Sunday, October 30, 2011

CBSE-10th 2012- Second Term March Sample Question Paper

Saturday, October 29, 2011

10th CBSE Question Banks - Ch. How do Organisms Reproduce ?

Chapter - 8 How Do Organisms Reproduce Points to remember:-

Q 1. What is the importance of DNA copying in reproduction?

Ans. DNA contains information for inheritance of features from parents to next generation. DNA 
present in the cell nucleus is the information source for making proteins. If the information is different,  different proteins will be made. Different proteins will eventually lead to altered body designs.

Q 2. Why is variation beneficial to the species but not necessarily for the individual?

Ans. Variations are useful for the survival of species in changed environmental situations. For 
example, if a population of reproducing organisms were suited to a particular niche (well-defined 
place of abode) and if the niche is drastically altered the population could be wiped out. However, if 
some variations were to be present in a few individuals in these populations, there would be some 
chance for them to survive. Thus if there were a population of bacteria living in temperate waters, and  if water temperature were to be increased by global warming , most of these bacteria would die,
but few variants resistant to heat would survive and grow further. Variation is thus useful for the 
survival of species over time

Q 3. How will an organism be benefited I sit reproduces through spores?

Ans. The spores are covered by thick walls that protect them until they come in to contact with 
suitable moist surface and can begin to grow.

Q 4. Can you think of reasons why more complex organisms cannot give rise to new individuals 
through regeneration?

Ans. The reason is that complex organisms are not merely random collection of cells. Specialised 
cells are organised in them as tissues and tissues are organised as organs. These organs have to be 
placed at definite positions in the body. In such a carefully organised situation, cell-by-cell division 
would be impractical. Complex multicellular, therefore, need to use more complex ways of 
reproduction.

Q 5. Why is vegetative propagation practiced for growing some types of plants?

Ans. (i) Plants raised by vegetative propagation can bear flower and fruits earlier than those 
produced from seeds.
(ii) Such methods also make possible the propagation of plants such as banana, orange, rose and 
jasmine that have lost the capacity to produce seeds.
(iii) All plants produced by this method are genetically similar enough to the parent plant to have its 
all characteristics

Q6. As a Goddess, she may be Lakshmi, Kali, Saraswati and Durga but s human she is an unwanted girl. In our highly prejudiced society girls are considered as a financial burden resulting in female foeticides. As a student, what three initiatives would you take in the common concern of "Save girl
child, save the mother Earth".[VALUE BASED]

Ans:  
- By inspiring people to give birth to a girl child.
- By motivating them to understand the fact that girls are not a burden but a boon as Kalpana
Chawala, Kiran Bedi and Indira Gadhi and many others proved to be.
- By making them aware of declining sex ratio and its impacts.

Q.7. Ritu's parent received a proposal for Ritu's marriage from a boy living abroad. Before everything was finalized, Reena, Ritu's friend asked Ritu's parents to ask the boy to get report of his blood test.
(a) Do you think it was right on the part of Ritu's parents to do so ?
(b) What moral value did Reena showed ?   [VALUE BASED]

Ans: 
(a) Yes, it is necessary to go through the blood reports of to be groom as it indicates about his being physically fit and free of STD.
(b) By doing so, Reena not only helped her friend but also made her parents aware of possibility of the groom being infected with STDs as he lived in an open society.

Download full Solved paper   HOW DO ORGANISMS REPRODUCE ?

Friday, October 28, 2011

10th Our Environment HIGHER ORDER THINKIN SKILLS (...

10th Our Environment HIGHER ORDER THINKIN SKILLS
Chapter - 15 Our Environment HIGHER ORDER THINKIN SKILLS (HOTS) QUESTIONS
1) Write any two ways of energy flow through an ecosystem.
2) Differentiate between biodegradable and non biodegradable with respect to the effect of biological processes on them and the way they affect our environment.
3) Which level shows the maximum biological magnification? Why?
4) Why is a pond self sustaining unit while an aquarium may not be? Justify the answer.

5) Arrange grasshopper, frog, grass, eagle and snake in the form of food chain. 6) If 1000 KJ energy is available at producer level, how much energy will be available at first carnival level?

7) Why do most food chains have 3-5 steps only?

