Answer : The concentration of NaOH is, 0.336 M
Explanation:
To calculate the concentration of base, we use the equation given by neutralization reaction:

where,
are the n-factor, molarity and volume of acid which is 
are the n-factor, molarity and volume of base which is NaOH.
We are given:

Putting values in above equation, we get:


Thus, the concentration of NaOH is, 0.336 M
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:
The Aufbau Principle simply helps us determine electron configuration of an atom by stating that in the ground state of an atom or ion, electrons fill subshells of the lowest available energy level, then they fill subshells of higher energy level. For example, the 1s subshell is filled before the 2s subshell is occupied. Now, when trying to figure out the electron configuration of a calcium, you need to know its atomic number to determine its amount of total electrons. Calcium has an atomic number of 20, which means it has 20 protons and 20 electrons. First remember that the "s" subshell only holds 2 electrons, the "p" subshell only hold 6 electrons, and the "d" subshell only holds up to 10 electrons. Using the Aufbau principle below, we can determine that the first two electrons will go in the 1s orbital. Since 1s can only hold two electrons the next 2 electrons go in the 2s orbital. The next six electrons will go in the 2p orbital. The p orbital can hold up to six electrons. We'll put six in the 2p orbital and then put the next two electrons in the 3s. Since the 3s is now full we'll move to the 3p where we'll place the next six electrons. We now go to the 4s orbital where we place the remaining two electrons. With this, the calcium electron configuration will be:

Hope that helps you understand!