Answer:
The volume will be 0.031 L
Explanation:
Since temperature is constant, Boyle's law is applied in this case.
Boyle's law States that at constant temperature, the volume of a given mass of a gas is inversely proportional to it's pressure.
From this statement it was deduced that P1V1 = P2V2
From the question, P1 = 1atm, V1 = 1.55L, P2 = 50 atm and V2 is to be calculated.
V2 = P1V1/P2
= 1×1.55÷50
=0.031 L
Answer:
Scheme is Attached.
When 3,5 dimethyl-4- octene is react with ozone two products are obtain.
2-Methyl butanal and 2 pentanon.
The complete reaction is given in attached file.
For a candle to burn, it requires a spark, which provides the activation energy for the oxidation reaction of the hydrocarbon making the candle.
It also requires oxygen to facilitate the oxidation of the hydrocarbon.
Therefore the two main requirements of combustion of a candle are oxygen and a spark (or an initial flame)
Answer:
34.3 g
Explanation:
Step 1: Write the balanced equation
2 CH₃CH₂OH ⇒ CH₃CH₂OCH₂CH₃ + H₂O
Step 2: Calculate the moles corresponding to 50.0 g of CH₃CH₂OH
The molar mass of CH₃CH₂OH is 46.07 g/mol.
50.0 g × 1 mol/46.07 g = 1.09 mol
Step 3: Calculate the theoretical moles of CH₃CH₂OCH₂CH₃ produced
The molar ratio of CH₃CH₂OH to CH₃CH₂OCH₂CH₃ is 2:1. The moles of CH₃CH₂OCH₂CH₃ theoretically produced are 1/2 × 1.09 mol = 0.545 mol.
Step 4: Calculate the real moles of CH₃CH₂OCH₂CH₃ produced
The percent yield of the reaction is 85%.
0.545 mol × 85% = 0.463 mol
Step 5: Calculate the mass corresponding to 0.463 moles of CH₃CH₂OCH₂CH₃
The molar mass of CH₃CH₂OCH₂CH₃ is 74.12 g/mol.
0.463 mol × 74.12 g/mol = 34.3 g
Answer:
A single compound is simultaneously oxidized and reduced.
Explanation:
In chemistry, disproportionation is a simultaneous oxidation and reduction of a single chemical specie.
What this means is that; in a disproportionation reaction, only one compound is both oxidized and reduced. This implies that two products are formed during disproportionation. One is the oxidized product while the other is the reduced product.
Consider the disproportionation of CuCl shown below;
2CuCl -----> CuCl2 + Cu
Here, CuCl2 is the oxidized product while Cu is the reduced product.
