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Vera_Pavlovna [14]
2 years ago
6

Fossils are the preserved remains or traces of an organism that used to be alive.

Chemistry
2 answers:
kotykmax [81]2 years ago
8 0

Answer:

This is True

Explanation:

A fossil is any remains or trace of an ancient organism. Fossils include body fossils, left behind when the soft parts have decayed away, as well as trace fossils

Evgesh-ka [11]2 years ago
3 0
True because the definition of a fossil is “the remains or impression of a prehistoric organism preserved in petrified form or as a mold or cast in rock.”
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Answer:

Explanation:

Figure 6. An instant cold pack consists of a bag containing solid ammonium nitrate and a second bag of water. When the bag of water is broken, the pack becomes cold because the dissolution of ammonium nitrate is an endothermic process that removes thermal energy from the water.

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How does fracking benefit humans?<br><br> (dont use go0gle please!)
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Explanation:

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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

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It is B nevermind I thought it was c but it wasn't.

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