Free Printable Atoms And Molecules _LS_ 2

1.

Raunak took 5 moles of carbon atoms in a container and Krish also took 5 moles of sodium atoms in another container of the same weight. (a) Whose container is heavier? what is the first given statement?

a)

The weights of the containers are equal.

b)

5 moles of carbon

c)

5 moles of sodium atoms

d)

5 moles x 6.02214076 × 10²³

^{Which is EQUAL TO 5 MOLES X}Avogadro's number

2.

Raunak took 5 moles of carbon atoms in a container and Krish also took 5 moles of sodium atoms in another container of the same weight. (a) Whose container is heavier? what does Raunak have in his container?

a)

Krish's container is heavier than Ronak's

b)

5 moles of carbon

c)

5 moles of sodium atoms

d)

5 moles x 6.02214076 × 10²³

^{Which is EQUAL TO 5 MOLES X}Avogadro's number

3.

Raunak took 5 moles of carbon atoms in a container and Krish also took 5 moles of sodium atoms in another container of the same weight. (a) Whose container is heavier? what does Krish have in his container?

a)

Krish's container is heavier than Ronak's

b)

The molar mass is the mass of a given chemical element or chemical compound (g) divided by the amount of substance (mol). n=m/M n = no of moles. m = given mass. M = molar mass.

c)

5 moles of sodium atoms

d)

5 moles x 6.02214076 × 10²³

^{Which is EQUAL TO 5 MOLES X}Avogadro's number

4.

Raunak took 5 moles of carbon atoms in a container and Krish also took 5 moles of sodium atoms in another container of the same weight. (a) Whose container is heavier? The number of atoms in Raunak's container?

a)

Krish's container is heavier than Ronak's

b)

The molar mass is the mass of a given chemical element or chemical compound (g) divided by the amount of substance (mol). n=m/M n = no of moles. m = given mass. M = molar mass.

c)

Mass of Raunak's container: 1mol.of C=12g. 5mol.of C= (12 x 5) =60g.

d)

5 moles x 6.02214076 × 10²³

^{Which is EQUAL TO 5 MOLES X}Avogadro's number

5.

Raunak took 5 moles of carbon atoms in a container and Krish also took 5 moles of sodium atoms in another container of the same weight. (a) Whose container is heavier? The number of atoms in Krish's container?

a)

Mass of Krish's container: 1 mol.of Na=23g. 5 mol of Na= (23 x 5) =115g

b)

The molar mass is the mass of a given chemical element or chemical compound (g) divided by the amount of substance (mol). n=m/M n = no of moles. m = given mass. M = molar mass.

c)

Mass of Raunak's container: 1mol.of C=12g. 5mol.of C= (12 x 5) =60g.

d)

5 moles x 6.02214076 × 10²³

^{Which is EQUAL TO 5 MOLES X}Avogadro's number

6.

Raunak took 5 moles of carbon atoms in a container and Krish also took 5 moles of sodium atoms in another container of the same weight. (a) Whose container is heavier? What is the formula for calculating molar mass?

a)

Mass of Krish's container: 1 mol.of Na=23g. 5 mol of Na= (23 x 5) =115g

b)

The molar mass is the mass of a given chemical element or chemical compound (g) divided by the amount of substance (mol). n=m/M n = no of moles. m = given mass. M = molar mass.

c)

Mass of Raunak's container: 1mol.of C=12g. 5mol.of C= (12 x 5) =60g.

d)

5 moles x 6.02214076 × 10²³

^{Which is EQUAL TO 5 MOLES X}Avogadro's number

7.

Raunak took 5 moles of carbon atoms in a container and Krish also took 5 moles of sodium atoms in another container of the same weight. (a) Whose container is heavier? What is the Mass of container containing 5 moles of C atoms?

a)

Mass of Krish's container: 1 mol.of Na=23g. 5 mol of Na= (23 x 5) =115g

b)

The molar mass is the mass of a given chemical element or chemical compound (g) divided by the amount of substance (mol). n=m/M n = no of moles. m = given mass. M = molar mass.

c)

Mass of Raunak's container: 1mol.of C=12g. 5mol.of C= (12 x 5) =60g.

d)

5 moles x 6.02214076 × 10²³

^{Which is EQUAL TO 5 MOLES X}Avogadro's number

8.

Raunak took 5 moles of carbon atoms in a container and Krish also took 5 moles of sodium atoms in another container of the same weight. (a) Whose container is heavier? Mass of container containing 5 moles of Na atoms?

a)

Mass of Krish's container: 1 mol.of Na=23g. 5 mol of Na= (23 x 5) =115g

b)

The molar mass is the mass of a given chemical element or chemical compound (g) divided by the amount of substance (mol). n=m/M n = no of moles. m = given mass. M = molar mass.

c)

Hence, Krish's container containing sodium atoms is heavier..

d)

5 moles x 6.02214076 × 10²³

^{Which is EQUAL TO 5 MOLES X}Avogadro's number

9.

