Which element requires the least amount of
energy to remove the most loosely held electron
from a gaseous atom in the ground state?
<h3>Answer-</h3><h3>Na</h3>
Answer:
B and C
Explanation:
When we have to do a buffer solution we always have to choose the reaction that has the <u>pKa closer to the desired pH value</u>. When we find the pKa values we will obtain:
![pKa_1=-Log[6.9x10^-^3]=2.16](https://tex.z-dn.net/?f=pKa_1%3D-Log%5B6.9x10%5E-%5E3%5D%3D2.16)
![pKa_2=-Log[6.2x10^-^8]=7.20](https://tex.z-dn.net/?f=pKa_2%3D-Log%5B6.2x10%5E-%5E8%5D%3D7.20)
![pKa_3=-Log[4.8x10^-^13]=12.31](https://tex.z-dn.net/?f=pKa_3%3D-Log%5B4.8x10%5E-%5E13%5D%3D12.31)
The closer value is pKa2 with a value of 7.2. Therefore we have to use the second reaction. In which 
is the <u>acid</u> and 
is the <u>base</u>. Therefore the answer for the first question is B and the answer for the second question is C.
Answer:
orbital
Explanation:
electrons are found in an orbital
Metals are all in the center of the periodic table and are all made up of metals
Answer:
ΔHr = -103,4 kcal/mol
Explanation:
<u>Using:</u>
<u>AH° (kcal/mol)
</u>
<u>Metano (CH)
</u>
<u>-17,9
</u>
<u>Cloro (CI)
</u>
<u>tetraclorometano (CCI)
</u>
<u>- 33,3
</u>
<u>Acido cloridrico (HCI)
</u>
<u>-22</u>
It is possible to obtain the ΔH of a reaction from ΔH's of formation for each compound, thus:
ΔHr = (ΔH products - ΔH reactants)
For the reaction:
CH₄(g) + Cl₂(g) → CCl₄(g) + HCl(g)
The balanced reaction is:
CH₄(g) + 4Cl₂(g) → CCl₄(g) + 4HCl(g)
The ΔH's of formation for these compounds are:
ΔH CH₄(g): -17,9 kcal/mol
ΔH Cl₂(g): 0 kcal/mol
ΔH CCl₄(g): -33,3 kcal/mol
ΔH HCl(g): -22 kcal/mol
The ΔHr is:
-33,3 kcal/mol × 1 mol + -22 kcal/mol× 4 mol - (-17,9 kcal/mol × 1 mol + 0kcal/mol × 4mol)
<em>ΔHr = -103,4 kcal/mol</em>
<em></em>
I hope it helps!