8) Select the biodegradable items from the list given below-
Polythene bags, old clothes, wilted flowers, pencil shavings, glass bangles, bronze statue, vegetable peels.
9) What will be impact on ecosystem if bacteria and fungi are removed from the Environment?
10) Express your feelings on the picture given down below. What will happen if all? Carnivores are eliminated from the environment? What measures will you take to save? Tiger?
More topics to read

X chemistry Carbon and its Compounds Sample guess ...

X chemistry Carbon and its Compounds Sample guess ...
Chapter – 4 Carbon and its Compounds - Points to remember and Important questions paper
1. Carbon forms a large variety of compouds because of its tetravalency and the property
of catenation.
2. Hydrocarbons are the compounds of carbon and hydrogen.
3. Hydrocarbons are of two types – saturated hydrocarbons (alkanes) and unsaturated
hydrocarbons (alkenes and alkynes)
4. Carbon forms covalent bonds with itself and other elements such as hydrogen, oxygen,
sulphur, nitrogen and chlorine.
5. The functional groups such as alcohols, aldehydes, ketones and carboxylic acids bestow
characteristic properties to the carbon compounds that contain them.
6. A group / series of hydrocarbons having similar structure and similar properties (i.e.
same functional group) is called a homologous series.
7. Carbon chains may be in the form of straight chains, branched chains or rings.
8. Carbon compounds with identical molecular formula but different structures are called
structural isomers.
9. Saturated hydrocarbons on combustion give carbondioxide and water with lots of heat.
10. Unsaturated hydrocarbons undergo addition reactions while saturated hydrocarbons
undergo substitution reactions.
11. Ethanol and Ethanoic acid (glacial acetic acid) are carbon compounds of importance in
our daily lives.
12. The molecules of soap are sodium or potassium salts of long chain carboxylic acids.
13. Detergents are ammonium or sulphonate salts of long chain carboxylic acids.
14. The action of soaps and detergents is based on the presence of both hydrophobic and
hydrophilic groups in the molecule and this helps to emulsify the oily dirt and hence its
removal.
QUESTIONS
1 mark questions:-
1. Name the type of bond formed when sharing of electrons occurs.
2. How many valence electrons are there in valence shell of carbon atom?
3. Define catenation.

Thursday, October 27, 2011

CBSE | NCERT - Explanation of The modern periodic Table Chemistry - E Notes


The modern version of the periodic law is stated as :“
The physical and chemical properties of the elements are the periodic functions of their atomic masses”.

LONG FORM OF THE PERIODIC TABLE

There are many forms of the periodic tale. The long form of the periodic table is the most convenient and the most widely used and is presented here. The horizontal rows are called PERIODS. 

Elements having similar chemical and physical properties appear in vertical columns and are known as GROUPS or FAMILIES. Altogether there are seven periods and 18 groups.

PERIODS
1ST Period Contain only 2 elements namely 1H, 2He and is the shortest period.

IInd Period Contains 8 elements namely 3Li, 4Be, 5B, 6C, 7N, 8O, 9F and 10Ne is known as the short period.

IIIrd Period Contains 8 elements : 11Na, 12Mg, 13Al, 14Si, 15S, 16S, 17Cl, and 18Ar. 

The IIIrd period is also known as short period.

IV Period starts with potassium and contains 18 elements :

19K, 20Ca, 21Sc, 22Ti, 23V, 24Cr, 25Mn, 26Fe, 27Co, 28Ni, 29Cu, 30Zn, 31Ga, 32Ge, 33As, 34Se, 35Br and 36Kr and is known as long period.


V Period Starts with rubidium and contains 18 elements : 37Rb, 38Sr, 39Y, 40Zr, 41Nb, 42Mo, 43Tc, 44Ru, 45Rh, 46PD, 47Ag, 48CD 49In, 50Sn, 51Sb, 52Te, 53I, 54Xe.The fifth


VI Period Consists of 32 elements, starting from cesium (55Sc) and ending with radon. It is called as longest period. This period includes the 4f shell that includes Lanthanides (58Ce, …..71Lu).



