Answer:
D
Explanation:
The electrons revolve around the nucleus and they contain negative charge
The question is incomplete, the complete question is
Which is NOT correct for when the silver and vanadium half-cells are connected via a salt bridge and a potentiometer? Ag^+ + 1 e^- rightarrow Ag Edegree = 0.7993 V V^2+ + 2e^- right arrow V E degree =-1.125 V Ag+ is reduced V is oxidized 1.924 V V2^+ is reduced Ag is oxidized I and II III, IV, and V I, II, and III III only IV and V
Answer:
only IV and V
Explanation:
If we look at the values of reduction potential for the two species, we will discover that vanadium has a negative reduction potential indicating its tendency towards oxidation.
On the other hand, solve has a positive reduction potential indicating a tendency towards reduction.
This implies that vanadium must be oxidized and silver reduced and not the not her way ground? Hence the answer above.
Answer: 2 moles
Explanation:
STP is Standard Temperature and Pressure. That means the pressure is 1.00 atm and the temperature is 273K. Since the oxygen is placed in the same container, we can use Ideal Gas Law to figure out what container the CO₂ used.
Ideal Gas Law: PV=nRT
P=1.00 atm
n=moles
R=0.08206 Latm/Kmol
T=273K
CO₂
Since we know that CO₂ has a 44.8 L container, we can use that to find the moles of oxygen.
There are 2 mol of oxygen.
Answer:
The shape of the BF3 molecule is best described as trigonal planar.
Explanation:
The Lewis Structure for BF3 is like this:
_ _
| F | | F |
\ /
B
|
| F |
---
It forms three angles of 120° each. The bonds are in the same planar that's why it is trigonal planar and they are exactly the same.
Boron and Fluorine have 3 covalent bonds, produced by electronic promotion that enables the 2py and 2pz orbitals, leaving an electron to pair in the 2px. So boron will have 3 possible electrons to pair in 2s1, 2px and 2py, remember that electronic configuration for B is 1s2, 2s2, 2p1
By hybridization between the orbitals 2s2 and 2p1, the electrons of F, can joined to make the covalent bond. The new B configuration is 1s2, 2s1, 2px1, 2py1 (these last three, hybrid orbitals)
