Potassium chlorate is an example of a binary ionic compound question 5 options 1 true 2 false

Write the full shape of IUPAC.

IUPAC stands for International Union of Pure and Applied Chemistry. It is a systematic employer which develops nomenclature for chemicals.

Name the scientist who gave:(a) law of conservation of mass(b) regulation of regular proportions.

(a) Law of conservation of mass changed into given via Antoine Lavoisier in 1744. According to this regulation, “Matter can neither be created nor destroyed in a chemical response”.

(b) Law of regular proportions was given by means of Joseph Proust in 1799. According to this regulation, “a chemical compound usually consists of equal elements in definite proportion via mass and it does now not depend upon the supply of compound.”

Name the choices regulation of chemical combination:(a) which was given by way of Lavoisier.(b) which turned into given by Proust.

(a) Law of conservation of mass become given through Antoine Lavoisier in 1744. According to this law, “Matter can neither be created nor destroyed in a chemical reaction”.

(b) Law of constant proportions changed into given by using Joseph Proust in 1799. According to this regulation, “a chemical compound constantly includes equal elements in particular proportion by means of mass and it does now not depend on the choices source of compound.”

Name the scientist who gave atomic idea of count.

Atomic idea of matter became given by John Dalton. It become the first entire try to describe all the depend within the form of atoms and their homes.

Which postulate of Dalton’s atomic idea is the choices end result of regulation of conservation of mass given by Lavoisier?

According to regulation of conservation of mass describes that, “Matter can neither be created nor destroyed in a chemical reaction”. This regulation gives result to the choices postulate of Dalton’s atomic theory which states that, “Atoms can neither be created and nor destroyed by using physical and chemical modifications.”

Which part of the choices Dalton’s atomic concept got here from the law of steady proportions given with the aid of Proust?

According to law of steady proportions describes that, “a chemical compound usually carries equal elements in definite share through mass and it does not depend on the choices supply of compound.”

This regulation offers result to the postulate of Dalton’s atomic concept which states that, “When two factors combine to form or greater than unique compounds then the choices special masses of 1 element B which combine with constant mass of the opposite element bear a easy ratio to one another.”

Which historical Indian truth seeker cautioned that all depend consists of very small debris? What name became given via him to these particles?

Maharshi Kanada was historic Indian scientist and sage who advised that each one count number consists of very small debris. He named the choices smallest particles as “Parmanu”.

Name any two laws of chemical combination.

Law of conservation of mass was given through Antoine Lavoisier in 1744. According to this regulation, “Matter can neither be created nor destroyed in a chemical response”.

Law of constant proportions became given with the aid of Joseph Proust in 1799. According to this regulation, “a chemical compound constantly carries identical factors in exact share with the aid of mass and it does no longer rely on the source of compound.”

‘If 100 grams of pure water taken from one of a kind assets is decomposed by using passing energy, 11 grams of hydrogen and 89 grams of oxygen are usually obtained’. Which chemical law is illustrated through this declaration?

The above statement justifies the choices Law of constant proportions which changed into given through Joseph Proust in 1799. According to this law, “a chemical compound usually includes identical elements in exact share via mass and it does no longer rely on the choices supply of compound.” This shows that something be the supply of water, it will always contain 1:eight ratio of Hydrogen and oxygen respectively.

‘If 100 grams of calcium carbonate (whether or not within the form of marble or chalk) are decomposed completely, then fifty six grams of calcium oxide and 44 grams of carbon dioxide are acquired’. Which regulation of chemical mixture is illustrated by this assertion?

The above announcement justifies the choices Law of consistent proportions which was given by Joseph Proust in 1799. According to this regulation, “a chemical compound continually consists of equal elements in exact proportion by way of mass and it does not rely upon the choices supply of compound.” This indicates that whatever be the supply of water, it will constantly comprise 1:eight ratio of Hydrogen and oxygen respectively.

What are the choices building blocks of rely?

Matter is made from indivisible particles known as Atoms. Atoms are the constructing blocks of matter.

How is the size of an atom indicated?

The size of an atom is indicated via its radius that’s referred to as atomic radius (radius of an atom). Atomic radius is measured in nanometre (nm) (1 metre = 109 nanometres or 1nm = 10_9m).

Name the choices unit in which the radius of an atom is commonly expressed.

Atomic radius is measured in nanometre (nm) (1 metre = 109 nanometres or 1nm = 10_9m).

Write the choices relation between nanometer and metre.

A nanometer is one billionth part of metre. 1 metre = 1/109 nanometres or 1nm = 10_9m).

The radius of an oxygen atom is zero.073 nm. What does the symbol ‘nm’ constitute?

The image ‘nm’ represents Nanometre. A nanometer is one billionth part of metre. 1 metre = 1/109 nanometres or 1nm = 10_9m).

Why is it no longer possible to see an atom despite the maximum powerful microscope?

Atoms are the choices building blocks of the problem. They are very small in length. Hence Why is it now not viable to peer an atom regardless of the most powerful microscope.

State whether or not the subsequent assertion is proper or fake:The image of detail cobalt is CO.

False, the symbol of detail cobalt is Co.

Define ‘molecular mass’ of a substance.

The molecular mass of a substance (an detail or a compound) can be defined as the choices common relative mass of a molecule of the choices substance compared with mass of an atom of carbon (C-12 isotope) taken as 12 atomic mass unit. The molecular mass of a substance can be determined with the aid of adding atomic hundreds of all the atoms present within the molecule of the substance.

What is meant by way of announcing that ‘the choices molecular mass of oxygen is 32’?

The image of oxygen is O2 because of this it carries atoms of oxygen. The molecular mass of 1 atom of oxygen is sixteen.

Hence the choices molecular mass of O2 is 32.

Fill inside the following blanks with appropriate phrases:(a) In water, the share of oxygen and hydrogen is …….. by mass.(b) In a chemical response, the choices sum of the choices loads of the reactants and the products stays unchanged. This is referred to as ………… .

(a) 8:1, anything be the choices source of water, it’ll usually contain 1:eight ratio of Hydrogen and oxygen respectively.

(b) Law of conservation of mass.

An detail A forms an oxide A2O5.(a) What is the choices valency of element A?(b) What might be the choices formulation of chloride of A?

(a) Let the choices valency of detail A be y, then 2y + 5(-2) = 0

So, y = valency of element A = five

(b) As valency of detail A is 5 and valency of chlorine is -1,

So, the components of chloride of A is ACl5.

An element X bureaucracy the subsequent compounds with hydrogen, carbon and oxygen:H2X, CX2, XO2, XO3

(i) In H2X: -2

(ii) In CX2: -2

If the choices aluminium salt of an anion X is Al2X3, what’s the valency of X? What may be the formulation of the magnesium salt of X?

Let the choices valency of X be y, then 2 x (+three) + three x y = zero

So, valency of X = y = -2

As valency of Mg is +2 and that of X is -2 so the method of Magnesium salt of X could be MgX.

The method of carbonate of a metallic M is M2CO3.(a) What could be the formula of its iodide?(b) What can be the formulation of its nitride?(c) What may be the system of its phosphate?

According to formula M2CO3, the metal M is univalent having valency of +1 and valency of carbonate is -2 i.e. bivalent.

(a) Iodine is univalent as valency of iodine is -1, hence method of iodide = MI

(b) Nitrogen is trivalent as valency of nitrogen is -three, for this reason system of nitride = M3N

(c) Phosphorus is trivalent as valency of phosphorus is -3, as a result method of phosphate = M3PO4

The atom of an detail X carries 17 protons, 17 electrons and 18 neutrons whereas the choices atom of an element Y contains 11 protons, eleven electrons and 12 neutrons.(a) What form of ion might be shaped by way of an atom of detail X? Write the image of ion fashioned.(b) What could be the choices variety of (i) protons (ii) electrons, and (iii) neutrons, in the ion shaped from X?(c) What kind of ion may be shaped through an atom of detail Y? Write the choices symbol of ion fashioned.(d) What may be the choices range of (i) protons (ii) electrons, and (iii) neutrons, within the ion shaped from Y?(e) What is the atomic mas of (i) X, and (ii) Y?(f) What could the factors X and Y be?

(a). Anion can be shaped with the aid of detail X; Symbol: X-

(b) (i). No. of protons in X = 17

(ii) No. of electrons in X = 18

(iii) No. of neutrons in X = 18

(c) Cation will be fashioned by means of element Y; Symbol: Y+

(d) (i). No. of protons in Y = 11

(ii) No. of electrons in Y = 10

(iii) No. of neutrons in Y = 12

(e) Atomic mass of X = No. of protons (17) + No. of neutrons (18) = 35 u

Atomic mass of Y = No. of protons (11) + No. of neutrons (12) = 23 u

(f) Element X is Chlorine (Cl).

Element Y is Sodium (Na).

What is a collection of 6.022 × 1023 debris referred to as?

The organization of 6.022 × 1023 debris is known as One mole of a substance.

What call is given to the quantity of substance containing 6.022 × 1023 particles (atoms, molecules or ions) of a substance?

A institution of 6.022 × 1023 debris (atom, molecules or ions) of a substance is referred to as a mole of that substance.

