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Ugo [173]
3 years ago
12

The following reactions can be used to prepare samples of metals. Determine the enthalpy change under standard state conditions

for each. (a) 2Ag2 O(s) ⟶ 4Ag(s) + O2 (g) (b) SnO(s) + CO(g) ⟶ Sn(s) + CO2 (g) (c) Cr2 O3 (s) + 3H2 (g) ⟶ 2Cr(s) + 3H2 O(l) (d) 2Al(s) + Fe2 O3 (s) ⟶ Al2 O3 (s) + 2Fe(s)
Chemistry
1 answer:
mamaluj [8]3 years ago
3 0

Answer:

a) 62.1 kJ/mol

b) 2.82 kJ/mol

c) 270.91 kJ/mol

d) -851.5 kJ/mol

Explanation:

The enthalpy change for a reaction in standard conditions (ΔH°rxn) can be calculated by:

ΔH°rxn = ∑n*ΔH°f, products - ∑n*ΔH°f, reagents

Where n is the number of moles in the stoichiometry reaction, and ΔH°f is the enthalpy of formation at standard conditions. ΔH°f = 0 for substances formed by only a single element. The values can be found in thermodynamics tables.

a) 2Ag₂O(s) → 4Ag(s) + O₂(g)

ΔH°f, Ag₂O(s) = -31.05 kJ/mol

ΔH°rxn = 0 - (2*(-31.05)) = 62.1 kJ/mol

b) SnO(s) + CO(g) → Sn(s) + CO₂(g)

ΔH°f,SnO(s) = -285.8 kJ/mol

ΔH°f,CO(g) = -110.53 kJ/mol

ΔH°f,CO₂(g) = -393.51 kJ/mol

ΔH°rxn = [-393.51] - [-110.53 - 285.8] = 2.82 kJ/mol

c) Cr₂O₃(s) + 3H₂(g) → 2Cr(s) + 3H₂O(l)

ΔH°f,Cr₂O₃(s) = -1128.4 kJ/mol

ΔH°f,H₂O(l) = -285.83 kJ/mol

ΔH°rxn = [3*(-285.83)] - [( -1128.4)] = 270.91 kJ/mol

d) 2Al(s) + Fe₂O₃(s) → Al₂O₃(s) + 2Fe(s)

ΔH°f,Fe₂O₃(s) = -824.2 kJ/mol

ΔH°f,Al₂O₃(s) = -1675.7 kJ/mol

ΔH°rxn = [-1675.7] - [-824.2] = -851.5 kJ/mol

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3 years ago
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V125BC [204]

Answer :

A = In

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Explanation :

Atomic number is defined as the number of protons or number of electrons.

Atomic number = number of protons = number of electrons

Mass number is defined as the sum of number of protons and number of neutrons.

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  Co                  27                     27                      31                    58

  In                    49                     49                     66                   115

  Ta                   73                     73                     108                  181

  Ba²⁺                56                    54                      81                   137

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4 0
3 years ago
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S_A_V [24]

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8 0
2 years ago
When a diprotic acid is titrated with a strong base, and the Ka1 and Ka2 are significantly different, then the pH vs. volume plo
dezoksy [38]

Complete question is;

When a diprotic acid is titrated with a strong base, and the Ka1 and Ka2 are significantly different, then the pH vs. volume plot of the titration will have

a. a pH of 7 at the equivalence point.

b. two equivalence points below 7.

c. no equivalence point.

d. one equivalence point.

e. two distinct equivalence points

Answer:

Option E - Two Distinct Equivalence points

Explanation:

I've attached a sample diprotic acid titration curve.

In diprotic acids, the titration curves assists us to calculate the Ka1 and Ka2 of the acid. Thus, the pH at the half - first equivalence point in the titration will be equal to the pKa1 of the acid while the pH at the half - second equivalence point in a titration is equal to the pKa2 of the acid.

Thus, it is clear that there are two distinct equivalence points.

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Thus, we can conclude that the given groups are placed in order of decreasing priority as follows.

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8 0
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