Chapter 2 of Class 9 Chemistry syllabus covers the topic of “Chemical Combination”. This chapter is important for students to understand the basic concepts of chemical bonding, types of chemical reactions, and the laws of chemical combination.
To help students prepare for their exams, there are various notes and study materials available online. These notes cover all the important topics related to Chapter 2 of the Class 9 Chemistry syllabus and include conceptual questions, short questions, long questions, multiple-choice questions, and numerical problems.
The conceptual questions cover important concepts related to chemical combination, such as the difference between elements, compounds, and mixtures, types of chemical reactions, and the law of conservation of mass. The short questions are designed to test the students’ understanding of the concepts in a concise manner, while the long questions require more in-depth understanding of the topics.
The multiple-choice questions help students to practice for their exams and improve their test-taking skills, while the numerical problems are designed to test their understanding of the mathematical calculations involved in chemical reactions.
Here are some tips for students to effectively study Chapter 2 of the Class 9 Chemistry syllabus:
- Read the textbook carefully and take notes.
- Understand the basic concepts of chemical bonding, types of chemical reactions, and the laws of chemical combination.
- Learn about balancing chemical equations and stoichiometry.
- Practice solving short and long questions, multiple-choice questions, and numerical problems to improve your understanding of the topics.
By following these tips and studying the notes and practice questions available online, students can effectively prepare for Chapter 2 of the Class 9 Chemistry syllabus and perform well in their exams.
Chemical Combination New Notes for Sindh Textbook in Pdf
Fermentation is chemical decomposition, in which glucose (C6H12O6) is converted into ethyl alcohol (C2H5-OH) and carbon dioxide (CO2).
(C6H12O6) → 2C2H5 -OH + 2CO2
What will be the amount of ethyl alcohol in grams and moles, which can be obtained by fermentation of 5000g of glucose.
Answer:
Data:
Mass of glucose = 5000 g
Mass of ethyl alcohol = ?
Moles of ethyl alcohol = ?
Solution:
Molecular mass of glucose = (6 × 12 + 12 × 1 + 6 × 16)
= 180 g
180 g of glucose contains = 1 mole
1 g of glucose contain = 1 / 180 moles
5000 g of glucose = 1 / 180 × 5000
= 27.72 moles
No. of moles:
1 mole of glucose on fermentation gives = 2 moles of ethyl alcohol
27.72 moles of glucose on fermentation gives = 2 × 27.72
No of moles of ethyl alcohol = 55.44 moles
So,
Molecular mass of ethyl alcohol = (2 × 12 + 5 × 1 + 16 +1)
= 46 g
1 mole of ethyl alcohol has mass = 46 g
54 moles of ethyl alcohol have mass = 46 × 54
No. of grams of ethyl alcohol = 2484 g
Silicon tetrachloride (SiCl4) can be prepared by heating (Si) in chlorine gas (Cl2).
Si+2CI2 → SiCl4
If we want to prepare 10 moles of (SiCl4) how many moles of molecular chlorine (Cl2) will be used in the reaction.
Answer:
Data:
Si + 2Cl2 → SiCl4
1 mole + 2 moles → 1 mole
Solution:
2 moles of chlorine is producing 1 mole of SiCl4. Then 20 moles are required to make 10 moles of SiCl4.
Calcium carbonate (CaCO3) on heating gives calcium oxide (CaO) and CO2 gas.
CaCO3 → CaO + CO2
Calculate how many grams of calcium oxide (CaO) can be obtained by heating 8 moles of CaCO3?
Answer:
Data:
CaCO3 → CaO + CO2
Number of moles of CaCO3 = 8 moles
Mass of CaO = ?
Solution:
CaCO3 → CaO + CO2
1 mole → 1 mole
8 mole → 8 mole
Molecular mass of CaO = 40 + 16
= 56 g
1 mole of CaO contains = 56 g
8 moles of CaO will contains = 56 × 8
Mass of CaO = 448 g
Consider the combination of (CO) with oxygen (O2) gas.
2CO + O2 → 2CO2?
Calculate the number of moles of (CO2) produced when 50 moles of oxygen are reacted with all of (CO)?
Answer:
Data:
2CO + O2 → 2CO2
Moles of oxygen = 50 moles
Number of moles of CO2 =?
Solution:
2CO + O2 → 2CO2
2 moles + 1mole → 2 moles
1 mole of oxygen produces = 2 moles of CO2
50 moles of oxygen will produce = 2 × 100
= 100 moles of CO2
When 1 mole of oxygen reacts with 2 moles of CO, 2 moles of CO2 is produced. Thus when 50 moles of oxygen will react with 2 moles of CO, 100 moles of CO2 is produced.