What is the choices numerical price of Avogadro wide variety?

The numerical fee of Avogadro variety is 6.022 × 1023.

How many atoms are found in one gram atomic mass of a substance?

One gram atomic mass = 6.022 × 1023 atoms.

How many molecules are found in one gram molecular mass of a substance?

One gram molecular mass = 6.022 × 1023 molecules.

What name is given to the choices quantity 6.022 × 1023?

Avagadro’s Number is the name given number 6.022 × 1023

Convert 12 g of oxygen fuel into moles.

Given Mass of oxygen gasoline(O2) = 12g

Molar mass of oxygen gas(O2) = 32g

As we realize that, Number of moles = mass of an detail/molar mass of an element

Hence, Number of moles = 12g /32g = zero.375.

How many moles are 3.6 g of water?

Given Mass of water(H2O) = three.6g

Molar mass of water(H2O) = 18g

As we recognize that, Number of moles = mass of an element/molar mass of an element

Hence, Number of moles = three.6g / 18g = 0.2 mole.

What is the mass of zero.2 mole of oxygen atoms?

Molar mass of oxygen atom(O) = 16g

Number of moles gift = 0.2 mole

As we recognise that, Number of moles = G mass of an detail/molar mass of an detail

Hence mass of an element = No. of moles X molar mass of an element.

Mass of oxygen atoms = 0.2×sixteen = 3.2g

Find the choices mass of two moles of oxygen atoms?

Molar mass of oxygen atom(O) = 16g

Number of moles gift = 2 mole

As we know that, Number of moles = G mass of an element/molar mass of an element

Hence mass of an element = No. of moles X molar mass of an detail.

Mass of oxygen atoms = 2×sixteen = 32g

Fill within the following blanks:(a) 1 mole carries …………. atoms, molecules or ions of a substance.(b) A mole represents an …….. number of particles of a substance.(c) 60 g of carbon detail are ……… moles of carbon atoms.(d) zero.five mole of calcium detail has a mass of ……….. .(e) 64 g of oxygen fuel consists of …………. Moles of oxygen atoms.

How many atoms are there in precisely 12 g of carbon -12 detail? (C = 12 u)

The number of atoms present in 12g of carbon-12 detail = 6.022 × 1023 atoms

What name is given to this number?

What name is given to the amount of substance containing this variety of atoms?

Calculate the choices mass of 12.044 × 1025 molecules of oxygen (O2).

One mole of O2 = 32 gm

6.022 × 1023 molecules of O2 have mass = 32 gm

So, 12.044 × 1025 molecules of O2 could have mass = 6400 gm = 6.4 Kg

What is the choices range of molecules in 1.5 moles of ammonia?

1 mole of ammonia includes 6.022 × 1023 molecules.

Therefore, 1.five mole of ammonia contains = 6.022 × 1023 × 1.5 = nine.03 × 1023 molecules of ammonia

How many moles of calcium carbonate (CaCO3) are present in 10 g of the substance? (Ca = forty u; C = 12 u; O = 16 u)

Molar mass of calcium carbonate (CaCO3) = forty+12+48 = 100g

Mass of the choices substance = 10g

As we know that, Number of moles = mass of an element/molar mass of an element

Hence Number of moles = 10/one hundred = zero.1 mole.

How many moles of O2 are there in 1.20 × 1022 oxygen molecules?

Since One mole of O2 contains = 6.022 × 1023 molecules of oxygen

So, 1 molecule of O2 has = 1/6.022 × 1023 moles of O2

Therefore, 1.2 × 1022 molecules of O2 may have = 1.2 × 1022 / 6.022 × 1023 moles of O2

If one mole of nitrogen molecules weights 28 g, calculate the choices mass of 1 molecule of nitrogen in grams.

Since 1 mol of nitrogen consists of 6.02 × 1023 molecules.

So in line with given information, 6.022 × 1023 molecules of N2 weigh = 28 gm

So, 1 molecule of N2 will weigh = 28 / 6.022 × 1023 grams of N2

= four.648 × 10-23 grams of N2

How many moles are there in 34.five g of sodium? (Atomic mass of Na = 23 u)

Molar mass of sodium(Na) = 23g

Given Mass of the sodium = 34.5g

As we recognise that, Number of moles= mass of an detail/molar mass of an detail

Hence Number of moles = 34.five/23 = 1.five moles.

What is the range of zinc atoms in a piece of zinc weighing 10 g?(Atomic mass of Zn = sixty five u)

Mole concept offers a dating among variety of debris(atoms) and their mass. Thus it is viable to calculate the choices wide variety of debris in a given mass.

Molar mass of Zinc(Zn) = 65g

Given Mass of the choices Zinc = 10g

As we realize that, Number of moles = mass of an element/molar mass of an element

Hence Number of moles= 10g/65g = zero.15 moles.

Since 1 mol of zinc consists of 6.02 × 1023 atoms

Therefore, zero.15 mol of zinc incorporates = 6.02 × 1023 × 0.15 mol = nine.03 × 1022 atoms of zinc

Calculate the choices mass of three.011 × 1024 atoms of carbon.

Mass of 6.022 × 1023 atoms of Carbon = 12 g

So, Mass of 1 Carbon atom = 12/6.022 × 1023 g

Hence, mass of three.011 × 1024 atoms of Carbon = three.011 × 1024 × 12 / 6.022 × 1023 = 60 g

If 6 g of oxygen includes 1 mole of oxygen atoms, calculate the mass of 1 atom of oxygen.

6.022 × 1023 atoms of Oxygen weigh = 16g

So, mass of one atom of Oxygen = sixteen/6.022 × 1023 = 2.656 × 10-23 g.

How many atoms are there in zero.25 mole of hydrogen?

1 mole of hydrogen contains = 6.022 × 1023 atoms of hydrogen

So, 0.25 moles of hydrogen will have = 6.022 × 1023 × 0.25 = 1.50 × 1023 atoms of hydrogen.

Calculate the variety of moles in 12.044 × 1025 atoms of phosphorus.

6.022 × 1023 atoms of phosphorus carries = 1 mole of phosphorus

So, 12.044 × 1025 atoms of phosphorus could have = 12.044 × 1025/6.022 × 1023 = two hundred moles

Calculate the wide variety of moles present in a drop of chloroform (CHCl3) weighing 0.0239 g. (Atomic hundreds: C = 12 u; H = 1 u; Cl = 35.five u)

Molar mass of chloroform (CHCl3) = (12+1+3×35.five) = 119.5g

Given Mass of the choices chloroform (CHCl3) = 0.0239g

As we realize that, Number of moles = mass of an element/molar mass of an detail

Hence Number of moles of chloroform (CHCl3) = 0.0239g/119.5g = 0.0002 moles.

What is the mass of 5 moles of sodium carbonate (Na2CO3)?(Atomic hundreds: Na = 23 u; C = 12 u; O = 16 u)

Molar mass of sodium carbonate (Na2CO3) = 2×23+12+three×sixteen = 106 g

Mass of one mole of sodium carbonate (Na2CO3) = 106g

Mass of 5 moles of sodium carbonate (Na2CO3) = five×106g = 530g.

Calculate the variety of molecules in 4 g of oxygen.

Mole concept affords a dating between number of debris and their mass. Thus it’s far possible to calculate the choices wide variety of particles in a given mass.

Molar mass of oxygen(O2) = 32g

As we recognise that, Number of moles = Given mass of an element/molar mass of an element

Hence, Number of moles of oxygen = 4g/32g = 0.12 moles

Since 1 mole of oxygen includes 6.022 × 1023 molecules.

Therefore, 0.12 mole of oxygen includes =

How many moles are represented via a hundred g of glucose, C6H12O6? (C = 12 u, H = 1 u, O = sixteen u)

Molar mass of glucose, C6H12O6 = 6×12+12×1+16×6 = 180g

Given Mass of the glucose, C6H12O6 = 100g

As we understand that, Number of moles = mass of an element/molar mass of an detail

Hence Number of moles glucose, C6H12O6 = 100g/180g = 0.55 moles.

Calculate the choices mass in grams of 0.17 mole of hydrogen sulphide, H2S.(Atomic masses: H = 1 u, S = 32 u)

For changing mole into mass in grams and vice-versa, we constantly need a courting among mass and mole.

Molar mass of hydrogen sulphide, H2S = 1×2+32 g /mol = 34g/mol

Therefore, range of grams of hydrogen sulphide, H2S in 0.17 mol of it = 0.17 mol of oxygen × 34 g /mol = 5.78 g of hydrogen sulphide, H2S.

Show by way of calculations that 5 moles of CO2 and five moles of H2O do not have the identical mass. How a whole lot is the difference of their loads?

One mole of CO2 weighs (12+2×16)g = 44g

5 moles of CO2 weighs 5×44g = 220g

One mole of H2O weighs (1×2+sixteen)g = 18g

5 moles of H2O weighs 5×18g = 90g

Hence from above calculation, it is able to be concluded that there is a difference of one hundred thirty g between five moles of CO2 and five moles of H2O.

