We use the equation for the standard enthalpy change of formation:
ΔHoreaction=∑ΔHof(products)−∑ΔHof(Reactants)
to calculate for the enthalpy for the reaction
2NaOH(s)+CO2(g)→Na2CO3(s)+H2O(l)
We now have
ΔHoreaction = { ΔHfo[Na2CO3(s)] + ΔHfo[H2O(l)] } - { ΔHfo[NaOH(s)] +
ΔHfo[CO2(g)] }
where we use the following Enthalpy of Formation (∆Hfo) values:
Substance ΔHf∘ (kJ/mol)
CO2(g) −393.509
H2O(l) −285.830
Na2CO3(s) −1130.68
NaOH(s) −425.609
and taking note of the coefficients of the products and the reactants,
ΔHoreaction = [1*(−1130.68) + 1*(−285.830)] − [2*(−425.609) + 1*(−393.509)]
= -1416.51 - (-1244.727)
= -171.783 kJ/mol
≈ -171.8 kJ/mol as our enthalpy for the given reaction.
Now considering the reaction
Na2CO3(s)→Na2O(s)+CO2(g) with enthalpy of reaction ΔHoreaction=321.5kJ/mol
we also use the equation for the standard enthalpy change of formation:
ΔHoreaction = ∑ΔHof(products)−∑ΔHof(Reactants)ΔHoreaction
= { ΔHfo[Na2O(s)] + ΔHfo[CO2(g)] } - { ΔHfo[Na2CO3(s)] }
to solve for the enthalpy of formation of Na2O(s):
ΔHfo[Na2O(s)] = ΔHoreaction - ΔHfo[CO2(g)] + ΔHfo[Na2CO3(s)]
Since the coefficients are all 1,
ΔHfo[Na2O(s)] = 321.5 - (-393.509) + (-1130.68)
ΔHfo[Na2O(s)] = -415.671 kJ/mol ≈ -415.7 kJ/mol
When the enthalpy change of the reaction is positive, the reaction is exothermic.
Answer:
C) The compound is largely ionic with A as the cation.
Explanation:
Pulings proposed the method to determine if the compound is ionic in nature or covalent in nature , by finding the difference between the electronegativity of the respective cation and anion .
The ion with higher electronegativity is the anion and the ion with lower electronegativity is the cation.
The electronegativity difference above 1.7 make the compound ionic in nature.
Hence, from the question ,
A is the cation and B is the anion.
And the electronegativity difference above 1.7 so the compound is ionic in nature.
The answer 1) Oxygen. And another smallest radius is fluorine, because it is the first in the periodic table.
True: due to
<span>- A stock solution is a solution that has been prepared with a greater concentration of solute than is needed for a procedure, so it must be diluted to the proper concentration(s) before use.
- It is very important that you master the concepts and techniques of preparing dilutions .
- A dilution is made by taking a measured portion of a stock solution, and adding to it a solvent (water or a buffer).
- This creates a new solution with a lower concentration of solute than was contained in the stock solution.
- Since the solute concentration of the stock solution is known as well as the volumes of stock and solvent, it is possible to calculate the concentration of solute in the dilute solution.
- The dilute solution is always less concentrated than the stock or starting solution.
</span>So the answer is True