Question

Please help me in questions for my final!! Can you tell me if they are correct?!

Answer 1

What is the electron configuration of an electrically neutral atom of sodium?
What is the electron configuration of an electrically neutral atom of neon?
What is the electron configuration of an electrically neutral atom of oxygen?
Answer:
Sodium is in period 3, so it is before the complications of the d orbital that arise in period 4. It is in group 1, so it will have an electron configuration of:
1s22s22p63s1
Neon comes before the complications of the d orbital as well, and it is the last element in period 2. This means it will complete the p orbital, so:
1s22s22p6
Oxygen is period 2 group 16. It will fill up 4 of the six spaces in the d orbital, so:
1s22s22p4

Answer 2

Methane (CH4), ammonia (NH3), and oxygen (O2) can react to form hydrogen cyanide (HCN) and water according to this equation:
CH4 + NH3 + O2  HCN + H2O
You have 8 g of methane and 10 g of ammonia in excess oxygen. Answer the following questions:
• What is the balanced equation for this reaction?
• Which reagent is limiting? Explain why.
• How many grams of hydrogen cyanide will be formed? Show your work.
Answer:
The balanced equation for this reaction is 2CH4 + 2NH3 + 3O2  2HCN + 6H2O. The reagent that is limiting is the CH4. This is because:
10 g of NH3 = 0.59 moles (10/17) and 8 g of CH4 is 0.5 moles (8/16). You have excess oxygen, so you need to find the limiting reagent. So, 0.588/2 = 0.294, and 0.5/2 = 0.25. Therefore, CH4 is your limiting reagent, and this is what you need to use to find the amount of HCN produced. (0.5 times 27) = 13.5 grams.

Answer 3

The little squares should be =