Calculate the choices mole ratio of 240 g calcium and 240 g of magnesium. (Ca = 40 u; Mg = 24 u)

Molar mass of calcium, Ca = 40g

Given Mass of the calcium = 240g

As we recognize that, Number of moles = mass of an detail/molar mass of an element

Hence Number of moles of Calcium = 240g/40g = 6 moles.

Molar mass of magnesium, Mg = 24g

Given Mass of the choices magnesium, Mg = 240g

As we recognize that, Number of moles = mass of an detail/molar mass of an element

Hence Number of moles of magnesium = 240g/24g = 10 moles.

From above calculation, the choices molar ratio can be inferred – 6:10 = 3:5

Define mole. What are the 2 matters that a mole represents?

Mole is described as the amount of substance that incorporates as many specific essential debris (inclusive of atoms, molecules) as the variety of atoms in 12g of carbon-12 isotope. One mole is also described as the amount of substance which incorporates Avogadro variety (6.023 x 1023) of debris.

What weight of each detail is found in 1.five moles of sodium sulphite, Na2SO3?(Atomic hundreds: Na = 23 u; S = 32 u; O = 16 u)

1.5 moles of Na2SO3 has 1.five×2 = 3 moles of Na

1.five moles of Na2SO3 has 1.five×1 = 1.five mole of S

1.five moles of Na2SO3 has 1.5×three = four.5 moles of O.

As we understand that Mass of an element= quantity of moles x atomic mass

Thus, mass of sodium = three × 23 g = 69g

Mass of sulphur = 1.5 × 32 = 48g

Mass of oxygen = 4.5 × 16 g = 72g

What is meant by way of ‘a mole of carbon atoms’?

A mole of carbon atoms means a carbon pattern measuring 12 g and containing 6.022 x 1023 carbon atoms.

Which has extra atoms, 50 g of aluminium or 50 g of iron? Illustrate your answer with the help of calculations.(Atomic masses: Al = 27 u; Fe = fifty six u)

Atomic mass of Al = 27u

1 mole of aluminium weighing 27g carries N = 6.022 x 1023 atoms of Al

So, 50g of Al contains = 50 × N(6.022 × 1023)/27 atoms of Al = eleven × 1023 atoms of Al

Atomic mass of Fe = 56 u

1 mole of iron weighing 56g includes N = 6.022 × 1023 atoms of Fe

So, 50g of Fe carries = 50 × N(6.022 × 1023)/fifty six atoms of Fe = five × 1023 atoms of Fe

Thus, 50g of Al has greater no. of atoms as compared to 50g of Fe. This is because Aluminium has a smaller atomic mass and is lighter.

Define gram atomic mass of a substance. How tons is the gram atomic mass of oxygen?

The atomic mass of an element expressed in grams is called gram atomic mass. It is determined with the aid of taking the choices atomic weight for an detail on the periodic desk and expressing it in gram. Weight in grams is numerically same to the atomic weight of the choices element.

The gram atomic mass of oxygen is equal to sixteen.

How many moles of oxygen atoms are found in one mole of the subsequent compounds?(i) Al2O3 (ii) CO2 (iii) Cl2O7(iv) H2SO4 (v) Al2(SO4)three

Moles of oxygen atom are –

(i). Al2O3 : three moles

(iii). Cl2O7 : 7 mole

(iv). H2SO4 : 4 mole

(v). Al2(SO4)three : 12 mole

Define gram molecular mass of a substance. How a good deal is the gram molecular mass of oxygen?

Gram molecular mass of a substance is referred to as the amount of the substance whose mass expressed in grams is numerically identical to its molecular mass. For instance: The molecular mass of CO2 is forty four u, its gram molecular mass is 44g.

The gram molecular mass of oxygen(O2) = 2×16g = 32g

If sulphur exists as S8 molecules, calculate the variety of moles in 100 g of sulphur. (S = 32 u)

Given mass of sulphur, S8 = 100g

Atomic mass of sulphur, S8 = 32 × eight g = 256g

Number of moles = Given mass of the element / Atomic mass of the choices detail

Number of moles = 100/ 256 = 0.39 moles.

What is supposed through the choices ‘molar mass’ of a substance? State the choices unit in which molar mass is normally expressed.

Molar mass can be described as the choices mass of 1 mole of a substance (may be an element or a compound). For example, One mole of oxygen(O2) weighs 32.0 g. Thus, Molar mass of one mole molecules of oxygen = 32 g /mol

Calculate the molar hundreds of the subsequent materials. Write the choices effects with proper devices.(i) Ozone molecule, O3(ii) Ethanoic acid, CH3COOH

Molar mass of ozone (O3) = three × gram atomic mass of O = three × sixteen g = forty eight g/ mole

(ii) Molar mass of Ethanoic acid (CH3COOH) = 2×C + 4×H + 2×O = (24 + four + 32)u = 60 g/ mole

Which of the following pair of elements represents a mole ratio of 1 : 1? A. 10 g of calcium and 12 g of magnesiumB. 12 g of magnesium and six g of carbonC. 12 g of carbon and 20 g of calciumD. 20 g of sodium and 20 g of calcium

Molar mass of magnesium = 24g

Given mass of magnesium = 12g

Number of moles = given mass of the detail/ atomic weight of detail

Number of moles of magnesium = 12/24 = zero.five moles

Molar mass of carbon = 12g

Given mass of carbon = 6g

Number of moles = given mass of the choices detail/ atomic weight of detail

Number of moles of carbon = 6/12 = zero.5 moles

Hence, 12 g of magnesium and six g of carbon represents a mole ratio of 1 : 1

Which of the subsequent efficiently represents 360 g water?(i) 2 moles of H2O(ii) 20 moles 0f water(iii) 6.022 × 1023 molecules of water(iv) 1.2044 × 1025 molecules of water A. (i)B. (i) and (iv)C. (ii) and (iii)D. (ii) and (iv)

One mole of water = 18 g

20 moles of water = 360g; it helps option (ii)

If 32g of sulphur has x atoms, then the wide variety of atoms in 32g of oxygen might be: A. x/2B. 2xC. xD. 4x

Atomic weight of sulphur = 32g

One atom of sulphur weighs 32g

Atomic weight of oxygen is 16g

Hence, 2 atoms of oxygen weighs 32g

A scholar desires to have three.011 × 1023 atoms each of magnesium and carbon factors. For this cause, he’s going to need to weigh: A. 24 g of magnesium and 6 g of carbonB. 12 g of carbon and 24 g of magnesiumC. 20 g of magnesium and 10 g of carbonD. 12 g of magnesium and 6 g of carbon

6.022 × 1023 atoms of magnesium = one mole of magnesium

3.011 × 1023 atoms of magnesium = 0.5 moles of magnesium

Mass of magnesium = atomic weight x wide variety of moles

Mass of magnesium = 24 × 0.five = 12g

Similarly 6.022 × 1023 atoms of carbon = one mole of carbon

3.011 × 1023 atoms of carbon = zero.five moles of carbon

Mass of carbon = atomic weight × number of moles

Mass of carbon = 12 × zero.five = 6g

The ratio of moles of atoms in 12g of magnesium and 16g of sulphur can be: A. 3 : 4B. four : 3C. 1 : 1D. 1 : 2

Molar mass of magnesium = 24g

Given mass of magnesium = 12g

Number of moles = given mass of the detail/ atomic weight of detail

Number of moles of magnesium = 12/24 = zero.five moles

Molar mass of sulphur = 32g

Given mass of sulphur = 16g

Number of moles = given mass of the element/ atomic weight of detail

Number of moles of sulphur = 16/32 = 0.5 moles

Hence, 12 g of magnesium and 16 g of sulphur represents a mole ratio of 1 : 1

If 12 gram of carbon has x atoms, then the number of atoms in 12 grams of magnesium can be: A. xB. 2xC. x/2D. 1.5x

Atomic weight of carbon = 12g

One atom of carbon weighs 12g

Atomic weight of magnesium is 24 g

Hence, half of atom of magnesium weighs 12g

Which of the following has the choices most variety of atoms? A. 18 g of H2OB. 18 g of O2C. 18 g of CO2D. 18 g of CH4

18 g of CH4 has the choices most number of atoms.

If 1 gram of sulphur dioxide consists of x molecules, what number of molecules will be found in 1 gram of oxygen?(S = 32 u; O = sixteen u)

1 mole of SO2 = Mass of S + 2 × Mass of O = 64 grams

64 g of SO2 = 1 mole

So, 1 g of SO2 = 1/sixty four mole

Now considering same moles of all the substances contain identical wide variety of molecules so, 1/sixty four mole of O2 may even incorporate x molecules like SO2.

32 g of oxygen = 1 mole

So, 1 g of oxygen = 1/32 mole

Now, 1/64 mole of oxygen consists of = x molecules

So, 1/32 mole of oxygen will incorporate = x × sixty four/32 = 2x molecules

The mass of one molecule of a substance is 4.sixty five × 10-23 g. What is its molecular mass? What should this substance be?

Mass of 1 molecule of substance = four.sixty five × 10-23 u

So, mass of one mole of substance = Mass of 6.022 × 1023 molecules of the choices substance

The substance is Nitrogen with molecular mass 28 u.