VII Period Like the sixth period would have a theoretical maximum of 32 elements. This period however is incomplete and at present contains 19 elements starting from 87Fr (francium) to 92U. All these elements are naturally occurring but rest are radioactive with very short half lives. These also include a part of inner transition elements, 90Th, …. 103Lr.


GROUPS

The atom of the element in a single vertical column have the same or very similar electronic configurations in the highest occupied orbitals and are therefore said to belong to the same GROUP or FAMILY of elements. According to the new IUPAC recommendations the groups are numbered form 1 to 18. Bases on the electronic configuration, we can classify elements in to four types.

1. Noble Gases                                                        2. Representative Elements                

3. Transition Elements and                                     4. Inner Transition Elements


1. Noble Gases
The noble gases are found at the end of each period in group 18. With the exception of helium, these elements have ns2 np6 electronic configuration in the outermost shell. Helium has 1s2 configuration. All the energy levels that are occupied by the electrons are completely filled and this stable arrangement of electrons cannot be easily altered. These elements have very low chemical reactivity.

2. Representative Elements (s and p block elements)
The elements of Group 1 (alkali metals), Group 2(alkaline earth metals) and Group 13,– 17constitute the Representative Elements. For the representative elements the period in which the element is located equal the principal quantum number of the differentiating electron, i.e., if an element is in nth period then the electronic configuration will be either ns1-2 or np1-5 .

Group 1 Consists of 1H(1s1), 3Li(2s1), 11Na(3s1), 19K(4s1), 37Rb(5s1), 55Cs(6s1), 87Fr(7s1).
The common outermost electronic configuration is ns-1.  Elements belonging to this group are known as Alkali Metals.

Group 2 Contains 4Be(2s2), 12Mg(3s2), 20Ca(4s2), 38Sr(5s2), 56Ba(6s2), 88Ra(7s2). The elements belonging to this group are known as Alkaline Earth Metals. The common outermost electronic configuration of the elements of this group is ns2.

Group 3 Starts with 5B(2s2 2p1). The general electronic configuration of the elements of this group is ns2 np1 . This group is also known as the Boron Family.

Group 4 Starts with 6C(2s2 2p2). The general electronic configuration is ns2 np2. This group is

also referred to as the carbon family.

Group 5 Starts with 7N(2S2 2P3).the general outermost electronic configuration of the elements

of this family ns2 np3. This group is also referred to as the Nitrogen Family.

Group 6 Starts with 8O(2S2 2p4) and is known as Oxygen Family. The general outermost electronic configuration of the elements of this family ns2 np4. The elements of this family are also known as Chalcogens.

Group 7 Contains 9F(2S2 2p5), 17Cl(3s2 3p5), 35Br(4s2 4p5), 53I(5s2 5p5) and 85At(6s2 6p5) The elements of this group are commonly known as Halogens.

Typical Elements
Elements of the third period are known as Typical elements, examples, 11Na, 12Mg, 13Al, 14Si, 15P, 16S and 17Cl Properties of all elements present in a particular group e.g. of group 1 resemble with the properties of 11Na and not with 3Li.

Bridge Elements
Elements of second period are known as Bridge Elements. Properties of the bridge elements resemble with the properties of the diagonal elements of the third period. For example, Li resembles Mg; Be resemble Mg; Be resembles Al; B resembles Si etc.

Note:- Noble gases are also grouped with representative P – block elements as they come at the end of each period.
The chemical and physical properties of the representative elements is determined by the number of electrons in the outer most shell called the Valence Shell. The number of valence electrons for groups 1 and 2 is the same as the group number for group 13 – 17, this number is obtained by subtracting 10 from the group number.

3. Transition Elements (d- block elements)
These are the elements of Groups 3 to 12 in the center of the period table. The elements in which the last electron enters the d sub-shell of the penultimate energy level are called d block elements. The outer most configuration of these elements is (n-1)d1 - 10 ns1 - 2 They are metals. They form colored ions and exhibit variable valency. However, Zn, Cd and Hg which too have (n – 1) d1 – 10 ns2 configuration in their outermost shell do not form colored ions and are not regarded as transition elements.

