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
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We know that the element Z = 119 would be placed right below the Fr, in the column of the alcaline metals.
We also know that the trend in the electronegativity is to decrease when you go up-down ia group.
The known electronegativities of the elements of this group are:
Li: 0.98
Na: 0.93
K: 0.82
Rb: 0.82
Cs: 0.79
Fr: 0.70
Then the hypotetical element Z = 119 would probably have an electronegativity slightly below 0.70, for sure in the range 0.60 - 0.70.
Cells divide in asexual reproduction!
I hope this helps
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Due to the variables that each inquiry has, Scientific inquiry follows a path of questioning and testing a hypothesis, however this changes in response to specific details.
I’m pretty sure the answer is “plant cell”
