Answer:
σ -> 2sp²
π -> 2p
Explanation:
The carbon has valence shell 2s 2p, and, both of them make 3 σ bonds and 1 π bond. The π bond only occurs in multiple bonds.
The σ bonds happen at the hybrids orbitals, which are orbitals formed by the association of the pure orbitals (s, p, d, f). The hybridization occurs to make possible to the atom to do the bonds because the electrons need to be isolated in it.
On the other hand, the π bonds only occur at pure orbitals. The subshell s only has 1 orbital, and the subshell p has 3 orbitals. So, because there are 3 σ bonds, it's necessary 3 hybrids orbitals (1 of s + 2 of p).
The σ bonds happen at the orbital 2sp² and the π bond at the 2p pure orbital.
CaCl2 +Na2CO3 ====>Nacl+CaCO3
Answer:
Yes
Explanation:
By definition, the equilibrium constanct, Kc, for the reaction A ⇒ 2B is
= [A]^1 / [B]^2
Substitute [A] = 4 and [B] = 2 in the equation,
[A]^1 / [B]^2
= 4^1 / 2^2
= 1
= Kc
So yes the reaction is at equilibrium.
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.