4. The Inner Transition elements ( f – Block Elements)
The rows of elements at the bottom of the periodic table are called the Lanthanide and  actinide series. These elements in which the last electron enters the f sub-shell of the antipenultimate  (third to the outermost shell) shell are called f block elements Their outer  electronic configuration is (n-2)f1 – 14 (n-1) d0-1 ns2. The differentiating electron is an felectron.
They are all metals. Within each series the properties of the elements are quite similar .

PERIODIC TRENDS IN PROPERTIES
1. VALENCE:-
An important chemical property of the elements exhibiting periodic trends is their Valence. It is defined as combining capacity of an element. It can also be defined it terms of valence electrons (electrons in the outermost shells). The valency is equal to number of valence electrons (or equal to 8 minus the number of valence electrons.

1. the valence of representative elements is usually equal to the number of electrons in the outermost orbitals and /or equal to eight minus the number of outermost electrons.

2. Transition elements do not exhibit any general trend. The reason for this that those elements have variable valencies due to availability of vacant d- subshells in them.

3. Inner transition elements also do not exhibit any general trend in the valency.

2. ATOMIC AND IONIC RADII
It is impossible to define the size of atoms as we know that atoms have no shop boundaries due to the delocalized picture of electron cloud. An estimate of he atomic size can be made by knowing the distance between the atoms in the combined state. 

There are three operational concepts of atomic radius.

a. If the bonding is covalent, the radius is called covalent radius.

b. If the bonding is ionic, the radius is called ionic radius.

c. If the two atoms are not bounded by a chemical bond (as in noble gases), the radius is called Vander Walls radius.

a. Covalent Radius : 

It is half of the distance between the nuclei of two like atoms bounded together by a single bond. For Example, the bond distance in hydrogen molecule (H2) is 74 pm and half of this distance is taken as the atomic radius of hydrogen. This radius is known as the covalent radius.

b. Ionic Radius:

It is the effective distance from the nucleus of an ion up to which it has its influence on its electron cloud.

c. van der Wall’s radius : 

It is one half of the distance between the nuclei of two adjacent atoms belonging to two neighboring molecules of an element in the solid state. The covalent radius is always smaller than the van der Wall’s radius because in the formation of chemical bond, the tow atoms have to come closer to each other. This is why the inert gases (where covalent radius is generally not possible) tend to have a larger size.

1. The size of atoms increases as we go down a column of periodic table. This increase is attributed to the increase in the number of shells around the nucleus.

2. The size of the atoms decrease as we go across the period from left to right except group 18 (Noble Gases). This decreases in the size is attributed to the increases in the nuclear charge and hence the attraction.

3. A positive ion is always smaller in size than the corresponding neutral atom.

4. A negative ion is always bigger in size than the corresponding neutral atom.

5. The size of ions increases as we go down a group provided that we are comparing ions of same charge.

6. Atoms or ions with the same electronic configurations are called as iso-electronic. If we consider a series of iso-electronic species (atoms or ions), the size decreases with the increasing atomic number. 

To illustrate the concept consider the radius of the following iso-electronic species, all having 10 electrons   N3- > O2- > F- > Ne > Na+ > Mg2+ > Al3+

Note that the successive increase in the values of Z/e ratio decreases the values of ionic (atomic ) radii.

3. IONIZATION ENERGY
The chemical nature of an element depends on the ability of its atoms to accept of donate

electrons. A quantitative measure of these tendencies is the Ionization Energy or the Electron

or the Electron Affinity.

The ionization Energy (IE) is defined as the energy required to remove an electron from an isolated gaseous atom (M) in its ground state.
M(g) + IE --------> M+(g) + e-

The ionization energy is expressed in units of kJ mol-1 or in ev / electron.

The energy required to remove the second electron from the same element is known as the second ionization energy. 

The second ionization energy is higher than that required for the removal of the firs electron because it is more difficult to remove an electron from a positively charged species than the second and so on. If the term ionization energy is not qualified (i.e., if the first, second and third is not motioned), it is taken as the first ionization energy.

M+(g) + second IE--------------> M+++(g) + e-

M++(g) + third IE--------------> M+++(g) + e-


FACTORS INFLUENCING IONIZATION ENERGY :
The ionization energy depends upon the following factors :

Nuclear Charge
Ionization energy increases with increase in the nuclear charge. With increase in the nuclear charge the force with which the electron is bound with the atom increases and hence it becomes more difficult to remove the outermost electron. For example the ionization energy of He is 567 kcal/mol, whereas that of H is only 314 kcal/mole. This increase in the ionization energy can be explained on the basis of the increase in the nuclear charge.

