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: The molar enthalpy change is 73.04 kJ/mol
Explanation:
moles of HCl= 
As NaOH is in excess 0.0415 moles of HCl reacts with 0.0415 moles of NaOH.
volume of water = 100.0 ml + 50.0 ml = 150.0 ml
density of water = 1.0 g/ml
mass of water = 
q = heat released
m = mass = 150.0 g
c = specific heat = 
= change in temperature = 
Thus 0.0415 mol of HCl produces heat = 3031.3 J
1 mol of HCL produces heat = 
Thus molar enthalpy change is 73.04 kJ/mol
Answer:
2Na⁺ (aq) and 2OH⁻(aq)
Explanation:
Spectator ions:
Spectator ions are those ions which are same on both side of chemical reaction. These ions are same in the reactant side and product side. Their presence can not effect the chemical equilibrium that's why when we write the net ionic equation these ions are neglect or omitted.
Given ionic equation:
Ba⁺²(aq) + 2OH⁻(aq) + 2Na⁺ (aq) + CO²⁻₃(aq) → BaCO₃(s) + + 2Na⁺ (aq) + 2OH⁻(aq)
In given ionic equation by omitting the spectator ions i.e, 2Na⁺ (aq) and 2OH⁻(aq) net ionic equation can be written as,
Net ionic equation:
Ba⁺²(aq) + CO²⁻₃(aq) → BaCO₃(s)
Answer:
A Planet
Explanation:
The earth for example, is a large body that orbits the sun, our local star