Raunak took 5 moles of carbon atoms in a container and Krish also took 5 moles of sodium atoms in another container of the same weight. (a) Whose container is heavier? Whose container is heavier?

a)

Mass of Krish's container: 1 mol.of Na=23g. 5 mol of Na= (23 x 5) =115g

b)

The molar mass is the mass of a given chemical element or chemical compound (g) divided by the amount of substance (mol). n=m/M n = no of moles. m = given mass. M = molar mass.

c)

Hence, Krish's container containing sodium atoms is heavier..

d)

5 moles x 6.02214076 × 10²³

^{Which is EQUAL TO 5 MOLES X}Avogadro's number

10.

Raunak took 5 moles of carbon atoms in a container and Krish also took 5 moles of sodium atoms in another container of the same weight. (a) Whose container is heavier? Whose container has more number of atoms?

a)

Mass of Krish's container: 1 mol.of Na=23g. 5 mol of Na= (23 x 5) =115g

b)

The molar mass is the mass of a given chemical element or chemical compound (g) divided by the amount of substance (mol). n=m/M n = no of moles. m = given mass. M = molar mass.

c)

Hence, Krish's container containing sodium atoms is heavier..

d)

Both containers of C and Na same no. of atoms which is 5 moles x 6.02214076 × 10²³

^{Which is EQUAL TO 5 MOLES X}Avogadro's number

11.

NaClrepresents

a)

Sodium bicarbonate

b)

Sodium carbonate

c)

Sodium Chloride

d)

Calcium Carbonate

12.

NaOHrepresents

a)

Sodium bicarbonate

b)

Sodium carbonate

c)

Sodium Chloride

d)

Sodium Hydroxide

13.

HCl represents

a)

Sulphuric Acid

b)

Hydrochloric Acid

c)

Nitric acid

d)

Chlorine gas

14.

MgO represents

a)

Magnesium Hydroxide

b)

Magnesium Chloride

c)

Magnesium Oxide

d)

Sodium Nitrate

15.

Chemical Formula used for Aluminium Chloride is

a)

AlCl₂

b)

Al₂Cl

c)

AlCl₃

d)

AlCl

16.

NaCl is

a)

Base

b)

Acid

c)

salt

17.

NaOH is

a)

Base

b)

Acid

c)

salt

18.

Chemical formula of copper sulphate

a)

CuSO₄

b)

Cu₂SO₄

c)

CuSO₃

d)

none of these

19.

CaCO₃represents

a)

Sodium bicarbonate

b)

Sodium carbonate

c)

Sodium Chloride

d)

Calcium Carbonate

20.

H₂SO₄ represents

a)

Sulphuric Acid

b)

Hydrochloric Acid

c)

Nitric acid

d)

Sulphurus Acid

21.

What is the total number of oxygen atoms in Ca₃(PO₄)₂?

a)

2

b)

4

c)

6

d)

8

e)

40

22.

Na₂CO₃represents

a)

Sodium bicarbonate

b)

Sodium carbonate

c)

Sodium Chloride

d)

Calcium Carbonate

23.

NaHCO₃represents

a)

Sodium bicarbonate

b)

Sodium carbonate

c)

Sodium Chloride

d)

Calcium Carbonate

24.

NaNO₃ represents

a)

Sodium Hydroxide

b)

Sodium Chloride

c)

Sodium Nitrite

d)

Sodium Nitrate

25.

Chemical Formula used for Aluminium oxide is

a)

AlO

b)

Al₂O₃

c)

AlO₄

d)

AlO₂

26.

Na₂Co₃represents

a)

Sodium bicarbonate

b)

Sodium carbonate

c)

Sodium Chloride

d)

Calcium Carbonate

27.

NaHCo₃represents

a)

Sodium bicarbonate

b)

Sodium carbonate

c)

Sodium Chloride

d)

Calcium Carbonate

28.

NaClrepresents

a)

Sodium bicarbonate

b)

Sodium carbonate

c)

Sodium Chloride

d)

Calcium Carbonate

29.

NaOHrepresents

a)

Sodium bicarbonate

b)

Sodium carbonate

c)

Sodium Chloride

d)

Sodium Hydroxide

30.

CaCo₃represents

a)

Sodium bicarbonate

b)

Sodium carbonate

c)

Sodium Chloride

d)

Calcium Carbonate

31.

CaCl₂represents

a)

Calcium bicarbonate

b)

Calcium Chloide

c)

Calcium Hydroxide

d)

Calcium Carbonate

32.

HCl represents

a)

Sulphuric Acid

b)

Hydrochloric Acid

c)

Nitric acid

d)

Chlorine gas

33.

H₂So₄ represents

a)

Sulphuric Acid

b)

Hydrochloric Acid

c)

Nitric acid

d)

Sulphurus Acid

34.

HNo₃ represents

a)

Sulphuric Acid

b)

Hydrochloric Acid

c)

Nitric acid

d)

Nitrous Acid

35.