Calcium carbonate (CaCO3) on heating gives calcium oxide (CaO) and CO2 gas.
CaCO3 → CaO + CO2
Calculate how many grams of calcium oxide (CaO) can be obtained by heating 8 moles of CaCO3?
Answer:
Data:
CaCO3 → CaO + CO2
Number of moles of CaCO3 = 8 moles
Mass of CaO = ?
Solution:
CaCO3 → CaO + CO2
1 mole → 1 mole
8 mole → 8 mole
Molecular mass of CaO = 40 + 16
= 56 g
1 mole of CaO contains = 56 g
8 moles of CaO will contains = 56 × 8
Mass of CaO = 448 g
Consider the combination of (CO) with oxygen (O2) gas.
2CO + O2 → 2CO2?
Calculate the number of moles of (CO2) produced when 50 moles of oxygen are reacted with all of (CO)?
Answer:
Data:
2CO + O2 → 2CO2
Moles of oxygen = 50 moles
Number of moles of CO2 =?
Solution:
2CO + O2 → 2CO2
2 moles + 1mole → 2 moles
1 mole of oxygen produces = 2 moles of CO2
50 moles of oxygen will produce = 2 × 100
= 100 moles of CO2
When 1 mole of oxygen reacts with 2 moles of CO, 2 moles of CO2 is produced. Thus when 50 moles of oxygen will react with 2 moles of CO, 100 moles of CO2 is produced.
Balance the following equation, which of them is a decomposition reaction or combination reaction.
a) CH4 + O2 → CO2 +H2O
b) NO2 → NO + O2
c) Na+O2 → Na2O
Answer:
a) CH4 + O2 → CO2 + H2O
CH4 + 2O2 → CO2 + 2H2O
It is not a combination reaction nor a decomposition reaction.
b) NO2 → NO + O2
2NO2 → 2NO + O2
Decomposition reaction
c) Na+O2 → Na2O
4Na+O2 → 2Na2O
Combination reaction
Balance the equation and decide which one is a single replacement reaction.
a) C2H2 + H2 → C4H8
b) Ca + H2O → Ca (OH)2 + H2
c) C2H5-OH+Na → C2H5ONa + H2
Answer:
a) C2H2 + H2 → C4H8
2C2H2 + 2H2 → C4H8
b) Ca + H2O → Ca (OH)2 + H2
Ca + 2H2O → Ca (OH)2 + H2
c) C2H5-OH + Na → C2H5ONa + H2
2C2H5-OH + 2Na → 2C2H5ONa + H2
b and c both are single replacement reactions.
Which of the following reaction is either a decomposition reaction or combination reaction?
a) MgCO3 → MgO + CO2
b) C2H4 + H2 → C2H6
c)BaCO3 → BaO + CO2
d) N2 + 3H2 → 2NH3
Answer:
a) MgCO3 → MgO + CO2
Decomposition reaction
b) C2H4 + H2 → C2H6
Combination reaction
c)BaCO3 → BaO + CO2
Decomposition reaction
d) N2 + 3H2 → 2NH3
Combination reaction
Balance the following equations by inspection method?
a) C+O2 → CO
b) CO+O2 → CO2
c) KNO3 → KNO2+O2
d) NaHCO3 →Na2CO3 + H2O + CO2
e) CaCO3 + HCI → CaCl2+H2O+CO2
f) NH3 + O2 → NO + H2O
Answer:
a) C+O2 → CO
Balance the number of atoms on each side.
Reactants Products
C (1) C (1)
O (2) O (1)
Here carbon element has the same number of atoms on both sides of the equation, but there are two oxygen atoms on left side and one oxygen atom on right side. So to balance the oxygen atom we will put 2 as co-effficient in front of CO.
C+O2 → 2CO
Reactants Products
C (1) C (2)
O (2) O (2)
Now to balance the number of carbon atoms on both side we will put 2 as co-efficient in front of C.
2C+O2 → 2CO
Reactants Products
C (2) C (2)
O (2) O (2)
Now the equation is balanced.
2C + O2 → 2CO
b) CO + O2 → CO2
Balance the number of atoms on each side.
Reactants Products
C (1) C (1)
O (3) O (2)
Here carbon element has the same number of atoms on both sides of the equation, but there are three oxygen atoms on left side and twq oxygen atom on right side. So to balance the oxygen atom we will put 2 as co-effficient in front of CO2.
