Answer:
b. CH₂Cl₂ is more volatile than CH₂Br₂ because of the large dispersion forces in CH₂Br₂
Explanation:
CH₂Cl₂ is more volatile than CH₂Br₂ (b.p of CH₂Cl₂ = 39,6 °C; b.p of CH₂Br₂ = 96,95°C). Thus, c. and d. are FALSE
Dipole-dipole interactions in CH₂Cl₂ are greater than the dipole-dipole interactions in CH₂Br₂ because Cl is more electronegative that Br (Cl = 3,16; Br = 2,96). But this mean CH₂Cl₂ is less volatile than CH₂Br₂ but it is false.
There are large dispersion forces in CH₂Br₂ because Br has more electrons and protons than Cl. Large disperson forces mean CH₂Br₂ is less volatile than CH₂Cl₂ and it is true.
I hope it helps!
They have same electronic configuration
1. NA = Sodium, NA was the chemicals symbols for the element would be NA because in Latin it is natrium for natron in which is sodium in English, Sodium was created by Humphry Davy in the year of 1807. Element name Sodium, Chemical symbol NA atomic number, 11 Atomic mass 23.
2. AG = Sliver, AG was the chemicals symbols for the element would be AG because in Latin it is argentum in which in English means bright and also sliver. Element name Sliver Chemical symbol AG atomic number, 47 Atomic mass 108.
Answer: 2.00 V
Explanation:
The balanced redox reaction is:
Here Al undergoes oxidation by loss of electrons, thus act as anode. Copper undergoes reduction by gain of electrons and thus act as cathode.
Where both 
are standard reduction potentials.
![E^0_{[Al^{3+}/Al]}=-1.66V](https://tex.z-dn.net/?f=E%5E0_%7B%5BAl%5E%7B3%2B%7D%2FAl%5D%7D%3D-1.66V)
![E^0_{[Cu^{2+}/Cu]}=0.340V](https://tex.z-dn.net/?f=E%5E0_%7B%5BCu%5E%7B2%2B%7D%2FCu%5D%7D%3D0.340V)
![E^0=E^0_{[Cu^{2+}/Cu]}- E^0_{[Al^{3+}/Al]}](https://tex.z-dn.net/?f=E%5E0%3DE%5E0_%7B%5BCu%5E%7B2%2B%7D%2FCu%5D%7D-%20E%5E0_%7B%5BAl%5E%7B3%2B%7D%2FAl%5D%7D)
Thus the standard cell potential is 2.00 V
