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3 years ago

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Calculate the percentage by mass of oxygen in Al2(SO4)3 (Al=27, S=32,O= 16) PLEASE SHOW FULL WORKING

2 answers:

White raven [17]3 years ago

6 0

Answer:

There is 61.538% oxygen in Al2(SO4)3.

Explanation:

Wt Of oxygen in the compound = 12*16 = 192 amu.

Total Wt. Of the compound = 2*12+3*32+12*16 = 312 amu.

Thus, percent of oxygen = Wt of oxygen/total Wt. Of compound *100

= 192/312 * 100=61.538 %

vlabodo [156]3 years ago

3 0

The percentage is 61.55%

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Given the two reactions PbCl2(aq)⇌Pb2+(aq)+2Cl−(aq), K3 = 1.84×10−10, and AgCl(aq)⇌Ag+(aq)+Cl−(aq), K4 = 1.14×10−4, what is the

Nimfa-mama [501]

Answer: The value of equilibrium constant for reaction is, $1.42\times 10^{-2}$

Explanation:

The given chemical equations are:

(1) $PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$ ; $K_3=1.84\times 10^{-10}$

(2) $AgCl(aq)\rightleftharpoons Ag^{+}(aq)+Cl^-(aq)$ ; $K_4=1.14\times 10^{-4}$

Now we have to calculate the equilibrium constant for chemical equation as:

$PbCl_2(aq)+2Ag^{+}(aq)\rightleftharpoons 2AgCl(aq)+Pb^{2+}(aq)$ ; $K=?$

We are reversing reaction 2 and multiplying reaction 2 by 2 and then adding both reaction, we get the final reaction.

The equilibrium constant for the reverse reaction will be the reciprocal of that reaction.

If the equation is multiplied by a factor of '2', the equilibrium constant of that reaction will be the square of the equilibrium constant.

If we are adding equations then the equilibrium constants will be multiplied.

The value of equilibrium constant for reaction is:

$K=(\frac{1}{K_4})^2\times K_3$

Now put all the given values in this expression, we get:

$K=(\frac{1}{1.14\times 10^{-4}})^2\times (1.84\times 10^{-10})$

$K=1.42\times 10^{-2}$

Hence, the value of equilibrium constant for reaction is, $1.42\times 10^{-2}$

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3 years ago

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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