The second one is more concentrated as they both times with the same thing but the second one (1.5) is bigger
Answer:
E° = 1.24 V
Explanation:
Let's consider the following galvanic cell: Fe(s) | Fe²⁺(aq) || Ag⁺(aq) | Ag(s)
According to this notation, Fe is in the anode (where oxidation occurs) and Ag is in the cathode (where reduction occurs). The corresponding half-reactions are:
Anode: Fe(s) ⇒ Fe²⁺(aq) + 2 e⁻
Cathode: Ag⁺(aq) + 1 e⁻ ⇒ Ag(s)
The standard cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.
E° = E°red, cat - E°red, an
E° = 0.80 V - (-0.44 V) = 1.24 V
Step 1-Light Dependent
CO2 and H2O enter the leaf
Step 2- Light Dependent
Light hits the pigment in the membrane of a thylakoid, splitting the H2O into O2
Step 3- Light Dependent
The electrons move down to enzymes
Step 4-Light Dependent
Sunlight hits the second pigment molecule allowing the enzymes to convert ADP to ATP and NADP+ gets converted to NADPH
Step 5-Light independent
The ATP and NADPH is used by the calvin cycle as a power source for converting carbon dioxide from the atmosphere into simple sugar glucose.
Step 6-Light independent
The calvin cycle converts 3CO2 molecules from the atmosphere to glucose
calvin cycle
The second of two major stages in photosynthesis (following the light reactions), involving atmospheric CO2 fixation and reduction of the fixed carbon into carbohydrate.
Explanation:
Mass of fructose = 33.56 g
Mass of water = 18.88 g
Total mass of the solution = Mass of fructose + Mass of water = M
M = 33.56 g + 18.88 g =52.44 g
Volume of the solution = V = 40.00 mL
Density =
a) Density of the solution:
b) Molar mass of fructose = 180.16 g/mol
Moles of fructose = 
Molar mass of water = 18.02 g/mol
Moles of water= 
Mole fraction of fructose in this solution:
Mole fraction of water = 
c) Average molar mass of of the solution:
=
d) Mass of 1 mole of solution = 42.50 g/mol
Density of the solution = 1.311 g/mL
d) Specific molar volume of the solution:
Answer:
E°(Ag⁺/Fe°) = 0.836 volt
Explanation:
3Ag⁺ + 3e⁻ => Ag°; E° = +0.800 volt
Fe° => Fe⁺³ + 3e⁻ ; E° = -0.036 volt
_________________________________
Fe°(s) + 3Ag⁺(aq) => Fe⁺³(aq) + 3Ag°(s) ...
E°(Ag⁺/Fe°) = E°(Ag⁺) - E°(Fe°) = 0.800v - ( -0.036v) = 0.836 volt
