1 mole of aluminium oxide (Al₂O₃) = 2 × 27 + 3 × 16 = 102g i.e., 102g of Al₂O₃ = 6.022 × 10²³ molecules of Al₂O₃ Then, 0.051 g of Al₂O₃ contains = 6.022×10²³/102×0.051 molecules = 3.011 × 10²⁰ molecules of Al₂O₃ The number of aluminium ions (Al³⁺) present in one molecules of aluminium oxide is 2. ThRead more
1 mole of aluminium oxide (Al₂O₃) = 2 × 27 + 3 × 16 = 102g
i.e., 102g of Al₂O₃ = 6.022 × 10²³ molecules of Al₂O₃
Then, 0.051 g of Al₂O₃ contains = 6.022×10²³/102×0.051 molecules
= 3.011 × 10²⁰ molecules of Al₂O₃
The number of aluminium ions (Al³⁺) present in one molecules of aluminium oxide is 2. Therefore, The number of aluminium ions (Al³⁺) present in
3.11 × 10²⁰ molecules (0.051g) of aluminium oxide (Al2O3) = 2 × 3.011 × 10²⁰
= 6.022 × 10²⁰
(a) 32 g of oxygen gas = 1 mole Then, 12g of oxygen gas = 12/32 mole = 0.375 mole (b) 18g of water = 1 mole Then, 20 g of water = 20/18 mole = 1.11 moles (approx.) (c) 44g of carbon dioxide = 1 mole Then, 22g of carbon dioxide = 22/44 mole = 0.5 mole For more answers visit to website: https://www.tiRead more
(a) 32 g of oxygen gas = 1 mole
Then, 12g of oxygen gas = 12/32 mole = 0.375 mole
(b) 18g of water = 1 mole
Then, 20 g of water = 20/18 mole = 1.11 moles (approx.)
(c) 44g of carbon dioxide = 1 mole
Then, 22g of carbon dioxide = 22/44 mole = 0.5 mole
(a) The mass of 1 mole of nitrogen atoms is 14g. (b) The mass of 4 moles of aluminium atoms is (4 × 27)g = 108g (c) The mass of 10 moles of sodium sulphite (Na2SO3) is 10 × [2 × 23 + 32 + 3 × 16]g = 10 × 126g = 1260g For more answers visit to website: https://www.tiwariacademy.com/ncert-solutions/clRead more
(a) The mass of 1 mole of nitrogen atoms is 14g.
(b) The mass of 4 moles of aluminium atoms is (4 × 27)g = 108g
(c) The mass of 10 moles of sodium sulphite (Na2SO3) is 10 × [2 × 23 + 32 + 3 × 16]g = 10 × 126g = 1260g
Compound Chemical formula Elements present Quick lime CaO Calcium, oxygen Hydrogen bromide HBr Hydrogen, bromine Baking powder NaHCO3 Sodium, hydrogen, carbon, oxygen Potassium sulphate K2SO4 Potassium, sulphur, oxygen For more answers visit to website: https://www.tiwariacademy.com/ncert-solutions/Read more
Compound Chemical formula Elements present
Quick lime CaO Calcium, oxygen
Hydrogen bromide HBr Hydrogen, bromine
Baking powder NaHCO3 Sodium, hydrogen, carbon, oxygen
Potassium sulphate K2SO4 Potassium, sulphur, oxygen
A polyatomic ion is a group of atoms carrying a charge (positive or negative). For example, ammonium ion (𝑁𝐻+/4), hydroxide ion (OH−), carbonate ion (𝐶𝑂2/3−), sulphateion (𝑆𝑂2/4−). For more answers visit to website: https://www.tiwariacademy.com/ncert-solutions/class-9/science/chapter-3/
A polyatomic ion is a group of atoms carrying a charge (positive or negative). For example, ammonium ion (𝑁𝐻+/4), hydroxide ion (OH−), carbonate ion (𝐶𝑂2/3−), sulphateion (𝑆𝑂2/4−).
Carbon + Oxygen → Carbon dioxide 3g of carbon reacts with 8 g of oxygen to produce 11g of carbon dioxide. If 3g of carbon is burnt in 50g of oxygen, then 3g 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 11g of carbon dioxide wRead more
Carbon + Oxygen → Carbon dioxide
3g of carbon reacts with 8 g of oxygen to produce 11g of carbon dioxide. If 3g of carbon is burnt in 50g of oxygen, then 3g 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 11g of carbon dioxide will be formed. The above answer is governed by the law of constant proportions.