Atomic Size
With the increase in the atomic size, the force with which the electron is bound to the atom decreases and hence with increase in the size, the ionization energy decreases. For example the ionization energy decreases as we don down the group.

Screening Effect
The force of attraction between the valence electrons and the nucleus is greatly shielded by the presence of core electrons. With the increase in the electrons in the inner sub shells the force of attraction between the outermost electron and the nucleus decreases and hence the ionization energy decrease.

Penetration of Electrons
The ionization energy also depends upon the penetrating power of the electrons. For example, the penetration of s-electrons. The f-electrons has the least penetration. The ionization energy increases with increasing penetration.

Stability of the Electronic Configuration
The half and fully filled orbitals are most stable as compared to their neighbours and hence the ionization energy of the fully filled or half filled orbitals is higher as compared to their neighbors. For instance, he ionization energy for N is higher that for C and O.

PERIODIC TRENDS
1. It is observed that the ionization energy of an element strongly depends on its electronic configuration and thus show periodic variations. The maxima are found at the noble gases which have completely filled electron shells. 

The high ionization energies of the noble gases can be connected with their extremely low chemical reactivity. Similarly, the high reactivity of alkali metals is reflected I their.


2. In a group : First ionization energy decreases as we go down a group in the table. It measures the ease of removing an electron from the outer shell. As we go down a group, this shell is farther away from the nucleus. As a result nuclear attraction decreases. Though the positive charge of nucleus increases, its effect is weakened due to the shielding supplied by the inner shells to the outermost shell.


3. In a period: 

As we go across a period from left to right, the atomic size decreases. As the number of shells in a particular period remain the same and the additional electrons are being continuously introduced in the same shell, the nuclear charge increases and hence the outer electrons are greatly attracted to the nucleus. Hence it becomes difficult to remove them and consequently ionization energy increases. The figure below shows the first ionization energy of elements of the second period as a function of the atomic number Z.

a. IE (B) < IE (Be). Be has its 2s orbital fully filled whereas B has one unpaired 2p electron and it is easier to remove a lone electron rather than that from a paired orbit. Hence extra-stability of fully filled sub- shell is the cause of this irregularity.

b. IE (O) < IE (N). nitrogen has an exactly half filled outermost electronic configuration and hence is extra stable. Thus extra-stability of half filled sub-shells is the cause of irregularity.

4. Whether all the outer shell electrons are removed, the next I.E. is much greater than the previous value of I.E. For the same element. Note that first I.E for the same is 72.64 eV.


4. ELECTRON AFFINITY
The electron affinity is the amount of energy releases when an isolated gaseous atom accepts an electron to form a monovalent gaseous ion.

X (g) + e- X- (g) + Energy

Electron affinities can be positive or negative. When energy is released in the process of attachment of an electron to an atom, the electron affinity is taken as positive and if energy is absorbed, electron affinity is taken to be negative like in inert gases. Thus the magnitude of electron affinity measures the tightness with which the atom can hold the additional electron. The larger value of E.A reflects the greater tendency of an atom to accept the electron.

Electron affinity values are influenced by

a. size of the atom b. Nuclear charge c. Electronic configuration

Fore more study points : Click On Links

9th chemistry Self-evaluation on Mole concept

chemistry adda: 9th chemistry Self-evaluation on Mole concept
Sample Problem 1. Calculate the number of mole in 52 g of Helium.

