I wrote the answer on this paper and here is the calculations step by step
very sry the "V" must be replaced with 285 and the 285 must be replaced with "V" and the answer is 231.09 cm3. sorry for the inconvenience.
Answer:
7 orbitals
Explanation:
An f sublevel has 7 orbitals
Answer:

Explanation:
Hello,
In this case, the undergoing chemical reaction is:

Next, we identify the limiting reactant by computing the available moles of ethane and the moles of ethane consumed by 60.0 grams of oxygen:

Thus, we notice there are less available moles, for that reason, the ethane is the limiting reactant. Finally, we can compute the produced moles of water by:

Best regards.