class9asciencekvnoida: 2016

Thursday 24 November 2016

shobha rollno. 16

atoms and molcules

exercise

1. A 0.24 g sample of compound of oxygen and boron was found by analysis to contain 0.096 g of boron and 0.144 g of oxygen. Calculate the percentage composition of the compound by weight.

Answer

Total mass of Compound = 0.24 g (Given)
Mass of boron = 0.096 g (Given)
Mass of oxygen = 0.144 g (Given)

Thus, percentage of boron by weight in the compound = 0.096 / 0.24 × 100%

= 40%

And, percentage of oxygen by weight in the compound = 0.144 / 0.24 × 100% = 60%

2. When 3.0 g of carbon is burnt in 8.00 g oxygen, 11.00 g of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.00 g of carbon is burnt in 50.00 g of oxygen? Which law of chemical combinations will govern your answer?

Answer

3.0 g of carbon combines with 8.0 g of oxygen to give 11.0 of carbon dioxide.

If 3 g of carbon is burnt in 50 g of oxygen, then 3 g of carbon will react with 8 g of oxygen. The remaining 42 g of oxygen will be left un-reactive.
In this case also, only 11 g of carbon dioxide will be formed.
The above answer is governed by the law of constant proportions.

Page No: 44

3. What are polyatomic ions? Give examples?

Answer

A polyatomic ion is a group of atoms carrying a charge (positive or negative). For example, Nitrate (NO₃^-), hydroxide ion (OH - ).

4. Write the chemical formulae of the following:

(a) Magnesium chloride

► MgCl₂

(b) Calcium oxide

► CaO

(c) Copper nitrate

► Cu (NO₃)₂

(d) Aluminium chloride

► AlCl₃

(e) Calcium carbonate

► CaCO₃

5. Give the names of the elements present in the following compounds:

(a) Quick lime
► Calcium and oxygen

(b) Hydrogen bromide
► Hydrogen and bromine

(c) Baking powder
► Sodium, hydrogen, carbon, and oxygen

(d) Potassium sulphate
► Potassium, sulphur, and oxygen

6. Calculate the molar mass of the following substances:

(a) Ethyne, C₂H₂
► Molar mass of ethyne, C₂H₂ = 2 × 12 + 2 × 1 = 26 g

(b) Sulphur molecule, S₈
►Molar mass of sulphur molecule, S₈ = 8 × 32 = 256 g

(c) Phosphorus molecule, P₄ (atomic mass of phosphorus = 31)
► Molar mass of phosphorus molecule, P₄ = 4 × 31 = 124 g

(d) Hydrochloric acid, HCl
► Molar mass of hydrochloric acid, HCl = 1 + 35.5 = 36.5 g

(e) Nitric acid, HNO₃
► Molar mass of nitric acid, HNO₃ = 1 + 14 + 3 × 16 = 63 g

7. What is the mass of-

(a) 1 mole of nitrogen atoms?

(b) 4 moles of aluminium atoms (Atomic mass of aluminium = 27)?

Answer

(a) The mass of 1 mole of nitrogen atoms is 14 g.

(b) The mass of 4 moles of aluminium atoms is (4 × 27) g = 108 g

(c) The mass of 10 moles of sodium sulphite (Na₂SO₃) is
10 × [2 × 23 + 32 + 3 × 16] g = 10 × 126 g = 1260 g

8. Convert into mole.

(a) 12 g of oxygen gas
(b) 20 g of water
(c) 22 g of carbon dioxide

Answer

(a) 32 g of oxygen gas = 1 mole
Then, 12 g of oxygen gas = 12 / 32 mole = 0.375 mole

(b) 18 g of water = 1 mole
Then, 20 g of water = 20 / 18 mole = 1.111 mole

(c) 44 g of carbon dioxide = 1 mole
Then, 22 g of carbon dioxide = 22 / 44 mole = 0.5 mole

9. What is the mass of:

(a) 0.2 mole of oxygen atoms?
(b) 0.5 mole of water molecules?