Solution. We know that, Atomic mass of He = 4u
So, its molar mass = 4 g  That is, 4 g of He contains 1 mole of He.
Or, 4 g of He = 1 mole of He
Or, 1 g of He = 1/4 mole of He
So, 52 grams of Helium =1/4 x 52 mole = 13 moles
Therefore, there are 13 moles in 52 g of He
Sample Problem 2. Calculate the number of moles for 12.044 X 1023 atoms of Helium.
Solution: 
6.022 X 1023 atoms of Helium           = 1 mole
1 atom of Helium                                 =1/6.022 X 1023 mole
12.044 X 1023 atoms of Helium         = [(1/6.022 X 1023) x  12.044 X 1023  ]moles
[ You may use  ----        Number of moles    = (given numbers of particles /Avogadro number)]
Sample Problem 3. How many moles are there in 5 grams of calcium?
Sample Problem 4. How many moles there in 12.044 X 1023 atoms of phosphorous?
Sample Problem 5. Calculate the mass of 0.5 mole of Nitrogen atoms.
Solution. We know that, Molar mass of of Nitrogen atoms 14 g
The mass of 1 mole of Nitrogen atoms = 14 g
The mass of 0.5 mole of Nitrogen atoms = 14x 0.5 g = 7 g
Alternate method,
                        Mass = Number of moles X Molar mass
                  Or,           = 0.5 X 14 = 7 g
Therefore, mass of 0.5 mole of N atoms is 7 grams.
Sample Problem 6. What is the mass of 3.011 X 1023 number of Nitrogen atoms?
Solution. We know that, Molar mass of of Nitrogen atoms 14 g
The mass of   6.022 X 1023 number of Nitrogen atoms= 14 g
The mass of  3.011 X 1023 number of Nitrogen atoms = [(14g/6.022 X 1023 )x 3.011 X 1023]g = 7 g

Sample Problem 7. Calculate the number of particles in each of the following:
(i) 0.1 mole of Carbon atoms
(ii) 46 grams of Sodium atoms.
Sample Problem 8 What is the mass of 4 moles of Aluminum atoms?
Determine the number of iron atoms in a piece of iron weighing 2.8 grams.
Sample Problem 9 If one mole of Carbon atoms weigh 12 grams, what is mass in grams of a single atom of Carbon?
Sample Problem 10. Calculate the mass of 0.5 mole of N2 gas.
Sample Problem 11. Convert 12.044 X 1022 molecules of Sulphur dioxide into moles?
Sample Problem 12. In which case the number of Hydrogen atoms is more- 2 mol of HCl
or 1 mol of NH3?
Sample Problem 13 . An ornament of silver contains 20 gram silver. Calculate the moles of silver present.
Sample Problem 14. If 1 g sulphur dioxide contains x molecules, what will be the number
of molecules in 1 g of methane?
Sample Problem 15. Calculate the number of aluminium ions present in 0.051 g of aluminium oxide.
Sample Problem 16. How many grams of neon will have the same number of atoms as 4 g of calcium?
Sample Problem 17. Calculate the number of iron atoms in a piece of iron weighing 2.8 grams.
Sample Problem 18. What is the mass of 3 moles of Zinc?
Sample Problem 19 : What is the ratio of molecules present in 6.6 grams of CO2 and 3.2 grams of SO2?
Sample Problem 20. If one gram of Sulphur contains x atoms, what will be the number of atoms in one gram of oxygen? (atomic mass of S = 32u)  

