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aleksandrvk [35]
3 years ago
12

Chlorine can bond with fluorine to form CIF. Chlorine can also bond with lithium to form LiCI Which compound will have a greater

partial charge?
Chemistry
1 answer:
Sergio [31]3 years ago
5 0

Answer: ClF will have a greater partial charge.

Explanation:

A polar covalent bond is defined as the bond which is formed when there is a low difference of electronegativities between the atoms, thus resulting in charge difference. Example: ClF

Non-polar covalent bond is defined as the bond which is formed when there is no difference of electronegativities between the atoms and thus there is no charge difference. Example: F_2

Ionic bond is formed when there is complete transfer of electron from a highly electropositive metal to a highly electronegative non metal. The electronegative difference between the elements is high. The charges on cation and anion neutralise each other. Example: LiCl

Thus as ClF will have greater partial charge.

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The standard reduction potentials of the following half-reactions are given in Appendix E in the textbook:
german

<u>Answer:</u>

<u>For 1:</u> The largest positive cell potential is of cell having 1st and 4th half reactions.

<u>For 2:</u> The standard electrode potential of the cell is 1.539 V

<u>For 3:</u> The smallest positive cell potential is of cell having 3rd and 4th half reactions. The standard electrode potential of the cell is 0.46 V

<u>Explanation:</u>

The substance having highest positive E^o potential will always get reduced and will undergo reduction reaction.

We are given:

Ag^++(aq.)+e^-\rightarrow Ag(s);E^o_{Ag^+/Ag}=0.799V\\\\Cu^{2+}+(aq.)+2e^-\rightarrow Cu(s);E^o_{Cu^{2+}/Cu}=0.337V\\\\Ni^{2+}(aq.)+2e^-\rightarrow Ni(s);E^o_{Ni^{2+}/Ni}=-0.28V\\\\Cr^{3+}(aq.)+3e^-\rightarrow Cr(s);E^o_{Cr^{3+}/Cr}=-0.74V

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

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

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

  • <u>Cell having 1st and 2nd half reactions:</u>

Silver has higher electrode potential. So, this will undergo reduction reaction and act as anode. Copper will undergo oxidation reaction and act as cathode.

E^o_{cell}=0.799-0.337=0.462V

  • <u>Cell having 1st and 3rd half reactions:</u>

Silver has higher electrode potential. So, this will undergo reduction reaction and act as anode. Nickel will undergo oxidation reaction and act as cathode.

E^o_{cell}=0.799-(-0.28)=1.079V

  • <u>Cell having 1st and 4th half reactions:</u>

Silver has higher electrode potential. So, this will undergo reduction reaction and act as anode. Chromium will undergo oxidation reaction and act as cathode.

E^o_{cell}=0.799-(-0.74)=1.539V

  • <u>Cell having 2nd and 3rd half reactions:</u>

Copper has higher electrode potential. So, this will undergo reduction reaction and act as anode. Nickel will undergo oxidation reaction and act as cathode.

E^o_{cell}=0.337-(-0.28)=0.617V

  • <u>Cell having 3rd and 4th half reactions:</u>

Nickel has higher electrode potential. So, this will undergo reduction reaction and act as anode. Chromium will undergo oxidation reaction and act as cathode.

E^o_{cell}=-0.28-(-0.74)=0.46V

Hence,

<u>For 1:</u> The largest positive cell potential is of cell having 1st and 4th half reactions.

<u>For 2:</u> The standard electrode potential of the cell is 1.539 V

<u>For 3:</u> The smallest positive cell potential is of cell having 3rd and 4th half reactions. The standard electrode potential of the cell is 0.46 V

8 0
3 years ago
Does anyone know what the formula would be for Rhenium V Biphosphate?
Georgia [21]

Answer:Re3(PO4)2 I think I'm wrong

Explanation:

8 0
3 years ago
In order to study hydrogen halide decomposition, a researcher fills an evacuated 1.11 L flask with 0.822 mol of HI gas and allow
Elza [17]

Answer:

Kc = 168.0749

Explanation:

  •           2HI(g)     ↔    H2(g) + I2(g)

initial mol:   0.822               0          0

equil. mol: 2(0.822 - x)         x           x

∴ [ HI ]eq = 0.055 mol/L = 2(0.822 - x) / (1.11 L )

⇒ 1.644 - 2x = 0.055 * 1.11

⇒ 1.644 = 2x + 0.06105

⇒ 2x = 1.583

⇒ x = 0.7915 mol equilibrium

⇒ [ H2 ] eq = 0.7915mol / 1.11L = 0.7130 M = [ I2 ] eq

⇒ Kc = ([ H2 ] * [ I2 ]) / [ HI ]²

⇒ Kc = ( 0.7130² ) / ( 0.055² )

⇒ Kc = 168.0749

 

4 0
3 years ago
The basic principle in balancing a chemical equation is to ______. View Available Hint(s) The basic principle in balancing a che
OlgaM077 [116]

Answer:

have the same number of atoms of each element in the reactants and in the products

Explanation:

<em>The basic principle in balancing a chemical equation would simply be to have the same number of atoms of each element in the reactants and in the products.</em>

<u>A balanced chemical equation is one that has the same number of atoms of each element on the reactant and the product's side of the equation.</u> For example, consider the equation below:

                     H_2 + O_2 --> H_2O

On the reactant's side, there are 2 atoms of H and O while there are 2 atoms of H and 1 atom of O on the product's side. This is an imbalanced equation. In order for it to be balanced, the number of atoms of H and O on the reactant side must be equal to the number of H and O on the product side as below.

                        2H_2 + O_2 --> 2H_2O

5 0
3 years ago
What is paper chromatography? Calculate the Rr value of a colored dye that traveled 52 mm on a chromatography strip while the so
Karo-lina-s [1.5K]

Answer:

Paper chromatography is a basic technique of chromatography. It consist in the separation of the mixe components using a solvent.

Explanation:

Paper chromatography is a basic technique of chromatography. It consists in the separation of the mixed components using a solvent.

Paper chromatography consists of put some dot of the mix using a glass capillary into a specialized paper, generally made of cellulose, this is called a stationary phase.

Then you put this paper into a camera of glass named, chromatography camera, where previously contain a solvent. The solvent also know as a mobile phase, the type can be defined before the test and involves a study of the kind of the mix, and the compound you want to separate.

The chromatography camera has to be closed all the time during the test, and you can't move at all because the movement of the solvent can alternate the result.

Very often, the solution of the solvent is a mix of different liquid substances with different polarities.

When the stationary phase put into the camera, the solvent starts to move up over the paper, until the separation of the compounds is observable.

the Rf is a value who relates the move of the mobile phase with the move of the distance traveled by the substance tested.

To undersant the paper chromatography, you can watch the images attached.

The first is an image of the chromatography camera.

The second one is an image of a cellulose paper after the chromatography is done. You can watch the dots who indicates the traveling of the compound across the paper.

The third one can show you the evolutions of paper chromatography, from the beginning to the end.

To calculate the Rf value you have to use the equation:

Rf = distance traveled by the substance/distance traveled by the solvent/

<u>Rf = 52mm/81mm =0.64</u>

<u></u>

3 0
3 years ago
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