Answer:
25.97oC
Explanation:
Heat lost by aluminum = heat gained by water
M(Al) x C(Al) x [ Temp(Al) – Temp(Al+H2O) ] = M(H2O) x C(H2O) x [ Temp(Al+H2O) – Temp(H2O) ]
Where M(Al) = 23.5g, C(Al) = specific heat capacity of aluminum = 0.900J/goC, Temp(Al) = 65.9oC, Temp(Al+H2O)= temperature of water and aluminum at equilibrium = ?, M(H2O) = 55.0g, C(H2O)= specific heat capacity of liquid water = 4.186J/goC
Let Temp(Al+H2O) = X
23.5 x 0.900 x (65.9-X) = 55.0 x 4.186 x (X-22.3)
21.15(65.9-X) = 230.23(X-22.3)
1393.785 - 21.15X = 230.23X – 5134.129
230.23X + 21.15X = 1393.785 + 5134.129
251.38X = 6527.909
X = 6527.909/251.38
X = 25.97oC
So, the final temperature of the water and aluminum is = 25.97oC
The given complex ion is as follow,
[Ru (CN) (CO)₄]⁻
Where;
[ ] = Coordination Sphere
Ru = Central Metal Atom = <span>Ruthenium
CN = Cyanide Ligand
CO = Carbonyl Ligand
The charge on Ru is calculated as follow,
Ru + (CN) + (CO)</span>₄ = -1
Where;
-1 = overall charge on sphere
0 = Charge on neutral CO
-1 = Charge on CN
So, Putting values,
Ru + (-1) + (0)₄ = -1
Ru - 1 + 0 = -1
Ru - 1 = -1
Ru = -1 + 1
Ru = 0
Result:
<span>Oxidation state of the metal species in each complex [Ru(CN)(CO)</span>₄]⁻ is zero.
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:
I think they are all correct