Which contains extra molecules, 10 g of sulphur dioxide (SO2) or 10 g of oxygen (O2)?(Atomic masses: S = 32 u; O = 16 u)

Molar mass of SO2 = (32 + 2×sixteen) g = 64g

Given mass of SO2 = 10g

1 mole of substance = 6.023 × 1023 debris of the choices substance

No. of moles of SO2 = 10g/64g = 0.15

Total no. of molecules of SO2 = 0.15 × 6.022 × 1023 = zero.90 × 1023 molecules of SO2

Molar mass of oxygen (O2) = 32g

Given mass of oxygen (O2) = 10g

1 mole of substance = 6.023 × 1023 particles of the substance

No. of moles of O2 = 10g/32g = 0.31

Total no. of molecules of O2 = zero.31 × 6.022 × 1023 = 1.88 × 1023 molecules of O2

Thus, 10g of O2 carries more no. of molecules.

What weight of oxygen gas will incorporate the choices identical variety of molecules as 56 g of nitrogen gas? (O = sixteen u; N = 14 u)

Given mass of nitrogen = 56g

Molar mass of nitrogen = 14g

No. of moles of nitrogen = given mass of the element/ molar mass of the detail = 56g/14g = 4 moles

Equal wide variety of moles of all the susbtances contain identical variety of molecules.

So, four moles of nitrogen and 4 moles of oxygen includes identical no. of molecules.

Hence, mass of 4 mole of oxygen = four × 16g = 64 g

What mass of nitrogen, N2, will include the choices same wide variety of molecules as 1.8 g of water, H2O? (Atomic masses: N = 14 u; H = 1 u; O = sixteen u)

Given mass of water = 1.8g

Molar mass of water = 18g

No. of moles of water = given mass of the element/ molar mass of the choices detail = 1.8g/18g = 0.1 moles

Equal quantity of moles of all of the substances contain identical number of molecules.

So, 0.1 moles of water and 0.1 moles of nitrogen carries equal no. of molecules.

Hence, mass of 0.1 mole of nitrogen = 0.1 × 28 g = 2.8 g.

If one gram of sulphur contains x atoms, calculate the choices number of atoms in one gram of oxygen detail. (Atomic loads: S = 32 u; O = sixteen u)

32 g of S = 1 mole

So, 1 g of S = 1/32 mole

Now considering the fact that identical moles of all of the materials comprise identical quantity of atoms so, 1/32 mole of oxygen may also incorporate x atoms like S.

sixteen g of oxygen = 1 mole

So, 1 g of oxygen = 1/sixteen mole

Now, 1/32 mole of oxygen contains = x atoms

So, 1/sixteen mole of oxygen will contain = x × 32/16 = 2x atoms.

How many grams of magnesium will have the choices equal wide variety of atoms as 6 grams of carbon? (Mg = 24 u; C = 12 u)

Given mass of carbon = 6g

Molar mass of carbon = 12g

No. of moles of carbon = Given mass of detail/ Molar mass of detail = 6g/12g = zero.5 moles

Equal variety of moles of all the substances incorporate identical range of molecules.

So, 0.5 moles of carbon and zero.5 moles of magnesium incorporates same no. of molecules.

Hence, mass of 0.five mole of magnesium = zero.5 × 24g = 12g

The mass of one atom of an element X is 2.0 × 10-23 g.(i) Calculate the choices atomic mass of detail X.(ii) What may want to element X be?

(i) Mass of 1 atom of element X = 2 × 10-23 g

Mass of one mole of atom of detail X = 2 × 10-23 × 6.022 × 1023 = 12.044 g

So, atomic mass of the detail X = mass of 1 mole of detail = 12 u

(ii) Element X is CARBON.

Name the choices element used as a popular for atomic mass scale.

Carbon is the choices element used as a general for atomic mass scale.

Which particular atom of the choices above detail is used for this cause?

C-12 atom is used for this purpose.

What cost has been given to the choices mass of this reference atom?

12 atomic mass unit (amu) is the choices mass of this reference atom.

Give one important downside of Dalton’s atomic idea of be counted.

According to one of the principal postulate of Dalton’s atomic concept, atoms were being taken into consideration as indivisible. But according to the choices current theory, it’s miles now known that that atoms may be further divided into still smaller/subatomic debris referred to as electrons, protons and neutrons.

Dalton’s atomic theory says that atoms are indivisible. Is this announcement still valid? Give motives on your solution.

According to one of the foremost postulate of Dalton’s atomic principle, atoms had been being taken into consideration as indivisible. But consistent with the cutting-edge theory, it’s far now regarded that that atoms can be similarly divided into still smaller/subatomic debris known as electrons, protons and neutrons.

Is it feasible to look atoms nowadays? Explain your solution.

To see atoms of different factors a unique type of microscope called Scanning Tunnelling microscope [STM] is used. It is an electron microscope for imaging surfaces at the choices atomic stage. It produces three-D pictures of the choices sample.

What is meant by the choices symbol of an element? Explain with examples.

Symbol can be described as the abbreviation/Sign used for the choices name of an element. The image of an element is commonly both the first letter or the first letters or the choices first and the choices 0.33 letters of the call of the choices detail.

For examples- Oxygen has image O, Aluminum has symbol Al.

Give symbols which have been derived from the “Latin names” of the choices factors.

(a) Aluminum- Al and Barium- Ba

Give two symbols which have been derived from the “Latin names” of the choices elements.

Iron(ferrum)- Fe and Copper(cuprum)- Cu.

Give the names and symbols of 5 familiar substances that you assume are factors.

Hydrogen- H, Helium-He, Lithium- Li , Beryllium-Be, Boron-B

Stat the chemical symbols for the subsequent factors:Sodium, Potassium, Iron, Copper, Mercury, Silver

Sodium- Na; Potassium- K; Iron- Fe; copper- Cu; Mercury- Hg; Silver- Ag.

Name the choices factors represented via the subsequent symbols:Hg, Pb, Au, Ag, Sn

Hg- Mercury; Pb- Lead; Au- Gold; Ag- Silver; Sn-Tin

What is supposed by way of atomicity? Explain with examples.

The number of atoms which constitutes a molecule of an element or a compound is known as atomicity of the element. For instance -The atomicity of the noble gases is 1, that of hydrogen, nitrogen, oxygen and so forth. is 2 each and of ozone is 3. Thus, noble gases, hydrogen and ozone are respectively monoatomic, diatomic and triatomic molecules.

What is the atomicity of the subsequent?(a) Oxygen(b) Ozone(c) Neon(d) Sulphur(e) Phosphorus(f) Sodium

What is meant with the aid of a chemical formula? Write the formulae of 1 detail and one compound.

The chemical system of a compound describes the composition of the choices molecules of the compound in phrases of the choices symbols of factors and the range of atoms of every element present in one molecule of the compound. In the choices chemical formula of a compound, the choices elements present are denoted with the aid of their symbols and the choices variety of atoms of each detail are shown with the aid of writing their range as subscripts to the choices symbols of the respective factors.

Formula of a compound- Water-H20

Formula of an detail-Sodium- Na.

Write the choices formulae of the following compounds. Also call the choices elements present in them.(a) Water(b) Ammonia(c) Methane(d) Sulphur dioxide(e) Ethanol.

(a) Water- H2O; As conveyed by the choices system, elements gift are hydrogen and oxygen.

(b) Ammonia-NH3; As conveyed by the choices method, factors present are nitrogen and hydrogen.

(c) Methane-Ch4; As conveyed by the choices components, elements gift are carbon and hydrogen.

(d) Sulphur dioxide-SO2; As conveyed by way of the formula, elements present are sulphur and oxygen.

(e) Ethanol-C2H5OH; As conveyed by way of the formulation, elements gift are carbon, hydrogen and oxygen.

Explain the difference among 2N and N2.

Both 2N and N2 are the styles of nitrogen. 2N denotes separate atoms of nitrogen whereas N2 denotes one molecule of nitrogen gas.

What do the following abbreviation stand for?(a) O (b) 2O (c) O2 (d) 3O2

(a) O – one atom of oxygen

(b) 2O – two separate atoms of oxygen

(c) O2 – one molecule of oxygen fuel

(d) 3O2 – 3 molecules of oxygen

What do the choices symbols, H2, S and O4 imply within the system H2SO4?

The above compound is Sulphuric acid (H2SO4). According to chemical formula of H2SO4, H2 denotes two atoms of hydrogen, S denotes one atom of sulphur and O4 denotes 4 atoms of oxygen.

In what form does oxygen fuel occur in nature?

Atoms of the majority of factors (besides noble gases) are chemically very reactive and does no longer exist inside the free country (as a single atom). Oxygen gasoline is diatomic and consequently its molecular formulation is O2.

In what form do noble gases occur in nature?

The atoms of only a few factors are chemically unreactive which are referred to as noble gases (which include helium, neon, argon and many others.). They exist within the free country (as single atom).

What is the choices distinction among 2H and H2?

Both 2H and H2 are the choices forms of hydrogen. 2H denotes two separate atoms of hydrogen whereas H2 denotes one molecule of hydrogen gas.

