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drek231 [11]
3 years ago
14

suppose you have a solution that might contain any or all of the following cations: Cu2+, Ag+,Ba2+ and Mn2+. The addition of HBr

causes a precipitate form. after the precipitate is separated by filtration, H2SO4 is added to the supernatant liquid, and another precipitate forms.
Chemistry
2 answers:
MAVERICK [17]3 years ago
8 0

The solution formed a precipitate when HBr is added. This indicates that the cation present would be silver ion(Ag^{+}) as silver forms an insoluble precipitate with bromide ion.

The equation representing the formation of silver bromide precipitate is:

Ag^{+}(aq)+HBr(aq)-->AgBr(s)+H^{+}(aq)

Bromides of all the other cations,Cu^{2+},Ag^{+},Ba^{2+},Mn^{2+}are soluble in aqueous solutions.

After removing the precipitate of AgBr by filtration, the supernatant solution is treated with H_{2}SO_{4}and another precipitate forms. This would be due to the presence of Ba^{2+}ions as barium sulfate is an insoluble precipitate.

The equation representing the formation of barium sulfate precipitate is:

Ba^{2+}(aq)+H_{2}SO_{4}(aq)-->BaSO_{4}(s)+2H^{+}(aq)

vitfil [10]3 years ago
5 0
Of the given cations, only Ag+ forms both an insoluble bromide and sulfate, leading to precipitatoin; therefore, the ion present must be Ag+.
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Irina-Kira [14]

Answer : The enthalpy of formation of H_2SO_4 is, -812.4 kJ/mole

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The formation of H_2SO_4 will be,

S+H_2+2O_2\rightarrow H_2SO_4    \Delta H_{formation}=?

The intermediate balanced chemical reaction will be,

(1) S+O_2\rightarrow SO_2     \Delta H_1=-298.2

(2) SO_2+\frac{1}{2}O_2\rightarrow SO_3    \Delta H_2=-98.2

(3) SO_3+H_2O\rightarrow H_2SO_4    \Delta H_3=-130.2

(4) H_2+\frac{1}{2}O_2\rightarrow H_2O    \Delta H_4=-285.8

Now adding all the equations, we get the expression for enthalpy of formation of H_2SO_4 will be,

\Delta H_{formation}=\Delta H_1+\Delta H_2+\Delta H_3+\Delta H_4

\Delta H=(-298.2)+(-98.2)+(-130.2)+(-285.8)

\Delta H=-812.4kJ/mol

Therefore, the enthalpy of formation of H_2SO_4 is, -812.4 kJ/mole

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