Chemistry term 2

Question 1

Describe the structure of silicon (IV) oxide.

Answer 1

Silicon(IV) oxide (SiO₂) has a macromolecular structure similar to that of diamond.

Question 2

State 3 properties which silicon (IV) oxide and diamond have in common.

Answer 2

1. High melting and boiling points
2. Hardness
3. Neither conducts electricity

Question 3

How would you show that silicon (IV) oxide is acidic and not basic or amphoteric?

Answer 3

it readily reacts with strong bases like sodium hydroxide (NaOH) to form a silicate salt, while it does not react with acids

Question 4

When ________ ions __________ electrons, reduction occurs.

Answer 4

When positive ions gain electrons, reduction occurs.

Question 5

When ……………. ions …………….. electrons, oxidation occurs.

Answer 5

When negative ions lose electrons, oxidation occurs.

Question 6

Where does reduction take place and where does oxidation take place?

Answer 6

Reduction takes place at the cathode and oxidation at the anode.

Question 7

Calculate the mass of Cu and O₂ required to produce 286 grams of Cu₂O.

Answer 7

Cu= 190 grams are required
O₂= 256 grams are required
wrong answer go through it

Question 8

When 72g of aluminum (Al) reacts completely with oxygen, 102g of aluminum oxide (Al₂O₃) is formed. How much aluminum oxide will be formed when 18g of aluminum reacts?

Answer 8

25.5g

Question 9

What is an exothermic reaction?

Answer 9

A reaction that transfers energy to the surroundings, increasing temperature.

Question 10

What is an endothermic reaction?

Answer 10

A reaction that absorbs energy from the surroundings, decreasing temperature.

Question 11

What is activation energy?

Answer 11

The minimum energy required for a reaction to occur.

Question 12

What is a reaction profile?

Answer 12

A diagram showing energy changes in a reaction.

Question 13

How does a catalyst affect a reaction?

Answer 13

It lowers activation energy by providing an alternative reaction pathway.

Question 14

In a reaction profile, where are the reactants and products in an exothermic reaction?

Answer 14

Products are lower than reactants because energy is released.

Question 15

In a reaction profile, where are the reactants and products in an endothermic reaction?

Answer 15

Products are higher than reactants because energy is absorbed.

Question 16

Is breaking bonds an exothermic or endothermic process?

Answer 16

Endothermic, as energy is absorbed.

Question 17

Is making bonds an exothermic or endothermic process?

Answer 17

Exothermic, as energy is released.

Question 18

How do you calculate the overall energy change in a reaction?

Answer 18

Total energy to break bonds minus total energy released forming bonds.

Question 19

What does a negative overall energy change mean?

Answer 19

The reaction is exothermic (more energy released than absorbed).

Question 20

What does a positive overall energy change mean?

Answer 20

The reaction is endothermic (more energy absorbed than released).

Question 21

What happens to energy in combustion reactions?

Answer 21

Energy is released, making combustion an exothermic reaction.

Question 22

What happens to energy in photosynthesis?

Answer 22

Energy is absorbed from sunlight, making it an endothermic reaction.

Question 23

What is bond energy?

Answer 23

The energy required to break or form a covalent bond.

Question 24

If a bond requires 500 kJ/mol to break, how much energy is released when it forms?

Answer 24

500 kJ/mol (bond making and breaking require the same energy).

Question 25

How can you tell if a reaction is exothermic or endothermic from bond energy values?

Answer 25

If energy released in bond making is greater, it's exothermic; if bond breaking requires more energy, it's endothermic.

Question 26

Why does electrolysis require energy input?

Answer 26

It is an endothermic reaction where energy is needed to break bonds in the reactants.

Question 27

A compound contains 55.85% carbon, 6.97% hydrogen, and 37.18% oxygen by mass. What is the empirical formula?

Answer 27

Step 1: Assume 100 g of the compound: - Carbon: 55.85 g - Hydrogen: 6.97 g - Oxygen: 37.18 g Step 2: Convert grams to moles: - Moles of C: 55.85 / 12.01 = 4.65 - Moles of H: 6.97 / 1.008 = 6.92 - Moles of O: 37.18 / 16.00 = 2.32 Step 3: Divide by the smallest number of moles (2.32): - C: 4.65 / 2.32 = 2.00 - H: 6.92 / 2.32 = 3.00 - O: 2.32 / 2.32 = 1.00 Step 4: Empirical formula is C₂H₃O.