Answer

(a) Mass of one mole of oxygen atoms = 16 g
Then, mass of 0.2 mole of oxygen atoms = 0.2 × 16g = 3.2 g

(b) Mass of one mole of water molecule = 18 g

Then, mass of 0.5 mole of water molecules = 0.5 × 18 g = 9 g

10. Calculate the number of molecules of sulphur (S8) present in 16 g of solid sulphur.

Answer

1 mole of solid sulphur (S8) = 8 × 32 g = 256 g
i.e., 256 g of solid sulphur contains = 6.022 × 10²³molecules
Then, 16 g of solid sulphur contains = 6.022 × 10²³/ 256 = 16 molecules
= 3.76375 × 10²² molecules

11. Calculate the number of aluminium ions present in 0.051 g of aluminium oxide.
(Hint: The mass of an ion is the same as that of an atom of the same element. Atomic mass of Al = 27 u)

Answer

mole of aluminium oxide (Al₂O₃) = 2 × 27 + 3 × 16
= 102 g
i.e., 102 g of Al₂O₃= 6.022 × 10²³molecules of Al₂O₃
Then, 0.051 g of Al₂O₃contains = 6.022 × 10²³/ 102 × 0.051 molecules
= 3.011 × 10²⁰ molecules of Al₂O₃

The number of aluminium ions (Al3+) present in one molecule of aluminium oxide is 2.

Therefore, the number of aluminium ions (Al³⁺) present in 3.011 × 10²⁰molecules (0.051 g ) of aluminium oxide (Al₂O₃) = 2 × 3.011 × 10²⁰
= 6.022 × 10²⁰

shobha rollno. 16

atoms and molecules

exercise

Thus, percentage of boron by weight in the compound = 0.096 / 0.24 × 100%

= 40%

Answer

3.0 g of carbon combines with 8.0 g of oxygen to give 11.0 of carbon dioxide.

Page No: 44

► MgCl₂

(b) Calcium oxide

► CaO

(c) Copper nitrate

► Cu (NO₃)₂

(d) Aluminium chloride

► AlCl₃

(e) Calcium carbonate

(b) 4 moles of aluminium atoms (Atomic mass of aluminium = 27)?

Answer

(a) The mass of 1 mole of nitrogen atoms is 14 g.

Then, mass of 0.5 mole of water molecules = 0.5 × 18 g = 9 g

10. Calculate the number of molecules of sulphur (S8) present in 16 g of solid sulphur.

Answer

shobha rollno.16

DIVERSITY IN LIVING ORGANISMS

EXERCISE

1. what are the advantages of classifiying organisms?
Answer

Following are the advantages of classifying organisms:
→ It makes us aware of and gives us information regarding the diversity of plants and animals.
→ It makes the study of different kinds of organisms much easier.
→ It tells us about the inter-relationship among the various organisms.
→ It helps us understanding the evolution of organisms.
→ It helps in the development of other life sciences easy.

→ It helps environmentalists to develop new methods of conservation of plants and animals.

2. How would you choose between two characteristics to be used for developing a hierarchy in classification?

Answer

We choose that characteristics which depends on the first characteristics and determines the rest variety.

3. Explain the basis for grouping organisms into five kingdoms.

Answer

The basis for grouping organisms into five kingdoms are:
→ Complexity of cell structure - There are two broad categories of cell structure: Prokaryotic and Eukaryotic. Thus, two broad groups can be formed, one having prokaryotic cell structure and the other having eukaryotic cell structure. Presence or absence of cell wall is another important characteristic.
→ Unicellular and multicellular organisms - This characteristic makes a very basic distinction in the body designs of organisms and helps in their broad categorizations.

→ Cell Wall: Presence and absence of cell wall leads into grouping.

→ Mode of nutrition -Organisms basically have two types of nutritions - autotrophic who can manufacture their own food and heterotrophic who obtain their food from external environment, i.e., from other organisms). Thus, organisms can be broadly classified into different groups on the basis of their mode of nutrition.

4. What are the major divisions in the Plantae? What is the basis for these divisions?

Answer

The major divisions in Kingdom Plantae are:
→Thallophyta
→ Bryophyta
→ Pteridophyta
→ Gymnosperms
→ Angiosperms

The following points constitute the basis of these divisions:
→ Whether the plant body has well differentiated, distinct components.
→ whether the differentiated plant body has special tissues for the transport of water and other substances.
→ The ability to bear seeds.
→ Whether the seeds are enclosed within fruits.

5. How are the criteria for deciding divisions in plants different from the criteria for deciding the subgroups among animals?

