Nascent oxygen has much higher reactivity than the oxygen bubbled through the reaction mixture. It doesn't stay nascent for long (you are right about it being converted quick to just O2), which is why it has to be generated in situ
P1 * V1 ÷ T1 = P2 * V2 ÷ T2
45 * 1.20 ÷ 314 = 96 * V2 ÷ 420
30,144 * V2 = 22,680
V2 = 22,680 ÷ 30,144
The new volume is approximately 0.75 liter.
I hope I helped
Answer:
2-Methylpentane is a branched-chain alkane with the molecular formula C6H14. It is a structural isomer of hexane composed of a methyl group bonded to the second carbon atom in a pentane chain.
Step 1 : Write balanced chemical equation
The balanced chemical equation for the reaction between iodine gas and chlorine gas is given below.
Step 2 : Set up ICE table
We will set up an ICE table for the above reaction
Following points are considered while drawing ICE table
- The initial concentration of product is assumed as 0
- The change in concentration (C) is assumed as x. Change (x) is negative for reactants and positive for products
- The coefficients in balanced equation are considered while writing C values
Check attached file for ICE table
Step 3 : Set up equilibrium constant equation
The equation for equilibrium constant can be written as
![K_{eq} = \frac{[ICl]^{2}}{[I_{2}][Cl_{2}]}](https://tex.z-dn.net/?f=%20K_%7Beq%7D%20%3D%20%5Cfrac%7B%5BICl%5D%5E%7B2%7D%7D%7B%5BI_%7B2%7D%5D%5BCl_%7B2%7D%5D%7D%20%20%20%20%20)
Step 4 : Solving for x
Keq at 298.15 K is given as 81.9
Let us plug in the equilibrium values (E) for I₂, Cl₂ and ICl from ICE table
81.9 = 
81.9 = 
![(2x)^{2} = 81.9 [ x^{2} -0.706x + 0.118]](https://tex.z-dn.net/?f=%20%282x%29%5E%7B2%7D%20%3D%2081.9%20%5B%20x%5E%7B2%7D%20-0.706x%20%2B%200.118%5D%20)
Solving the above equation using quadratic formula we get
x = 0.488 or x = 0.254
The value 0.488 cannot be used because the change (C) cannot be greater that initial concentration of the reactants.
Therefore the change in concentration of the gases during the reaction is 0.254 M
Hence, x = 0.254 M
From the ICE table, we know that the equilibrium concentration of ICl is 2x
![[ICl]_{eq} = 2 ( 0.254) = 0.508 M](https://tex.z-dn.net/?f=%20%5BICl%5D_%7Beq%7D%20%3D%202%20%28%200.254%29%20%3D%200.508%20M%20)
The concentration of ICl when the reaction reaches equilibrium is 0.508 M
This question is testing to see how well you understand the "half-life" of radioactive elements, and how well you can manipulate and dance around them. This is not an easy question.
The idea is that the "half-life" is a certain amount of time. It's the time it takes for 'half' of the atoms in any sample of that particular unstable element to 'decay' ... their nuclei die, fall apart, and turn into nuclei of other elements.
Look over the table. There are 4,500 atoms of this radioactive substance when the time is 12,000 seconds, and there are 2,250 atoms of it left when the time is ' y ' seconds. Gosh ... 2,250 is exactly half of 4,500 ! So the length of time from 12,000 seconds until ' y ' is the half life of this substance ! But how can we find the length of the half-life ? ? ?
Maybe we can figure it out from other information in the table !
Here's what I found:
Do you see the time when there were 3,600 atoms of it ?
That's 20,000 seconds.
... After one half-life, there were 1,800 atoms left.
... After another half-life, there were 900 atoms left.
... After another half-life, there were 450 atoms left.
==> 450 is in the table ! That's at 95,000 seconds.
So the length of time from 20,000 seconds until 95,000 seconds
is three half-lifes.
The length of time is (95,000 - 20,000) = 75,000 sec
3 half lifes = 75,000 sec
Divide each side by 3 : 1 half life = 25,000 seconds
There it is ! THAT's the number we need. We can answer the question now.
==> 2,250 atoms is half of 4,500 atoms.
==> ' y ' is one half-life later than 12,000 seconds
==> ' y ' = 12,000 + 25,000
y = 37,000 seconds .
Check:
Look how nicely 37,000sec fits in between 20,000 and 60,000 in the table.
As I said earlier, this is not the simplest half-life problem I've seen.
You really have to know what you're doing on this one. You can't
bluff through it.
