Surface tension under water results from greater attraction of liquid molecules to each other, due to a process called cohesion, than to molecules in the air, due to a process called adhesion.
Answer:
water was added to powdered rock
Explanation:
Answer:
163.2g
Explanation:
First let us generate a balanced equation for the reaction. This is shown below:
4Al + 3O2 —> 2Al2O3
From the question given, were were told that 3.2moles of aluminium was exposed to 2.7moles of oxygen. Judging by this, oxygen is excess.
From the equation,
4moles of Al produced 2moles of Al2O3.
Therefore, 3.2moles of Al will produce = (3.2x2)/4 = 1.6mol of Al2O3.
Now, let us covert 1.6mol of Al2O3 to obtain the theoretical yield. This is illustrated below:
Mole of Al2O3 = 1.6mole
Molar Mass of Al2O3 = (27x2) + (16x3) = 54 + 48 =102g/mol
Mass of Al2O3 =?
Number of mole = Mass /Molar Mass
Mass = number of mole x molar Mass
Mass of Al2O3 = 1.6 x 102 = 163.2g
Therefore the theoretical of Al2O3 is 163.2g
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
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