Electrochemical cell representation for above reaction is,
Br-/Br2//I2/I-
Reaction at Anode: Br2 + 2e- → 2Br- (1)
Reaction at Cathode: 2I- → I2 + 2e- (2)
Standard reduction potential for Reaction 1 = Ered(anode) = 1.066 v
Standard reduction potential for Reaction 2 = Ered(cathode) = 0.535 v
Eo cell = Ered(cathode) - Ered(anode)
= 0.535 - 1.066
= -0.531v
Now, we know that ΔGo = -nF (Eo cell) ..............(3)
Also, ΔGo = RTln(K) ..........(4)
Equation 3 and 4 we get,
ln (K) = nF (Eo cell) / RT
= 2 X 96500 X (-0.531)/ (8.314 X 298)
∴ K = 1.085 X 10^-18.
Answer : The half-life at this temperature is, 3.28 s
Explanation :
To calculate the half-life for second order the expression will be:
![t_{1/2}=\frac{1}{k\times [A_o]}](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%3D%5Cfrac%7B1%7D%7Bk%5Ctimes%20%5BA_o%5D%7D)
When,
= half-life = ?
![[A_o]](https://tex.z-dn.net/?f=%5BA_o%5D)
= initial concentration = 0.45 M
k = rate constant = 
Now put all the given values in the above formula, we get:
Therefore, the half-life at this temperature is, 3.28 s
Absorbance measures the ability of the substance to absorb light at a specific wavelength.
Absorbance is also equal to the product of molar absorptivity, path length and molar concentration.
The mathematical expression is given as:
(1)
where, A = absorbance
= molar absorptivity
l = path length
c = molar concentration.
The above formula is said to Beer's Law.
Absorptivity of protein x = 
Path length = 1 cm
Molar concentration = 
Put the values in formula (1)
= 
Thus, absorbance at 280 nm =
gas to liquid
Explanation:
The change of state indicated by this analogy is from gas to liquid.
Cylinder to the left is filled with gases
Cylinder to the right is made up of liquid.
- Gases occupy the volumes of containers they are introduced into.
- They are random and possess a high kinetic energy.
- Liquids have definite volume and flow with one another.
- The gases in A are dispersed and in random motion.
- This phase change is called condensation
learn more:
Phase change brainly.com/question/1875234
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Answer:
3 regions of high electronic density
2 simple links
1 double bond
Explanation:
Hello!
First we calculate the total valence electrons that make up the molecule (6 + 6 * 3 = 24)
We locate six electrons forming the bonds and complete the octecs with the remaining 18 electrons.
We move 2 electrons of an oxygen to complete the sulfur octet.
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