Answer:
Explanation:
1. the 1/2 reaction that occurs at the cathode
3Cl2(g) +6e^- -------------> 6Cl^- (aq)
2 the 1/2 reaction that occurs at the anode
2MnO2(s) + 8OH^-(aq) ----------> 2MnO4^- (aq) + 4H2O(l) +6e^-
2MnO2(s) + 8OH^-(aq) ----------> 2MnO4^- (aq) + 4H2O(l) +6e^-
E0 = -0.59v
3Cl2(g) +6e^- -------------> 6Cl^- (aq)
E0 = 1.39v
3Cl2 (g) + 2MnO2 (s) + 8OH^(−) (aq)---------> 6Cl^(−) (aq) + 2MnO4^(−) (aq) + 4H2O (l)
E0cell = 0.80v
Answer:me being a tanned ginger
Explanation: i cant date u
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.
Answer:
Helum (He)g will escape faster
Explanation:
the phenomemenon can be explained by the Graham's law of diffusion.
Graham's law of difussion states that the rate of difussion is inversely proportional to the square root of the molecular mass,which means the gas with lower molecular mass will escape faster.
Helium gas has a molecular mass of 4 while Neon has a molecular mass of 10.
rate of diffusion of He/rate of difussion of Ne=√4/10=√0.4=0.63
It means He(g) will move 0.63 times faster than Ne(g) under the same condition
