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MAXImum [283]
3 years ago
7

Compare and contrast and give the definitions of: Partial + charge Partial - charge

Chemistry
2 answers:
sineoko [7]3 years ago
5 0
When two elements of different electronegativity forms a compound, -ve charge is shifted to more electronegative element & +ve towards less electronegative.

So, the charge on less electronegative element in polar compound is known as "Partial +ve charge" and charge on more electronegative element in Polar compound is known as "Partial -ve charge"

Hope this helps!
Aloiza [94]3 years ago
3 0
Partial charges are formed in molecules when electronegativity difference between is small. 

Partial charges formed can be permanent as well temporary. Inductive effect creates permanent partial charge in the molecule, whereas electromeric effect creates temporary partial charge in the molecule.

Inductive effect: when a covalent bond is formed between the atoms of different electronegatives, the electron density is more towards the more electronegative(EN) atom of the bond which results in the formation of a polar covalent bond. This polarity developed in molecule creates dipole moment.


Electromeric effect: It is temporary delocalisation of 
π-electron in compound containing multiple covalent bond( double and triple bond).

Partial negative charge is formed when more EN element is attached to the molecule, whereas partial negative charge is formed when less EN element is attached to the molecule.

Definitions:

1.) Partial positive charge: The positive charge developed in a molecule which is non-integral is called partial positive charge. Denoted by +
δ.
2.) Partial negative charge: The negative charge developed in a molecule which is non-integral is called partial negative charge. Denoted by -δ.
Hope this helps!






 


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Viktor [21]

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9.8×10^-4...... is the answer

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2 years ago
What is the name of the the following formula Cd(HOOCCOO)2
Andreyy89

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4 0
2 years ago
What type of compound is formed when a secondary (2°) alcohol is treated with Jones' reagent?
EastWind [94]

Answer:

Option D, ketone

Explanation:

Since Jones reagent (CrO3/H2SO4) is a strong oxidizing agent and oxidize the secondary alcohol to ketone.

Example , isopropanol gets oxidized to propanone.

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6 0
3 years ago
It took 55 days for a radioactivity of 1.75 x 1012 Bq to remain 0.135 Ci. What is the half-life of this radioactivity?
Mnenie [13.5K]

From the calculations, the half life of the material is 6.5 days.

<h3>What is radioactivity?</h3>

The term radioactivity has to do with the spontaneous disintegration of a specie.

Uisng the formula;

N=Noe^-kt

N= amount at time t = 0.135 Ci or 4.995 ×10^9 Bq

No = amount initially present =  1.75 x 10^12 Bq

k = rate constant = ?

t = time taken = 55 days

Hence;

4.995 ×10^9  = 1.75 x 10^12e^-55k

4.995 ×10^9/1.75 x 10^12 = e^-55k

2.85 * 10^-3 = e^-55k

ln2.85 * 10^-3 = -55k

k = ln2.85 * 10^-3/-55

k = 0.1066 day-1

Half life = 0.693/ 0.1066 day-1

= 6.5 days

Learn more about radioactivity:brainly.com/question/1770619

#SPJ1

4 0
1 year ago
In the activity, click on the E∘cell and Keq quantities to observe how they are related. Use this relation to calculate Keq for
inessss [21]

Answer:

The standard cell potential of the reaction is 0.78 Volts.

Explanation:

Cu^{2+}(aq)+Fe(s)\rightarrow Cu(s)+Fe^{2+}(aq)

Reduction at cathode :

Cu^+(aq)+2e^-\rightarrow Cu(s)

Reduction potential of  Cu^{2+} to Cu=E^o_{1}=0.34 V

Oxidation at anode:

Fe(s)\rightarrow Fe^{2+}(aq)+2e^-

Reduction potential of  Fe^{2+} to Fe=E^o_{2}=-0.44 V

To calculate the E^o_{cell} of the reaction, we use the equation:

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

Putting values in above equation, we get:

E^o_{cell}=0.34V -(-0.44 V)=0.78 V

The standard cell potential of the reaction is 0.78 Volts.

3 0
3 years ago
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