Answer:
The first high part is Q4, then the low part is Q7, the following high part is Q6, and the energy moving from the next two high points is Q5.
Explanation:
The first high part is Q4, then the low part is Q7, the following high part is Q6, and the energy moving from the next two high points is Q5 because of the diagram.
Explanation:
Formula to calculate hybridization is as follows.
Hybridization =
where,
V = number of valence electrons present in central atom
N = number of monovalent atoms bonded to central atom
C = charge of cation
A = charge of anion
So, hybridization of
is as follows.
Hybridization =
=
= 2
Hybridization of
is sp. Therefore,
is a linear molecule. There will be only two electron groups through which Be is attached.
Similarly, hybridization of
is calculated as follows.
Hybridization =
=
= 5
Therefore, hybridization of
is
is also a linear molecule. Though there are three lone pair of electrons present on a xenon atom and it is further attached with fluorine atoms through two electron pairs. Hence, there are in total five electron groups.
Thus, we can conclude that out of the given options
is the correct examples of linear molecules for five electron groups.
P=18000000/6 zeros. not sure how to do rest
Answer:
0.189 g.
Explanation:
- This problem is an application on <em>Henry's law.</em>
- Henry's law states that the solubility of a gas in a liquid is directly proportional to its partial pressure of the gas above the liquid.
- Solubility of the gas ∝ partial pressure
- If we have different solubility at different pressures, we can express Henry's law as:
<em>S₁/P₁ = S₂/P₂,</em>
S₁ = 0.0106/0.792 = 0.0134 g/L and P₁ = 0.321 atm
S₂ = ??? g/L and P₂ = 5.73 atm
- So, The solubility of the gas at 5.73 atm (S₂) = S₁.P₂/P₁ = (0.0134 g/L x 5.73 atm) / (0.321 atm) = 0.239 g/L.
<em>The quantity in (g) = S₂ x V = (0.239 g/L)(0.792 L) = 0.189 g.</em>
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Answer:
true
covalent bonds are between non metals and nonmetals. and they are sharing electrons.