I hope you understand my working:
1) Finding the mol of NH3 to find the mol of (NH4)2SO4 (ammonium sulfate)
2) Mr of (NH4)2SO4
3) Theoretical yield: The actual grams of (NH4)2SO4 produced when reacting 0.514 mol of NH3 to 0.514 mol H2SO4
4) Using formula of (given grams)/(theoretical grams or actual grams) * 100 = 73%
5) Basic algebra
The mass would be same because of the law of conservation which states that the mass of the reactants must equal to the mass of products
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:
Explanation:
Partial pressure of oil = mole fraction of oil x total pressure
mole fraction of oil = mole of oil / mole of water + mole of oil
= mole of oil = mass of oil / molecular weight of oil
= 20 / 100 = .2
mole of water = 80 / 18
= 4.444
mole fraction of oil = .2 / .2 + 4.444
= .2 / 4.644
Partial pressure of oil = mole fraction of oil x total pressure
= (.2 / 4.644 ) x 760 mm
= 32.73 mm Hg .
