The balanced chemical equation is written as:
<span>CsF(s) + XeF6(s) ------> CsXeF7(s)
We are given the amount of </span>cesium fluoride and <span>xenon hexafluoride used for the reaction. We need to determine first the limiting reactant to proceed with the calculation. From the equation and the amounts, we can say that the limiting reactant would be cesium fluoride. We calculate as follows:
11.0 mol CsF ( 1 mol </span>CsXeF7 / 1 mol CsF ) = 11.0 mol <span>CsXeF7</span>
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q = mC∆T
q = (30.0g)(0.900J/goC)(50oC)
q = 1350 J
So, the right answer is 1350 J
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Answer:
12 moles of F₂
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
N₂ + 3F₂ —> 2NF₃
From the balanced equation above,
3 moles of F₂ reacted to produce 2 moles of NF₃.
Finally, we shall determine the number of mole of F₂ needed to produce 8 moles of NF₃. This can be obtained as illustrated below:
From the balanced equation above,
3 moles of F₂ reacted to produce 2 moles of NF₃.
Therefore, Xmol of F₂ will react to produce 8 moles of NF₃ i.e
Xmol of F₂ = (3 × 8)/2
Xmol of F₂ = 12 moles
Thus, 12 moles of F₂ is needed for the reaction.
It is A since the tails are hydrophobic.
Answer:
2.78 moles of water are produced.
Explanation:
Given data:
Number of moles of H₂O produced = ?
Number of moles of oxygen react = 3.25 mol
Solution:
Chemical equation:
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
Now we will compare the moles of water with oxygen.
O₂ : H₂O
7 : 6
3.25 : 6/7×3.25 = 2.78 mol