Answer:
Cu(s) in Cu(NO₃)₂(aq)
Explanation:
The standard reduction potential (E°) is the energy necessary to reduce the atom in a redox reaction. When an atom reduces it gains electrons from other than oxides. As higher is E°, easily it will reduce. The substance that reduces is at the cathode of a cell, where the electrons go to, and the other that oxides are at the anode of the cell.
The standard reduction potentials from Al(s) and Cu(s) are, respectively, -1.66V and +0.15V, so the half-cell of Cu(s) in Cu(NO₃)₂(aq) is the cathode.
Since the question manages to include moles, pressure, volume, and temperature, then it is evident that in order to find the answer we will have to use the Ideal Gas Equation: PV = nRT (where P = pressure; V = volume; n = number of moles; R = the Universal Constant [0.082 L·atm/mol·K]; and temperature.
First, in order to work out the questions, there is a need to convert the volume to Litres and the temperature to Kelvin based on the equation:
250 mL = 0.250 L
58 °C = 331 K
Also, based on the equation P = nRT ÷ V
⇒ P = (2.48 mol)(0.082 L · atm/mol · K)(331 K) ÷ 0.250 L
⇒ P = (67.31 L · atm) ÷ 0.250 L
⇒ P = 269.25 atm
Thus the pressure exerted by the gas in the container is 269.25 atm.
Answer:
a) 
b) 
d) 
d) 
Explanation:
From the question we are told that:
Moles of N2 
Atmospheric pressure 
Temperature 

Initial heat 
a)
Generally the equation for change in temperature is mathematically given by

Where




b)
Generally the equation for ideal gas is mathematically given by

For v double


Therefore



Total Work-done 



c)
Generally the equation for amount of heat added is mathematically given by



d)
Generally the equation for change in internal energy of the gas is mathematically given by


