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Ask QuestionPosted by Manjushree Hiremat 7 years, 11 months ago
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Gurvinder Kaur 7 years, 11 months ago
{tex}PbS(s)+4H_2O_2(aq) \to PbSO_4(s)+4H_2O(l){/tex}
Posted by Mihir Makwane 7 years, 11 months ago
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Mihir Makwane 7 years, 11 months ago
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Gurvinder Kaur 7 years, 11 months ago
| One is heated in the presence of excess of oxygen or air. | one is heated in the absence or limited supply of oxygen or air |
| This method is employed for sulphide ores. | This method is employed for carbonate ores. |
| Sulphur dioxide is produced along with metal oxide. | Carbon dioxide is produced along with metal oxide. |
| <div>Example: For the ores ZnS (sphalerite) and Cu2S (chalcocite), balanced equations for the roasting are:</div> <div>{tex}2ZnS+3O_2\to2 ZnO+ 2SO_2{/tex}</div> <div>{tex}2 Cu_2S + 3O_2\to 2 Cu_2O + 2 SO_2{/tex}</div> |
<div>Example: {tex} ZnCO_3\to ZnO+ CO_2{/tex}</div> |
Posted by Uday Rajvansh 7 years, 11 months ago
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Posted by Pragati Singh 7 years, 11 months ago
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Posted by Pragati Singh 7 years, 11 months ago
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Posted by Pragati Singh 7 years, 11 months ago
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Gurvinder Kaur 7 years, 11 months ago
From the pH we can calculate the hydrogen ion concentration. Knowing hydrogen ion concentration and the ionic product of water we can calculate the concentration of hydroxyl ions. Thus we have:
[H+] = antilog (-pH)
= antilog [-9.95]
= 1.12 X 10-10
So, [OH-] = Kw / [H+]
= (1 X 10-14) / (8.91 X 10-5)
= 8.91 X 10-5 M
The concentration of the corresponding codenium ion is also the same as that of hydroxyl ion. Let the codeine ion be denoted by [M+]. Thus
Kb = [M+][OH-] / [MOH]
= (8.91 X 10-5)2 / 0.005
= 1.59 X 10-6
pKb = -log Kb = - log(1.59 X 10-6)
pKb = 5.80
Posted by Pragati Singh 7 years, 11 months ago
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Posted by Pragati Singh 7 years, 11 months ago
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Gurvinder Kaur 7 years, 11 months ago
Enthalpy of neutralization of is the heat change taking place when 1 gram equivalent of an acid (or base) is neutralized by 1 gram equivalent of base (or acid) in a dilute solution.
Heat of neutralization of strong acid-strong base is always constant, i.e. ΔH = -57.1 kj/mole.
HCl (aq) + NaOH (aq){tex}\to{/tex} NaCl (aq) + {tex}H_2O{/tex} (l)...ΔH = -57.1 kj/mole
Enthalpy of neutralization of any strong acid (like HCl, HNO3, H2SO4) with a strong base (like LiOH, NaOH, KOH) or vice versa is always the same i.e. 57.1 kj/mol. This is because strong acids, strong bases and salt that they form are all completely ionized in dilute aqueous solutions.
Thus the reaction between any strong acid and strong base for example in the above case may be written as :
NaOH (aq) + HCl(aq) {tex}\to{/tex} NaCl (aq) + {tex}H_2O{/tex} (l)... ΔH = -57.1 kj/mole
they will dissociate as :
Na(+) (aq) + OH(-) (aq) + H(+) (aq) + Cl(-) (aq) {tex}\to{/tex} Na(+) (aq) + Cl(-) (aq) + {tex}H_2O{/tex} (l)
common ions will cancel out..
H(+) (aq) + OH(-) (aq){tex}\to H_2O(l){/tex}
Therefore, neutralization is simply a reaction between H(+) ions given by acids and OH(-) ions given by base to form one mole of {tex}H_2O{/tex}.Since strong acid and strong base completely ionize in aqueous solution, the number of H(+) and OH(-) produced by 1 gram equivalent of strong acid and strong base is always the same, hence enthalpy of neutralization between a strong acid and strong base is always constant i.e. ΔH = -57.1 kj/mole.
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Posted by Sangam Tiwari 7 years, 11 months ago
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Gurvinder Kaur 7 years, 11 months ago
Its a type of redox reaction in which a species reduced and oxidized simultaneously i.e. same element reduce and oxidise , to form two different products.
Example: Chlorine gas reacts with dilute sodium<a href="https://en.wikipedia.org/wiki/Sodium_hydroxide" title="Sodium hydroxide"> </a>hydroxide to form sodium chloride, sodium chlorate and water.
{tex}3 Cl_2 + 6 OH^− \to 5 Cl^− + ClO_3^ − + 3 H_2O{/tex}
The chlorine gas reactant is in oxidation state 0. In the products, the chlorine in the Cl− ion has an oxidation number of −1, having been reduced, whereas the oxidation number of the chlorine in the ClO3 − ion is +5, indicating that it has been oxidized.
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Manjushree Hiremat 7 years, 11 months ago
1Thank You