## Friday, 5 January 2018

### What produces more severe burns, boiling water or steam?

Answer : Steam produces more severe burns as compare to boiling water as it has extra high heat energy.

### Why is ice at 273 K more effective in cooling than water at the same temperature?

Answer : The heat energy  absorbed by ice to change to liquid state, without showing any rise in temperature is known as the latent heat. We already know, the  heat can change the physical state of a matter by overcoming the forces of attraction between the particles. For water freezing point and melting point is 0°C. So water has extra amount of hidden energy as compare to ice.

### Give two reasons to justify Question 6 from Chapter 1. MATTER IN OUR SURROUNDINGS

(a) water at room temperature is a liquid.
(b) an iron almirah is a solid at room temperature.

Answer :  (a) water at room temperature is a liquid because its melting temperature is with in range of room temperature. As we know force of molecular attraction of a substance becomes high at low temperature and weak  at high temperature. Due to this  water at 0°C becomes solid due to increased molecular particles attraction.At room temperature arround 20 °C, due to increase in tempreture,  kinetic energy of the water particles increases. Due to the increase in kinetic energy, the particles start vibrating with greater speed. The energy supplied by heat overcomes the forces of attraction between the water particles. The
particles leave their fixed positions and start moving more freely. A stage is reached when the solid ice melts and is converted to water a liquid.

(b) an iron almirah is a solid at room temperature.This is because the melting point of iron is  1536 °C or 2797 °F or 1809 K where as room temperature is around 20°C. At this temperature, kinetic energy of the iron molecules is not enough to  overcomes the forces of attraction between the water particles. There for iron almirah is solid at room temperature.

### What is the physical state of water at. (a) 25°C (b) 0°C (c) 100°C ?

Answer : Physical states of water at given temperatures :
(a) 25°C - Liquid as familiar water
(b) 0°C  - Solid  as ice
(c) 100°C- Gaseous as water vaporour

### Arrange the following substances in increasing order of forces of attraction between the particles. water, sugar, oxygen.

Answer : Substances in increasing order of forces of attraction between the particles.
• Oxygen
• Water
• Sugar

### Give reason for the following observations Question3 from Chapter 1. MATTER IN OUR SURROUNDINGS

(a) Naphthalene balls disappear with time without leaving any solid.
(b) We can get the smell of perfume sitting several metres away.

Answer : (a) Naphthalene balls disappear with time without leaving any solid. This is due to the property of sublimation. Some solids do not exist in their liquid form or changes from solid to liquid; instead they directly evaporate from their solid form to vapour forms. Naphthalene is a sublime substance with very high vapors pressure,  that show the property of sublimation. So naphthalene balls disappear after some time due to the property of sublimation.

(b) We can get the smell of perfume sitting several metres away. This is due to high rate of  diffusion of perfume particles. Liquid generally vaporize at normal temperature. Perfume particles on vaporizing, intermix on their own with each other and air molecules around at a very fast rate. They do so by getting into the spaces between the particles. This intermixing of particles of two different types of matter on their own is called diffusion.Therefor the smell of perfume reach us sitting metres away.

### Convert the following temperatures to the Kelvin scale. (a) 25°C (b) 373°C

Answer : Kelvin is the SI unit of temperature, 0° C =273.16 K. For convenience, we take 0° C = 273 K
after rounding off the decimal. To  convert a temperature on the Celsius scale to the Kelvin scale we have to add 273 to the given temperature.

(a) 25°C = 25 + 273 = 298 K

(b) 373°C = 373 + 273 = 646 K

### Convert the following temperatures to the celsius scale. (a) 293 K (b) 470 K.

Answer : As we know 0° C =273.16 K. For convenience, we take 0° C = 273 K after rounding off the decimal.
To change a temperature on the Kelvin scale to the Celsius scale we have to subtract 273 from the given temperature

(a) 293 K =  293-273  = 20° C
(b) 470 K = 470-273   = 197° C

### Which of the following are chemical changes?Question 11 from Chapter 2. IS Matter Around Us Pure ?

(a) Growth of a plant
(b) Rusting of iron
(c) Mixing of iron filings and sand
(d) Cooking of food
(e) Digestion of food
(f) Freezing of water
(g) Burning of a candle

(a) Growth of a plant
(b) Rusting of iron
(d) Cooking of food
(e) Digestion of food
(g) Burning of a candle

### Classify the following into elements, compounds and mixtures Question 10 from Chapter 2. IS Matter Around Us Pure ?

(a) Sodium
(b) Soil
(c) Sugar solution
(d) Silver
(e) Calcium carbonate
(f) Tin
(g) Silicon
(h) Coal
(i) Air
(j) Soap
(k) Methane
(l) Carbon dioxide
(m) Blood

Answer : Classification of materials as elements, compounds and mixtures.

