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3 years ago

Definiton of atomic mass unit and the name of the atom mass it is on.

1 answer:

7 0

Atomic mass is the total mass of neutrons and protons in a certain isotope. It is usually called: AMU or atomic mass unit.

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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, $6.093\times 10^{32}M$

6 0

3 years ago

Katyanochek1 [597]

Answer:

The answer to your question is V2 = 4.97 l

Explanation:

Data

Volume 1 = V1 = 4.40 L Volume 2 =

Temperature 1 = T1 = 19°C Temperature 2 = T2 = 37°C

Pressure 1 = P1 = 783 mmHg Pressure 2 = 735 mmHg

Process

1.- Convert temperature to °K

T1 = 19 + 273 = 292°K

T2 = 37 + 273 = 310°K

2.- Use the combined gas law to solve this problem

P1V1/T1 = P2V2/T2

-Solve for V2

V2 = P1V1T2 / T1P2

-Substitution

V2 = (783 x 4.40 x 310) / (292 x 735)

-Simplification

V2 = 1068012 / 214620

-Result

V2 = 4.97 l

6 0

3 years ago

If parents wanted to pick their child's traits like hair color, gender, eye color, what would doctors have to do?

kap26 [50]

Answer:

Nothing

Explanation:

I would say this because the parents are picking the baby's traits and they make all the decisions. And i guess the doctor would have to follow the traits, if i'm right i tried

5 0

3 years ago

Which shows an isomer of the molecule below?

kobusy [5.1K]

Answer:

Explanation:

3 0

2 years ago

Read 2 more answers

Carbondioxide gas is't collectable in water. why?

Lisa [10]

Answer:

carbon dioxide is a gas so it isn't collected over water.

Explanation:

it works for insoluble gases such as hydrogen,or gases that do not dissolve easily in water such as ammonia and chlorine are readily soluble in water and are not collected this way.

hope it is helpful for you.

4 0

3 years ago