H₃Po4 represents

a)

Sulphuric Acid

b)

Hydrochloric Acid

c)

Nitric acid

d)

Phosphoric acid

36.

NaNo₃ represents

a)

Sodium Hydroxide

b)

Sodium Chloride

c)

Sodium Nitrite

d)

Sodium Nitrate

37.

MgO represents

a)

Magnesium Hydroxide

b)

Magnesium Chloride

c)

Magnesium Oxide

d)

Sodium Nitrate

38.

Chemical Formula used for Ferrous oxide is

a)

FeO

b)

Fe₂O₃

c)

FeSo₄

39.

Chemical Formula used for Ferric oxide is

a)

FeO

b)

Fe₂O₃

c)

FeSo₄

40.

Chemical Formula used for Aluminium oxide is

a)

AlO

b)

Al₂O₃

c)

AlO₄

d)

AlO2

41.

Chemical Formula used for Aluminium Chloride is

a)

AlCl₂

b)

Al₂Cl

c)

AlCl₃

d)

AlCl

42.

Chemical Formula used for Aluminium Sulphate is

a)

AlSo₂

b)

AlSo₄

c)

Al(So₄₎₃

d)

Al₂(So₄₎₃

43.

Chemical Formula used for Zinc Nitrate is

a)

ZnNo₃

b)

Zn(No₃)₂

c)

ZnNo

d)

ZnN

44.

NaCl is

a)

Base

b)

Acid

c)

salt

45.

NaOH is

a)

Base

b)

Acid

c)

salt

46.

What is the symbol of iron?

a)

Fe

b)

FE

c)

I

d)

Ir

47.

The symbol for Potassium is

a)

P

b)

K

c)

Po

d)

Ka

48.

(a) is the symbol for Gold.

49.

Silver derived its symbol from the Latin word (a) .

50.

(a) is used to conduct heat in a thermometer.

51.

What is the symbol for the element which is considered as fool's gold?

a)

Co

b)

Ca

c)

C

d)

Cu

52.

What is the lightest element?

a)

Helium

b)

Neon

c)

Oxygen

d)

Hydrogen

53.

What is the symbol of the element which is considered as the wonder metal?

a)

Cu

b)

Ni

c)

Fe

d)

Al

54.

The complete list of the names of the elements together with their symbols and other characteristics is found in the (a) .

55.

Which among the following is not a metallic element?

a)

gold

b)

silver

c)

mercury

d)

carbon

56.

Who was the first scientist to give the symbol of Elements.

a)

John Dalton

b)

Joseph Proust

c)

Bohr

d)

Rutherford

57.

Lithium is found in most igneous rocks. Its element symbol is:

a)

LI

b)

L

c)

LT

d)

Li

58.

Beryllium is said to have a sweet taste. The symbol for beryllium is:

a)

B

b)

b

c)

Be

d)

By

59.

Most of the earth's atmosphere consists of nitrogen gas. The symbol for nitrogen is:

a)

ni

b)

N

c)

Nt

d)

No

60.

Dalton used _________________ in his symbols of elements

a)

Circle

b)

Square

c)

Triangle

d)

Rectangle

61.

The English name of the element is ________

a)

Silver

b)

Copper

c)

Gold

d)

Aurum

62.

The Latin name of Iron is

a)

Iron

b)

France

c)

Franc

d)

Ferrum

63.

What is English name of the element B

a)

Barium

b)

Boron

c)

Beryllium

d)

Bromine

64.

Ni is the symbol of _________

a)

Nitrogen

b)

Neon

c)

Nickel

d)

Tin

65.

Chemical Symbols represent the ____________________

a)

Name and number of atoms

b)

Initial of Person

c)

Chemical properties of elements

d)

Name of the discover

66.

Atomicity is ......

a)

the number of atoms present in an element.

b)

the number of atoms present in a molecule.

c)

atomic mass

d)

none of the above

67.

The Latin name of Potassium is......

a)

Natrium

b)

Kalium

c)

Plum bum

d)

None of the above

68.

Hydragyrum is the Latin name of ......

a)

Iron

b)

Mercury

c)

Tin

d)

Tungsten

69.

a)

diatomic

b)

monoatomic

70.

a)

diatomic

b)

monoatomic

71.

a)

diatomic

b)

monoatomic

72.

a)

diatomic

b)

monoatomic

73.

a)

diatomic

b)

monoatomic

74.

a)

diatomic

b)

monoatomic

75.

a)

diatomic

b)

monoatomic

76.

a)

diatomic

b)

monoatomic

77.

a)

diatomic

b)

monoatomic

78.

a)

diatomic

b)

monoatomic

79.

The correct way to write an ionic compound is first _______ and then ________ ( separate your answers with a comma)

(a)

80.

According to the number of elements in the ion, it is classified as:

S^-2

a)

Monoatomic cation

b)

Polyatomic cation

c)

Monoatomic anion

d)

Polyatomic anion

81.

