In the lake that the rivers lead, water molecules evaporate into the sky and form clouds. In the sky, these water droplets condense and form clouds that will eventually rain.
According to the balanced equation of the reaction:
2C2H2 + 5O2 → 4CO2 + 2H2O
So we can mention all as liters,
A) as we see that 2 liters of C2H2 react with 5 liters of oxygen to produce 4 liters of CO4 and 2 liters of H2O
So, when we have 75L of CO2
and when we have 2 L of C2H2 reacts and gives 4 L of CO2
2C2H2 → 4CO2
∴ The volume of C2H2 required is:
= 75L / 2
= 37.5 L
B) and, when we have 75 L of CO2
and 4CO2 → 2H2O
∴ the volume of H2O required is:
= 75 L /2
= 37.5 L
C) and from the balanced equation and by the same way:
when 5 liters O2 reacts to give 4 liters of CO2
and we have 75 L of CO2:
5 O2 → 4 CO2
?? ← 75 L
∴ the volume of O2 required is:
= 75 *(5/4)
= 93.75 L
D) about the using of the number of moles the answer is:
no, there is no need to find the number of moles as we called everything in the balanced equation by liters and use it as a liter unit to get the volume, without the need to get the number of moles.
Answer:
Mass = 99.8 g
Explanation:
Given data:
Mass of potassium nitride = ?
Mass of nitrogen produced = 10.65 g
Solution:
Chemical equation:
2K₃N→ 6K + N₂
Moles of nitrogen:
Number of moles = mass/ molar mass
Number of moles = 10.65 g / 28 g/mol
Number of moles = 0.38 mol
Now we will compare the moles of nitrogen with potassium nitride.
N₂ ; K₃N
1 : 2
0.38 : 2×0.38 =0.76
Mass of potassium nitride:
Mass = molar mass × number of moles
Mass = 131.3 g/mol × 0.76 mol
Mass = 99.8 g
3.
protium (A = 1), deuterium (A = 2), and tritium (A = 3).
Answer:The product formed on reaction with hydroxide ion as nucleophile is 2R-hexane-2-ol.
The product formed on reaction with water would be a 50:50 mixture of
2S-hexane-2-ol. and 2R-hexane-2-ol.
Explanation:
2S-iodohexane on reactiong with hydroxide ion would undergo SN² substitution reaction that is substitution bimolecular. Hydroxide ion has a negative charge and hence it is a quite good nucleophile .
The rate of a SN² reaction depends on both the substrate and nucleophile . Here the substrate is a secondary carbon center having Iodine as a leaving group.SN² reaction takes place here as hydroxide ion is a good nucleophile and it can attack the secondary carbon center from the back side leading to the formation of 2R-hexane-2-ol.
In a SN² reaction since the the nucleophile attacks from the back-side so the product formation takes place with the inversion of configuration.
When the same substrate S-2-iodohexane undergoes a substitution reaction with water as a nucleophile then the reaction occurs through (SN¹) substitution nucleophilic unimolecular mechanism .
The rate of a SN¹ reaction depends only on the nature of substrate and is independent of the nature of nucleophile.
The SN¹ reaction is a 2 step reaction , in the first step leaving group leaves leading to the formation of a carbocation and once the carbocation is formed then any weaker nucleophile or even solvent molecules can attack leading the formation of products.
In this case a secondary carbocation would be generated in the first step and then water will attack this carbocation to form the product in the second step.
The product formed on using water as a nucleophile would be a racemic mixture of R and S isomers of hexane -2-ol in 50:50 ratio. The two products formed would be 2R-hexane-2-ol and 2S-hexane-2-ol.
Kindly refer the attachment for reaction mechanism and structure of products.