Asexual reproduction.<span>
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Hey there!
No of hybrid orbitals , H = ( V +S - C + A ) / 2
Where H = no . of hybrid orbitals
V = Valence of the central atom = 5
S = No . of single valency atoms = 4
C = No . of cations = 1
A = No . of anions = 0
For PCl4 +
Plug the values we get H = ( 5+4-1+0) / 2
H = 4 ---> sp3 hybridization
sp3 hybrid orbitals are used by phosphorous in the PCl4+ cations
Answer C
Hope that helps!
Answer:
1s2 2s2 2p2
Explanation:
it has 6 electrons in two energy levels so the sub levels are 1s, 2s and 2p
The solution would be like this for this specific problem:
<span>Given:
H2 = </span><span>2.6 atm
CL2 = 3.14 atm</span>
<span>
pressure H2 = 2.6 - x
pressure Cl2 = 3.14 - x
<span>pressure HBr = 2x = 1.13
x = 1.13 / 2 = 0.565
<span>pressure H2 = 2.6 - 0.565 = 2.035
pressure Br2 = 3.14 - 0.565 = 2.575
Kp = (1.13)^2 / 2.035 x 2.575</span></span></span>
= 1.2769 / (5.240125)
= 0.24367739319195629875241525726963
= 0.244
<span>Therefore, the Kp for the reaction at the given temperature
is 0.244.
To add, </span>the hypothetical pressure of a gas if
it alone occupied the whole volume of the original mixture at the same
temperature is called the partial pressure or Kp.
The most stable isotope would be lead-82.