What do the following denote?(i) N (ii) 2N (iii) N2 (iv) 2N2

All of the above options are the sorts of nitrogen.

(i) N denotes one atom of nitrogen.

(ii) 2N denotes two separate atoms of nitrogen

(iii) N2 denotes one molecule of nitrogen gas.

(iv) 2N2 denotes molecules of nitrogen fuel.

What is the significance of the system of a substance?

With the assist of molecular method of a compound, we will calculate its percent composition by way of mass. First, we calculate the molecular mass of the compound. From this we will find out mass of one mole of the choices compound, that is identical to its gram molecular mass.

What is the importance of the choices components H2O?

The molecular components of glucose is C6H12O6. Calculate its molecular mass. (Atomic loads: C = 12 u; H = 1 u; O = 16 u)

Molecular system of glucose is C6H12O6

Molecular mass of glucose = (6xC) + (12xH) + (6xO) = seventy two + 12 + 96 = 180u.

Calculate the molecular hundreds of the subsequent:(a) Hydrogen, H2(b) Oxygen, O2(c) Chlorine, Cl2(d) Ammonia, NH3(e) Carbon dioxide, CO2(Atomic masses: H = 1 u; O = sixteen u; Cl = 35.5 u; N = 14 u; C = 12 u)

(a) Molecular mass of Hydrogen (H2) = 2 x H = 2 x 1 u = 2 u

(b) Molecular mass of oxygen (O2) = 2 x O = 2 x 16 u = 32 u

(c) Molecular mass of chlorine (Cl2) = 2 x Cl = 2 x 35.5 = seventy one u

(d) Molecular mass of Ammonia (NH3) = 1 x N + 3 x H = 14 + three = 17 u

(e) Molecular mass of carbon dioxide (CO2) = 1 x C + 2 x O = 12 + 32 = 44 u

Calculate the choices molecular hundreds of the subsequent compounds:(a) Methane, CH4(b) Ethane, C2 H6(c) Ethene, C2 H4(d) Ethyne, C2 H2(Atomic masses: C = 12 u; H = 1 u)

(a). Molecular mass of methane (CH4) = 12 + four = 16 u

(b). Molecular mass of ethane (C2H6) = 2 x 12 + 6 x 1 = 30 u

(c). Molecular mass of ethane (C2H4) = 2 x 12 + four x 1 = 28 u

(d). Molecular mass of ethyne (C2H2) = 2 x 12 + 2 x 1 = 26 u

Calculate the molecular hundreds of the following compounds:(a) Methanol, CH3 OH(b) Ethanol, C2 H5 OH

(a) Molecular mass of Methanol (CH3OH) = 1 x C + 3xH + 1xO +1xH = (12+3+16+1)u = 32u

(b) Molecular mass of Ethanol (C2H5OH) = 2xC + 5xH + 1xO + 1xH = (24 + 5 + sixteen + 1) = 46u

Calculate the molecular mass of ethanoic acid, CH3 COOH.(Atomic loads: C = 12 u; H = 1 u; O = sixteen u)

Molecular mass of ethanoic acid (CH3COOH) = 1xC + 3xH + 1xC + 2xO + 1xH = 12+three+12+32+1 = 60u

Calculate the choices molecular mass of nitric acid, HNO3. (Atomic loads: H = 1 u; N = 14 u; O = sixteen u)

Molecular mass of Nitric acid (HNO3) = 1xH + 1xN + 3xO = (1 + 14 + forty eight) u = sixty three u

Calculate the choices molecular mass of chloroform (CHCl3). (Atomic loads: C = 12 u; H = 1 u; Cl = 35.five u)

Molecular mass of chloroform (CHCl3) = 1 x C + 1 x H + 3 x Cl = (12 + 1 + 106.5)u = 119.five u

Calculate the molecular mass of hydrogen bromide (HBr). (Atomic loads: H = 1 u; Br = 80 u)

Molecular mass of hydrogen bromide (HBr) = 1 x H + 1 x Br = (1 + 80) u = 81u

Calculate the choices molecular hundreds of the following compounds:(a) Hydrogen sulphide, H2S(b) Carbon disulphide, CS2

(a) Molecular mass of hydrogen sulphide (H2S) = 2xH + 1xS = (2+32)u = 34u

(b) Molecular mass of Carbon disulphide (CS2) = 1xC + 2xS = (12+2 x 32)u = 76 u

State the choices law of conservation of mass. Give one instance to illustrate this law.

Law of conservation of mass was given by means of Antoine Lavoisier in 1744. According to this law, “Matter can neither be created nor destroyed in a chemical response”. This law states that mass of an remoted device will remain consistent over time. It way that when mass is enclosed in a gadget and nothing is authorized in or out, its amount will in no way alternate. That is mass will be conserved, and hence that is known as Law of Conservation of Mass. This means total mass of products is usually identical to the entire mass of reactants.

In a chemical reaction, the choices substances that combine or react are referred to as reactants and the new substance/materials shaped are called product or merchandise. A chemical response can be represented in standard as follows:

Example: When calcium oxide is dissolved in water, then calcium hydroxide is fashioned. The reaction involved in this could be written as:

Calcium oxide + Water → Calcium hydroxide

In this reaction calcium oxide and water are reactants at the same time as calcium hydroxide is product.

In this reaction 74 g of calcium hydroxide is acquired when 56 g of calcium oxide reacts with 18 g of water, that’s proved by way of test.

Calcium oxide (56 g) + Water (18 g) → Calcium hydroxide (seventy four g)

Here the total mass of reactants, i.e. calcium oxide and water is same to seventy four g. And the choices mass of product, i.e. calcium hydroxide is also same to 74g. This proves that the whole mass of reactants is continually equal to the whole mass of product, which proves the choices Law of Conservation of Mass.

State the law of steady proportions. Give one example to demonstrate this regulation.

Law of constant proportions was given via Joseph Proust in 1799. According to this law, “a chemical compound continually consists of same factors in exact share with the aid of mass and it does now not depend on the choices source of compound.” Compounds are fashioned via the aggregate of or greater factors. In a compound the choices ratio of the atoms or element by mass stays usually equal no matter the choices supply of compound. This means a sure compound always shaped through the combination of atoms in equal ratio by means of mass. If the ratio of mass of constituent atoms can be altered the new compound is fashioned.

Example: Water is fashioned by the choices mixture of hydrogen and oxygen. The ratio of loads of hydrogen and oxygen is usually in 1:8 in water regardless of source of water. Whether you gather the choices water from a properly, river, pond or from anywhere the ratio in their constituent atoms by means of mass will constantly equal.

State the choices numerous postulates of Dalton’s atomic theory of matter.

Main postulates of Dalton’s atomic principle

1). Elements are fabricated from extremely small/ indivisible particles referred to as atoms.

2). Atoms of a given element are same in size, mass, and other homes.

three). Atoms of various factors differ size-wise, mass, and different properties.

4). Atoms cannot be subdivided, created, or destroyed.

five). Atoms of various factors combine in easy ratios to shape chemicals.

6). In chemical reactions, atoms are mixed, separated, or rearranged.

Which postulate of Dalton’s atomic concept can provide an explanation for the choices regulation of conservation of mass?

According to regulation of conservation of mass which describes that, “Matter can neither be created nor destroyed in a chemical response”. This law offers result to the choices postulate of Dalton’s atomic concept which states that, “Atoms can neither be created and nor destroyed by way of physical and chemical modifications.”

Which postulate of Dalton’s atomic theory can give an explanation for the law of steady proportions?

According to regulation of consistent proportions which describes that, “a chemical compound constantly consists of identical factors in particular proportion by means of mass and it does no longer depend on the source of compound.”

This law gives end result to the choices postulate of Dalton’s atomic theory which states that, “When two factors combine to form two or more than specific compounds then the choices exceptional loads of 1 detail B which integrate with fixed mass of the other detail endure a easy ratio to each other.”

What is the choices importance of the choices symbol of an detail? Explain with the choices assist of an instance.

Symbol may be defined as the abbreviation used for the choices call of an detail. The image of an element are normally both the first letter or the choices first letters of the name of element or the first and the 0.33 letters of the choices name of the choices element.

For example, the symbol of a few factors are the first letter of the choices name of that element consisting of Hydrogen- H, Oxygen- O.

Some symbols are derived from the choices first letters of the names of the choices element inclusive of Calcium- Ca, Aluminum- Al.

Some symbol derived from the choices first and the choices 0.33 letter of the names of the factors together with Arsenic-As

Symbol of many factors are taken from their English call, whilst symbol of many factors are taken from their Greek or Latin names together with the latin call of sodium is Natrium, hence the choices symbol is Na.

Explain the importance of the symbol H.

Significance of image H

● The image H refers to the detail hydrogen.

● The image H refers to one atom of hydrogen.

● The image H refers to at least one mole of hydrogen i.e., 6.022 × 1023 atoms.

● The symbol H refers to 1 gram atomic mass of hydrogen.

What is an atom? How do atoms normally exist?

