According to TABC laws on alcohol use, it is illegal to sell alcohol in many places in texas thus no you can't discount alcohol in Texas.
Alcohol is prohibited in many parts of Texas thus anyone under the age of 21 may not be sold alcohol, including adults. Beer and wine may be purchased from licensed establishments Monday through Friday from 7:00 a.m. to midnight. On Saturday, they may sell them from 7:00 a.m. until 1:00 p.m. Additionally, they may sell on Sundays from noon to midnight. Only liquor stores, nevertheless, are allowed to sell distilled alcohol. On Sundays before noon, restaurants are allowed to serve alcoholic beverages with food. On the other hand, it is prohibited to sell any alcoholic beverages with a volume of more than 17% on Sunday (ABV). On Thanksgiving Day, New Year's Day, or Christmas Day, no establishment may offer any alcoholic beverages. Alcohol from a single producer cannot be sold by a restaurant or other retailer. And alcohol cannot be sold by food trucks.
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The answer is Three
!!!!!!
To know this you pretty much do have to kind of memorize a few electronegativities. I don't recall ever getting a table of electronegativities on an exam.
From the structure, you have:
I remember the following electronegativities most because they are fairly patterned:
EN
H
=
2.1
EN
C
=
2.5
EN
N
=
3.0
EN
O
=
3.5
EN
F
=
4.0
EN
Cl
=
3.5
Notice how carbon through fluorine go in increments of
~
0.5
. I believe Pauling made it that way when he determined electronegativities in the '30s.
Δ
EN
C
−
Cl
=
1.0
Δ
EN
C
−
H
=
0.4
Δ
EN
C
−
C
=
0.0
Δ
EN
C
−
O
=
1.0
Δ
EN
O
−
H
=
1.4
So naturally, with the greatest electronegativity difference of
4.0
−
2.5
=
1.5
, the
C
−
F
bond is most polar, i.e. that bond's electron distribution is the most drawn towards the more electronegative compound as compared to the rest.
When the electron distribution is polarized and drawn towards a more electronegative atom, the less electronegative atom has to move inwards because its nucleus was previously favorably attracted to the electrons from the other atom.
That means generally, the greater the electronegativity difference between two atoms is, the shorter you can expect the bond to be, insofar as the electronegative atom is the same size as another comparable electronegative atom.
However, examining actual data, we would see that on average, in conditions without other bond polarizations occuring:
r
C
−
Cl
≈
177 pm
r
C
−
C
≈
154 pm
r
C
−
O
≈
143 pm
r
C
−
F
≈
135 pm
r
C
−
H
≈
109 pm
r
O
−
H
≈
96 pm
So it is not necessarily the least electronegativity difference that gives the longest bond.
Therefore, you cannot simply consider electronegativity. Examining the radii of the atoms, you should notice that chlorine is the biggest atom in the compound.
r
Cl
≈
79 pm
r
C
≈
70 pm
r
H
≈
53 pm
r
O
≈
60 pm
So assuming the answer is truly
C
−
C
, what would have to hold true is that:
The
C
−
F
bond polarization makes the carbon more electropositive (which is true).
The now more electropositive carbon wishes to attract bonding pairs from chlorine closer, thereby shortening the
C
−
Cl
bond, and potentially the
C
−
H
bond (which is probably true).
The shortening of the
C
−
Cl
bond is somehow enough to be shorter than the
C
−
C
bond (this is debatable).
A stable arrangement of eight valence electrons : ³⁵Cl⁻¹
<h3>Further explanation</h3>
Chlorine is a halogen gas, located in group 17, p block
Chlorine has an atomic number of 17 and an atomic mass of 35
Electron configuration: [Ne] 3s²3p⁵
If we look at the electron configuration, then Cl will bind 1 more electron so that the configuration is stable like Argon (atomic number 18)
So by binding this one electron, chlorine forms negative ions (anions)
³⁵Cl⁻¹
B. Cl⁻² binds 2 electrons, exceeding the octet rule
C. Cl⁺¹, releases 1 electron, remains unstable
D. Cl, the neutral form of Cl, is still unstable with a 7-electron valence configuration
Answer:
In a long channel MOSFET, the width of the pinch-off region is assumed small relative to the length of the channel. Thus, neither the length nor the voltage across the inversion layer change beyond the pinch-off, resulting in a drain current independent of drain bias. Consequently, the drain current saturates.
Explanation:
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