CO + O2 → 2CO2
Reactants Products
C (1) C (2)
O (3) O (4)
Now to balance the number of carbon atoms and oxygen atoms on both side we will put 2 as co-efficient in front of CO.
2CO + O2 → 2CO2
Reactants Products
C (2) C (2)
O (4) O (4)
Now the equation is balanced.
2CO + O2 → 2CO2
c) KNO3 → KNO2+ O2
Balance the number of atoms on each side.
Reactants Products
K (1) K (1)
N (1) N (1)
O (3) O (4)
Here both potassium and nitrogen elements have the same number of atoms on both sides of the equation, but there are three oxygen atoms on left side and four oxygen atom on right side. So to balance the oxygen atom we will put 2 as co-effficient in front of KNO3.
2 KNO3 → KNO2+O2
Reactants Products
K (2) K (1)
N (2) N (1)
O (6) O (4)
Now to balance the number of potassium, nitrogen and oxygen atoms on both side we will put 2 as co-efficient in front of CO.
2 KNO3 → 2KNO2+O2
Reactants Products
K (2) K (2)
N (2) N (2)
O (6) O (6)
Now the equation is balanced.
2 KNO3 → 2KNO2+O2
d) NaHCO3 → Na2CO3 + H2O + CO2
Balance the number of atoms on each side.
Reactants Products
Na (1) Na (1)
H (1) H (1)
C (1) C (2)
O (3) O (6)
Here only hydrogen element has the same number of atoms on both sides of the equation, but there are three oxygen atoms and one carbon atom on left side and six oxygen atoms and two carbon atoms on right side. And two sodium atoms on right side and one sodium atom on left side of equation. So to balance all these we will put 2 as co-effficient in front of NaHCO3.
2NaHCO3 → Na2CO3 + H2O + CO2
Reactants Products
Na (2) Na (2)
H (2) H (2)
C (2) C (2)
O (6) O (6)
Now the equation is balanced.
2NaHCO3 → Na2CO3 + H2O + CO2
e) CaCO3 + HCl → CaCl2 + H2O + CO2
Balance the number of atoms on each side.
Reactants Products
Ca (1) Ca (1)
C (1) C (1)
O (3) O (3)
H (1) H (2)
Cl (1) Cl (2)
Here calcium carbon and oxygen elements have the same number of atoms on both sides of the equation, but there are two hydrogen atoms on right side and two chlorine atoms on right side. So to balance the hydrogen and chlorine atoms we will put 2 as co-effficient in front of HCl.
CaCO3 + 2HCl → CaCl2+H2O+CO2
Reactants Products
Ca (1) Ca (1)
C (1) C (1)
O (3) O (3)
H (2) H (2)
Cl (2) Cl (2)
Now the equation is balanced.
CaCO3 + 2HCl → CaCl2+H2O+CO2
f) NH3 + O2 → NO + H2O
Balance the number of atoms on each side.
Reactants Products
N (1) N (1)
H (3) H (2)
O (2) O (2)
Here sodium element has the same number of atoms on both sides of the equation, but there are two hydrogen atoms on right side and three hydrogen atoms on left side. So to balance the hydrogen atoms we will put 4 as co-effficient in front of NH3.
4NH3 + O2 → NO + H2O
Reactants Products
N (4) N (1)
H (12) H (2)
O (2) O (2)
Now to balance the nitrogen atom and hydrogen atoms on the right side we will put 4 in front of NO and 6 in front of H2O as co-efficient.
4NH3 + O2 → 4NO + 6H2O
Reactants Products
N (4) N (4)
H (12) H (12)
O (2) O (10)
Now to balance the oxygen atoms on both sides we will put 5 in front of O2.
4NH3 + 5O2 → 4NO + 6H2O
Reactants Products
N (4) N (4)
H (12) H (12)
O (5) O (10)
Now the equation is balanced.
4NH3 + 5O2 → 4NO + 6H2O
Explain double displacement reaction with examples?
Answer:
Double-displacement reaction:
It is a reaction in which two compounds exchange their partners, so that two new compounds are formed. In double displacement reaction usually there is an exchange of ionic radicals. For example,
Sodium sulfide reacts with hydrochloric acid to form sodium chloride and hydrogen sulfide gas:
Na2S ( aq ) + 2HCl ( aq ) → 2NaCl ( aq ) + H2S ( g )
Another example, Potassium hydroxide reacts with nitric acid to form water and potassium nitrate:
KOH ( aq ) + HNO3 ( aq ) → H2O ( l ) + KNO3 ( aq )
What is a single replacement reaction? Give an example?
