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DENIUS [597]
3 years ago
13

Help please if its right i'll give brainiest

Chemistry
1 answer:
Mkey [24]3 years ago
5 0
I believe it is spontaneous generation
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Which example does not involve knowledge of chemistry?
Trava [24]

I would honestly say C. All others have ingredients in them that makes them work, the combination of stuff to put out the fire, film is made from a certain material and drugs are an ovbious mixture. Looking at the stars is the only thing you don't need chemistry for.

6 0
3 years ago
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What’s the bladder wall made up of?
AleksandrR [38]

Answer:

The bladder wall is made of many layers, including: Urothelium or transitional epithelium. This is the layer of cells that lines the inside of the kidneys, ureters, bladder, and urethra. Cells in this layer are called urothelial cells or transitional cells

Explanation:

8 0
3 years ago
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svetoff [14.1K]
The answer is C: How long does it take distilled water to evaporate from a 5” diameter container?
5 0
3 years ago
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For the Zn - Cu^2+ voltaic cell Zn(s) + Cu^2+(aq, 1M) + Cu(s) E degree _cell = 1.10 V Given that the standard reduction potentia
Fittoniya [83]

Answer : The value of E^o_{(Cu^{2+}/Cu)} is, 0.34 V

Explanation :

Here, copper will undergo reduction reaction will get reduced. Zinc will undergo oxidation reaction and will get oxidized.

The oxidation-reduction half cell reaction will be,

Oxidation half reaction:  Zn\rightarrow Zn^{2+}+2e^-

Reduction half reaction:  Cu^{2+}+2e^-\rightarrow Cu

Oxidation reaction occurs at anode and reduction reaction occurs at cathode. That means, gold shows reduction and occurs at cathode and chromium shows oxidation and occurs at anode.

The overall balanced equation of the cell is,

Zn+Cu^{2+}\rightarrow Zn^{2+}+Cu

To calculate the E^o_{(Cu^{2+}/Cu)} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

E^o_{cell}=E^o_{(Cu^{2+}/Cu)}-E^o_{(Zn^{2+}/Zn)}

Putting values in above equation, we get:

1.10V=E^o_{(Cu^{2+}/Cu)}-(-0.76V)

E^o_{(Cu^{2+}/Cu)}=0.34V

Hence, the value of E^o_{(Cu^{2+}/Cu)} is, 0.34 V

8 0
3 years ago
How can you determine which bond in a structure is more polar without using an electronegativity table?
UkoKoshka [18]
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).
5 0
3 years ago
Read 2 more answers
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