Q.31. To prepare sulfur from hydrogen sulfide extracted from the natural gas, the hydrogen sulfide gas with sulfur dioxide gaseous: we thus obtain liquid sulfur and water steam. 1) Write the equation of this reaction and calculate, using the data, its standard enthalpy at 298K. 2) Explain why the standard enthalpy of formation of liquid sulfur isn't zero? (Explain in one sentence only). Given: Standard molar enthalpies of formation at 298.15K Substance and State AH° (kJ/mol) SO, (g) -298.8 S(1) S(s) +11.8 H,S(g) -20.6 H,O(g) -241.8 H,O(1) -286

(a) Magnesium reacts with hydrochloric acid to produce hydrogen gas. AH, for HCI(aq) = -167.16 kJ/mol %3D AH", for MgCl(aq) = -801.17 kJ/mol Calculate the standard enthalpy change for this reaction. For this reaction, explain the sign of AT for the surroundings (1) (i) Write the net ionic equation for this reaction What type of reaction is this? In our everyday lives, we observe what happens around us using five biological senses. Give the names of the biological senses. (v) (vi) If you reacted magnesium with hydrochloric acid in the laboratory, specifically what would you observe? Give four observations (b) In the laboratory, what do we wear in different situations to protect ourselves? (c) Give four reasons for working in the fume hood in the laboratory.

Johnson and Steele determined the standard enthalpy of combustion of UO₂(s) in gaseous fluorine. The products of combustion are UF(s) and O2(g). The measured value of the enthalpy of combustion is -1112.6 kJ mol-¹ (per mole of UO₂), and the standard enthalpy of formation of UF(s) is -2197.7 kJ mol-¹. Calculate the standard enthalpy of formation of UO2(s) in kJ mol-¹.

Q.30. To prepare sulfur from hydrogen sulfide extracted from the natural gas, the hydrogen sulfide gas with sulfur dioxide gaseous: we thus obtain liquid sulfur and water steam. 1) Write the equation of this reaction and calculate, using the data, its standard enthalpy at 298K. 2) Explain why the standard enthalpy of formation of liquid sulfur isn't zero? (Explain in one sentence only) Given: Standard molar enthalpies of formation at 298.15K AH (kl/mol) -298.8 Substance and State SO2 (g) S(1) S(s) H-S(g) H2O(g) HO(1) +11.8 -20.6 -241.8 -286

give answer with detail explanation

Background-info: https://drive.google.com/file/d/1G7sPTuESIgWk9wpRdqAnFcsBmOlbBfLv/view?usp=sharing Question: Select each of the following substances for which the standard enthalpy of formation, AH°f, is equal to zero. Assume that all conditions are at standard pressure and temperature. CaO (s) O2 (8) H (aq) Uc(s, graphite) Na e)

Use the standard of enthalpies table to determine the change in enthalpy for: C6H6 (l) + 3/2 O2 (g) --> CO2 (g) + H2O (l) (just balance this equation before you compute for enthalpy change.)

Use the standard of enthalpies table to determine the change in enthalpy for: SO 2(g) + ½ O 2(g) --> SO 3(g)

The standard molar enthalpy of formation for oxygen gas, ?2(?), is zero. The enthalpy of formation for atomic oxygen, ?(?), is +250??/???. Why is there a difference in these values?

Standard Enthalpy of Formations in kJ/mol at 298.15 K   ΔHof   ΔHof AlCl3(s)       -704.2.0     Al2O3(s)         -1676 H2O(l)    -285.8             H2O(g)            -241.8            NH4ClO3(s)           -265.2                    N2O(g)        81.6    Use data above to calculate the enthalpy change for the reaction:                                6Al(s)  +  6NH4ClO3(s)  =>   2Al2O3(s)   + 2AlCl3(s)  +  3 N2(g)  +  12H2O(g)               ΔH = ?