Mass of boron = 0.096g (Given) Mass of oxygen = 0.144g (Given) Mass of sample = 0.24g (Given) Thus, percentage of boron by weight in the compound = 0.096×1000. / 24% = 40% Thus, percentage of oxygen by weight in the compound = 0.144×1000./24% = 60 % For more answers visit to website: https://www.tiwRead more
Mass of boron = 0.096g (Given)
Mass of oxygen = 0.144g (Given)
Mass of sample = 0.24g (Given)
Thus, percentage of boron by weight in the compound = 0.096×1000. / 24% = 40%
Thus, percentage of oxygen by weight in the compound = 0.144×1000./24%
= 60 %
Calculate the number of aluminum ions present in 0.051 g of aluminum 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 = 102g i.e., 102g of Al₂O₃ = 6.022 × 10²³ molecules of Al₂O₃ Then, 0.051 g of Al₂O₃ contains = 6.022×10²³/102×0.051 molecules = 3.011 × 10²⁰ molecules of Al₂O₃ The number of aluminium ions (Al³⁺) present in one molecules of aluminium oxide is 2. ThRead more
1 mole of aluminium oxide (Al₂O₃) = 2 × 27 + 3 × 16 = 102g
i.e., 102g of Al₂O₃ = 6.022 × 10²³ molecules of Al₂O₃
Then, 0.051 g of Al₂O₃ contains = 6.022×10²³/102×0.051 molecules
= 3.011 × 10²⁰ molecules of Al₂O₃
The number of aluminium ions (Al³⁺) present in one molecules of aluminium oxide is 2. Therefore, The number of aluminium ions (Al³⁺) present in
3.11 × 10²⁰ molecules (0.051g) of aluminium oxide (Al2O3) = 2 × 3.011 × 10²⁰
= 6.022 × 10²⁰
For more answers visit to website:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-9/science/chapter-3/
Calculate the number of molecules of Sulphur (S8) present in 16 g of solid Sulphur.
1 mole of solid sulphur (S₈) = 8 × 32g = 256g i.e., 256g of solid sulphur contains = 6.022 × 10²³ molecules Then, 16g of solid sulpur contains 6.022×10²³ /256×16 molecules = 3.76 × 10²² molecules (approx) For more answers visit to website: https://www.tiwariacademy.com/ncert-solutions/class-9/sciencRead more
1 mole of solid sulphur (S₈) = 8 × 32g = 256g
i.e., 256g of solid sulphur contains = 6.022 × 10²³ molecules
Then, 16g of solid sulpur contains 6.022×10²³ /256×16 molecules
= 3.76 × 10²² molecules (approx)
For more answers visit to website:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-9/science/chapter-3/
Convert into mole.
(a) 32 g of oxygen gas = 1 mole Then, 12g of oxygen gas = 12/32 mole = 0.375 mole (b) 18g of water = 1 mole Then, 20 g of water = 20/18 mole = 1.11 moles (approx.) (c) 44g of carbon dioxide = 1 mole Then, 22g of carbon dioxide = 22/44 mole = 0.5 mole For more answers visit to website: https://www.tiRead more
(a) 32 g of oxygen gas = 1 mole
Then, 12g of oxygen gas = 12/32 mole = 0.375 mole
(b) 18g of water = 1 mole
Then, 20 g of water = 20/18 mole = 1.11 moles (approx.)
(c) 44g of carbon dioxide = 1 mole
Then, 22g of carbon dioxide = 22/44 mole = 0.5 mole
For more answers visit to website:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-9/science/chapter-3/
What is the mass of:
(a) The mass of 1 mole of nitrogen atoms is 14g. (b) The mass of 4 moles of aluminium atoms is (4 × 27)g = 108g (c) The mass of 10 moles of sodium sulphite (Na2SO3) is 10 × [2 × 23 + 32 + 3 × 16]g = 10 × 126g = 1260g For more answers visit to website: https://www.tiwariacademy.com/ncert-solutions/clRead more
(a) The mass of 1 mole of nitrogen atoms is 14g.
(b) The mass of 4 moles of aluminium atoms is (4 × 27)g = 108g
(c) The mass of 10 moles of sodium sulphite (Na2SO3) is 10 × [2 × 23 + 32 + 3 × 16]g = 10 × 126g = 1260g
For more answers visit to website:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-9/science/chapter-3/
Calculate the molar mass of the following substances.