According to the number of elements in the compound, it is classified as: NaHCO₃

a)

Binary compound

b)

Ternary compound

c)

Polyatomic compound

82.

According to the number of elements in the ion, it is classified as:

CO₃^-2

a)

Monoatomic cation

b)

Polyatomic cation

c)

Monoatomic anion

d)

Polyatomic anion

83.

According to the number of elements in the ion, it is classified as:

Al⁺³

a)

Monoatomic cation

b)

Polyatomic cation

c)

Monoatomic anion

d)

Polyatomic anion

84.

According to the number of elements in the ion, it is classified as:

N^-3

a)

Monoatomic cation

b)

Polyatomic cation

c)

Monoatomic anion

d)

Polyatomic anion

85.

According to the number of elements in the ion, it is classified as:

NH₄⁺

a)

Monoatomic cation

b)

Polyatomic cation

c)

Monoatomic anion

d)

Polyatomic anion

86.

According to the number of elements in the ion, it is classified as:

OH^-

a)

Monoatomic cation

b)

Polyatomic cation

c)

Monoatomic anion

d)

Polyatomic anion

87.

According to the number of elements in the ion, it is classified as:

Ag⁺

a)

Monoatomic cation

b)

Polyatomic cation

c)

Monoatomic anion

d)

Polyatomic anion

88.

According to the number of elements in the compound, it is classified as:

KCl

a)

Binary compound

b)

Ternary compound

c)

Polyatomic compound

89.

According to the number of elements in the compound, it is classified as:

AgBrO₃

a)

Binary compound

b)

Ternary compound

c)

Polyatomic compound

90.

According to the number of elements in the compound, it is classified as:

NaNO₃

a)

Binary compound

b)

Ternary compound

c)

Polyatomic compound

91.

According to the number of elements in the compound, it is classified as:

H₂O

a)

Binary compound

b)

Ternary compound

c)

Polyatomic compound

92.

The name of the ionic compound is:

CaSO₄(gypsum)

a)

Calcium sulfate

b)

calcium sulfite

93.

The name of the ionic compound is:

HCl(muriatic acid)

a)

Hydrochloric acid

b)

Hydrochlorate acid

94.

The name of the ionic compound is:

Li₂O(rechargeable batteries)

a)

Lithium oxide

b)

Oxide lithium

95.

The name of the ionic compound is:

Al(OH)₃(magnesia milk)

a)

Hydroxide aluminum

b)

Aluminum hydroxide

96.

The name of the ionic compound is:

NaCl(salt)

a)

Sodium chloride

b)

Sodium chlorate

97.

Calcium chloride when dissolved in water dissociates into its ions according to the following equation.

CaCl₂ (aq) ‑> Ca₂+ (aq) + 2Cl– (aq)

Calculate the number of ions obtained from CaCl₂ when 222 g of it is dissolved in water. What is the first step in deriving the formula of CaCl_2?

a)

Calcium chloride is a chemical compound of calcium and chlorine. The elements in calcium chloride are calcium and chlorine. The valency of calcium is 2 and chlorine is 1. Therefore, we write the symbols as Ca²and Cl¹.

b)

The valencies are exchanged and written as subscripts to the elements. Therefore, the formula of calcium chloride would be CaCl₂.

c)

Calcium is an Alkaline Earth Metal in the second column of the periodic table. This means that calcium has 2 valence electrons it readily gives away in order to seek the stability of the octet. This makes calcium a Ca+2 cation.

d)

Chlorine is a Halogen in the 17th column or p5 group.Chlorine has 7 valence electrons. It needs one electron to make it stable at 8 electrons in its valence shells. This makes chlorine a Cl−1 anion.

e)

Calcium chloride, when dissolved in water, dissociates into its ions, Ca+2 cation and Cl−1 anion.

98.

Calcium chloride when dissolved in water dissociates into its ions according to the following equation.

CaCl₂ (aq) ‑> Ca₂+ (aq) + 2Cl– (aq)

Calculate the number of ions obtained from CaCl₂ when 222 g of it is dissolved in water. What is the second step in deriving the formula of CaCl_2?

a)

No of moles =Given mass/molar mass. The molar mass of 1 mole of calcium chloride = 111g

b)

The valencies are exchanged and written as subscripts to the elements. Therefore, the formula of calcium chloride would be CaCl₂.

c)

Calcium is an Alkaline Earth Metal in the second column of the periodic table. This means that calcium has 2 valence electrons it readily gives away in order to seek the stability of the octet. This makes calcium a Ca+2 cation.

d)

Chlorine is a Halogen in the 17th column or p5 group.Chlorine has 7 valence electrons. It needs one electron to make it stable at 8 electrons in its valence shells. This makes chlorine a Cl−1 anion.

e)

Calcium chloride, when dissolved in water, dissociates into its ions, Ca+2 cation and Cl−1 anion.