The building blocks of all varieties of be counted is referred to as atoms. An atom is the smallest and indivisible part of an detail that may take take part in a chemical reaction. Atoms of majority of the choices factors are very reactive and do now not exist in the unfastened nation (as single atom). They exist in mixture with the atoms of the choices same elements or another detail.

What is a molecule? Explain with an example.

A molecule is the smallest part of an detail or compound that has unbiased lifestyles. A molecule include one or more than one atoms. The molecules of factors incorporate atoms of most effective one kind. For instance Oxygen gasoline; it incorporates atoms of oxygen. Hence it is a molecule.

What is the difference among the molecule of an element and the molecule of a compound? Give one example of each.

Molecules of factors: The molecules of an detail include two or extra similar atom chemically bonded together, for example oxygen fuel has 2 oxygen atoms combined together whereas ozone gasoline has 3 oxygen atoms combined together, so ozone exists within the form of O3. In comparison to it, molecules of compounds include two or extra extraordinary kinds of atoms chemically bonded collectively. For instance: the choices molecule sulphur dioxide (SO2) include one atom of sulphur chemically bonded with two atom of oxygen.

Define atomic mass unit. What is its image?

‘amu’ is the choices symbol of Atomic mass unit, however in recent times it’s miles denoted 1 through ‘u’.

Define atomic mass of an element.

Atomic mass of an element may be defined as the choices common relative mass of an atom of the element as compared with mass of an atom of carbon (C-12 isotope) taken as 12 amu

Carbon-12 is taken into consideration as unit to calculate atomic mass. Carbon-12 is an isotope of carbon. The relative mass of all atoms are discovered with recognize to C-12.

One atomic mass = 1/12 of the mass of 1 atom of C-12.

This means atomic mass unit = 1/twelfth of carbon-12.

What is meant with the aid of announcing that ‘the choices atomic mass of oxygen is sixteen’?

The atomic mass of oxygen is 16u, this indicates one atom of oxygen is sixteen times heavier than 1/12th of carbon atom.

The Latin language name of an element is natrium. The English call of this element is: A. sodiumB. potassiumC. magnesiumD. sulphur

Natrium is the latin call of Sodium.

The atomicities of ozone, sulphur, phosphorus and argon are respectively: A. eight, three, 4 and 1B. 1, three, four and 8C. four, 1, eight and 3-d. 3, 8, 4 and 1

The atomicity of ozone(O3) is three, sulphur(S)- 8, phosphorus(P)- four, Argon(noble gasoline, Ar)- 1.

The symbol of a metal element that is used in making thermometers is: A. AgB. HgC. MgD. Sg

Mercury is used within the making of thermometers. Its symbol is Hg.

The atomic theory of matter changed into proposed via: A. John KennedyB. LavoisierC. ProustD. John Dalton

John Dalton gave atomic theory of matter.

One of the subsequent factors has an atomicity of ‘one’. This element is: A. heliumB. hydrogenC. sulphurD. ozone

The atomicity of helium is 1. It is a noble fuel.

The English name of an element is potassium, its Latin call may be: A. plumbumB. cuprumC. kaliumD. natrium

the choices latin call of potassium is kalium.

The regulation of conservation of mass changed into given by using: A. DaltonB. ProustC. LavoisierD. Berzelius

The law of conservation of mass became given by means of Lavoisier.

The detail having atomicity ‘four’ is maximum probable to be: A. argonB. fluorineC. phosphorusD. francium

The atomicity of phosphorus is four.

If 1.4 g of calcium oxide is formed via the complete decomposition of calcium carbonate, then the quantity of calcium carbonate taken and the amount of carbon dioxide fashioned might be respectively: A. 2.2 g and 1.1 gB. 1.1 g and a couple of.5 gC. 2.five g and 1.1 gD. five.zero g and 1.1 g

CaCO3 gets decomposed into calcium oxide(CaO) and carbondioxide(CO2). According to law of conservation of mass

The regulation of regular proportions was given via: A. ProustB. LavoisierC. DaltonD. Berzelius

The law of steady proportions turned into given by way of proust.

The law of regular phosphorus, sulphur and krypton, the choices elements having the bottom and maximum atomicities are respectively: A. sulphur and kryptonB. krypton and ozoneC. phosphorus and sulphurD. krypton and sulphur

Krypton has the bottom atomicity and Sulphur has highest atomicity i.e. eight.

One nm is identical to: A. 10-9 mmB. 10-7 cmC. 10-9 cmD. 10-6 m

1nm = 10-nine m, whereas 1m = 100cm, hence 1nm = 10-7 cm.

The scientist who proposed the first letter (or first letter and some other letter) of the choices Latin or English name of an element as its image, was: A. DaltonB. ProustC. LavoisierD. Berzelius

Berzelius proposed the choices first letter (or first letter and another letter) of the choices Latin or English call of an detail as its symbol.

The atoms of which of the following pair of factors are maximum in all likelihood to exist in free kingdom? A. hydrogen and heliumB. argon and carbonC. neon and nitrogenD. helium and neon

helium and neon are noble gases.

Which of the subsequent elements has the identical molecular mass as its atomic mass? A. nitrogenB. neonC. oxygenD. chlorine

Neon has the same molecular mass as its atomic mass.

In water, the proportion of oxygen and hydrogen by way of mass is: A. 1 : 4B. 1 : 8C. four : 1D. 8 : 1

The percentage of oxygen and hydrogen by means of mass in water may be calculated as-

(H2O) = 1xO(sixteen): 2xH(1) = 16:2 = 8:1.

In hydrogen peroxide (H2O2), the share of hydrogen and oxygen via mass is: A. 1 : 8B. 1 : 16C. 8 : 1D. sixteen : 1

(b); The share of hydrogen and oxygen via mass in hydrogen peroxide can be calculated as-

(H2O2) = 2xH(1): 2xO(16) = 2:32 = 1:16.

The symbols of the factors cobalt, aluminium, helium and sodium respectively written with the aid of a pupil are as follows. Which symbol is the perfect one? A. COB. ALC. HeD. So

It is the best symbol of Helium; the appropriate symbols of cobalt, aluminium, and sodium are Co, Al and Na respectively.

Copper sulphate reacts with sodium hydroxide to form a blue precipitate of copper hydroxide and sodium sulphate. In an test, 15.95 g of copper sulphate reacted with 8.0 g of sodium hydroxide to shape 9.seventy five g of copper hydroxide and 14.2 g of sodium sulphate. Which law of chemical combination is illustrated by means of this data? Give purpose for your preference.

Here are the choices equations for copper sulphate answer reacting with sodium hydroxide solution:

copper sulfate + sodium hydroxide → copper hydroxide + sodium sulfate

CuSO4 + 2NaOH → Cu(OH)2 + Na2SO4

(15.95g + 8g → nine.75g + 14.2g)

As all the reactants and merchandise stay in the sealed response container then it is straightforward to expose that the full mass is unchanged. This statistics indicates the regulation of conservation of mass.

Potassium chlorate decomposes, on heating, to shape potassium chloride and oxygen. When 24.five g of potassium chlorate is decomposed completely, then 14.nine g of potassium chloride is shaped. Calculate the mass of oxygen shaped. Which regulation of chemical combination have you used in solving this hassle?

According to the above question-

According to the choices law of conservation of mass-

Mass of potassium chloride + mass of oxygen = Mass of potassium chlorate

Mass of oxygen(x) = Mass of potassium chlorate- mass of potassium chloride

Hence mass of oxygen = nine.6 g

In an test, 4.90 g of copper oxide turned into acquired shape 3.92 g of copper. In any other experiment, 4.fifty five g of copper oxide gave, on reduction, 3.sixty four g of copper. Show with the choices assist of calculations that these figures verify the law of steady proportions.

So, 1 equivalent of Cu reacts with zero.25 equivalent of O2 to shape 1.25 equivalent of copper oxide.

Here again, you can see that 1.25 equal of CuO decomposed to shape 1 equivalent of Cu and zero.25 equal of oxygen.

Hence, regulation of consistent proportion is established.

Magnesium and oxygen integrate in the ratio of 3 : 2 through mass to shape magnesium oxide. What mass of oxygen gasoline might be required to react completely with 24 h of magnesium?

i.e. 3 equivalents of Mg reacts with 2 equivalents of O2 to shape 1 equivalent of MgO.

When mass of Mg = 3x = 24 gm

Then, mass of oxygen required = 2x = 16 gm

When five g of calcium is burnt in 2 g of oxygen, then 7 g of calcium oxide is produced. What mass of calcium oxide will be produced while five g of calcium is burnt in 20 g of oxygen? Which law of chemical combination will govern your answer?

When five gm of calcium is burnt in 2gm of oxygen, then 7 gm of calcium oxide is fashioned. So, calcium and oxygen combine in the constant proportion of five:2 by mass. Now, whilst 5 gm of calcium is burnt in 20 gm of oxygen, then also 7 gm of calcium oxide can be shaped due to the fact chemical reactions follows law of constant proportion. As a result, 18 gm of oxygen may be left unreacted.

