Answer:
131.5 kJ
Explanation:
Let's consider the following reaction.
CaCO₃(s) → CaO(s) + CO₂(g)
First, we will calculate the standard enthalpy of the reaction (ΔH°).
ΔH° = 1 mol × ΔH°f(CaO(s)) + 1 mol × ΔH°f(CO₂(g)
) - 1 mol × ΔH°f(CaCO₃(s)
)
ΔH° = 1 mol × (-634.9 kJ/mol) + 1 mol × (-393.5 kJ/mol) - 1 mol × (-1207.6 kJ/mol)
ΔH° = 179.2 kJ
Then, we calculate the standard entropy of the reaction (ΔS°).
ΔS° = 1 mol × S°(CaO(s)) + 1 mol × S°(CO₂(g)
) - 1 mol × S°(CaCO₃(s)
)
ΔS° = 1 mol × (38.1 J/mol.K) + 1 mol × (213.8 J/mol.K) - 1 mol × (91.7 J/mol.K)
ΔS° = 160.2 J/K = 0.1602 kJ/K
Finally, we calculate the standard Gibbs free energy of the reaction at T = 25°C = 298 K.
ΔG° = ΔH° - T × ΔS°
ΔG° = 179.2 kJ - 298 K × 0.1602 kJ/K
ΔG° = 131.5 kJ
Answer:
We also need the nitrogen intake
Explanation:
The nitrogen balance is the difference between the nitrogen intake and nitrogen excreted. If this calculation is positive, the body grow; if it is negative, the body is decreased.
Answer:
magnesium= +2
aluminum= +3
phosphorous= -3
lithium= +1
fluorine= -1
Explanation:
As fluorine having seven valance electrons in its outer most shell with atomic number nine. And for completing its outer most shell it needs one more electron that is why it form ion of -1 value.
Now Aluminum has three electrons in its outer most shell with atomic number 13. So it need to lose these three electrons in order to be in stable state that is why it have ion in +3 state.
Same rule apply for other elements too.
The reaction is a double displacement one which means the cation of one reactant is substituted to the cation of the other reactant to identify the products. Hence in this reaction, the products are silver ammonium nitrate (Ag(NH3)2NO3) and hydrochloric acid (HCl).
Answer : The ions present in the solution of 
are 
and 
in aqueous state.
Explanation :
When is in aqueous solution then they dissociates into their ions.
The reaction in aqueous medium is,
The charge on potassium ion is +1 and on carbonate ion is -2. To neutralize the charge on carbonate ion, two potassium ion must be used.
Therefore, the ions present in solution of are and [tex]CO^{2-}_3[tex] in aqueous state.