Single replacement reaction:
In a single-displacement reaction, a free element displaces another element from a compound to produce a different compound and a different free element. A more active element displaces a less active element from its compound.
For example, zinc replaces hydrogen in hydrogen chloride to give zinc chloride.
Zn + 2HCl → ZnCl2 + H2
What is decomposition reaction? Will two or more elements always be the products of this type of reaction? Explain with examples.
Decomposition Reactions:
A reaction in which a chemical substance breaks down to form two or more simpler substances is called a decomposition reaction. These reactions require energy for decomposition.
Yes, it is necessary that two or more elements always be the products of this type. For example,
CaCO3 → CaO + CO2
What is a combination reaction? Give an example.
Combination reaction:
In a chemical reaction, when two or more reactants combine to form a new product, it is called combination reaction. For example, calcium oxide (CaO) reacts with carbon dioxide (CO2) to form calcium carbonate (CaCO3).
CaO + CO2 → CaCO3
What is chemical equation? What is a co-efficient? Give an example of balanced equation.
Answer:
Chemical equation:
A chemical equation is the shorthand method or symbolic representation of a chemical reaction in the form of symbols and formulae. In chemical equation, the reactants are given on the left-hand side and the products on the right-hand side. The reactants and products are separated by the single arrow or double arrow depending upon the type of reaction. For example, hydrogen gas (H2) can react with oxygen gas (O2) to form water (H2O). The chemical equation for this reaction is written as:
2H2 + O2 → 2H2O
The ‘+’ is read as ‘reacts with’ and the arrow ” means ‘produces’. The chemical formulas on the left represent the starting substances, called reactants. The substances produced by the reaction are shown on the right and are called products.
Co-efficient:
The numbers in front of the formulas are called coefficients. They show the number of molecules that react with each other. Where no coefficient is given, there only one number is considered.
2H2 + O2 → 2H2O
Example of a balanced chemical equation:
Consider the example when hydrogen burns in air to form water.
H2 + O2 → H2O
Balance the number of atoms on each side.
Reactants Products
H (2) H (2)
O (2) O (1)
Here hydrogen element has the same number of atoms on both sides of the equation, but there are two oxygen atoms on left side and one oxygen atom on right side. So to balance the oxygen atom we will put 2 as co-effficient in front of water.
H2 + O2 → 2H2O
Reactants Products
H (2) H (4)
O (2) O (2)
Now to balance the number of hydrogen atoms on both side we will put 2 as co-efficient in front of H2.
2H2 + O2 → 2H2O
Reactants Products
H (4) H (4)
O (2) O (2)
Now the equation is balanced.
2H2 + O2 → 2H2O
Define the following terms:
a) Chemical reaction
b) Reactants
c) Products
Answer:
a) Chemical reaction:
Any change which alters the composition of a substance is called chemical change. For example, when coal burns, it forms smoke and ashes. The burning of coal is a chemical change.
b) Reactants:
A reactant is a substance that reacts or initiates the chemical reaction to form product. For example in the burning of coal, coal is reactant.
c) Products:
Products are the species formed from chemical reactions. During a chemical reaction reactants are converted into products. For example, in the reaction of burning of coal, it forms smoke and ashes. These are products.
The formula for rust is Fe2O3. How many moles of (Fe) are present in 30g of rust?
Answer:
Data:
Formula of rust = Fe2O3
Mass of rust = 30 g
Number of moles of Fe = ?
Solution:
Molecular mass of Fe2O3 = (2 × 55) + (3 × 16)
= 110 + 48
= 158 g
158 g of Fe2O3 contains number of moles of Fe = 2 moles
1 g of Fe2O3 contains number of moles of Fe = 2 / 158 moles
30 g of Fe2O3 contains number of moles of Fe = 2 / 158 × 30
Number of moles of Fe = 0.378 moles
Calculate the molar mass of the following substances.
a) S8
b) CS2
c) CHCl3 (Chloroform).
d) CH3 -COOH (Acetic acid)
Answer:
a) S8
Molar mass of S8
= (8 × atomic mass of S)
= (8 g × 32 g)
= 256 g/mol
b) CS2
Molar mass of CS2
= (1 × atomic mass of C) + (2 × atomic mass of S)
= (1× 12 g) + (2 × 32 g)
= 12 g + 64 g = 76 g/mol
c) CHCl3 (Chloroform)
Molar mass of CHCl3
= (1 × atomic mass of C) + (1 × atomic mass of H) + (3 atomic mass of Cl)
= (1× 12 g) +(1g) + ( 3 × 35 g)
= 12 g + 1 g + 105g = 118 g/mol
d) CH3 -COOH (Acetic acid)
Molar mass of CHCl3
= (2 × atomic mass of C) + (3 × atomic mass of H) + (2 × atomic mass of O)
= (2× 12 g) + (3 × 1 g) + ( 2 ×16 g)
= 24 g + 3 g + 32g = 59 g/mol
State the law of definite proportions in your own words.
