Answer: D. Potassium bromide
I think it’s the third option but I’m not entirely sure
The theoretical yield of NaBr given that 2.36 moles of FeBr₃ reacts is 7.08 moles
<h3>Balanced equation </h3>
2FeBr₃ + 3Na₂S → Fе₂S₃ + 6NaBr
From the balanced equation above,
2 moles FeBr₃ reacted to produce 6 moles of NaBr
<h3>How to determine the theoretical yield of NaBr</h3>
From the balanced equation above,
2 moles FeBr₃ reacted to produce 6 moles of NaBr
Therefore,
2.36 moles FeBr₃ will react to produce = (2.36 × 6) / 2 = 7.08 moles of NaBr
Therefore,
Thus, the theoretical yield of NaBr is 7.08 moles
Learn more about stoichiometry:
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Answer:
N - 1s²2s²2p³
Explanation:
Nitrogen is located in the p-block of the periodic table (groups 13-18) and is on the 2nd period.
The 2nd period tells us the principal energy level (a quantum number) is n = 2. Therefore, it must have already filled up the 1s sublevel.
The groups 13-18 on period 2 tells us that the 2s sublevel is also filled.
Nitrogen is located in Group 15. That means that there are 3 electrons that have filled the 2p sublevel, out of a possible 6.
Therefore, our electron configuration is 1s²2s²2p³
2p³ (Shorthand Config)
[He] 2s²2p³ (Noble Gas Config)