(a) Molar mass of ethyne, C2H2 = 2 × 12 + 2 × 1 = 28g (b) Molar mass of sulphur molecule, S8 = 8 × 32 = 256g (c) Molar mass of phosphorus molecule, P4 = 4 × 31 = 124g (d) Molar mass of hydrochloric acid, HCl = 1 + 35.5 = 36.5g (e) Molar mass of nitric acid, HNO3 = 1 + 14 + 3 × 16 = 63g For more answRead more
(a) Molar mass of ethyne, C2H2 = 2 × 12 + 2 × 1 = 28g
(b) Molar mass of sulphur molecule, S8 = 8 × 32 = 256g
(c) Molar mass of phosphorus molecule, P4 = 4 × 31 = 124g
(d) Molar mass of hydrochloric acid, HCl = 1 + 35.5 = 36.5g
(e) Molar mass of nitric acid, HNO3 = 1 + 14 + 3 × 16 = 63g
For more answers visit to website:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-9/science/chapter-3/
Give the names of the elements present in the following compounds.
Compound Chemical formula Elements present Quick lime CaO Calcium, oxygen Hydrogen bromide HBr Hydrogen, bromine Baking powder NaHCO3 Sodium, hydrogen, carbon, oxygen Potassium sulphate K2SO4 Potassium, sulphur, oxygen For more answers visit to website: https://www.tiwariacademy.com/ncert-solutions/Read more
Compound Chemical formula Elements present
Quick lime CaO Calcium, oxygen
Hydrogen bromide HBr Hydrogen, bromine
Baking powder NaHCO3 Sodium, hydrogen, carbon, oxygen
Potassium sulphate K2SO4 Potassium, sulphur, oxygen
For more answers visit to website:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-9/science/chapter-3/
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 For more answers visit to website: https://www.tiwariacademy.com/ncert-solutions/class-9/science/chapter-3/
(a) Magnesium chloride →MgCl2
(b) Calcium oxide →CaO
(c) Copper nitrate →Cu(NO3)2
(d) Aluminium chloride →AlCl3
(e) Calcium carbonate →CaCO3
For more answers visit to website:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-9/science/chapter-3/
What are polyatomic ions? Give examples.
A polyatomic ion is a group of atoms carrying a charge (positive or negative). For example, ammonium ion (𝑁𝐻+/4), hydroxide ion (OH−), carbonate ion (𝐶𝑂2/3−), sulphateion (𝑆𝑂2/4−). For more answers visit to website: https://www.tiwariacademy.com/ncert-solutions/class-9/science/chapter-3/
A polyatomic ion is a group of atoms carrying a charge (positive or negative). For example, ammonium ion (𝑁𝐻+/4), hydroxide ion (OH−), carbonate ion (𝐶𝑂2/3−), sulphateion (𝑆𝑂2/4−).
For more answers visit to website:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-9/science/chapter-3/
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 combination will govern your answer?
Carbon + Oxygen → Carbon dioxide 3g of carbon reacts with 8 g of oxygen to produce 11g of carbon dioxide. If 3g of carbon is burnt in 50g of oxygen, then 3g 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 11g of carbon dioxide wRead more
Carbon + Oxygen → Carbon dioxide
3g of carbon reacts with 8 g of oxygen to produce 11g of carbon dioxide. If 3g of carbon is burnt in 50g of oxygen, then 3g 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 11g of carbon dioxide will be formed. The above answer is governed by the law of constant proportions.
For more answers visit to website:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-9/science/chapter-3/
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 = 0.096g (Given) Mass of oxygen = 0.144g (Given) Mass of sample = 0.24g (Given) Thus, percentage of boron by weight in the compound = 0.096×1000. / 24% = 40% Thus, percentage of oxygen by weight in the compound = 0.144×1000./24% = 60 % For more answers visit to website: https://www.tiwRead more
Mass of boron = 0.096g (Given)
Mass of oxygen = 0.144g (Given)
Mass of sample = 0.24g (Given)
Thus, percentage of boron by weight in the compound = 0.096×1000. / 24% = 40%
Thus, percentage of oxygen by weight in the compound = 0.144×1000./24%
= 60 %
For more answers visit to website:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-9/science/chapter-3/