NH₂-CH₂-COOH + HNH-CH₂-COOH → NH₂-CH₂-CO-NH-CH₂-COOH + H₂O
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Answer:
82.08 %
Explanation:
- <em>The percent yield of the reaction = [(actual yield)/(calculated yield)] x 100.
</em>
Actual yield = 26.80 kg.
- <em><u>To get the calculated yield:
</u></em>
- The balanced equation of reacting N2 with H2 to produce NH3 is:
N₂ + 3H₂ → 2NH₃
- It is clear that 1.0 mole of N₂ reacts with 3.0 moles of H₂ to produce 2.0 moles of NH₃.
- N₂ is present in excess and H₂ is the limiting reactant.
- We need to convert the mass of H₂ added (5.79 kg) to moles using the relation:
n = mass /molar mass = (5790 g) / (2.01 g/mol) = 2880.6 mol.
- We can get the no. of moles of NH₃ produced.
<u><em>Using cross multiplication:
</em></u>
3.0 mole of H₂ produce → 2.0 moles off NH₃, from the stichiometry.
2880.6 mol of H₂ produces → ??? moles of NH₃.
- The no. of moles of NH₃ produced = (2880.6 mol)(2.0 mol) / (3.0 mol) = 1920.4 mol.
- We can know get the calculated yield of NH₃ = no. of moles x molar mass = (1920.4 mol) (17.00 g/mol) = 32646.76 g ≅ 32.65 kg.
∴ <em>The percent yield of the reaction = [(actual yield)/(calculated yield)] x 100 = </em>[(26.8 kg) / (32.65 kg)] x 100 <em>= 82.08 %.</em>
Answer: 6.02 x 10^25
Explanation:
In order to find the number of molecules you simply multiply the number of moles by 6.022 x 10^23.
1 mole is always equal to 6.022 x 10^23 no matter what element.
100.0 x 6.022 x 10^23 = 6.02 x 10^25
The answer only has three significant figures because 100.0 only has three.
Answer:
A will be much less than 0 degrees
Explanation:
See the graph
https://prnt.sc/10ahuzc
