Question

1.When HCl(aq) and NaOH(aq) are mixed in a beaker, the beaker feels warm to the touch. What is known about the enthalpy of this reaction?a. The reaction is exothermic.b. The reaction is endothermic.c. Heat is absorbed from the surroundings.d. \DeltaH is positive.Part B 2. Consider this combination reaction: 2Mg(s)+O2(g)→2MgO(s) ΔH=−1204 kJ What is the enthalpy for the decomposition of 1 mole of MgO(s) into Mg(s) and O2(g)? View Available Hint(s) A. -602 kJ/mol B. 602 kJ/mol C. -1204 kJ/mol D. 1204 kJ/mol Part C 3. The enthalpy for the formation of 1 mole of NH3(aq) is -80.29 kJ/mol. What is the enthapy for the formation of 3 moles of NH3(aq)? View Available Hint(s) The enthalpy for the formation of 1 mole of is -80.29 . What is the enthapy for the formation of 3 moles of ? A -83.29 kJ B. -518×103 kJ C. -240.87 kJ D. -26.76 kJ

Answer 1

1. The correct answer is option a.

2. The correct answer is option B.

3. The correct answer is option C.

Step-by-step explanation:

1. When acid reacts with base heat is generated along with formation of salt and water.

`HCl+NaOH\rightarrow NaCl+H_2O,\Delta H=Negative`

Those reaction in which heat released as a product is called exothermic reaction.Exothermic reaction have negative value of enthalpy of reaction

2.
`2Mg(s)+O_2(g)→2MgO(s), \Delta H=-1204 kJ`

If we reveres the equation we will have the reaction in which MgO is getting decomposed into Mg and oxygen gas.

`2MgO\rightarrow 2Mg+O_2(g),\Delta H=1204 kJ`

Divide the whole equation by 2.

`MgO\rightarrow Mg+(1)/(2)O_2(g),\Delta H=602 kJ`

602 kJ is the enthalpy for the decomposition of 1 mole of MgO(s).

3.

The enthalpy for the formation of 1 mole of liquid ammonia = -80.29 kJ/mol

So, enthalpy of formation of 3 moles of liquid ammonia :

`3 mol* (-80.29 kJ/mol)=-240.87 kJ`

-240.87 kJ is the enthapy for the formation of 3 moles of liquid ammonia.