 Elements Compounds Mixture (a) Sodium (d) Silver  (f) Tin (g) Silicon (e) Calcium carbonate (k) Methane (l) Carbon dioxide (b) Soil (c) Sugar solution (h) Coal (i) Air (j) Soap (m) Blood

### Which of the following will show "Tyndall effect"? Question 9 from Chapter 2. IS Matter Around Us Pure ?

(a) Salt solution
(b) Milk
(c) Copper sulphate solution
(d) Starch solution.

Answer : Scattering of a beam of light passing through a suspension or a colloidal solution is called the Tyndall effect. Both a suspension and  a colloidal solution is a heterogeneous mixture in which the particles which may or may not be seen with naked eyes,  can scatter a beam of light passing through it.
As per the description above, following will show "Tyndall effect"

(b) Milk
(d) Starch solution.

### Identify the solutions among the following mixtures Question 8 from Chapter 2. IS Matter Around Us Pure ?

(a) Soil
(b) Sea water
(c) Air
(d) Coal
(e) Soda water.

Answer : As we know a solution is a homogeneous mixture of two or more substances. The component  of the solution which is present in larger amount and dissolves the other is called solvent and other component with lesser quantity which gets dissolved in solvent is called the solute.
As per the description above, the following mixtures are solutions:

(b) Sea water
(c) Air
(e) Soda water

### Which of the following materials fall in the category of a "pure substance" ? Question 7 from Chapter 2. IS Matter Around Us Pure ?

(a) Ice
(b) Milk
(c) Iron
(d) Hydrochloric acid
(e) Calcium oxide
(f) Mercury
(g) Brick
(h) Wood
(i) Air.

Answer : A pure substance contains only one kind of pure matter and its composition is the same throughout. Mixtures are constituted by more than one kind of pure substance.A pure substance cannot be separated into other kinds of matter by any physical process.Whatever the source of a substance may be, it will always have the same characteristic properties. Therefor as per the description of pure substance above, the following materials fall in the category of a “pure substance”:
(a) Ice
(c) Iron
(d) Hydrochloric acid
(e) Calcium oxide
(f) Mercury

### How would you confirm that a colourless liquid given to you is pure water?

Answer :  For a given atmospheric pressure, every liquid has a distinct boiling point. Water in its pure form  boils at 100° C or 273 K at 1 atmospheric pressure. If given colourless liquid starts boiling below or above 100° C or 273 K, it is either not pure water or some other liquid. So boiling point of a liquid, is an indicator of its purity  at a particular atmospheric pressure.

### Classify each of the following as a homogeneous or heterogeneous mixture. soda water, wood, air, soil, vinegar, filtered tea.

 Homogeneous Mixture Heterogeneous Mixture Soda water Air Vinegar Wood Soil Filtered tea

### Explain the following giving examples Question 4 from Chapter 2. IS Matter Around Us Pure ?

(a) saturated solution
(b) pure substance
(c) colloid
(d) suspension

(a) Saturated solution : A solution at any particular temperature, having maximum amount of solute which  can be dissolve in it, is called a saturated solution. In other words, at a given temperature no more solute can be dissolved in a saturated solution.
(b) Pure substance : A pure substance contains only one kind of pure matter and its composition is the same throughout. Mixtures are constituted by more than one kind of pure substance.A pure substance cannot be separated into other kinds of matter by any physical process.Whatever the source of a substance may be, it will always have the same characteristic properties. Examples Sugar, sodium chloride.
(c) colloid : A colloid or a colloidal solution is a heterogeneous mixture in which the particles of a colloid are uniformly spread throughout the solution. Due to the relatively smaller size of particles,  the mixture appears to be almost homogeneous,  for example, milk.
(d) Suspension : A suspension is a heterogeneous mixture in which the solute particles do not dissolve but remain suspended throughout the bulk of the medium. Particles of a suspension are visible to the naked eye.
The particles of a suspension scatter a beam of light passing through it and make its path visible.The solute particles settle down when a suspension is left undisturbed, that is, a suspension is unstable. They can
be separated from the mixture by the process of filtration. When the particles settle down, the suspension breaks and it does not scatter light any more.

### Pragya tested the solubility of three different substances at different temperatures and collected the data as given below Question 3 from Chapter 2. IS Matter Around Us Pure ?

