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torisob [31]
3 years ago
12

Need help on stoichiometry asap please

Chemistry
1 answer:
lbvjy [14]3 years ago
8 0
1. Convert 135g Fe2O3 to moles, divide by 2, and multiply the resulting number by the molar mass of Al
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__________ ___________ can release different forms of energy such as electricity or light, in addition to the release of heat en
Stells [14]

Answer:

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Explanation:

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8 0
3 years ago
Read 2 more answers
An electrochemical cell at 25°C is composed of pure copper and pure lead solutions immersed in their respective ionis. For a 0.6
ExtremeBDS [4]

Answer :

(a) The concentration of Pb^{2+} is, 0.0337 M

(b) The concentration of Pb^{2+} is, 6.093\times 10^{32}M

Solution :

<u>(a) As per question, lead is oxidized and copper is reduced.</u>

The oxidation-reduction half cell reaction will be,

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

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

The balanced cell reaction will be,  

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

Here lead (Pb) undergoes oxidation by loss of electrons, thus act as anode. Copper (Cu) undergoes reduction by gain of electrons and thus act as cathode.

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

E^o_{[Pb^{2+}/Pb]}=-0.13V

E^o_{[Cu^{2+}/Cu]}=+0.34V

E^o=E^o_{[Cu^{2+}/Cu]}-E^o_{[Pb^{2+}/Pb]}

E^o=0.34V-(-0.13V)=0.47V

Now we have to calculate the concentration of Pb^{2+}.

Using Nernest equation :

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Pb^{2+}]}{[Cu^{2+}]}

where,

n = number of electrons in oxidation-reduction reaction = 2

E_{cell} = 0.507 V

Now put all the given values in the above equation, we get:

0.507=0.47-\frac{0.0592}{2}\log \frac{[Pb^{2+}]}{(0.6)}

[Pb^{2+}]=0.0337M

Therefore, the concentration of Pb^{2+} is, 0.0337 M

<u>(b) As per question, lead is reduced and copper is oxidized.</u>

The oxidation-reduction half cell reaction will be,

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

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

The balanced cell reaction will be,  

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

Here Copper (Cu) undergoes oxidation by loss of electrons, thus act as anode. Lead (Pb) undergoes reduction by gain of electrons and thus act as cathode.

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

E^o_{[Pb^{2+}/Pb]}=-0.13V

E^o_{[Cu^{2+}/Cu]}=+0.34V

E^o=E^o_{[Pb^{2+}/Pb]}-E^o_{[Cu^{2+}/Cu]}

E^o=-0.13V-(0.34V)=-0.47V

Now we have to calculate the concentration of Pb^{2+}.

Using Nernest equation :

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cu^{2+}]}{[Pb^{2+}]}

where,

n = number of electrons in oxidation-reduction reaction = 2

E_{cell} = 0.507 V

Now put all the given values in the above equation, we get:

0.507=-0.47-\frac{0.0592}{2}\log \frac{(0.6)}{[Pb^{2+}]}

[Pb^{2+}]=6.093\times 10^{32}M

Therefore, the concentration of Pb^{2+} is, 6.093\times 10^{32}M

6 0
3 years ago
H2, N2, O2 molecules. . . A)must be polar. must be nonpolar. . B)can be polar or nonpolar depending on geometric . C)configurati
LenKa [72]

Answer: The correct option is A.

Explanation: The given molecules are the molecules of same element.

These molecules are considered as diatomic species.

Polar molecules are the molecules in which some polarity is present in the bond. These molecules are formed when there is some difference in the electronegativities of the elements. Example: HCl

Non-polar molecules are the molecules where no polarity is present in the bond. These molecules are formed when there is no difference in the electronegativities of the elements. Example: H_2, O_2

The given molecules are non-polar in nature.

Hence, these molecules must be non-polar. So, the correct option is A.

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3 years ago
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What is the pH of a 4.7 x 10-9 M HCl solution? *
ValentinkaMS [17]
11.21 or 2.93 I think
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3 years ago
Can someone please help and list what each type of thing they are? (Element, Compound, Mixture of elements, Mixture of compounds
enot [183]
1. B
2. C
3. E
4. D
5. A
6. B
7. A
8. E
9. A

Hope this helps!
5 0
3 years ago
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