A liquid compound X of molecular mass 18 u can be acquired from some of herbal assets. All the choices animals and plants want liquid X for their survival. When an electric powered contemporary is surpassed via 200 grams of natural liquid X below appropriate conditions, then 178 grams of gas Y and 22 grams of fuel Z are produced. Gas Y is produced at the high-quality electrode whereas fuel Z is acquired at the bad electrode. Moreover, fuel Y supports combustion while gas Z burns itself causing explosions.(a) Name (i) liquid X (ii) gasoline Y, and (iii) gasoline Z(b) What is the ratio to the choices mass of detail Z to the mass of detail Y in the liquid X?(c) Which law of chemical aggregate is illustrated through this example?(d) Name two sources of liquid X.(e) State an essential use of Y in our lifestyles.

(a) (i) Water (ii) Oxygen (iii) Hydrogen

(c) Law of Constant proportions

(d) Rivers and Wells

(e) Gas Y (oxygen) is vital For breathing

One of the choices sorts of a certainly going on solid compound P is usually used for making the flooring of homes. On including some drops of dilute hydrochloric acid to P, brisk effervescence are produced. When 50 g of reactant P become heated strongly, than 22 g of a fuel Q and 28 g of a solid R had been produced as merchandise. Gas Q is the choices equal which produced brisk effervescence on adding dilute HCl to P. Gas Q is stated to motive global warming while solid R is used for white-washing.(a) What is (i) strong P (ii) gas Q, and (iii) strong R.(b) What is the entire mass of Q and R acquired from 50 g of P?(c) How does the total mass of Q and R formed compare with the choices mass of P taken?(d) What end do you get from the choices contrast of masses of merchandise and reactant?(e) Which regulation of chemical mixture is illustrated by using the instance given in this hassle?

(a) Solid P – Calcium Carbonate (CaCO3)

Gas Q – Carbon dioxide (CO2)

Solid R – Calcium oxide (CaO)

(b) Total mass of Q and R = 22gm + 28gm = 50gm

(c) Total mass of Q and R (50gm) is equal to mass of reactant (50gm).

(d) The regulation of conservation of mass is observed, i.e. general mass of product is identical to mass of reactant.

(e) Law of conservation of mass is illustrated by way of the instance.

What can we call those particles which have extra or much less electrons than the choices everyday atoms?

Those particles which have more or less electrons than the choices regular atoms are called Ions.

What do we name those particles that have:(a) greater electrons than the choices everyday atoms?(b) less electrons than the normal atoms?

(a) Those particles which have extra electrons than the regular atoms are known as Anions.

(b)Those particles that have less electrons than the regular atoms are called Cations.

Define ‘system mass’ of a compound.

Formula mass of an ionic compound is acquired by way of adding atomic hundreds of all of the atoms in a system unit of the compound. For example: Formula mass of sodium chloride (NaCl) = Atomic mass of sodium + atomic mass of chlorine = 39 + 35.five = seventy four.5

What do we call the ones particles which are fashioned:(a) through the benefit of electrons by way of atoms?(b) by means of the choices loss of electrons via atoms?

(a) Anions are shaped via the advantage of electrons via atoms

(b) Cations are formed by means of the lack of electrons via atoms.

State whether or not the following statements are true or false:(a) A sodium ion has high quality price as it has more protons than a impartial atom(b) A chloride ion has bad rate as it has extra electrons than a impartial atom

(a) A sodium ion has high quality price as it has greater protons than a impartial atom

(b) A chloride ion has negative price as it has extra electrons than a impartial atom

Write down the formulae for the following compounds:(a) Calcium oxide(b) Magnesium hydroxide

(a) The method of Calcium oxide is CaO

(b) The components of Magnesium hydroxide is Mg(OH)2.

An detail Z has a valency of 3. What is the components of oxide of Z?

Valency of detail Z = three

So, formula of oxide of element = Z2O3

What is the call of a particle which includes 10 electrons, 11 protons and 12 neutrons?

The valence electrons found in sodium atom are 11. In the above announcement, there is loss of one electron proven which makes it Sodium ion(Cation) denoted by Na+.

Name the choices particle which has 18 electrons, 18 neutrons and 17 protons in it.

The valence electrons found in chlorine atom are 17. In the above announcement, there may be gain of 1 electron shown which makes it chloride ion(anion) denoted with the aid of Cl-.

Fill inside the following blanks with suitable words:(a) The particle that’s formed by means of the choices advantage of electrons by an atom is known as ………….(b) The particle that’s shaped with the aid of the lack of electrons by way of an atom is called ………….(c) The particle that’s shaped by the loss or benefit of electrons by using an atom is called ……….(d) A potassium ion has superb alternate because it contains much less …………. than ………….(e) A sulphide ion has terrible charge as it carries less …………… than ……………

(a) Anion is shaped through the choices attractiveness/advantage of electrons through an atom.

(b) Cation is formed by donating/lack of electrons by an atom.

Name the elements water is made of. What are the choices valencies of those elements? Work out the choices chemical formulation for water.

Water(H2O) is made up of varieties of factors known as hydrogen and oxygen.

The valency of hydrogen is 1 as simplest one valence electron is found in its outermost shell and one is wanted. The valency of oxygen is two as six valence electrons are found in its outermost shell and are wished.

So consistent with the valencies of both the factors present in water, the choices formulation will become H2O.

If the choices valency of hydrogen is 1 and that of nitrogen is 3, work out the choices components for ammonia.

The valency of hydrogen is 1 and that of nitrogen is 3. So in line with the choices valencies of each the elements found in ammonia, The system turns into NH3.

Work out the choices formula for sulphur dioxide. (Valencies: S = four; O = 2)

The valency of sulphur is four and that of oxygen is two. So in step with the choices valencies of both the choices factors present in sulphur dioxide, The components becomes SO4.

If the choices valency of carbon is 4 and that of sulphur is 2, exercise session the choices components of the choices compound formed by the choices aggregate of carbon with sulphur. What is the choices name of this compound?

The valency of carbon is four and that of sulphur is two. So in keeping with the choices valencies of each the factors gift, the choices components will become CS2. The name of compound is carbon disulphide.

An detail X has a valency of four whereas another detail Y has a valency of one. What might be the method of the choices compound fashioned between X and Y?

The valency of X is 4 and that of Y is 1. So consistent with the choices valencies of each the elements gift, the choices formulation becomes XY4. The name of compound is carbon disulphide.

An element B indicates valencies of 4 and 6. Write the choices formulae of its oxides.

When detail B indicates valency of 4 and oxygen has valency of two. The formulation will become BO2.

When detail B suggests valency of 6 and oxygen has valency of 2. The formulation will become BO3.

An detail X of valency 3 combines with some other detail Y of valency 2. What can be the method of the choices compound shaped?

The method of the choices compound formed is X2Y3.

Work out the choices system for magnesium hydrogencarbonate.

The valency of magnesium is two and that of hydrogen carbonate(HCO3) is 1. So according to the choices valencies of both the choices factors present in magnesium hydrogencarbonate, the method becomes Mg(HCO3)2.

An detail X has a valency of 2. Write the choices simplest system for:(a) bromide of the choices detail(b) oxide of the detail

(a) The valency of detail X is 2 and that of bromine is -1. So according to the choices valencies of each the elements, the system turns into XBr2.

(b) The valency of detail X is two and that of oxide is -2. So in step with the valencies of both the choices factors, the choices components will become XO.

Work out the choices formulae for the following compounds:(a) Sodium oxide(b) Calcium carbonate

(a) The valency of sodium is 1 and that of oxide is -2. So in line with the choices valencies of each the factors, the choices formula becomes Na2O.

(b) The valency of calcium is 2 and that of carbonate is -2. So consistent with the choices valencies of each the factors, the formulation will become CaCO3.

Calculate the choices formulae loads of the subsequent compounds:(i) Sodium oxide, Na2O(ii) Aluminium oxide, Al2O3(Given: Atomic loads: Na = 23 u; O = 16 u; Al = 27 u)

(a) Molecular mass of Na2O = (2 x Na) + (1 x O) = (2 x 23) + (1 x 16) = 62u

(b) Molecular Mass of Al2O3 = (2 x Al) + (three x O) = (2 x 27) + (3 x sixteen) = 102u

Name the subsequent compounds. Also write the choices symbols/formulae of the choices ions present in them:(a) CuSO4 (b) (NH4)2SO4(c) Na2O (d) Na2CO3(e) CaCl2

(a) Copper sulphate; Cu2+ and

(c) Sodium oxide; Na+ and O2-

(d) Sodium carbonate; Na+ and

(e) calcium chloride; Ca2+ and Cl-

Write the cations and anions gift, if any, inside the following:(a) CH3COONa (b) NaCl(c) H2 (d) NH4NO3

(a) Cation: Na+; Anion: CH3COO-

(b) Cation: Na+; Anion Cl-

(c) H2 is a covalent molecule. It has no cation and anion

Give the formulate of the choices compounds shaped from the following sets of factors:(a) calcium and fluorine(b) hydrogen and sulphur(c) nitrogen and hydrogen(d) carbon and chlorine(e) sodium and oxygen(f) carbon and oxygen

(a) The valency of calcium is 2 and that of fluorine is -1. So in line with the choices valencies of both the factors, the components will become CaF2.

