Some Basic Concepts Of Chemistry Test 9

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Some Basic Concepts Of Chemistry Test 9

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Weekly Quiz Competition

Question 1
1 / -0

For the reaction 2NaN₃(s) → 2Na(s) + 3N₂ (g), how many moles of sodium azide (NaN₃) must react in order to produce 8.75 moles of nitrogen gas?

A
13.1 moles NaN₃

B
2.50 moles NaN₃

C
2.00 moles NaN₃

D
5.83 moles NaN₃

Solution

2NaN₃ → 2Na + 3N₂
As above reaction is already balanced we can write:
2 moles of NaN₃ ≡ 3 moles of N₂
So, moles of NaN₃ ≡ 8.75 moles of N₂
8.75 x 2/3 = 5.83 moles of NaN₃.
Question 2
1 / -0

Amount of water produced by the combustion of 16g of methane is:

A
34 g

B
30 g

C
32 g

D
36 g

Solution

C H₄ + 2O₂ => CO₂ + 2 H₂O complete combustion with out CO.
1 mole of methane 16 grams reacts with 2 moles of oxygen = 64 grams . to produce 2 moles of water = 36 grams.
So 16 grams of methane produces 36 grams of water.
Question 3
1 / -0

Branch of chemistry in which quantities of reactants and products of a chemical reaction are determined is known as:

A
Thermodynamics

B
Stoichiometry

C
Chemical kinetics

D
Stereochemistry

Solution

Stoichiometry comes from the Greek “stoiechion” (element) and “metron” (to measure). As such, stoichiometry deals with determining the amounts of reactants and products that are consumed and produced within a given chemical reaction.
The stoichiometric coefficient of any species that does not participate in a given chemical reaction is zero.
The principles of stoichiometry are based upon the law of conservation of mass. Matter can neither be created nor destroyed, so the mass of every element present in the product(s) of a chemical reaction must be equal to the mass of each and every element present in the reactant(s).
Question 4
1 / -0

The mass of 1 Χ 10²² molecules of CuSO₄.5H₂O is:

A
41.59 g

B
415.9 g

C
4.14 g

D
None of these

Solution
Question 5
1 / -0

The mass of one mole of NaOH will be (At. mass of Na = 23, O = 16, H = 1)

A
16

B
40

C
80

D
23

Solution

Mass of 1 mole of NaOH = 23 + 16 + 1 = 40
Question 6
1 / -0

If 2 moles of H₂ and O₂ are available for the reaction, identify the limiting reagent for the reaction,
2H₂ + O₂ → 2H₂O

A
H₂

B
O₂

C
H₂O

D
Both H₂ and O₂

Solution

As we know that the limiting reagent is the substance that is totally consumed in reaction. So when we take 2 moles of H₂ and O₂ then the H₂ molecules are totally consumed in the reaction. So H₂ is the limiting reagent.
Question 7
1 / -0

Which of the following has more number of atoms?

A
1 mole of Neon gas

B
1 mole of Ozone gas

C
1 mole of Oxygen gas

D
1 mole of Carbon monoxide gas

Solution

Neon's atomicity is 1 Ozone's atomicity is 3 Oxygen's atomicity is 2 Carbon monoxide's atomicity is 2 Now Ozone has more atom in a molecule than any other gas in the options. So in 1 mole or 6.022×10²³ Ozone has more atom.
Question 8
1 / -0

Molecular mass of glucose molecule (C₆H₁₂O₆) is

A
172 g

B
182 g

C
180 g

D
192 g

Solution

Mass of C=12g
Mass of H =1g
Mass of O =16g
Thus,
Mass of C₆H₁₂O₆= 12×6+1×12+16×6
=72+12+96
=180g
Question 9
1 / -0

If 0.5 mole of BaCl₂ is mixed with 0.2 mole of Na₃PO₄ ,then maximum number of moles of Ba₃(PO₄)₂ that can be formed

