When ignited, solid ammonium dichromate decomposes in a fiery display. This is the reaction for a "volcano" demonstration. The decomposition produces nitrogen gas, water vapor, and chromium(III) oxide. The temperature is constant at 25°C.

Substance ΔHf° (kJ/mol) S° (kJ/mol K)

Cr2O3(g) –1147 0.08115

H2O(l) –242 0.1187

N2(g) 0 0.1915

(NH4)2Cr2O7(s) –22.5 0.1137

Determine ΔSuniv° (in kJ/mol K).

a) 6.39 b) 7.66 c) 6.03 d) 84.3 e) 5.22

Option B is correct.

ΔSuniv° = 7.66 kJ/mol.K

Explanation:

The decomposition of (NH₄)₂Cr₂O₇ is represented as

(NH₄)₂Cr₂O₇ (s) → N₂ (g) + 4H₂O (l) + Cr₂O₃ (g)

Substance | ΔHf° (kJ/mol) | S° (kJ/mol K)

Cr₂O₃(g) | -1147 | 0.08115

H₂O(l) | -242 | 0.1187

N₂(g) | 0 | 0.1915

(NH₄)₂Cr₂O₇(s) | -22.5 | 0.1137

Note that

ΔSuniv = ΔSsurr + ΔSsys = (-ΔHrxn /Tsurr) + ΔSsys

ΔHrxn = Hf°(products) - Hf°(reactants)

Hf°(products) = (1×0) + (4×-242) + (1×-1127) = -2,095 kJ

Hf°(reactants) = (1×-22.5) = -22.5 kJ

ΔHrxn = -2095 - (-22.5) = -2,072.5 kJ/mol

ΔSsys = S°(products) - S°(reactants)

S°(products) = (1×0.1915) + (4×0.1187) + (1×0.08115) = 0.74745 kJ/K

S°(reactants) = (1×0.1137) = 0.1137 kJ/K

ΔSsys = 0.74745 - (0.1137) = 0.63375 kJ/mol.K

ΔSuniv = ΔSsurr + ΔSsys = (-ΔHrxn /Tsurr) + ΔSsys

Tsurr = 25°C = 298.15 K

ΔHrxn = -2,072.5 kJ/mol

ΔSsys = 0.63375 kJ/mol.K

ΔSuniv = (2,072.5/298.15) + 0.63375 = 7.60 kJ/K. The closest answer amongst the options is 7.66 kJ/K.

Hope this Helps!!!

8

explanation:

each principal energy level above the first contains one s orbital and three p orbitals. a set of three p orbitals, called the p sublevel, can hold a maximum of six electrons. therefore, the second level can contain a maximum of eight electrons - that is, two in the s orbital and 6 in the three p orbitals.

mark brainliest and have a great day!

b.

the accuracy of the data.