Some Basic Concepts of Chemistry Test

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Some Basic Concepts of Chemistry Test - 32

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Question 1
1 / -0

Mass can neither be created nor destroyed. This is in agreement with which law of chemical combination?

A
Law of definite proportion

B
Law of multiple proportion

C
Law of conservation of mass

D
Law of conservation of momentum

Solution

The Law of conservation of mass states that when elements and compounds react to form new products, mass cannot be lost or gained.
Question 2
1 / -0

Which scientist proposed the concept of atomic mass?

A
Avogadro

B
Gay Lussac

C
Proust

D
Dalton

Solution

The concept of atomic mass was proposed by William Prout. Early atomic mass theory was proposed by the English chemist William Prout in a series of published papers in 1815 and 1816. Known as Prout's Law, Prout suggested that the known elements had atomic weights that were whole number multiples of the atomic mass of hydrogen.

Option C is correct.
Question 3
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1.008 grams of hydrogen combines with 35.453 g of chlorine to form 36.463 g of a pure compound (hydrogen chloride). Dalton's explanation for this experimental fact might be:

A
One atom of hydrogen combines with 35.453 atoms of chlorine in this reaction

B
Hydrogen and chlorine atoms always combine in a 1:35 ratio.

C
Chlorine accepts hydrogen's electron to form polar covalent hydrogen chloride

D
Hydrogen and chlorine atoms aren't created or destroyed in the process so the reactant mass is the same as the product mass.

Solution

Dalton Atomic theory postulates-
All matter is made up of atoms and atoms are indestructible cannot be created or destroyed so the total mass of reactants remains equal to the total mass of products.
Question 4
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Which of the following statement is true about mole?

A
It is used to calculate number of particles.

B
1 mole is equal to $6.023\times10^{23}$ particles.

C
It is the chemist's counting unit

D
All of the above

Solution

Mole is used to find number of particles. 1 mole is equal to Avogadro number of atoms/molecules. It is chemist's counting unit.
Question 5
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The formula for % composition of a compound is:

A
molar mass of a compound/mass due to specific component $\times$ 100

B
mass due to specific component $\times$ 100

C
mass due to specific component/molar mass of a compound $\times$ 100

D
molar mass of a specific component/temperature $\times$ 100

Solution

To calculate the per cent composition of a component in a compound: Find the molar mass of the compound by adding up the masses of each atom in the compound using the periodic table or a molecular mass calculator. Calculate the mass due to the component in the compound you are for which you are solving by adding up the mass of these atoms. Divide the mass due to the component by the total molar mass of the compound and multiply by 100.

% $\text{composition of a compound}=\cfrac{\text{ Mass due to specific component}}{\text{molar mass of a compound}}\times 100$

Hence, the correct option is $\text{C}$
Question 6
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Which is not one of the laws of chemical combinations?

A
Law of multiple proportion

B
Law of conservation of mass

C
Law of conservation of energy

D
Law of definite proportion

Solution

The Law of conservation of energy is not one of the laws of chemical combinations.

Hence, the correct option is C.
Question 7
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Which of these would support the idea that mass is conserved in a reaction that produces gas as a product?

A
Heating the reactants to ensure the reaction occurs in a gaseous state.

B
Subtracting the mass of the gas from the mass of the solid and liquid products.

C
Mixing the reactants and measuring their total mass.

D
Trapping the gas and measuring its mass.

Solution

As gas can escape after reaction so after its release trap the gas and finding its mass so that it will be in accordance with the law of conservation of mass.
Question 8
1 / -0

A solution contains 100 g of urea in 400 g of water. Calculate the concentration in terms of mass by mass percentage of the solution.

A
20

B
25

C
10

D
80

Solution

Mass of solute = 100 g

Mass of solvent = 400 g

Mass of solution = 400 + 100 = 500 g

Mass % of solution $= \dfrac{\text{Mass of solute} \times 100}{\text{Mass of solution}}$

$= \dfrac{100 \times 100}{500}$

$= 20$ %

Hence, the correct option is $\text{A}$
Question 9
1 / -0

Find the mass percentages (mass %) of Na, H, C, and O in sodium hydrogen carbonate.

A
30, 20, 45, 5

B
28, 1, 14, 57

C
24, 23, 12, 1

D
None of above

Solution

22.99 g (1 mol) of Na
12.01 (1 mol) of H
48.0 (1 mol) of C
48.0 (3 mole $\times$ 16.00 gram per mole) of O
The mass of one mole of $NaHCO_3$ is $22.99 g + 1.01 g + 12.01 g + 48.00 g = 84.01 g$
And the mass percentages of the elements are
mass % $Na = \dfrac{22.99 g}{84.01 g} \times 100 = 27.36$ %
mass % $H = \dfrac{1.01 g}{84.01 g} \times 100 = 1.20$ %
mass % $C = \dfrac{12.01 g}{84.01 g} \times 100 = 14.30$ %
mass % $O = \dfrac{48.00 g}{84.01 g} \times 100 = 57.14$ %
Question 10
1 / -0

Mass number is denoted by:

A
D

B
S

C
A

D
Z

Solution

The mass number (A) is the number of nucleons, which is the total number of protons and neutrons in the nucleus of an atom.

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Some Basic Concepts of Chemistry Test - 32

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Some Basic Concepts of Chemistry Test - 32

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Question 1

Law of definite proportion

Law of multiple proportion

Law of conservation of mass

Law of conservation of momentum

Solution

Question 2

Avogadro

Gay Lussac

Proust

Dalton

Solution

Question 3

One atom of hydrogen combines with 35.453 atoms of chlorine in this reaction

Hydrogen and chlorine atoms always combine in a 1:35 ratio.

Chlorine accepts hydrogen's electron to form polar covalent hydrogen chloride

Hydrogen and chlorine atoms aren't created or destroyed in the process so the reactant mass is the same as the product mass.

Solution

Question 4

It is used to calculate number of particles.

1 mole is equal to 6.023×10236.023\times10^{23}6.023×1023 particles.

It is the chemist's counting unit

All of the above

Solution

Question 5

molar mass of a compound/mass due to specific component ×\times× 100

mass due to specific component ×\times× 100

mass due to specific component/molar mass of a compound ×\times× 100

molar mass of a specific component/temperature ×\times× 100

Solution

Question 6

Law of multiple proportion

Law of conservation of mass

Law of conservation of energy

Law of definite proportion

Solution

Question 7

Heating the reactants to ensure the reaction occurs in a gaseous state.

Subtracting the mass of the gas from the mass of the solid and liquid products.

Mixing the reactants and measuring their total mass.

Trapping the gas and measuring its mass.

Solution

Question 8

20

25

10

80

Solution

Question 9

30, 20, 45, 5

28, 1, 14, 57

24, 23, 12, 1

None of above

Solution

Question 10

D

S

A

Z

Solution

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1 mole is equal to $6.023\times10^{23}$ particles.

molar mass of a compound/mass due to specific component $\times$ 100

mass due to specific component $\times$ 100

mass due to specific component/molar mass of a compound $\times$ 100

molar mass of a specific component/temperature $\times$ 100