The relative volumes of chloroform and water that should be used is 9:10
Concentration of solution in chloroform = 
( moles of chloroform )
Concentration of solution in water = 
( moles of water )
Dissociation constant at 
; 
Concentration of solution in chloroform / Concentration of solution in water
Meaning;
Since mole is present in chloroform and mole is present in water, Total mole of Caffeine present is 
Now, we substitute our given values into the equation
Therefore, the relative volumes of chloroform and water that should be used is 9:10
Learn more; brainly.com/question/11060225
The electrons and the nuclei will settle into positions that minimize repulsion and maximize attraction.
SOLVENT- A substance (usually a liquid) capable of dissolving one or more pure substances. SOLUTE- Solid, liquid or gas that is dissolved in a solvent. SOLUTION- A homogeneous (looks the same throughout) mixture of a solvent and one or more solutes. AQUEOUS SOLUTION- Solution in which water is the solvent.
Answer
it raises the boiling point substance is dissolved in water
Answer : The correct option is, (b) +0.799 V
Solution :
The values of standard reduction electrode potential of the cell are:
![E^0_{[H^{+}/H_2]}=+0.00V](https://tex.z-dn.net/?f=E%5E0_%7B%5BH%5E%7B%2B%7D%2FH_2%5D%7D%3D%2B0.00V)
![E^0_{[Ag^{+}/Ag]}=+0.799V](https://tex.z-dn.net/?f=E%5E0_%7B%5BAg%5E%7B%2B%7D%2FAg%5D%7D%3D%2B0.799V)
From the cell representation we conclude that, the hydrogen (H) undergoes oxidation by loss of electrons and thus act as anode. Silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.
The half reaction will be:
Reaction at anode (oxidation) : 
Reaction at cathode (reduction) : 
The balanced cell reaction will be,
Now we have to calculate the standard electrode potential of the cell.
![E^o_{cell}=E^o_{[Ag^{+}/Ag]}-E^o_{[H^{+}/H_2]}](https://tex.z-dn.net/?f=E%5Eo_%7Bcell%7D%3DE%5Eo_%7B%5BAg%5E%7B%2B%7D%2FAg%5D%7D-E%5Eo_%7B%5BH%5E%7B%2B%7D%2FH_2%5D%7D)
Therefore, the standard cell potential will be +0.799 V
Answer:
D.) Products are weakly favored
Explanation:
For the reaction:
2SO₃ ⇄ O₂ + 2SO₂ + 198kJ/mol
The kc is defined as:
kc = [O₂] [SO₂]² / [SO₃]²
As the kc is 8,1:
8,1 [SO₃]² = [O₂] [SO₂]²
The products are favored 8,1 times. This is a weakly favored because the usual kc are in the order of 1x10⁴. Thus, right answer is:
D.) Products are weakly favored
I hope it helps!
