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kogti [31]
3 years ago
7

The electron pair in a C-F bond could be considered Question 3 options: closer to C because carbon has a larger radius and thus

exerts greater control over the shared electron pair closer to C because carbon has a lower electronegativity than fluorine an inadequate model since the bond is ionic centrally located directly between the C and F closer to F because fluorine has a higher electronegativity than carbon
Chemistry
1 answer:
Leokris [45]3 years ago
4 0

Answer:

closer to F because fluorine has a higher electronegativity than carbon

Explanation:

Electronegativity refers to the ability of an atom in a bonding situation to draw the shared electrons of the bond closer to itself.

Electronegativity increases across the period and decreases down the group. A highly electronegative atom draws the shared electron pair of a bond towards itself.

When two atoms are bonded together, the electron pair is always drawn closer to the atom that has a higher electronegativity.

Hence, the electron pair in a C-F bond could be considered closer to F because fluorine has a higher electronegativity than carbon.

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Calculate for the following electrochemical cell at 25°C, Pt H2(g) (1.0 atm) H (0.010 M || Ag (0.020 M) Ag if E (H) - +0.000 V a
viva [34]

Answer : The correct option is, (b) +0.799 V

Solution :

The values of standard reduction electrode potential of the cell are:

E^0_{[H^{+}/H_2]}=+0.00V

E^0_{[Ag^{+}/Ag]}=+0.799V

From the cell representation we conclude that, the hydrogen (H) undergoes oxidation by loss of electrons and thus act as anode. Silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.

The half reaction will be:

Reaction at anode (oxidation) : H_2\rightarrow 2H^{+}+2e^-    

Reaction at cathode (reduction) : Ag^{+}+1e^-\rightarrow Ag    

The balanced cell reaction will be,  

H_2+2Ag^{+}\rightarrow 2H^{+}+2Ag

Now we have to calculate the standard electrode potential of the cell.

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

E^o_{cell}=E^o_{[Ag^{+}/Ag]}-E^o_{[H^{+}/H_2]}

E^o_{cell}=(+0.799V)-(+0.00V)=+0.799V

Therefore, the standard cell potential will be +0.799 V

4 0
3 years ago
How many milliliters of a 0.40%(w/v) solution of nalorphine must be injected to obtain a dose of 1.5 mg?
ozzi
The number  of Ml  of  a  0.40 %w/v solution  of   ,nalorphine  that must  be injected  to  obtain  a  dose  of 1.5 mg is  calculated as  below


since M/v%   is  mass  of solute  in  grams per 100  ml

convert Mg to  g
1 g = 1000 mg  what  about  1.5 mg =?  grams
=   1.5 /1000 = 0.0015 grams


volume is therefore =  100 (  mass/ M/v%)

= 100  x(  0.0015/ 0.4) =  0.375  ML
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4 years ago
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Answer:

2.173 moles of ethanol is presented in a 100.0g sample of ethanol .

Explanation:

The amount of substance that contains as many Particles as there are atoms in exactly 12g of carbon- '12 isotope is called 1 mole '= 46 u.

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3 years ago
If stomach acid has a ph of 1.3, what is the pOH
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Which of these elements does not exist as a diatomic molecule?
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Answer:

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