Answer:
The heat gain by the system,
q
=
−
250
kJ
.
The work done on the system ,
w
=
−
500
kJ
.
The First Law of Thermodynamics state that
Δ
U
=
q
+
w
=
−
750
kJ
Explanation:
Answer:
<h3>A.</h3>
Explanation:
If A does not work, please use the alteritive choice, which is D!
が機能しない場合は、別の選択肢であるDを使用してください。 ^ 3 ^
<h3>Answer:</h3>
Excess Reagent = NBr₃
<h3>Solution:</h3>
The Balance Chemical Equation for the reaction of NBr₃ and NaOH is as follow,
2 NBr₃ + 3 NaOH → N₂ + 3 NaBr + 3 HBrO
Calculating the Limiting Reagent,
According to Balance equation,
2 moles NBr₃ reacts with = 3 moles of NaOH
So,
40 moles of NBr₃ will react with = X moles of NaOH
Solving for X,
X = (40 mol × 3 mol) ÷ 2 mol
X = 60 mol of NaOH
It means 40 moles of NBr₃ requires 60 moles of NaOH, while we are provided with 48 moles of NaOH which is Limited. Therefore, NaOH is the limiting reagent and will control the yield of products. And NBr₃ is in excess as some of it is left due to complete consumption of NaOH.
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Answer:
1) acetylide
2) enol
3) aldehydes
4) tautomers
5) alkynes
6) Hydroboration
7) Keto
8) methyl ketones
Explanation:
Acetylide anions (R-C≡C^-) is a strong nucleophile. Being a strong nucleophile, we can use it to open up an epoxide ring by SN2 mechanism. The attack of the acetylide ion occurs from the backside of the epoxide ring. It must attack at the less substituted side of the epoxide.
Oxomercuration of alkynes and hydroboration of alkynes are similar reactions in that they both yield carbonyl compounds that often exhibit keto-enol tautomerism.
The equilibrium position may lie towards the Keto form of the compound. Usually, if terminal alkynes are used, the product of the reaction is a methyl ketone.
