Answer:
B. 2 Na(s) + O₂(g) → Na₂O₂(s); ΔG∘f=−451.0 kJ/mol
D. 2 SO(g) + O₂(g) → 2 SO₂(g); ΔG°f=−600.4 kJ/mol
Explanation:
The spontaneity of a reaction is given by the value of the standard Gibbs free energy of the reaction (ΔG°rxn). The more negative is the ΔG°rxn, the more spontaneous is a reaction.
The ΔG°rxn can be calculated using the following expression:
ΔG°rxn = ∑np × ΔG°f(products) − ∑nr × ΔG°f(reactants)
By definition, the standard Gibbs free energy of formation of simple substances in their most stable state is zero. That is why, in the reaction of formation of a compound ΔG°rxn = ΔG°f(product).
<em>Based on the standard free energies of formation, which of the following reactions represent a feasible way to synthesize the product?
</em>
<em> A. N₂(g) + H₂(g) → N₂H₄(g); ΔG°f=159.3 kJ/mol. </em>
<em> </em>Not feasible. ΔG°rxn = ΔG°f(product) > 0.
<em>B. 2 Na(s) + O₂(g) → Na₂O₂(s); ΔG°f=−451.0 kJ/mol</em>
Feasible. ΔG°rxn = ΔG°f(product) < 0.
<em>C. 2 C(s) + 2 H₂(g) → C₂H₄(g); ΔG°f=68.20 kJ/mol</em>
Not feasible. ΔG°rxn = ΔG°f(product) > 0.
<em>D. 2 SO(g) + O₂(g) → 2 SO₂(g); ΔG°f=−600.4 kJ/mol</em>
Feasible. ΔG°rxn = ΔG°f(product) < 0.
The answer is 0.59 M.
Molar mass (Mr) of MgCl₂ is the sum of the molar masses of its elements.
So, from the periodic table:
Mr(Mg) = 24.3 g/l
Mr(Cl) = 35.45 g/l
Mr(MgCl₂) = Mr(Mg) + 2Mr(Cl) = 24.3 + 2 · 35.45 = 24.3 + 70.9 = 95.2 g/l
So, 1 mol has 95.2 g/l.
Our solution contains 55.8g in 1 l of solution, which is 55.8 g/l
Now, we need to make a proportion:
1 mole has 95.2 g/l, how much moles will have 55.8 g/l:
1 M : 95.2 g/l = x : 55.8 g/l
x = 1 M · 55.8 g/l ÷ 95.2 g/l ≈ 0.59 M
A) Using paper chromatography to separate pigments by density
Answer:
Explanation:
Hello,
For the given chemical reaction:
We first must identify the limiting reactant by computing the reacting moles of Al2S3:
Next, we compute the moles of Al2S3 that are consumed by 2.50 of H2O via the 1:6 mole ratio between them:
Thus, we notice that there are more available Al2S3 than consumed, for that reason it is in excess and water is the limiting, therefore, we can compute the theoretical yield of Al(OH)3 via the 2:1 molar ratio between it and Al2S3 with the limiting amount:
Finally, we compute the percent yield with the obtained 2.10 g:
Best regards.
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