Question 28

A compound contains 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. What is the molecular formula if the molar mass is 180 g/mol?

Answer 28

Step 1: Assume 100 g of the compound: - Carbon: 40.0 g - Hydrogen: 6.7 g - Oxygen: 53.3 g Step 2: Convert grams to moles: - Moles of C: 40.0 / 12.01 = 3.33 - Moles of H: 6.7 / 1.008 = 6.64 - Moles of O: 53.3 / 16.00 = 3.33 Step 3: Divide by the smallest number of moles (3.33): - C: 3.33 / 3.33 = 1.00 - H: 6.64 / 3.33 = 2.00 - O: 3.33 / 3.33 = 1.00 Step 4: Empirical formula is CH₂O. Step 5: Calculate empirical formula mass (CH₂O): - Mass of C: 12.01 g/mol - Mass of H₂: 2 \* 1.008 = 2.016 g/mol - Mass of O: 16.00 g/mol - Empirical formula mass = 12.01 + 2.016 + 16.00 = 30.026 g/mol Step 6: Find the ratio (n): - n = 180 / 30.026 = 6 Step 7: Multiply the empirical formula by 6: - Molecular formula = C₆H₁₂O₆.

Question 29

A compound contains 30.0% carbon, 5.0% hydrogen, and 65.0% oxygen by mass. What is the empirical formula?

Answer 29

Step 1: Assume 100 g of the compound: - Carbon: 30.0 g - Hydrogen: 5.0 g - Oxygen: 65.0 g Step 2: Convert grams to moles: - Moles of C: 30.0 / 12.01 = 2.50 - Moles of H: 5.0 / 1.008 = 4.96 - Moles of O: 65.0 / 16.00 = 4.06 Step 3: Divide by the smallest number of moles (2.50): - C: 2.50 / 2.50 = 1.00 - H: 4.96 / 2.50 = 1.98 ≈ 2.00 - O: 4.06 / 2.50 = 1.62 ≈ 1.62 Step 4: Empirical formula is C₁H₂O₁.6, which can be rounded to C₁H₂O₂ (simplified).

Question 30

A compound contains 53.0% carbon, 7.0% hydrogen, and 40.0% oxygen by mass. What is the molecular formula if the molar mass is 132 g/mol?

Answer 30

Step 1: Assume 100 g of the compound: - Carbon: 53.0 g - Hydrogen: 7.0 g - Oxygen: 40.0 g Step 2: Convert grams to moles: - Moles of C: 53.0 / 12.01 = 4.42 - Moles of H: 7.0 / 1.008 = 6.94 - Moles of O: 40.0 / 16.00 = 2.50 Step 3: Divide by the smallest number of moles (2.50): - C: 4.42 / 2.50 = 1.77 - H: 6.94 / 2.50 = 2.78 ≈ 3 - O: 2.50 / 2.50 = 1.00 Step 4: Empirical formula is C₂H₃O. Step 5: Calculate empirical formula mass (C₂H₃O): - Mass of C: 12.01 \* 2 = 24.02 g/mol - Mass of H₃: 3 \* 1.008 = 3.024 g/mol - Mass of O: 16.00 g/mol - Empirical formula mass = 24.02 + 3.024 + 16.00 = 43.044 g/mol Step 6: Find the ratio (n): - n = 132 / 43.044 = 3 Step 7: Multiply the empirical formula by 3: - Molecular formula = C₆H₉O₃.

Question 31

A compound contains 60.0% carbon, 10.0% hydrogen, and 30.0% oxygen by mass. What is the empirical formula?

Answer 31

Step 1: Assume 100 g of the compound: - Carbon: 60.0 g - Hydrogen: 10.0 g - Oxygen: 30.0 g Step 2: Convert grams to moles: - Moles of C: 60.0 / 12.01 = 5.00 - Moles of H: 10.0 / 1.008 = 9.92 - Moles of O: 30.0 / 16.00 = 1.88 Step 3: Divide by the smallest number of moles (1.88): - C: 5.00 / 1.88 = 2.66 - H: 9.92 / 1.88 = 5.28 - O: 1.88 / 1.88 = 1.00 Step 4: Empirical formula is C₃H₅O.