Answer:
The Law of Definite Proportions:
The Law of definite proportions states that any chemical compound will always contain a fixed ratio of elements by mass.
A French scientist named Joseph Proust conducted experiments that demonstrated this law.
Examples of the Law of Definite Proportions:
Let us take, for example, the compound water. Whatever the source of water, its composition is that of two atoms of hydrogen and one atom of oxygen. This figure shows that water, from any source, is always made up of two atoms of hydrogen and one atom of oxygen. If we calculate the molecular weight of water, we come up with 18 g/mol. In 1 mole of water, there are 2 grams of hydrogen and 16 grams of oxygen. By weight, we have a percentage of 11% hydrogen and 89% oxygen in one mole of water. This translates into a ratio of 1:8 of hydrogen to oxygen in water.
What is law of multiple proportions? Explain with examples.
Answer:
Law of Multiple Proportion:
The law of multiple proportions states that when two elements combine to form more than one compound, the mass of one element, which combines with a fixed mass of the other element, will always be ratios of whole numbers
Example of the Law of Multiple Proportion:
Let us demonstrate the law of multiple proportions with these two elements: nitrogen monoxide and nitrogen dioxide.
Nitrogen monoxide is made up of one nitrogen (N) atom and one oxygen (O) atom, forming (NO), and nitrogen dioxide is made of one nitrogen (N) atom and two oxygen (O) atoms, forming (NO2).
Nitrogen has an atomic mass of 14 and oxygen has an atomic mass of 16. So, for nitrogen monoxide (NO), we can say that it is made of 14 parts by mass nitrogen and 16 parts by mass oxygen – the ratio of nitrogen to oxygen is therefore 14:16.
For nitrogen dioxide (NO2), we can say it is made of 14 parts by mass nitrogen and 32 parts by mass oxygen – the ratio of nitrogen to oxygen is therefore 14:32.
For nitrogen trioxide (N2O3), we can say it is made of 28 parts by mass nitrogen and 48 parts by mass oxygen – the ratio of nitrogen to oxygen is therefore 28:48.
For nitrogen tetra oxide (N2O4), we can say it is made of 28 parts by mass nitrogen and 64 parts by mass oxygen – the ratio of nitrogen to oxygen is therefore 28:64.
For nitrogen penta oxide (N2O5), we can say it is made of 28 parts by mass nitrogen and 80 parts by mass oxygen – the ratio of nitrogen to oxygen is therefore 28:80.
By fixing the mass of (N), the mass of oxygen (O) in different oxides varies. The oxygen varies ratios are 1 : 2 : 3 : 4 : 5.
Q.4) State the law of reciprocal proportion and illustrate it with examples .
Answer:
Law of reciprocal proportion
This law was proposed by Ritcher in the year 1799. It states that “when two different elements separately combine with the fixed mass of third element, the proportions in which they combine with one another shall be either in the same ratio or some simple multiple of it.”
Example
When two elements C and O combine separately with H to form methane (CH4) and water (H2O) respectively it is very clear that in methane 3 g of C combine with 1 g of hydrogen, an in water 8 g of O combine with the same (fixed) mass i.e. 1 g of H. Now when C and O combine with each other to form carbon dioxide, they do so in the same proportion i.e. 12:32 = 3:8 parts by mass.
What is empirical formula? Give an example?
Answer:
Empirical formula:
Empirical formula is the simplest whole number ratio of atoms present in a compound.
For example, the molecular formula of glucose is C6H12O6. If we divide it by least common factor (6 in this case) we get CH2O which is the empirical formula of glucose.
Mathematically,
Molecular formula = (empirical formula)n
Where n is any number 1,2,3,…etc. Empirical formula multiplied by n gives the molecular formula.
What is molecular formula? Give an example?
