The enthalpy<span> of </span>solution<span>, </span>enthalpy<span> of dissolution, or heat of </span>solution<span> is the</span>enthalpy<span> change associated with the dissolution of a substance in a solvent at constant pressure resulting in infinite dilution. The </span>enthalpy<span> of </span>solution<span> is most often expressed in kJ/mol at constant temperature. </span>
Answer:
1.00 M
Explanation:
Sn^2+ reacts with KMNO4 as follows;
5Sn^2+(aq) + 2MnO4^-(aq) + 16H^+(aq) ----> 5Sn^4+(aq) + 2Mn^+(aq) + 8H2O(l)
The number of moles of MnO4^- reacted = 42.1/1000 L × 0.145 mol/L
= 0.0061 moles
If 5 moles of Sn^2+ reacts with 2 moles of MnO4^-
x moles of Sn^2+ reacts with 0.0061 moles of MnO4^-
x= 5 × 0.0061/2
x= 0.015 moles
Since the volume of the Sn^2+ solution is 15.00mL or 0.015 L
number of moles = concentration × volume
Concentration = number of moles/volume
Concentration= 0.015 moles/0.015 L
Concentration = 1 M
Answer:
Yes, a precipitate of PbS forms
Explanation:
The equation of the reaction is given as:
Pb(NO₃)₂ + Na₂S → PbS + 2NaNO₃
The lead sulfide forms a precipitate in the aqeous solution.
Precipitation is a form of reaction in which ions combines to form a solid precipiate. Most double displacement reactions in which ionic compounds are the reactants results in formation of a precipitate as the product.
There are rules of solubility which guides a reaction that would lead to the formation of a precipitate. The mos applicable of the rules to the reaction stated above is that "carbonates, phosphates, sulfides, oxides and hydroxides are insolube". The sulfide of lead formed in the product is therefore insoluble.