99.

Calcium chloride when dissolved in water dissociates into its ions according to the following equation.

CaCl₂ (aq) ‑> Ca₂+ (aq) + 2Cl– (aq)

Calculate the number of ions obtained from CaCl₂ when 222 g of it is dissolved in water. What is the characteristic of calcium_?

a)

No of moles =Given mass/molar mass.

b)

Molar mass of CaCl₂ = 40+2x35.5 = 111g

111g of CaCl₂ represent = 1 mol

c)

Calcium is an Alkaline Earth Metal in the second column of the periodic table. This means that calcium has 2 valence electrons it readily gives away in order to seek the stability of the octet. This makes calcium a Ca+2 cation.

d)

Chlorine is a Halogen in the 17th column or p5 group.Chlorine has 7 valence electrons. It needs one electron to make it stable at 8 electrons in its valence shells. This makes chlorine a Cl−1 anion.

e)

Calcium chloride, when dissolved in water, dissociates into its ions, Ca+2 cation and Cl−1 anion.

100.

Calcium chloride when dissolved in water dissociates into its ions according to the following equation.

CaCl₂ (aq) ‑> Ca₂+ (aq) + 2Cl– (aq)

Calculate the number of ions obtained from CaCl₂ when 222 g of it is dissolved in water. What is the characteristic of Chlorine?

a)

No of moles =Given mass/molar mass.

b)

Molar mass of CaCl₂ = 40+2x35.5 = 111g

111g of CaCl₂ represent = 1 mol

c)

222g of CaCl₂ represent = 222/111 x 1 = 2 mol

d)

Chlorine is a Halogen in the 17th column or p5 group.Chlorine has 7 valence electrons. It needs one electron to make it stable at 8 electrons in its valence shells. This makes chlorine a Cl−1 anion.

e)

Calcium chloride, when dissolved in water, dissociates into its ions, Ca+2 cation and Cl−1 anion.

101.

Calcium chloride when dissolved in water dissociates into its ions according to the following equation.

CaCl₂ (aq) ‑> Ca₂+ (aq) + 2Cl– (aq)

Calculate the number of ions obtained from CaCl₂ when 222 g of it is dissolved in water. What happens when Calcium chloride dissolves in water?

a)

No of moles =Given mass/molar mass.

b)

Molar mass of CaCl₂ = 40+2x35.5 = 111g

111g of CaCl₂ represent = 1 mol

c)

222g of CaCl₂ represent = 222/111 x 1 = 2 mol.

d)

Therefore 222g of CaCl2 is equivalent to 2 moles of CaCl2. Since 1 formula unit CaCl2 gives 3 ions.

e)

Calcium chloride, when dissolved in water, dissociates into its ions, Ca+2 cation and Cl−1 anion.

102.

Calcium chloride when dissolved in water dissociates into its ions according to the following equation.

CaCl₂ (aq) ‑> Ca₂+ (aq) + 2Cl– (aq)

Calculate the number of ions obtained from CaCl₂ when 222 g of it is dissolved in water. How are moles calculated?

a)

No of moles =Given mass/molar mass.

b)

Molar mass of CaCl₂ = 40+2x35.5 = 111g

111g of CaCl₂ represent = 1 mol

c)

222g of CaCl₂ represent = 222/111 x 1 = 2 mol.

d)

Therefore 222g of CaCl2 is equivalent to 2 moles of CaCl2. Since 1 formula unit CaCl2 gives 3 ions.

e)

1 molecule of CaCl₂ form ions = 3. Therefore, 1 mol of CaCl2 will give 3 moles of ions 2 moles of CaCl2 would give 3×2=6 moles of ions.

103.

Calcium chloride when dissolved in water dissociates into its ions according to the following equation.

CaCl₂ (aq) ‑> Ca₂+ (aq) + 2Cl– (aq)

Calculate the number of ions obtained from CaCl₂ when 222 g of it is dissolved in water. What is the molar mass of CaCl₂?

a)

No. of ions = No. of moles of ions × Avogadro number

= 6 × 6.022 ×10*23

b)

Molar mass of CaCl₂ = 40+2x35.5 = 111g

111g of CaCl₂ represent = 1 mol

c)

222g of CaCl₂ represent = 222/111 x 1 = 2 mol.

d)

Therefore 222g of CaCl2 is equivalent to 2 moles of CaCl2. Since 1 formula unit CaCl2 gives 3 ions.

e)

1 molecule of CaCl₂ form ions = 3. Therefore, 1 mol of CaCl2 will give 3 moles of ions 2 moles of CaCl2 would give 3×2=6 moles of ions.

104.

Calcium chloride when dissolved in water dissociates into its ions according to the following equation.