A
0.5

B
0.1

C
0.3

D
0.7

Solution

Balanced equation for reaction between BaCl₂ and Na₃PO₄ is as follows:
3BaCl₂ + 2Na₃PO₄ → Ba₃(PO₄)₂ + 6NaCl
3 moles of BaCl₂ react with 2 moles of Na₃PO₄ to give 1 mole of Ba₃(PO₄)₂
0.5 moles of BaCl₂ will react with (2/3) x 0.5 = 0.33 moles of Na₃PO₄
Available moles of Na₃PO₄ = 0.2
So, Na₃PO₄ is the limiting reagent
Now, 2 moles of Na₃PO₄ give 1 mole of Ba₃(PO₄)₂
So, 0.2 moles of Na₃PO₄ will give ½ x 0.2 = 0.1 mole of Ba₃(PO₄)₂
Hence, maximum number of moles of Ba₃(PO₄)₂ formed = 0.1
Question 10
1 / -0

The atomic mass of an element is usually fractional because:

A
Elements contain impurities

B
Elements are mixture of isotopes

C
Elements are mixture of allotropes

D
Elements are mixture of isobars

Solution

The atomic masses of most elements are fractional because they exist of a mixture of Isotopes of different masses. the fractional atomic masses arise because of this mixture.
Mass =total mass of all atoms/number of atom.
Question 11
1 / -0

If you have 2 moles of A and 3 moles of B, Identify the limiting reagent in the following reaction.
A + B → AB

A
A

B
B

C
A as well as B

D
None

Solution

A + B → AB divide given moles of each reactant by their stoichiometric coefficient , the one with least ratio is limit reagents divide with their co efficient A=2/1 =2 B=3/1 =3. So A is the limiting reagent.
Question 12
1 / -0

According to the chemical equation,
Mg(s) + 2HCl → MgCl₂(aq) + H₂(g)
How many moles of magnesium are needed to produce 12.34 moles of hydrogen gas?

A
24.68 mol

B
3.09 mol

C
6.17 mol

D
12.34 mol

Solution

1mole of Mg gives 1mole water so, to produce 12.34 moles of water we required 12.34mole of Mg.
Question 13
1 / -0

Two different atoms A and B combine together to form AB. What will be AB?

A
Atom

B
Ion

C
Element

D
Molecule

Solution

A molecule is an electrically neutral group of two or more atoms held together by chemical bonds. Molecules are distinguished from ions by their lack of electrical charge.
Question 14
1 / -0

20 L of H₂ gas at S.T.P weigh about:

A
20 g

B
1.8 g

C
12.2 g

D
448 g

Solution

1 mole of hydrogen (about 2 g) occupies a volume of 22.4 L at S.T.P
Hence, weight of 20 L of H2 at S.T.P = (2 x 20)/22.4 = 1.8g
Question 15
1 / -0

One type of anaerobic respiration converts glucose, C₆H₁₂O₆, to ethanol C₂H₅OH and carbon dioxide. If the molecular weight of glucose is 180 grams/mol and the molar mass of ethanol is 50 g/mol how many grams of carbon dioxide are produced when 1 mol of glucose is digested via respiration?

A
88 grams

B
0 grams

C
40 grams

D
None of these

Solution

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Some Basic Concepts Of Chemistry Test 9

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Question 1

13.1 moles NaN3

2.50 moles NaN3

2.00 moles NaN3

5.83 moles NaN3

Solution

Question 2

34 g

30 g

32 g

36 g

Solution

Question 3

Thermodynamics

Stoichiometry

Chemical kinetics

Stereochemistry

Solution

Question 4

41.59 g

415.9 g

4.14 g

None of these

Solution

Question 5

16

40

80

23

Solution

Question 6

H2

O2

H2O

Both H2 and O2

Solution

Question 7

1 mole of Neon gas

1 mole of Ozone gas

1 mole of Oxygen gas

1 mole of Carbon monoxide gas

Solution

Question 8

172 g

182 g

180 g

192 g

Solution

Question 9

0.5

0.1

0.3

0.7

Solution

Question 10

Elements contain impurities

Elements are mixture of isotopes

Elements are mixture of allotropes

Elements are mixture of isobars

Solution

Question 11

A

B

A as well as B

None

Solution

Question 12

24.68 mol

3.09 mol

6.17 mol

12.34 mol

Solution

13.1 moles NaN₃

2.50 moles NaN₃

2.00 moles NaN₃

5.83 moles NaN₃

H₂

O₂

H₂O

Both H₂ and O₂