Answer:
A
Explanation:
In a galvanic cell, energy is produced by spontaneous chemical processes.
The cathode and anode of this cell will depend on the relative position of the two metals in the electrochemical series.
Aluminium is higher in the electrochemical series so aluminium will be the anode. Silver is lower in the electrochemical series so silver will be the cathode.
Recall that oxidation (electron loss) occurs at the anode while reduction (electron gain) occurs at the cathode.
I think the correct answer from the choices listed above is option 3. It is the compound Silver bromide that is insoluble to water. It has a very low solubility which is approximately 0.14 milligram per liter of water. Hope this answers the question. Have a nice day.
Answer:
there is 2% of hydrogen and 98% of nitrogen (mass percent)
Explanation:
assuming ideal gas behaviour
P*V=n*R*T
n= P*V/(R*T)
where P= pressure=1.02 atm , V=volume=7.47 L , T=absolute temperature= 296 K and R= ideal gas constant = 0.082 atm*L/(mole*K)
thus
n= P*V/(R*T) = 1.02 atm*7.47 L/( 296 K * 0.082 atm*L/(mole*K)) = 0.314 moles
since the number of moles is related with the mass m through the molecular weight M
n=m/M
thus denoting 1 as hydrogen and 2 as nitrogen
m₁+m₂ = mt (total mass)
m₁/M₁+m₂/M₂ = n
dividing one equation by the other and denoting mass fraction w₁= m₁/mt , w₂= m₂/mt , w₂= 1- w₁
w₁/M₁+w₂/M₂ = n/mt
w₁/M₁+(1-w₁) /M₂ = n/mt
w₁*(1/M₁- 1/M₂) + 1/M₂ = n/mt
w₁= (n/mt- 1/M₂) /(1/M₁- 1/M₂)
replacing values
w₁= (n/mt- 1/M₂) /(1/M₁- 1/M₂) = (0.314 moles/3.48 g - 1/(14 g/mole)) /(1/(1 g/mole)-1/(14 g/mole))= 0.02 (%)
and w₂= 1-w₁= 0.98 (98%)
thus there is 2% of hydrogen and 98% of nitrogen
