Answer: A pattern of same atomic orbitals can be seen about elements in the same period with respect to electron structures.
Explanation:
The horizontal rows in a period table are called periods.
Elements present in the same period will have same atomic orbitals.
For example, electronic distribution of Na is 2, 8, 1 and it is a third period element.
Similarly, electronic distribution of Cl is 2, 8, 7 and it is also a third period element.
Hence, both Na and Cl will have K, L, M shells, that is, they have three atomic orbitals.
Thus, we can conclude that a pattern of same atomic orbitals can be seen about elements in the same period with respect to electron structures.
Answer:
The activation energy was reached was 10:45 a.m. The additional energy did not affect the reaction.
Explanation:
Answer: Option (b) is the correct answer.
Explanation:
The elements which have excess or deficiency of electrons will react readily.
Atomic number of Mn is 25 and electronic configuration of 
is [Ar]
. This configuration is stable.
Atomic number of Cr is 24 and electronic configuration of 
is [Ar]
. This configuration is not stable.
Atomic number of Fe is 26 and electronic configuration of 
is [Ar]
. This configuration is stable.
Atomic number of Cu is 29 and electronic configuration of 
is [Ar]
. This configuration is not stable.
Atomic number of Al is 13 and electronic configuration of Al is 
. This configuration is not stable.
Atomic number of Ba is 56 and electronic configuration of 
is [Kr]
. This configuration is stable.
Atomic number of Mg is 12 and electronic configuration of 
is 
. This configuration is stable.
Atomic number of Sn is 50 and electronic configuration of Sn is [Kr]
. This configuration is stable.
Thus, we can conclude that out of the given options, only Fe and reactants would lead to a spontaneous reaction as they have incomplete sub-shells. Therefore, in order to gain stability they will readily react.
A. It’s mass has changed if not that’s then it’s B.
Answer:
0.45M
Explanation:
The following were obtained from the question:
C1 = 0.75M
V1 = 100mL
V2 = 165mL
C2 =?
Applying the dilution formula C1V1 = C2V2, the concentration of the diluted solution can be calculated for as follows:
0.75 x 100 = C2 x 165
Divide both side by 165
C2 = (0.75 x 100) /165
C2 = 0.45M
The concentration of the diluted solution is 0.45M
