The answer that would best complete the given statement above would be option A. Both "Ode to the West wind" and "Ode for Melancholy" praise <span>something non-human. Hope this answers your question. Have a great day ahead!</span>
5. B air is a mixture of many elements, but is not a chemically fused.
6. B Beef stew, composition varies throughout.
7. A. They can be chemically separated into their component elements, but they are all homogenous, and as such, have constant composition, which differs from the components properties, as the components must undergo a chemical change to become compounds.
Answer:
Laboratories use both distilled water and deionized water as controls in experiments. Deionization removes only non-charged organic matter from the water.
Explanation:
Distilled water removes even more impurities than deionization does, if the water undergoes a filtering process before boiling and distillation.
Answer:
1. Options A and B
2. Options B and C
3.. B. Net ∆G = -16.7 KJ/mol; C. Net ∆G = -14.2 KJ/mol
Explanation:
1. The spontaneity of a chemical reaction depends on its standard free energy change, ∆G. If ∆G is negative, the reaction is favourable, but when it is positive, the reaction is unfavorable.
Therefore, since reaction A and B have ∆G to be positive, they are unfavorable
2. Coupling an unfavorable reaction to a favourable reaction can help the reaction to proceed in the forward direction as long as the net free energy change is negative.
Coupling reaction A and C, as well as reaction B and C will make the reactions to become favourable as net ∆G is negative in both instances.
3. A and C: net ∆G = 13.8 - 30.5 = -16.7 KJ/mol
B and C: net ∆G = 16.3- 30.5 = -14.2 KJ/mol
1) Answer is: molar mas of ammonia is 17.031 g/mol.
M(NH₃) = Ar(N) + 3 · Ar(H) · g/mol.
M(NH₃) = 14.007 + 3 · 1.008 · g/mol.
M(NH₃) = 17.031 g/mol.
2) Answer is: molar mas of lead(II) chloride is 278.106 g/mol.
M(PbCl₂) = Ar(Pb) + 2 · Ar(Cl) · g/mol.
M(PbCl₂) = 207.2 + 2 · 35.453 · g/mol.
M(PbCl₂) = 278.106 g/mol.
3) Answer is: molar mas of acetic acid is 60.052 g/mol.
M(CH₃COOH) = 2 · Ar(C) + 2 · Ar(O) + 4 · Ar(H) · g/mol.
M(CH₃COOH) = 2 · 12.0107 + 2 · 15.9994 + 4 · 1.008 · g/mol.
M(CH₃COOH) = 60.052 g/mol.