Answer:
Molecular formula:
Molecular formula is the actual number of atoms of each element present in a molecule of that compound. Molecular formula is derived from empirical formula. Like,
Molecular formula = (Empirical formula)n
For example, the molecular formula of benzene is C6H6 which is derived from the empirical formula CH where the value of n is 6.
Q.7) Can one substance have the same empirical formula and molecular formula? Explain with examples.
Answer:
One substance can have the same empirical and molecular formula. For example, formaldehyde (CH2O), ammonia (NH3) and methane (CH4) all have same empirical and molecular formula.
What is the difference between empirical formula and molecular formula?
Answer:
Empirical formula is the simplest whole number ratio of atoms present in a compound. For example, the molecular formula of glucose is C6H12O6. If we divide it by least common factor (6 in this case) we get CH2O which is the empirical formula of glucose.
On the other hand, Molecular formula is the actual number of atoms of each element present in a molecule of that compound. Molecular formula is derived from empirical formula. For example, molecular formula of benzene is C6H6 which is derived from the empirical formula (CH)n where the value of n is 6.
What is atomic mass unit?
Answer:
Atomic mass unit:
Atomic mass unit is the mass equal to the 1/12th of the mass of a carbon-12 (12C). In 1961, by an international agreement, an atom of C-12 that has 6 protons and 6 neutrons has a mass of exactly 12 atomic mass unit is taken as a standard. Thus it is represented as a.m.u. For example,
Atomic mass of oxygen (O) = 16 a.m.u
The value of atomic mass of carbon in periodic table is 12.011 a.m.u, rather than 12.00 a.m.u? Explain?
Answer:
Most of the elements consist of a mixture of isotopes. For example, carbon has three isotopes C-12 (98.88%), C-13 (1.11%) and C-14 (traces). Thus average atomic mass of C is 12.011. So it is taken as 12.011 a.m.u in the periodic table.
the atomic masses of 17CI35 (75%) and 17CI37 (25%). Calculate the average atomic mass of chlorine.
Answer:
Data:
Mass of Cl = 35
Percentage composition of Cl = 75%
Mass of Cl = 37
Percentage composition of Cl = 25%
Solution:
Average atomic mass = mixture of isotopes / 100
By putting the values in equation,
Average atomic mass = 35 ×75 + 37 ×25 / 100
Average atomic mass of Cl = 35.5
Hence the average atomic mass of Cl is 35.5 amu.
How many atoms are there in 5 moles of sulphur?
Answer:
Data:
Number of moles of sulphur = 5 moles
Number of atoms of sulphur = ?
Solution:
1 mole of sulphur contain = 6.022 × 1023
5 moles of sulphur contain = 6.022 × 1023 × 5
= 30.11 × 1023
= 3.011 × 1024 atoms.
So, 5 moles of sulphur will contain 3.011 × 1024 atoms.
Sindh Class 9th Chemistry Notes Chapter 2 Multiple Choice Questions
5 moles of H2O are equal to:
A. 80g
B. 90g
C. 100g
D. 90 a.m.u
The sum of the atomic masses of all atoms in a formula unit of substance is called:
A. Empirical Formula
B. Molecular Formula
C. Molecular Mass
D. Formula Mass
44 a.m.u. of CO2 is equal to:
A. Molar Mass
B. Atomic Mass
C. Molecular Mass
D. Mass Number
The mass of (1) mole of a substance expressed in grams, is called:
A. Empirical Formula
B. Molecular Formula
C. Molecular Mass
D. Molar Mass
The sum of the atomic masses of all atoms in a molecule is called:
A. Empirical Formula
B. Molecular Formula
C. Molecular Mass
D. Formula Mass
A formula that indicates the actual number and type of atoms in a molecule is called:
A. Empirical Formula
B. Molecular Formula
C. Molecular Mass
D. Formula Mass
A formula that gives only the relative number of each type of atom in a molecule, is called:
A. Empirical Formula
B. Molecular Formula
C. Molecular Mass
D. Formula Mass
The average mass of the natural mixture of isotopes, which is compared to the mass of one atom of C-12 a.m.u, is called:
A. Atomic Number
B. Mass Number
C. Atomic Mass
D. None of these
A given compound always contains exactly the same proportion of elements, by mass, is the statement of:
A. Law of conservation of mass
B. Law of definite proportions
C. Law of multiple proportions
D. Law of reciprocal proportions
Mass is neither created nor destroyed during a chemical change, is the statement of:
A. Law of conservation of mass
B. Law of definite proportions
C. Law of multiple proportions
D. Law of reciprocal proportions
