Chemistry term 2 Flashcards
Describe the structure of silicon (IV) oxide.
Silicon(IV) oxide (SiO₂) has a macromolecular structure similar to that of diamond.
State 3 properties which silicon (IV) oxide and diamond have in common.
- High melting and boiling points
- Hardness
- Neither conducts electricity
How would you show that silicon (IV) oxide is acidic and not basic or amphoteric?
it readily reacts with strong bases like sodium hydroxide (NaOH) to form a silicate salt, while it does not react with acids
When ________ ions __________ electrons, reduction occurs.
When positive ions gain electrons, reduction occurs.
When ……………. ions …………….. electrons, oxidation occurs.
When negative ions lose electrons, oxidation occurs.
Where does reduction take place and where does oxidation take place?
Reduction takes place at the cathode and oxidation at the anode.
Calculate the mass of Cu and O₂ required to produce 286 grams of Cu₂O.
Cu= 190 grams are required
O₂= 256 grams are required
wrong answer go through it
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?
25.5g
What is an exothermic reaction?
A reaction that transfers energy to the surroundings, increasing temperature.
What is an endothermic reaction?
A reaction that absorbs energy from the surroundings, decreasing temperature.
What is activation energy?
The minimum energy required for a reaction to occur.
What is a reaction profile?
A diagram showing energy changes in a reaction.
How does a catalyst affect a reaction?
It lowers activation energy by providing an alternative reaction pathway.
In a reaction profile, where are the reactants and products in an exothermic reaction?
Products are lower than reactants because energy is released.
In a reaction profile, where are the reactants and products in an endothermic reaction?
Products are higher than reactants because energy is absorbed.
Is breaking bonds an exothermic or endothermic process?
Endothermic, as energy is absorbed.
Is making bonds an exothermic or endothermic process?
Exothermic, as energy is released.
How do you calculate the overall energy change in a reaction?
Total energy to break bonds minus total energy released forming bonds.
What does a negative overall energy change mean?
The reaction is exothermic (more energy released than absorbed).
What does a positive overall energy change mean?
The reaction is endothermic (more energy absorbed than released).
What happens to energy in combustion reactions?
Energy is released, making combustion an exothermic reaction.
What happens to energy in photosynthesis?
Energy is absorbed from sunlight, making it an endothermic reaction.
What is bond energy?
The energy required to break or form a covalent bond.
If a bond requires 500 kJ/mol to break, how much energy is released when it forms?
500 kJ/mol (bond making and breaking require the same energy).
How can you tell if a reaction is exothermic or endothermic from bond energy values?
If energy released in bond making is greater, it’s exothermic; if bond breaking requires more energy, it’s endothermic.
Why does electrolysis require energy input?
It is an endothermic reaction where energy is needed to break bonds in the reactants.
A compound contains 55.85% carbon, 6.97% hydrogen, and 37.18% oxygen by mass. What is the empirical formula?
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.
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?
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.
A compound contains 30.0% carbon, 5.0% hydrogen, and 65.0% oxygen by mass. What is the empirical formula?
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).
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?
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.
A compound contains 60.0% carbon, 10.0% hydrogen, and 30.0% oxygen by mass. What is the empirical formula?
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.
