Answer:
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Explanation:
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Answer : The enthalpy of formation of 
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 will be,
The intermediate balanced chemical reaction will be,
(1) 
(2) 
(3) 
(4) 
Now adding all the equations, we get the expression for enthalpy of formation of will be,
Therefore, the enthalpy of formation of is, -812.4 kJ/mole
Answer:
Explanation:
THE QUESTION SHOULD BE
Write the equilibrium constant expression, Kc, for the following reaction: If either the numerator or denominator is 1, please enter 1 CaCO3(s).
CaCO3(s) <=> Ca2+(aq) + CO32-(aq)
Write the mathematical form of the concentration of the products, then divide it by the concentration of the reactants as follows
Kc =[Ca2+]^1 . [CO32-]^1 / [CaCO3]^1
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.