CaCl₂ (aq) ‑> Ca₂+ (aq) + 2Cl– (aq)

Calculate the number of ions obtained from CaCl₂ when 222 g of it is dissolved in water. How many moles are present in 222 g of CaCl_2?

a)

No. of ions = No. of moles of ions × Avogadro number

= 6 × 6.022 ×10*23

b)

= 6 × 6.022 ×10^23

= 36.132×10^23

= 3.6132 × 10^24 ions

c)

222g of CaCl₂ represent = 222/111 x 1 = 2 mol.

d)

Therefore 222g of CaCl2 is equivalent to 2 moles of CaCl2. Since 1 formula unit CaCl2 gives 3 ions.

e)

1 molecule of CaCl₂ form ions = 3. Therefore, 1 mol of CaCl2 will give 3 moles of ions 2 moles of CaCl2 would give 3×2=6 moles of ions.

105.

Calcium chloride when dissolved in water dissociates into its ions according to the following equation.

CaCl₂ (aq) ‑> Ca₂+ (aq) + 2Cl– (aq)

Calculate the number of ions obtained from CaCl₂ when 222 g of it is dissolved in water. How many moles are present in 222 g of CaCl_2?

a)

No. of ions = No. of moles of ions × Avogadro number

= 6 × 6.022 ×10*23

b)

= 6 × 6.022 ×10^23

= 36.132×10^23

= 3.6132 × 10^24 ions

c)

222g of CaCl₂ represent = 222/111 x 1 = 2 mol.

d)

Therefore 222g of CaCl2 is equivalent to 2 moles of CaCl2. Since 1 formula unit CaCl2 gives 3 ions.

e)

1 molecule of CaCl₂ form ions = 3. Therefore, 1 mol of CaCl2 will give 3 moles of ions 2 moles of CaCl2 would give 3×2=6 moles of ions.

106.

Calcium chloride when dissolved in water dissociates into its ions according to the following equation.

CaCl₂ (aq) ‑> Ca₂+ (aq) + 2Cl– (aq)

Calculate the number of ions obtained from CaCl₂ when 222 g of it is dissolved in water. 1 formula unit CaCl2 gives how ions_?

a)

No. of ions = No. of moles of ions × Avogadro number

= 6 × 6.022 ×10*23

b)

= 6 × 6.022 ×10^23

= 36.132×10^23

= 3.6132 × 10^24 ions

c)

222g of CaCl₂ represent = 222/111 x 1 = 2 mol.

d)

Therefore 222g of CaCl2 is equivalent to 2 moles of CaCl2. Since 1 formula unit CaCl2 gives 3 ions.

e)

1 molecule of CaCl₂ form ions = 3. Therefore, 1 mol of CaCl2 will give 3 moles of ions 2 moles of CaCl2 would give 3×2=6 moles of ions.

107.

Calcium chloride when dissolved in water dissociates into its ions according to the following equation.

CaCl₂ (aq) ‑> Ca₂+ (aq) + 2Cl– (aq)

Calculate the number of ions obtained from CaCl₂ when 222 g of it is dissolved in water. 1 formula unit CaCl2 gives how ions_?

a)

No. of ions = No. of moles of ions × Avogadro number

= 6 × 6.022 ×10*23

b)

= 6 × 6.022 ×10^23

= 36.132×10^23

= 3.6132 × 10^24 ions

c)

222g of CaCl₂ represent = 222/111 x 1 = 2 mol.

d)

Therefore 222g of CaCl2 is equivalent to 2 moles of CaCl2. Since 1 formula unit CaCl2 gives 3 ions.

e)

1 molecule of CaCl₂ form ions = 3. Therefore, 1 mol of CaCl2 will give 3 moles of ions 2 moles of CaCl2 would give 3×2=6 moles of ions.

108.

Calcium chloride when dissolved in water dissociates into its ions according to the following equation.

CaCl₂ (aq) ‑> Ca₂+ (aq) + 2Cl– (aq)

Calculate the number of ions obtained from CaCl₂ when 222 g of it is dissolved in water. 2 moles of CaCl2 would give how many moles of ions?

a)

No. of ions = No. of moles of ions × Avogadro number

= 6 × 6.022 ×10*23

b)

= 6 × 6.022 ×10^23

= 36.132×10^23

= 3.6132 × 10^24 ions

c)

222g of CaCl₂ represent = 222/111 x 1 = 2 mol.

d)

Therefore 222g of CaCl2 is equivalent to 2 moles of CaCl2. Since 1 formula unit CaCl2 gives 3 ions.

e)

1 molecule of CaCl₂ form ions = 3. Therefore, 1 mol of CaCl2 will give 3 moles of ions 2 moles of CaCl2 would give 3×2=6 moles of ions.

109.

Calcium chloride when dissolved in water dissociates into its ions according to the following equation.

