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3 years ago

Hydrogen sulfide (H2S) can be produced by

1 answer:

8 0

that can produce the gas include petroleum/natural gas drilling and refining, wastewater treatment, coke ovens, tanneries, and kraft paper mills.

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For a certain chemical reaction, the standard Gibbs free energy of reaction at 30.0 °C is 110. kJ. Calculate the equlilibriunm c

mestny [16]

Answer:

Explanation:

The equation that relates standard Gibbs free energy, ΔG, with equilibrium constant, K, is:

ΔG = -RT ln K

Where R is gas constant, 8.314J/molK, and T is absolute temperatue (30.0°C + 273.15 = 303.15K).

Replacing (110kJ = 110000J):

110000J/mol = -8.314J/molK*303.15K ln K

-43.644 = lnK

1.11x10⁻¹⁹ = K

7 0

3 years ago

If the solid copper starts off at the normal melting temperature of copper and the liquid copper is put into the mold at the cry

Agata [3.3K]

Heat of what? This is an unclear question.

3 0

3 years ago

1. Which phase change is accompanied by the release of heat? A) H20(s)--> H2O(g) B) H20(l) ->H2O(s) C) H20(l)→ H2O(g) D) H

docker41 [41]

Answer:

B, liquid to solid.

Explanation: Since heat is being released, the particles for H2O would clump up. Heat is basically being taken out.

5 0

3 years ago

What type of bond exists between the atoms in a molecule of N2?

vivado [14]

Answer:

covalent bonds

Explanation:

Nitrogen atoms will form three covalent bonds (also called triple covalent) between two atoms of nitrogen because each nitrogen atom needs three electrons to fill its outermost shell.

(if it is still confusing i have another way of explaining so jus let me know :)

6 0

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 :

(a) As per question, lead is oxidized and copper is reduced.

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

(b) As per question, lead is reduced and copper is oxidized.

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.

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

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

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