Tuesday, October 25, 2011

Numerical Problems based on mole concept 9th Chemistry

CBSE CHEMISTRY CLASS 9TH CHAPTER - 3 ATOMSAND MOLECULES


Download solved paper based on mole concept
9th Mole concept numerical problems solved-CBSE SET- 1
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9th Mole concept numerical problems solved-CBSE SET- 2
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practice your learning 
1. Calculate the number of moles of magnesium present in a magnesium ribbon weighing 12 g. Molar atomic mass of magnesium is 24g/ mol
2. Verify by calculating that
(a) 5 moles of CO2 and 5 moles of H2O do not have the same mass.
(b) 240 g of calcium and 240 g magnesium elements have a mole ratio of 3:5.
3. Find the ratio by mass of the combining elements in the following compounds.
 (a) CaCO3 (b) MgCl2 (c) H2SO4 (d) C2H5OH  (e) NH3  (f) Ca(OH)2
4. Calcium chloride when dissolved in water dissociates into its ions according to the following equation.
CaCl2 (aq) → Ca2+ (aq) + 2Cl– (aq)
Calculate the number of ions obtained from CaCl2 when 222 g of it is dissolved in water.
5. The difference in the mass of 100 moles each of sodium atoms and sodium ions is 5.48002 g. Compute the mass of an electron.
6. Cinnabar (HgS) is a prominent ore of mercury. How many grams of mercury are present in 225 g of pure HgS? Molar mass of Hg and S are 200.6 g /mol and 32 g/ mol respectively.
7. The mass of one steel screw is 4.11g. Find the mass of one mole of these steel screws. Compare this value with the mass of the Earth (5.98 × 1024kg). Which one of the two is heavier and by how many times?
8. A sample of vitamin C is known to contain 2.58 ×1024 oxygen atoms. How many moles of oxygen atoms are present in the sample?
9. Compute the difference in masses of 103 moles each of magnesium atoms and magnesium ions. (Mass of an electron = 9.1×10–31 kg)
10. Compute the number of ions present in 5.85 g of sodium chloride.
11. A gold sample contains 90% of gold and the rest copper. How many atoms of gold are present in one gram of this sample of gold?
12. Compute the difference in masses of one mole each of aluminium atoms and one mole of its ions. (Mass of an electron is 9.1×10–28 g). Which one is heavier?
13. A silver ornament of mass ‘m’ gram is polished with gold equivalent to 1% of the mass of silver. Compute the ratio of the number of atoms of gold and silver in the ornament.
14. A sample of ethane (C2H6) gas has the same mass as 1.5 ×1020 molecules of methane (CH4). How many C2H6 molecules does the sample of gas contain?
15. Fill in the blanks
(a) In a chemical reaction, the sum of the masses of the reactants and products remains unchanged. This is called ————.
(b) A group of atoms carrying a fixed charge on them is called ————.
(c) The formula unit mass of Ca3 (PO4)2 is ————.
(d) Formula of sodium carbonate is ———— and that of ammonium sulphate is ————.
Problems based on mole concept

9th Atoms and Molecules - Mole Concept and Problems based on mole concept o

Sunday, October 23, 2011

Work, Power, Energy MCQ for physics class 9

WORK AND ENERGY   9th physics Term-II
1. When a body falls freely towards the earth, then its total energy
(a) increases
(b) decreases
(c) remains constant
(d) first increases and then decreases
2. A car is accelerated on a levelled road and attains a velocity 4 times of its initial velocity. In this process the potential energy of the car
(a) does not change
(b) becomes twice to that of initial
(c) becomes 4 times that of initial
(d) becomes 16 times that of initial
3. In case of negative work the angle between the force and displacement is
(a) 00 (b) 450 (c) 900 (d ) 1800
4. An iron sphere of mass 10 kg has the same diameter as an aluminium sphere of mass is 3.5 kg. Both spheres are dropped simultaneously from a tower. When they are 10 m above the ground, they have the same
(a) acceleration
(b) momenta
(c) potential energy
(d) kinetic energy
5. A girl is carrying a school bag of 3 kg mass on her back and moves 200 m on a levelled road. The work done against the gravitational force will be (g =10 m s–2)
(a) 6 ×103 J
(b) 6 J
(c) 0.6 J
(d) zero
6. Which one of the following is not the unit of energy?
(a) joule
(b) newton metre
(c) kilowatt
(d) kilowatt hour
7. The work done on an object does not depend upon the
(a) displacement
(b) force applied
(c) angle between force and displacement
(d) initial velocity of the object
8. Water stored in a dam possesses
(a) no energy
(b) electrical energy
(c) kinetic energy
(d) potential energy
9. A body is falling from a height h. After it has fallen a height h/2 , it will possess
(a) only potential energy
(b) only kinetic energy
(c) half potential and half kinetic energy
(d) more kinetic and less potential energy
 10. How are Joule (J) and ergs (erg) related?
(a) 1J = 107 erg (b) 1erg = 10 -7J (c) 1J = 10-7 erg (d) None
Solution
1. (c) 2. (a) 3. (d) 4. (a) 5. (d) 6. (c) 7. (d) 8. (d) 9. (c) 10.(b)
10. A rocket is moving up with a velocity v. If the velocity of this rocket is suddenly tripled, what will be the ratio of two kinetic energies?
Ans: Initial velocity = u, then v = 3 u 
Initial kinetic energy = 1/2 m u2 
Final kinetic energy (K.E.) =1/2 m v2 =1/2m (3u)2 =9 x (1/2 m u2 ) 
(K.E) initial : (K.E) final=1:9
11. Avinash can run with a speed of 8 m s–1 against the frictional force of 10 N, and Kapil can move with a speed of 3 m s–1 against the frictional force of 25 N. Who is more powerful and why?
Ans:            Power of Avinash      PA = FA x vA  =  10 × 8 = 80 W                
 The power of Kapil            Pk = Fk x vk    =  25 × 3 = 75 W
So, Avinash is more powerful than Kapil.
12. A boy is moving on a straight road against a frictional force of 5 N. After travelling a distance of 1.5 km he forgot the correct path at a round about of radius 100 m. However, he moves on the circular path for one and half cycle and then he moves forward upto 2.0 km. Calculate the work done by him.

