The answer to this question is theory
I think the answer is D no change. Though you add more CO2, but the pressure is not mentioned. If the pressure is constant and the reaction is already balanced, the H2O is also saturation and can not absorb more CO2.
The δs∘rxn for the reaction 
→ 
will be -146 J/K.
Entropy would be a measurable physical characteristic and a scientific notion that is frequently connected to a condition of disorder, unpredictability, or uncertainty.
Entropy would be a measurement of the system's unpredictability or disorder. The entropy increases as randomness do. It has broad properties as well as a state function. It has the unit 
.
Entropy of the reaction can be calculated by the reaction.
Δ
= 2 mol × 
× 
- 1 mol × 
Δ = 2 mol × 240 J/mol.K - 2 mol × 210 J/mol.K-1 mol ×205.2 J/mol.K
Δ = -146.8 J/K
Therefore, the δs∘rxn for the reaction → will be -146 J/K.
To know more about reaction
brainly.com/question/20305735
#SPJ4
Answer:
C2H3Br + O2 → CO2 + H2O + HBr
Explanation:
The term balancing of chemical reaction equation has a unique meaning in chemistry. What it actually means is to ensure that the number of atoms of each element on the left hand side of reaction equation becomes equal to the number of atoms of the same element on the right hand side of the reaction equation.
When we look at the equation; C2H3Br + O2 → CO2 + H2O + HBr, the number of atoms of each element on the left and right hand sides of the given equation are not the same hence the equation is unbalanced.
If we look at the equation; 2C2H3Br + 5O2 → 4CO2 + 2H2O + 2HBr, the number of atoms of each element on both sides of the reaction equation are now equal, thus the later equation is the balanced version of the former.
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