CaCl₂ (aq) ‑> Ca₂+ (aq) + 2Cl– (aq)

Calculate the number of ions obtained from CaCl₂ when 222 g of it is dissolved in water. 2 moles of CaCl2 would give how many moles of ions?

a)

No. of ions = No. of moles of ions × Avogadro number

= 6 × 6.022 ×10*23

b)

= 6 × 6.022 ×10^23

= 36.132×10^23

= 3.6132 × 10^24 ions

c)

222g of CaCl₂ represent = 222/111 x 1 = 2 mol.

d)

Therefore 222g of CaCl2 is equivalent to 2 moles of CaCl2. Since 1 formula unit CaCl2 gives 3 ions.

e)

1 molecule of CaCl₂ form ions = 3. Therefore, 1 mol of CaCl2 will give 3 moles of ions 2 moles of CaCl2 would give 3×2=6 moles of ions.

110.

Calcium chloride when dissolved in water dissociates into its ions according to the following equation.

CaCl₂ (aq) ‑> Ca₂+ (aq) + 2Cl– (aq)

Calculate the number of ions obtained from CaCl₂ when 222 g of it is dissolved in water. What is the first step to calculate the number of ions obtained from CaCl₂ ?

a)

No. of ions = No. of moles of ions × Avogadro number

= 6 × 6.022 ×10*23

b)

= 6 × 6.022 ×10^23

= 36.132×10^23

= 3.6132 × 10^24 ions

c)

222g of CaCl₂ represent = 222/111 x 1 = 2 mol.

d)

Therefore 222g of CaCl2 is equivalent to 2 moles of CaCl2. Since 1 formula unit CaCl2 gives 3 ions.

e)

1 molecule of CaCl₂ form ions = 3. Therefore, 1 mol of CaCl2 will give 3 moles of ions 2 moles of CaCl2 would give 3×2=6 moles of ions.

111.

Calcium chloride when dissolved in water dissociates into its ions according to the following equation.

CaCl₂ (aq) ‑> Ca₂+ (aq) + 2Cl– (aq)

Calculate the number of ions obtained from CaCl₂ when 222 g of it is dissolved in water. What is the second step to calculate the number of ions obtained from CaCl₂ ?

a)

No. of ions = No. of moles of ions × Avogadro number

= 6 × 6.022 ×10*23

b)

= 6 × 6.022 ×10^23

= 36.132×10^23

= 3.6132 × 10^24 ions

c)

222g of CaCl₂ represent = 222/111 x 1 = 2 mol.

d)

Therefore 222g of CaCl2 is equivalent to 2 moles of CaCl2. Since 1 formula unit CaCl2 gives 3 ions.

e)

1 molecule of CaCl₂ form ions = 3. Therefore, 1 mol of CaCl2 will give 3 moles of ions 2 moles of CaCl2 would give 3×2=6 moles of ions.

112.

If one mole of carbon atoms weighs 12 gram, what is the mass (in gram) of 1 atom of carbon? What is the given information?

a)

1 mole of carbon atoms weighs 12 gram

b)

1 mole of carbon atoms weighs = 6.022 x 10²³

c)

Therefore, 6.022 x 10²³ atoms of carbon = 12 g

d)

1 atom of carbon = 12/ (6.022 x 10²³)

= 1.993 x 10^-23

113.

If one mole of carbon atoms weighs 12 gram, what is the mass (in gram) of 1 atom of carbon? What is the second step of the calculation?

a)

1 mole of carbon atoms weighs 12 gram

b)

1 mole of carbon atoms weighs = 6.022 x 10²³

c)

Therefore, 6.022 x 10²³ atoms of carbon = 12 g

d)

1 atom of carbon = 12/ (6.022 x 10²³)

= 1.993 x 10^-23

114.

If one mole of carbon atoms weighs 12 gram, what is the mass (in gram) of 1 atom of carbon? What is the third step of the calculation?

a)

1 mole of carbon atoms weighs 12 gram

b)

1 mole of carbon atoms weighs = 6.022 x 10²³

c)

Therefore, 6.022 x 10²³ atoms of carbon = 12 g

d)

1 atom of carbon = 12/ (6.022 x 10²³)

= 1.993 x 10^-23

115.

If one mole of carbon atoms weighs 12 gram, what is the mass (in gram) of 1 atom of carbon? What is the fourth step of the calculation?

a)

1 mole of carbon atoms weighs 12 gram

b)

1 mole of carbon atoms weighs = 6.022 x 10²³

c)

Therefore, 6.022 x 10²³ atoms of carbon = 12 g

d)

1 atom of carbon = 12/ (6.022 x 10²³)

= 1.993 x 10^-23

116.

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? What is given to you in the question?

a)

When 3.0 g of carbon is burnt in 8.00 g oxygen, 11.00 g of carbon dioxide is produced.

b)

To find: What mass of carbon dioxide will be formed when 3.00 g of carbon is burnt in 50.00 g of oxygen?

c)

A combustion reaction occurs when carbon (C) burns in oxygen (O₂) to give Carbon dioxide CO₂).

d)

As per the given condition, when 3.0 g of carbon is burnt in 8.00 g oxygen, 11.00 g of carbon dioxide is produced.