(results are given in the following table, as grams of substance dissolved in 100 grams of water to form a saturated solution).
 Substance Dissolved Temperature in K 283   293    313    333    353 Solubility Potassium nitrate Sodium chloride Potassium chloride Ammonium chloride 21 36 35 24 32 36 35 37 62 36 40 41 106 37 46 55 167 37 54 66

(a) What mass of potassium nitrate would be needed to produce a saturated solution of potassium nitrate in 50 grams of water at 313 K?

Answer :  Mass of potassium nitrate at 313 K per 100 grams of water in given saturated solution of potassium nitrate = 62 Grams
Required mass of potassium nitrate at 313 K per 50 grams of water for a saturated solution of potassium nitrate = 62 / 100 ×50 = 31 grams

(b) Pragya makes a saturated solution of potassium chloride in water at 353 K and leaves the solution to cool at room temperature. What would she observe as the solution cools? Explain.
Answer :  A saturated solution of potassium chloride in water at 353 K has a specific solubility.If it is allowed to cool down at room temperature around 293 K, with the decrease in temperature, solubility  of potassium chloride in the solution will also become low. It will dissolve less in water and undissolved potassium chloride result in  formation of  a layer of potassium chloride crystals at the bottom.

(c) Find the solubility of each salt at 293 K. Which salt has the highest solubility at this temperature?

Answer : The  solubility of a saturated solution for a given temperature is expressed as  Mass of solute in gm per 100 gm of solvent
1. The solubility of Potassium nitrate in a saturated solution with water at 293 K = 32 gm of Potassium nitrate per 100 gm of water
2. The solubility of Sodium chloride in a saturated solution with water at 293 K = 36 gm of Sodium chloride per 100 gm of water
3. The solubility of Potassium chloride in a saturated solution with water at 293 K = 37 gm Potassium chloride per 100 gm of water
4. The solubility of Ammonium chloride in a saturated solution with water at 293 K = 35 gm Ammonium chloride per 100 gm of water
Ammonium chloride has  the highest solubility at the given temperature of 293 K

(d) What is the effect of change of temperature on the solubility of a salt?
Answer : Solubility of salt gets  gets affected with change of temperature. It increases with rise in temperature and decreases with fall in temperature.

### Write the steps you would use for making tea. Use the words : solution, solvent, solute, dissolve, soluble, insoluble, filtrate and residue.

Answer : For making tea, first of all we will take desired cups of  water as solvent in a tea pan. Then it is allowed to boil on stove. Tea leaves, as salute is added to it and  brewed. While brewing , the colour of water as solvent gets changed to tea colour as soluble part of tea as salute gets dissolve in heated water  as solvent. There after,   milk is added to it as solute and is further allowed to boil for some times. Insoluble tea leaves as residue  are removed by passing the mixture through a tea strainer. Sugar as solute according to need is added to filtrate so obtained and stirred with spoon. Sugar gets dissolve in filtrate. Resulting solution is in the form of  Tea, ready for use.

### Which separation techniques will you apply for the separation of the following?Question 1`Chapter 2. IS Matter Around Us Pure ? from

(a) Sodium chloride from its solution in water.
(b) Ammonium chloride from a mixture containing sodium chloride and ammonium chloride.
(c) Small pieces of metal in the engine oil of a car.
(d) Different pigments from an extract of flower petals.
(e) Butter from curd.
(f) Oil from water.
(g) Tea leaves from tea.
(h) Iron pins from sand.
(i) Wheat grains from husk.
(j) Fine mud particles suspended in water.

Answer :Separation techniques for separation of given substance mixtures :-
 Mixtures of Substance Separation techniques (a) Sodium chloride from its solution in water. By Evaoparation (b) Ammonium chloride from a mixture containing sodium chloride and ammonium chloride By Sublimation (c) Small pieces of metal in the engine oil of a car. By Filtration (d) Different pigments from an extract of flower petals. Using Chromatography (e) Butter from curd. By Centrifugation (f) Oil from water. Using separating funnel (g) Tea leaves from tea. By Filtration (h) Iron pins from sand. By a Magnet (i) Wheat grains from husk. By Winnowing (j) Fine mud particles suspended in water. By Centrifugation

