a. W = 0 J
b. W = - 308.028 J
<h3>Further explanation</h3>
Given
Nitrogen gas expands in volume from 1.6 L to 5.4 L
Required
The work done
Solution
Isothermal :
W = -P . ΔV
Input the value :
a. At a vacuum, P = 0
So W = 0
b. At pressure = 0.8 atm
W = - 0.8 x ( 5.4 - 1.6)
W = -3.04 L.atm ( 1 L.atm = 101.325 J)
W = - 3.04 x 101.325
W = - 308.028 J
<u>Answer:</u> The correct answer is Option B.
<u>Explanation:</u>
Decomposition is a type of chemical reaction in which larger compound breaks down into two or more smaller compounds.
Double displacement reactions is defined as the chemical reaction in which exchange of ions takes place.
Synthesis reaction is a type of reaction in which two or more smaller compounds combines to form a single large compound.
Single displacement reaction is a type of reaction in which a more reactive metal displaces a less reactive metal from its chemical reaction.
In stomach, an acid is present known as hydrochloric acid and to neutralize its effect, antacid is taken which has 
as a component.
The reaction between HCl and 
is a type of neutralization reaction and it is a type of double displacement reaction.
The equation between the two follows:
Hence, the correct answer is Option B.
2 Al+ 3 CuO-> 1 Al2O3+ 3Cu
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
