Answer:
None of the alpha particles fired at the foil are being repelled back, like they were in the Rutherford atom simulation.I hope this is correct.
Answer:
When the volume of product increases.
When the weight of the product decreases.
Option (a) and (d) are correct.
Explanation:
The overall density of the product can be decreased:
a. Increase the volume of the product (and keep the material same)
d. Decrease the weight of the product ( and keep the same material ).
Density is calculated as the ratio of mass to the volume.
Density is inversely related to volume and directly related to mass.
So, when the mass of the product is increased then the density will increase keeping the material same. Density will also increase when the volume of the product decreased.
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:
O ice melts at 0°C that is the answer
A or D I think but go with A it makes more since