Substitution Reactions are those reactions in which one nucleophile replaces another nucleophile present on a substrate. These reactions can take place via two different mechanism i.e SN¹ or SN². In SN¹ substitution reactions the leaving group leaves first forming a carbocation and nucleophile attacks carbocation in the second step. While in SN² reactions the addition of Nucleophile and leaving of leaving group take place simultaneously.
Example:
OH⁻ + CH₃-Br → CH₃-OH + Br⁻
In above reaction,
OH⁻ = Incoming Nucleophile
CH₃-Br = Substrate
CH₃-OH = Product
Br⁻ = Leaving group
Organic reactions are typically slower than ionic reactions because in organic compounds the covalent bonds are first broken, this breaking of bonds is a slower step, while, in ionic compounds no bond breakage is required as it consists of ions, so only bond formation takes place which is a quicker and fast step.
Answer:
= 9.28 g CO₂
Explanation:
First write a balanced equation:
CH₄ + 2O₂ -> 2H₂O + CO₂
Convert the information to moles
7.50g CH₄ = 0.46875 mol CH₄
13.5g O₂ = 0.421875 mol O₂
Theoretical molar ratio CH₄:O₂ -> 1:2
Actual ratio is 0.46875 : 0.421875 ≈ 1:1
If all CH₄ is used up, there would need to be more O₂
So O₂ is the limiting reactant and we use this in our equation
Use molar ratio to find moles of CO₂
0.421875 mol O₂ * 1 mol CO₂/2 mol O₂=0.2109375 mol CO₂
Then convert to grams
0.2109375 mol CO₂ = 9.28114 g CO₂
round to 3 sig figs
= 9.28 g CO₂
Answer:
A) if the system is isothermal then all the heat added to the system will be used to do work (since none is used to raise the temperature of the gas). The heat added will be equal to the work done = 340 J
B) change in internal energy of the system of the process is isothermal will be zero, since there is no rise in temperature.
C) an adiabatic process is one involving no heat loss or gain through the system, Therefore heat gain will be zero
D) if the process is adiabatic then there is no heat loss or gain through the system and hence there is no change in temperature. Change in internal energy will be zero
E) if the process is isobaric then, there is no work done and the total heat to the system is equal zero
F) if there is no work done, and no heat added, then the internal energy will be equal zero.
C. Of the products is equal to the reactants.
Good luck out there! :)
