1)C. 2)A. 3)D. 4)C. 5)A. 6)A. 7)B. 8)B. 9)A. 10)B.
Answer:
25.08 grams of O₂ are needed to react with 8.15 g of C₂H₂.
Explanation:
The balanced reaction is:
2 C₂H₂ + 5 O₂ → 4 CO₂ + 2 H₂O
By reaction stoichiometry, the following amounts of moles of each compound participate in the reaction:
- C₂H₂: 2 moles
- O₂: 5 moles
- CO₂: 4 moles
- H₂O: 2 moles
The molar mass of each compound is:
- C₂H₂: 26 g/mole
- O₂: 32 g/mole
- CO₂: 44 g/mole
- H₂O: 18 g/mole
Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
- C₂H₂: 2 moles* 26 g/mole= 52 g
- O₂: 5 moles* 32 g/mole= 160 g
- CO₂: 4 moles* 44 g/mole= 176 g
- H₂O: 2 moles* 18 g/mole= 36 g
Then you can apply the following rule of three: if by stoichiometry 52 grams of C₂H₂ react with 160 grams of O₂, 8.15 grams of C₂H₂ react with how much mass of O₂?

mass of O₂= 25.08 grams
<u><em>25.08 grams of O₂ are needed to react with 8.15 g of C₂H₂.</em></u>
Answer:
The answer to your question is given below.
Explanation:
Aluminium has atomic number of 13. Thus, the electronic configuration of aluminium can be written as:
Al (13) —› 1s² 2s²2p⁶ 3s²3p¹
The orbital diagram is shown on the attached photo.
D.
If we take a sample of 100g of the given compound, we should find that it contains 74.1g of oxygen and 25.9g of nitrogen. The molar masses of oxygen and nitrogen are approximately 16.0g/mol and 14.0g/mol respectively. To find the appropriate number of moles of oxygen and nitrogen found in one mole of the given compound, we divide the mass of the sample by the molar mass of each substance.
For oxygen: 74.1g/(16.0g/mol) = 4.63125 mol (round up to 5 mol)
For nitrogen: 25.9g/(14.0g/mol) = 1.85 mol (round up to 2 mol)
Therefore one mole of this compound contains 2 moles of nitrogen and 5 moles of oxygen.