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wlad13 [49]
1 year ago
10

Potassium carbonate (K2CO3) has one potassium atom bonded to carbon, and that carbon is also bonded to three oxygens. Their elec

tronegativities are shown in the list:
K 0.82
C 2.55
O 3.44
How would you describe the bonds (K-C and C-O) in this molecule?
A: K-C is polar covalent, and C-O is nonpolar covalent.
B: K-C is ionic, and C-O is nonpolar covalent.
C: K-C is ionic, and C-O is polar covalent.
D: K-C is polar covalent, and C-O is polar covalent.
Chemistry
1 answer:
dybincka [34]1 year ago
3 0

Answer:

I'm not all the way sure but I think it is C

Explanation:

The reason is because an ionic bond essentially donates an electron to the other atom participating in the bond. Also, the C-O I think is polar since it is evenly distributed.

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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
What is the role of blood in the transportation of materials throughout the body
scZoUnD [109]
Blood carries hormones to specific organs.
7 0
3 years ago
How would these cells appear under a microscope at a higher magnification.
gulaghasi [49]

they would be more detailed? what type of cells though?


5 0
3 years ago
Study this chemical reaction: (aq)(s)(s)(aq) Then, write balanced half-reactions describing the oxidation and reduction that hap
disa [49]

The question is incomplete, complete question is:

Study this chemical reaction:

FeSO_4 (aq) + Zn (s)\rightarrow Fe (s) + ZnSO_4 (aq)

Then, write balanced half-reactions describing the oxidation and reduction that happen in this reaction.

Oxidation:

Reduction:

Answer:

Oxidation taking place in given reaction :

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

Reduction taking place in given reaction;

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

Explanation:

Redox reaction is defined as the reaction in which oxidation and reduction reaction occur side by side.

Oxidation reaction is defined as the chemical reaction in which an atom looses its electrons. The oxidation number of the atom gets increased during this reaction.

X\rightarrow X^{n+}+ne^-

Reduction reaction is defined as the chemical reaction in which an atom gains electrons. The oxidation number of the atom gets reduced during this reaction.

X^{n+}+ne^-\rightarrow X

FeSO_4 (aq) + Zn (s) \rightarrow Fe (s) + ZnSO_4 (aq)

In the given reaction, iron(II) ions are getting reduced and zinc metal is getting oxidized to zinc(II) ions.

Oxidation :

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

Reduction ;

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

6 0
2 years ago
Helppppppppppppppppppp me
masya89 [10]
The answer is A.
A pure substance is pure, so it cannot be separated in most cases.
8 0
3 years ago
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