Answer

The characteristics used to classify plants is different from animals because the basic design are different, based on the need to make their own food (plants) or acquire food (animals).
Criteria for deciding divisions in plants are:
→ Differentiated/ Undifferentiated plant body
→ Presence/ absence of vascular tissues
→With/without seeds
→ Naked seeds/ seeds inside fruits

But the animals can't be divided into groups on these criteria. It is because the basic designs of animals are very different from plants. They are divided on the basis of their body structure.

6. Explain how animals in Vertebrata are classified into further subgroups.

Answer

Animals in Vertebrata are classified into five classes:

(i) Class Pisces: This class includes fish such as Scoliodon, tuna, rohu, shark, etc. These animals mostly live in water. Hence, they have special adaptive features such as a streamlined body, presence of a tail for movement, gills, etc. to live in water.

(ii) Class Amphibia: It includes frogs, toads, and salamanders. These animals have a dual mode of life. In the larval stage, the respiratory organs are gills, but in the adult stage, respiration occurs through the lungs or skin. They lay eggs in water.

(iii) Class Reptilia: It includes reptiles such as lizards, snakes, turtles, etc. They usually creep or crawl on land. The body of a reptile is covered with dry and cornified skin to prevent water loss. They lay eggs on land.

(iv) Class Aves: It includes all birds such as sparrow, pigeon, crow, etc. Most of them have feathers. Their forelimbs are modified into wings for flight, while hind limbs are modified for walking and clasping. They lay eggs.

(v) Class Mammalia: It includes a variety of animals which have milk producing glands to nourish their young ones. Some lay eggs and some give birth to young ones. Their skin has hair as well as sweat glands to regulate their body temperature.

Thursday 8 September 2016

DATE-8/09/16 ABHISHEK SINGH ROLL.NO.25

QUESTION AND ANSWER

Q 1) AN OBJECT EXPERIENCES A NET ZERO EXTERNAL UNBALANCED FORCE. IS IT POSSIBLE FOR OBJECT TO BE TRAVELLING WITH A NON ZERO VELOCITY? IF YES, STATE THE CONDITIONS THAT MUST BE PLACED ON THE MAGNITUDE AND DIRECTION OF THE VELOCITY. IF NO, PROVIDE A REASON.

ANS 1) YES AN OBJECT MAY TRAVEL WITH A NON ZERO VELOCITY EVEN WHEN THE NET EXTERNAL FORCE ON IT IS ZERO. A RAIN DROP FALLS DOWN WITH A CONSTANT VELOCITY. THE WEIGHT OF THE DROP IS BALANCED BY THE UPTHRUST AND THE VISCOSITY OF AIR. THE NET FORCE DROP FROM THE ZERO.

Q 2) WHEN WE BEAT THE CARPET WITH A STICK, IT DUST COMES INTO THE OF IT. EXPLAIN, WHY?
WHEN WE BEAT A CARPET WITH STICK , IT COMES TO MOTION. B UT THE DUST PARTICLES CONTINUE TO AT REST DUE TO THE INERTIA AND GET DETACHED FROM THE CARPET.

Q 3) WHY IS IT ADVISED TO TIE ANY LUGGAGE KEPT ON THE ROOF OF A BUS WITH A ROPE?
DUE TO SUDDEN JERKS OR DUE TO THE BUS TAKING SHARP TURN TO THE ROAD, THE LUGGAGE MAY FALL DOWN FROM THE ROOF BECAUSE OF ITS TENDENCY TO CONTINUE MOVING IN THE ORIGINAL DIRECTION. TO AVOID THIS., THE LUGGAGE IS TIED WITH THE A ROAP ON THE ROOF.

Q 4) A BATSMAN HITS A CRICKET BALL WHICH THEN ROLLS ON A LEVEL GROUND. AFTER COVERING A SHORT DISTANCE , THE BALL COMES TO REST. THE BNALL HARD ENOUGH
ANS) THERE IS THE FORCE ON THE BALL OPPOSING THE MOTION.
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SOUBHAGYA , ROLL NO 18 CHAPTER= MOTION

NAME-SOUBHAGYA SINGH ROLL=18

CHAPTER MOTION

QUESTION AND ANSWER

Wednesday 7 September 2016

SHUBHAM KUMAR ROLL NO 56

TISSUES

QUESTION AND ANSWER

Q1) DEFINE THE TERM TISSUES?

