Potassium 23.5g/39.0983g/mol = 0.601mol
The Ratio of reactants is 2 to 1 so (0.601mol)/2 = 0.3005mol
Therefore 0.3005mol of F2 is needed to find liters use
formula V = nRT/P (V)Volume = 22.41L
(T)Temperature = 273K or 0.0 Celsius
(P)Pressure = 1.0atm
<span>(R)value is always .08206 with atm n = 0.3005moles
(273)(.08206)(0.3005)/1 = V V = 6.7319 Liters</span>
Answer:
Part A. The half-cell B is the cathode and the half-cell A is the anode
Part B. 0.017V
Explanation:
Part A
The electrons must go from the anode to the cathode. At the anode oxidation takes place, and at the cathode a reduction, so the flow of electrons must go from the less concentrated solution to the most one (at oxidation the concentration intends to increase, and at the reduction, the concentration intends to decrease).
So, the half-cell B is the cathode and the half-cell A is the anode.
Part B
By the Nersnt equation:
E°cell = E° - (0.0592/n)*log[anode]/[cathode]
Where n is the number of electrons being changed in the reaction, in this case, n = 2 (Sn goes from S⁺²). Because the half-reactions are the same, the reduction potential of the anode is equal to the cathode, and E° = 0 V.
E°cell = 0 - (0.0592/2)*log(0.23/0.87)
E°cell = 0.017V
Answer:
A. Theoretical yield = 3.51g
B. %yield = 75%
Explanation:
The balanced equation for the reaction is given below:
C2H4 + 3O2 —> 2CO2 + 2H2O
Molar Mass of O2 = 16 x 2 = 32g/mol
Mass of O2 from the balanced equation = 3 x 32 = 96
Molar Mass of H20 = (2x1) + 16 = 18g/mol
Mass of H2O from the balanced equation = 2 x 18 = 36g
A. From the equation,
96g of O2 produced 36g of H2O
Therefore, 9.35g of O2 will produce = (9.35 x 36)/96 = 3.51g of H2O
Therefore,theoretical yield of water (H2O) = 3.51g
B. Theoretical yield = 3.51
Actual yield = 2.63g
%yield =?
%yield = Actual yield/Theoretical yield x 100
%yield = 2.63/3.51
%yield = 75%
There are 2 full orbitals and 2 electrons left over.