Ans: F = 5 N 
 W = F.S = 5 × [1500 + 200 + 2000] = 18500 J.
13. Can any object have mechanical energy even if its momentum is zero? Explain.


Ans: Yes, mechanical energy comprises both potential energy and kinetic energy. Momentum is zero which means velocity is zero. Hence, there is no kinetic energy but the object may possess potential energy.
14. Can any object have momentum even if its mechanical energy is zero? Explain.
Ans: No. Since mechanical energy is zero, there is no potential energy and no kinetic energy. Kinetic energy being zero, velocity is zero. Hence, there will be no momentum.
15. The power of a motor pump is 2 kW. How much water per minute the pump can raise to a height of 10 m? (Given g = 10 m s–2)
p=w/Dt=mgh/Dt = [(m x10 x10)/60]       Þ m = 1200kg

Thursday, October 20, 2011

Power and LAW OF CONSERVATION OF ENERGY(9th physics)

CBSE PHYSICS: LAW OF CONSERVATION OF ENERGY(9th physics):
LAW OF CONSERVATION OF ENERGY


Energy can neither be created nor destroyed, but it
is transformed from one form to another. Alternatively,
whenever energy gets transformed, the total energy
remains unchanged.
Proof – Freely falling body
It may be shown that in the absence of external frictional force the total mechanical energy of a body remains constant.
Let a body of mass m falls from a point A, which is at a height h from the ground as shown in fig.
At A,
Kinetic energy kE = 0
Potential energy Ep = mgh
Total energy E = Ep + Ek = mgh + 0= mgh
During the fall, the body is at a position B. The body has moved a distance x from A.
At B,
velocity v2 = u2 + 2as
applying, v2 = 0 + 2ax = 2ax
Kinetic energy Ek = 1/2 mv2 = 1/2 m x 2gx = mgx
Potential energy Ep = mg (h – x)
Total energy E = Ep + Ek = mg (h-x) + mgx = mgh – mgx + mgx= mgh
If the body reaches the position C.
At C,
Potential energy Ep = 0
Velocity of the body C is
v2 = u2 + 2as
u = 0, a = g, s = h
applying v2 = 0 + 2gh = 2gh
kinetic energy Ek =1/2 mv2=1/2 m x 2gh= mgh
Total energy at C
E = Ep + Ek
E = 0 + mgh
E = mgh
Thus we have seen that sum of potential and kinetic energy of freely falling body at all points remains same. Under the force of gravity, the mechanical energy of a body remains constant.
RATE OF DOING WORK (OR) POWER
Power is defined as the rate of doing work or work done per unit time.
Power =work done/time taken
P = w/t
UNIT OF POWER
The unit of power is J/S known as watt, its symbol is W.
1 watt =1 joule/1 second
1 W = 1 J/ S
Commercial unit of energy is kilo watt hour
We pay electricity bill in terms of unit or kWh. It is a commercial unit of electric energy consumed by the user.
Watt hour = power in watt x time in hour.
Example : How much energy will be used when a hundred watt bulb is used for 10 hour?
Energy = 100 watt x 10 hour = 1000 w h = 1kw h
I kwh is known as 1 unit.
One kilowatt hour means thousand watt of power is consumed in one hour.
1 kWh = 1 kW x 1 h
= 1000 W x 60 x 60 s
= 1000 Js-1 x 3600 s
= 3.6 x 106 J
1 unit = 1 kilowatt hour = 3.6x106 J