3g + 8g →11 g ( from the above reaction)

e)

3 g of carbon must also combine with 8 g of oxygen only.

This means that (50−8)=42g of oxygen will remain unreacted.

117.

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? What are we required to find?

a)

Then, it also depicts that the carbon dioxide contains carbon and oxygen in a fixed ratio by mass, which is 3:8.

b)

To find: What mass of carbon dioxide will be formed when 3.00 g of carbon is burnt in 50.00 g of oxygen?

c)

A combustion reaction occurs when carbon (C) burns in oxygen (O₂) to give Carbon dioxide CO₂).

d)

As per the given condition, when 3.0 g of carbon is burnt in 8.00 g oxygen, 11.00 g of carbon dioxide is produced.

3g + 8g →11 g ( from the above reaction)

e)

3 g of carbon must also combine with 8 g of oxygen only.

This means that (50−8)=42g of oxygen will remain unreacted.

118.

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? What is the chemical reaction?

a)

Then, it also depicts that the carbon dioxide contains carbon and oxygen in a fixed ratio by mass, which is 3:8.

b)

In this case also, only 11 g of carbon dioxide will be formed. The remaining 42 g of oxygen will be left un-reactive. The above answer is governed by the law of constant proportions.

c)

A combustion reaction occurs when carbon (C) burns in oxygen (O₂) to give Carbon dioxide CO₂).

d)

As per the given condition, when 3.0 g of carbon is burnt in 8.00 g oxygen, 11.00 g of carbon dioxide is produced.

3g + 8g →11 g ( from the above reaction)

e)

3 g of carbon must also combine with 8 g of oxygen only.

This means that (50−8)=42g of oxygen will remain unreacted.

119.

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? What happens to the above reaction when excessive oxygen is supplied?

a)

Then, it also depicts that the carbon dioxide contains carbon and oxygen in a fixed ratio by mass, which is 3:8.

b)

In this case also, only 11 g of carbon dioxide will be formed. The remaining 42 g of oxygen will be left un-reactive. The above answer is governed by the law of constant proportions.

c)

A combustion reaction occurs when carbon (C) burns in oxygen (O₂) to give Carbon dioxide CO₂).

d)

As per the given condition, when 3.0 g of carbon is burnt in 8.00 g oxygen, 11.00 g of carbon dioxide is produced.

3g + 8g →11 g ( from the above reaction)

e)

3 g of carbon must also combine with 8 g of oxygen only.

This means that (50−8)=42g of oxygen will remain unreacted.

120.

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? DOES THIS chemical reaction display a condition?

a)

Then, it also depicts that the carbon dioxide contains carbon and oxygen in a fixed ratio by mass, which is 3:8.

b)

In this case also, only 11 g of carbon dioxide will be formed. The remaining 42 g of oxygen will be left un-reactive. The above answer is governed by the law of constant proportions.

c)

A combustion reaction occurs when carbon (C) burns in oxygen (O₂) to give Carbon dioxide CO₂).

d)

As per the given condition, when 3.0 g of carbon is burnt in 8.00 g oxygen, 11.00 g of carbon dioxide is produced.

3g + 8g →11 g ( from the above reaction)

e)

3 g of carbon must also combine with 8 g of oxygen only.

This means that (50−8)=42g of oxygen will remain unreacted.

121.

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? How does this reaction prove the law of constant proportions?

a)

Then, it also depicts that the carbon dioxide contains carbon and oxygen in a fixed ratio by mass, which is 3:8.

b)

In this case also, only 11 g of carbon dioxide will be formed. The remaining 42 g of oxygen will be left un-reactive. The above answer is governed by the law of constant proportions.

c)

A combustion reaction occurs when carbon (C) burns in oxygen (O₂) to give Carbon dioxide CO₂).

d)

As per the given condition, when 3.0 g of carbon is burnt in 8.00 g oxygen, 11.00 g of carbon dioxide is produced.

3g + 8g →11 g ( from the above reaction)

e)

3 g of carbon must also combine with 8 g of oxygen only.

This means that (50−8)=42g of oxygen will remain unreacted.

122.

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?

a)

Then, it also depicts that the carbon dioxide contains carbon and oxygen in a fixed ratio by mass, which is 3:8.

b)

In this case also, only 11 g of carbon dioxide will be formed. The remaining 42 g of oxygen will be left un-reactive. The above answer is governed by the law of constant proportions.

c)

A combustion reaction occurs when carbon (C) burns in oxygen (O₂) to give Carbon dioxide CO₂).

d)

As per the given condition, when 3.0 g of carbon is burnt in 8.00 g oxygen, 11.00 g of carbon dioxide is produced.

3g + 8g →11 g ( from the above reaction)

e)

3 g of carbon must also combine with 8 g of oxygen only.

This means that (50−8)=42g of oxygen will remain unreacted.

Atoms And Molecules _LS_ 2_2

Total questions: 122

Worksheet time: 1hrs 1mins

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