The balanced chemical reaction is:
<span>2 I2 + KIO3 + 6 HCl ---------> 5 ICl + KCl + 3 H2O
</span>
We are given the amount of the product to be produced from the reaction. This will be the starting point of our calculations.
28.6 g ICl (1 mol / 162.35 g ICl ) ( 2 mol I2 / 5 mol ICl ) ( 253.81 g I2 / 1 mol I2 ) = 17.88 g I2
<u>Answer:</u> The concentration of solute is 0.503 mol/L
<u>Explanation:</u>
To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

where,
= osmotic pressure of the solution = 24 atm
i = Van't hoff factor = 2 (for NaCl)
c = concentration of solute = ?
R = Gas constant = 
T = temperature of the solution = ![25^oC=[273+25]=298K](https://tex.z-dn.net/?f=25%5EoC%3D%5B273%2B25%5D%3D298K)
Putting values in above equation, we get:

Hence, the concentration of solute is 0.503 mol/L
Answer:
x = 2+
Explanation:
1) FADH2 + Q => FAD + QH2
Since H is added to Q
=> Reactant reduced is Q
(2) Balancing charges on both sides of the equation gives:
QH2 + 2 cyt c(Fe3+) => Q + 2 cyt c(Fe2+) + 2 H+
Thus x = 2+
It moves to a higher energy level<span />
Answer:
The major product formed is a benzyl allyl compound, namely 2-methylene propylbenzene
Explanation:
<u>Part 1.</u>
The product formed from this Wittig reaction with phosphonium ylide is a phenylallyl compound, 2-methylene propylbenzene. Structure and reaction synthesis is attached in file.
<u>Part 2</u>
The types of transformation involved are:
- alkylation (as there is a transfer of alkyl group)
- addition (reaction of ylide)