(b) The valency of hydrogen is 1 and that of sulphur is -2. So consistent with the choices valencies of each the factors, the choices components will become H2S.

(c) The valency of nitrogen is -three and that of hydrogen is 1. So in keeping with the choices valencies of each the choices factors, the method will become NH3.

(d) The valency of carbon is 4 and that of chlorine is -1. So consistent with the choices valencies of each the factors, the components turns into CCl4.

(e) The valency of sodium is 1 and that of oxygen is -2. So according to the choices valencies of both the choices elements, the choices components will become Na2O.

(f) The valency of carbon is 4 and that of oxygen is -2. So in step with the choices valencies of each the elements, the choices formulation will become CO2.

What are (i) ionic compounds, and(ii) molecular compounds? Give examples of every sort of compounds.

(i) Ionic compounds- The compounds which might be formed by means of the choices aggregate of metals and non-metals are known as ionic compounds. For instance- NaCl, CaCl2.

(i) Molecular Compounds- These compounds are formed by means of the choices combination of non-metallic factors. For ex. HCl and H2SO4.

What is an ion? How is an ion formed? Explain with the choices help of examples of various ions.

An ion is a charged particle and may be negatively or undoubtedly charged.

A negatively charged ion is referred to as an „anion‟ and the undoubtedly charged ion, a „cation‟. For example, sodium chloride (NaCl). Its constituent particles are definitely charged sodium ions (Na+) and negatively charged chloride ions (Cl–). Ions may also include a unmarried charged atom or a set of atoms that have a net charge on them. A organization of atoms sporting a price is called a polyatomic ion e.g. Calcium oxide (Ca+2 O-2)

The valencies (or costs) of some of the ions are given beneath:Using this records, write down the choices formulate of the following compounds:(i) Sodium phosphate(ii) Ammonium sulphate(iii) Calcium hydroxide(iv) Lead bromide

i. Sodium phosphate – Na3PO4

ii. Ammonium sulphate – (NH4)2SO4

iii. Calcium Hydroxide – Ca(OH)2

iv. Lead bromide – PbBr2

What is the distinction among a cation and an anion? Explain with examples.

A definitely charged ion is known as cation. For example : Sodium ion:Na , Magnesium ion: Mg2 A cation is shaped by way of the choices loss of one or extra electrons by means of an atom For example : sodium atom, loses one electron to shape a sodium ion Na Sodium atom (A cation)

A negatively charged ion is referred to as anion. Cl- (chloride ion), O-2 (oxide ion) and many others. An anions is fashioned through the choices gain of one or more electrons with the aid of an atom. For instance a chlorine atom profits one electron to shape a chloride ion Cl-. Chlorine atom Chloride ion (An anion).

The valencies (or fees) of a number of the choices ions are given under:Using this statistics, write down the formulae of:(i) Sodium sulphide(ii) Copper nitrate

Explain the choices formation of (i) sodium ion, and (ii) chloride ion, form their respective atoms giving the choices range of protons and range of electrons in every one in all them. What is the reason for high quality fee on a sodium ion and a poor charge on a chloride ion?

Formation of Sodium ion- The atomic variety of sodium is eleven. The preparations of electrons within the shells of the sodium atom is within the combination of two,8,1. The valence electron present within the outermost shell of sodium atom is 1. The sodium atom can lose one electron to shape sodium ion(Na+). There may be discount in number of electrons, E=10. Number of protons continue to be same as 11. Due to the choices lower in one electron, there is fine price on a sodium ion.

Formation of Chloride ion- The atomic wide variety of chlorine is 17. The arrangements of electrons within the shells of the chlorine atom is within the aggregate of two,eight,7. The valence electron gift in the outermost shell of chlorine atom is 1. The chlorine atom can lose one electron to form chlorine ion(Cl-). There might be growth in wide variety of electrons, E=18. Number of protons remain same as 17. Due to the boom in a single electron, there is poor rate on a chloride ion.

Write the symbols/formulae of two easy ions and compound ions (or polyatomic ions).

Symbol of simple ions− Na+, Al3+

Symbol of two compound ions- NH4+, NO32-

An element Y has a valency of 4. Write the choices method for its:(i) chloride(ii) oxide(iii) sulphate(iv) carbonate(v) nitrate

(i) The valency of chloride ion is -1. Hence the choices formulation turns into YCl4

(ii)The valency of oxide ion is -2. Hence the formulation turns into YO2

(iii) The valency of sulphate ion is -4. Hence the choices components will become Y(SO4)2

(iv) The valency of carbonate ion is -4. Hence the choices components becomes Y(CO3)2

(v) The valency of nitrate ion is -1. Hence the formulation will become Y(NO3)four

Define ‘system unit’ of an ionic compound. What is the method unit of (i) sodium chloride, and (ii) magnesium chloride?

Formula unit is defined as the choices empirical method of a compound which gives the choices best entire quantity ratio of atoms of the various factors present within the molecule of the choices compound.

Formula Unit of Sodium Chloride- NaCl

Formula Unit of Magnesium Chloride- MgCl2.

Calculate the components loads of the following compounds:(i) Calcium chloride(ii) Sodium carbonate(Given: Atomic masses: Ca = 40 u; Cl = 35. five u; Na = 23 u; C = 12 u; O = 16 u)

(i) Formula Mass of Calcium chloride (CaCl2) = 1xCa + 2xCl = (forty+71) u = 111 u

(ii) Formula Mass of Sodium carbonate (Na2CO3) = 2xNa + 1xC + 3xO = (2×23 + 1×12 + 3×16) u = 106 u

The atomic wide variety of an element X is 13. What might be the variety of electrons in its ion X3+? A. 11B. 15B. 16D. 10

X3+ indicates loss of 3 electrons. Hence the ultimate no. of electrons gift are 10.

Which of the following represents a accurate chemical method? A. CaClB. Na3NC. NaSO4D. NaS

according to the choices valencies of sodium(1) and nitrogen(3).

If the choices number of electrons in an ion Z3- is 10, the choices atomic range of element Z can be: A. 7B. 5C. 10D. 8

as the image suggests the gain of 3 electrons, the choices atomic number of Z is in all likelihood to be 7.

The anion of an detail has: A. extra electrons than the everyday atomB. less electrons than the choices regular atomC. greater protons than the choices regular atomD. identical wide variety of electrons as regular atom

An anion is shaped via the choices gain of electrons.

A particle X has 17 protons, 18 neutrons and 18 electrons. This particle is maximum possibly to be: A. a cationB. an anionC. a moleculeD. a compound

An anion is formed by way of the choices advantage of electrons.

An detail that can show off valencies of 2, 4 and 6 can be: A. CopperB. ironC. mercuryD. sulphur

Sulphur has 2 valence electrons in its outermost shells. Hence it famous valencies of 2, 4 and 6.

The atomic variety of an element E is 16. The number of electrons in ion E2- may be: A. 16B. 18C. 15D. 14

as the symbol suggests the choices benefit of two electrons, the choices atomic number is likely to be 18.

The cation of an element has: A. the same quantity of electrons as its neutral atomB. extra electrons than a impartial atomC. much less protons than a neutral atomD. much less electrons than a neutral atom

A cation is formed by way of the choices lack of electrons.

Two elements X and Y have valencies of 5 and 3, and 3 and a couple of, respectively. The factors X and Y are most probably to be respectively: A. copper and sulphurB. sulphur and ironC. phosphorus and nitrogenD. nitrogen and iron

Nitrogen have five electrons in the outermost shell of its atom, consequently its valencies are five and 3. Iron can lose 2 and three electrons from its outermost shell, so its valencies are 2 and three.

The wide variety of electrons in an ion Y2+ is 10. The atomic variety of detail Y is maximum likely to be: A. 8B. 12C. 10D. 14

as the symbol shows the choices loss of electrons, the choices atomic variety is possibly to be 12.

A particle P has 18 electrons, 20 neutrons and 19 protons. This particle ought to be: A. a moleculeB. a binary compoundC. an anionD. a cation

The lack of an electron in the above facts proves that it’s far a cation.

An ionic compound will be formed by means of the combination of one of the following pairs of factors. This pair of elements is: A. chlorine and calciumB. calcium and sodiumC. sulphur and carbonD. chlorine and chlorine

An ionic compound will always formed through the choices combination of an anion and a cation.

Molecular compounds are commonly fashioned via the choices aggregate among: A. a metallic and a non-metalB. two special non-metalsC. two distinct metalsD. any two gaseous factors

Molecular compounds are commonly fashioned by means of the aggregate among two special non-metals

The formulation of a compound is X3Y. The valencies of factors X and Y could be respectively: A. 1 and 3B. 3 and 1C. 2 and 3D. 3 and a pair of

As the choices system suggests, the valencies of factors X and Y may be 1 and 3 respectively.

The method of the choices sulphate of an element X is X2(SO4)three. The system of nitride of detail X might be: A. X2NB. XN2C. XND. X2N3

Nitrogen is trivalent as valency of nitrogen is -3, as a result components of nitride = XN