### 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)

1 mole of Aluminium Oxide (Al2O3) = 2 × 27 + 3 × 16  = 102 g

We know, 102 g of
Aluminium Oxide (Al2O3) = 6.022 × 1023 molecules of Aluminium Oxide (Al2O3)
We know, The number of atoms present in given mass = (Given mass ÷ molar mass) × Avogadro number
Then, 0.051 g of Aluminium Oxide (Al2O3) contains
= (0.051 ÷ 102 ) × 6.022 × 1023 molecules of Aluminium Oxide (Al2O3)
= 3.011 × 1020 molecules of Aluminium Oxide (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 × 1020 molecules (0.051 g ) of Aluminium Oxide (Al2O3)
= 2 × 3.011 × 1020

= 6.022 ×
1020

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

1 mole of Sulphur (S8 = 8 × 32 g = 256 g
also as
1 mole = 6.022 × 1023 in number of particles of that substance
So 256 g of  sulphur contains = 6.022 × 1023  molecules of sulphur
Therefor 16 g of sulphur will contains = 3.76 ×
1022  molecules (approx)

### What is the mass of Question 9 from Chapter 3. Atoms And Molecules

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

Answer : Mass of atoms (m) in given moles (n) = Molar Mass of substance (M)× Given Moles of substance (n)

(a) ∴  Mass of 0.2 mole of oxygen atoms = Molar Mass of oxygen × Given Moles  =  16g × 0.2   = 3.2 g

(b)
∴ Mass of 0.5 mole of water molecules =Molar Mass of water × Given Moles = 18 g × 0.5  = 9 g

### Convert into mole Question 8 from Chapter 3. Atoms And Molecules

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

Answer: To get the mass of 1 mole of atom of that element, that is, molar mass, we have to take the same numerical value but change the units from 'u' to 'g'.
∴ 1 mole substance = Total Atomic mass of a substance (u)       Or
= Total gram atomic mass of substance (g)  Or
= Molar Mass
= 1 Mole of substance                                Or
=  6.022 × 1023 in number of particles of that substance
Here required mole of substance  =  Given mass of substance in 'g' ÷ Molar Mass

(a) 12 g of oxygen gas =  Given mass of oxygen in 'g' ÷ Molar Mass of Oxygen
= 12 ÷ 32
= 0.375 mole of oxygen
(b) 20 g of water = Given mass of water in 'g' ÷ Molar Mass of Water
= 20 ÷ 18
= 1.11 mole of water

(c) 22 g of carbon dioxide.Given mass of Carbon Dioxide in 'g' ÷ Molar Mass of Carbon Dioxide
= 22 ÷ 44
= 0.5 mole of carbon dioxide

### What is the mass of Question 7 from Chapter 3. Atoms And Molecules

(a) 1 mole of nitrogen atoms?
(b) 4 moles of aluminium atoms (Atomic mass of aluminium= 27)?
(c) 10 moles of sodium sulphite (Na2SO3)?

Answer :The mass of 1 mole of a substance is equal to its relative atomic or molecular mass in grams. The atomic mass of an element gives us the mass of one atom of that element in atomic mass units (u). To get the mass of 1 mole of atom of that element, that is, molar mass, we have to take the same numerical value but change the units from 'u' to 'g'.
Molar mass of atoms is also known as gram atomic mass. For example, atomic mass of hydrogen=1u. So, gram atomic mass of
hydrogen = 1 g.

(a) The mass of 1 mole of Nitrogen atoms = 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) =10 × [2 × 23 + 32 + 3 × 16] g = 10 × 126 g = 1260 g

### Calculate the molar mass of the following substances Question 6 from Chapter 3. Atoms And Molecules

(a) Ethyne, C2H2
(b) Sulphur molecule, S8
(c) Phosphorus molecule, P4 (Atomic mass of phosphorus = 31)
(d) Hydrochloric acid, HCl
(e) Nitric acid, HNO3

(a) Molar Mass of Ethyne, C2H2; =     2 × 12 + 2 × 1 =  26 g
(b) Molar Mass of Sulphur molecule, S8 =   8 × 32  =  256 g
(c) Molar Mass of Phosphorus molecule, P4 (Atomic mass of phosphorus = 31)  =   4 × 31  =  124 g
(d) Molar Mass of Hydrochloric acid, HCl  =   1 + 35.5 =  36.5 g
(e) Molar Mass of Nitric acid, HNO3 =   1 + 14 + 3 × 16  =  63 g

### Give the names of the elements present in the following compounds Question 5 from Chapter 3. Atoms And Molecules

(a) Quick lime
(b) Hydrogen bromide
(c) Baking powder
(d) Potassium sulphate.

