Answer:
Equilibrium constant is 0.4
Explanation:
We propose the equilibrium:
PCl₅(g) ⇄ PCl₃(g) + Cl₂(g)
Initially 0.72 mol
We have the 0.72 moles of the PCl₅.
React x x x
X amount has reacted, so in the equilibrium we have:
0.72 - x 0.4 0.4
If we initially have 0.72 moles and we have 0.4 moles at the equilibrium, it means that 0.3 moles has been reacted.
Let's make the expression for Kc:
Kc = [PCl₃] . [Cl₂] / [PCl₅]
Kc = 0.4 . 0.4 / 0.4
Kc = 0.4
The amount of solute needed is concentration.
Answer:
It leads to more frequent collisions, which increase reaction rate
Explanation:
An increase in reactant concentration leads to an increase in reactant particles which increases the number of collision in a reaction and then causes a reaction to proceed faster.
This belief was first suggested by the collision theory which states that for a reaction to occur, the reacting particles must collide with one another. However, further studies show that not all collisions lead to a reaction, and only effective collisions lead to a reaction.
Hence, an increase in reactant concentration increases the frequency of collision which in turn increases the chances of effective collision and subsequently sincreases the reaction rate.
The sign of ΔSsys for a solid explosive converts to a gas is positive.
The term used for the measurement of randomness and disorderliness in a system is known as Entropy (S). ΔS calculates the change in entropy and it is positive when entropy increase and negative when entropy decreases.
Entropy increases when heat is added to the system. Addition of heat increases the randomness of the molecules and hence the entropy increases.
When a solid explosive converts to a gas the randomness increases. Explosive causes heat and the solid is converted into gas. The randomness of molecules is maximum in gases thus the entropy increases. The sign of ΔS is positive for this process.
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Answer:
See explanation.
Explanation:
I highly suggest you watch OChem Tutor's videos on IUPAC nomenclature because the actual naming would take a lot of time to teach in text-based format. But here is how to name them:
1) I think there are two seperate pictures for number 1. The molecule on the left is 1-pentene and the one on the right is 4-methyl-1-pentene. If the whole thing is one molecule but there is just a bond missing where the red marker numbers are, that molecule would be 9-methyl-1,6-decadiene.
2) 4-methyl-2-pentene
3) 2,4-octadiene
4) 1,5-nonadiene
5) 2,5-dimethyl-3-hexene
6) 3,6-dimethyl-2,4-heptadiene
7) 2,5,5-trimethyl-2-hexene