equals A C comma space t h e r e f o r e space t h e space t r i a n g l e space i s space a n space i s o s c e l e s space t r i a n g l e.

Mon September 29, 2014  

Ans: Dear kalitapadmanath@yahoo.in

Thanks for asking us a question in Ask the Expert section of TopperLearning.com.
Answer to 1st MCQ is colourless. As gases released during the process are SO 2 and SO 3 which are colourless.
In case of second MCQ: One of the options must be an acid or the compound which will lower down the pH. Option (a): Sodium hydroxide is basic in nature, if it is present in water then pH will be more than 7. Option (b): Sodium chloride is neutral in nature which also will NOT affect the pH. Option (c) Sodium bicarbonate and (d) Sodium carbonate are basic in nature so here also pH will be more than 7.
Regards

Topperlearning Team.

Mon September 29, 2014  

Ans: Carbon has an electronic arrangement of 2, 4. In diamond, each carbon shares electrons with four other carbon atoms forming four single bonds.  In graphite, each carbon atom uses three of its electrons to form simple bonds to its three close neighbours. That leaves a fourth electron in the bonding level. These spare electrons in each carbon atom become delocalized over the whole of the sheet of atoms in one layer. They bare no longer associated directly with any particular atom or pair of atoms, but are free to wander throughout the whole sheet. The structures of diamond and graphite are as follows:

Sun September 28, 2014  

Ans: L e t space x space a n d space y space b e space t h e space s u m space o f space a m o u n t space i n v e s t e d space i n space t w o space t y p e s space o f space b o n d s. G i v e n space t h a t comma x plus y equals 20000... left parenthesis 1 right parenthesis S i n c e space S e e m a space e a r n s space 12 space p e r c e n t space o n space t h e space f i r s t space t y p e space a n d space 15 space p e r c e n t space o n space t h e space s e c o n d space t y p e space a n d space h e r space t o t a l e a r n i n g space i s space R s.2850 comma space w e space h a v e comma 0.12 x plus 0.15 y equals 2850... left parenthesis 2 right parenthesis rightwards double arrow 12 x plus 15 y equals 285000.... left parenthesis 3 right parenthesis S o l v i n g space e q u a t i o n s space left parenthesis 1 right parenthesis space a n d space left parenthesis 3 right parenthesis comma space w e space h a v e comma y equals R s.15 comma 000 space a n d space x equals R s.5 comma 000

Sun September 28, 2014  

Ans:

Sun September 28, 2014  

Ans: L e t space t h e space r a t e space o f space b l a c k space s o c k s space equals R s. x L e t space t h e space r a t e space o f space b l u e space s o c k s equals R s. y G i v e n space t h a t space x equals 2 y L e t space k space b e space t h e space n u m b e r space o f space b l u e space s o c k s space J o h n space i n t e n d space t o space b u y. T h e space t o t a l space a m o u n t space h e space i n t e n d e d space t o space p a y equals 4 x plus k y equals 4 x plus k open parentheses x over 2 close parentheses I n t e r c h a n g e d space b i l l equals k x plus 4 y equals k x plus 4 open parentheses x over 2 close parentheses equals k x plus 2 x G i v e n space t h a t space t h e space b i l l space h a s space b e e n space i n t e r c h a n g e d space a n d space h e n c e space t h e space cos t space h a s space b e e n space i n c r e a s e space b y space 50 %. rightwards double arrow k x plus 2 x equals 3 over 2 open square brackets 4 x plus k open parentheses x over 2 close parentheses close square brackets rightwards double arrow 2 k x plus 4 x equals 12 x plus fraction numerator 3 k x over denominator 2 end fraction rightwards double arrow 2 k minus fraction numerator 3 k over denominator 2 end fraction equals 12 minus 4 rightwards double arrow k over 2 equals 8 rightwards double arrow k equals 16 T h u s comma space t h e space r a t i o space o f space n u m b e r space o f space p a i r s space o f space b l a c k space s o c k s space t o space t h e space n u m b e r space o f space p a i r s space o f space b l u e space s o c k s equals 4 over 16 equals 1 fourth.
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