A TISSUES IS A GROUP OR COLLECTION OF SIMILAR OR DISIMILAR CELLS WHICH WORK TOGETHER TO ACHIEVE A PARTICULAR FUNCTION CELLS OF A TISSUES HAVE COMMON ORIGIN.

Q2) HOW MANY TYPES OF ELEMENTS TOGETHER MAKE UP THE XYLEM TISSUES? NAME THEM.

XYLEM IS A COMPLEX TISSUE. IT MADE UP THE FOLLOWING FOUR KINDS OF CELL

1. TRACHEIDS

2. VESSELS

3. XYLEM PARENCHYMA

4.XYLEM FIBRES.

Q3)HOW ARE SIMPLE TISSUE IS DIFFERENT FROM COMPLEX TISSUE IN PLANT?

A SIMPLE TISSUE IS MADE UP OF ONE TYPE OF CELL AND COMPLEX TISSUE IS MADE UP OF MORE THAN ONE TYPE OF CELL.

Q4) DIFFERENTIATE BETWEEN PARENCHYMA, COLLENCHYMA AND SCLERENCHYMA ON THE BASIS OF THIER CELL WALL?

THE DIFFERENCE BETWEEN PARENCHYMA , COLLENCHYMA AND SCERENCHYMA ON THE BASIS ON IT THE CELL WALL.

PARENCHYAM CELL

1. PARENCHYMA CELL. WALL IS PRIMARY

2. PARENCHYAM ARE THIN AND MADE UP OF CELLULOS.

-f 1.CELL WALL IS PRIMARY.

2. CELL WALL WAS LOCALISED. THICKENING OF CELLULOSE.

SCERENCHYMA CELL

1.CELL WALL IS SECONDARY.

2.CELL WALL IS VERY THICK.

Q5)WHAT ARE THE FUNCTION OF STOMATA?

THE IMPORTANT FUNCTION OF STOMATA ARE:

1. EXCHANGE OF GASES WITH ATMOSPHERE, E.G., ENTRY OF CARBON DIOXIDE FOR PHOTOSYNTHESIS AND OF OXYGEN FOR RESPIRATION.

2. TRANSPIRATION,I.E., LOSS OF WATER IN THE FORM OF WATER VAPOUR

Q7) WHAT ARE THE SPECIFIC FUNCTION OF THE CARDIAC MUSCLE?

CARDIAC MUSCLE IS PRESENT IN THE HEART. IT CONTRACTS AND RELAXES RAPIDLY AND CONTINUOUSLY WITH A RHYTHM, BUT IT NEVER GETS FATIGUED.

Q8) DIFFERENTIATE BETWEEN STRIATED, UNSTRIATED AND MUSCLE ON THE BASIS OF THEIR STRUCTURE AND SITE/ LOCATION IN THE BODY?
THE DIFFERENCE BETWEEN THESE ARE:
STRIATED MUSCLE

1. THE CELLS OF STRIATED MUSCLES ARE LONG, CYLINDRICAL UNBRANCHED AND MULTINUCLEATED.
2. STRIATED MUSCLES ARE PRESENT IN OUR LIMBS AND JOIN THE BONES.
UNSTRIATED MUSCLE
1. THE CELLS OF UNSTRIATED MUSCLE ARE LONG, POINTED AT THE ENDS AND UNNUCLEATED.
2.THESE MUSCLES ARE PRESENT IN ALIMENTARY CANAL, BLOOD VESSELS, IRIS OF THE EYE, URETER AND BRONCHI.
CARDIAC MUSCLE
1. THE CELLS OF CARDIAC MUSCLEWS ARE CYLINDRICAL BRANCHED AND UNINUCLEATED.
2. CARDIAC MUSCLE PRESENT IN THE HEART.

Q10) NAME THE FOLLOWING:
A)TISSUE THAT FORMS INNER LINING OF OUR MONTH.
B)TISSUE THAT CONNECTS MUSCLE TO BONE IN HUMANS.
C)TISSUE THAT TRANSPORTS FOOD IN THE PLANTS.
D)TISSUE THAT STORES FAT IN OUR BODY.
E)CONNECTIVE TISSUE WITH A FLUID MARTIX.
F)TISSUE PRESENT IN THE BRAIN.
A)SQUAMOUS EPITHELIUM
B)TENDON
C)PHLOEM
D)ADIPOSE TISSUE
E)VASCULAR TISSUE
F)NERVOUS TISSUE