 Compound Chemical Formula Elements (a) Quick lime (b) Hydrogen bromide (c) Baking powder (d) Potassium sulphate. CaO HBr NaHCO3  K2SO4 Calcium, Oxygen Hydrogen, Bromine Sodium, Hydrogen, Carbon, Oxygen Potassium, Sulphur, Oxygen

### Write the chemical formulae of the following Question 4 from Chapter 3. Atoms And Molecules

(a) Magnesium chloride
(b) Calcium oxide
(c) Copper nitrate
(d) Aluminium chloride
(e) Calcium carbonate.

(a) Magnesium chloride     => MgCl2
(b) Calcium oxide         => CaO
(c) Copper nitrate        => Cu(NO3)2.
(d) Aluminium chloride    => AlCl3
(e) Calcium carbonate.    => CaCO3

### What are polyatomic ions? Give examples

Answer : Polyatomic ions are clusters or group of atoms that always have a  fixed  charge attach to them. This charge may be positive or negative .
Ammonium  NH4+
Hydroxide OH-
Nitrate NO3 -
Carbonate CO32-
Sulphate  SO42-

### 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 willbe formed when 3.00 g of carbon is burnt in 50.00 g of oxygen? Which law of chemical combination will govern your answer?

Answer : We know that in burning process carbon utilises oxygen to form Carbon Dioxide .
Carbon + Oxygen => Carbon Dioxide + Heat Energy

Here as given,  in burning, 3.0 g of carbon utilises 8.00 g of oxygen to form 11.00 g of carbon dioxide.When 3.00 g of carbon is burnt in 50.00 g of oxygen, it will use just 8.00 g of oxygen from  the total of 50 gm of Oxygen to produce 11 gm of Carbon Dioxide. 42 gm of Oxygen will be left behind without any change.
The above answer is governed by the following two laws of chemical combination:

1. Law of conservation of mass which states that mass can neither be created nor destroyed in a chemical reaction.
Here in terms of mass 3g of Carbon and 50 gm of Oxygen combines to give 11 gm of Carbon Dioxide and 42 gm of unused Oxygen keeping the total mass of combining

chemicals 53 gm (3+50 gm)  by weight constant before and after the reaction.

2. Law of constant proportions which states that In a chemical substance the elements are always present in definite proportions by mass.Here respective of the excess quantity of Oxygen available for burning of same weight of Carbon, the amount of carbon dioxide will remain same.

### 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.

Mass of boron as given = 0.096 g
Mass of oxygen as given  = 0.144 g
Mass of sample as given   = 0.24 g
Percentage of boron by weight in the compound = (0.096 ÷ 0.24 )  x 100 = 40%

Percentage of oxygen by weight in the compound = (0.144 ÷ 0.24)  x 100 = 60%

### Complete the following table Question 19 from Chapter 4. Structure Of The Atom

 Atomic Number Mass Number Number of Neutrons Number of Protons Number of Electrons Name of the Atomic Species 9 - 10 - - - 10 32 - - - Sulphure - 24 - 12 - - - 2 - 1 - - - 1 0 1 0 -

 Atomic Number Mass Number Number of Neutrons Number of Protons Number of Electrons Name of the Atomic Species 9 19 10 9 9 Fluorine 16 32 16 16 16 Sulphure 12 24 12 12 12 Magnesium 1 2 1 1 1 Deuterium 1 1 0 1 1 Protium

(b)  8 (√)
(c) 17 (×)
(d) 18 (×)

### Isotopes of an element have (a) the same physical properties (b) different chemical properties (c) different number of neutrons (d) different atomic numbers.

Answer : (a) the same physical properties (×)
(b) different chemical properties (×)
(c) different number of neutrons (√)
(d) different atomic numbers. (×)

### Rutherford’s alpha-particle scattering experiment was responsible for the discovery of (a) Atomic Nucleus (b) Electron (c) Proton (d) Neutron

Answer : (a) Atomic Nucleus (√)
(b) Electron (×)
(c) Proton (×)
(d) Neutron (×)

### For the following statements, write T for True and F for False Question 14 from Chapter 4. Structure Of The Atom

(a) J.J. Thomson proposed that the nucleus of an atom contains only nucleons.
(b) A neutron is formed by an electron and a proton combining together. Therefore, it is neutral.
(c) The mass of an electron is about 1 2000 times that of proton.
(d) An isotope of iodine is used for making tincture iodine, which is used as a medicine.
(a) J.J. Thomson proposed that the nucleus of an atom contains only nucleons.(F)
(b) A neutron is formed by an electron and a proton combining together. Therefore, it is neutral.(F)
(c) The mass of an electron is about 1/ 2000 times that of proton.(T)
(d) An isotope of iodine is used for making tincture iodine, which is used as a medicine.(T)