Answer:
342.8 kJ are absorbed
Explanation:
In the reaction:
4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(l) ΔH° = 1168 kJ
<em>As ΔH > 0, the heat is absorbed. Also, when 4 moles of NH3 are involved in the reaction, there are absorbed 1168 kJ</em>.
Having this in mind, moles of NH3 in 20.00g are:
20.00g × (1mol / 17.0307g) = <em>1.174 moles</em>
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Thus, 1.174 moles of NH3 absorbed:
1.174 moles × (1168 kJ / 4 moles) = <em>342.8 kJ are absorbed</em>.
Answer: The value of enthalpy of the given reaction is 44.000 kJ/mol , that is energy is supplied to water to change into water vapors
Explanation:
Enthalpy of formation of water in liquid state,
Enthalpy of formation of water in gaseous state,
The value of enthalpy of the given reaction is 44.000 kJ/mol, that is energy is supplied to water to change into water vapors.
Crazy stuff, it will bubble up and flow over
Answer:
Approximately 2000 J.
General Formulas and Concepts:
<u>Thermodynamics</u>
Specific Heat Formula: q = mcΔT
- <em>q</em> is heat (in J)
- <em>m</em> is mass (in g)
- <em>c</em> is specific heat (in J/g °C)
- ΔT is change in temperature (in °C or K)
Explanation:
<u>Step 1: Define</u>
<em>Identify variables</em>
[Given] <em>c</em> = 0.897 J/g °C
[Given] <em>m</em> = 79 g
[Given] ΔT = 28°C
[Solve] <em>q</em>
<em />
<u>Step 2: Solve for </u><em><u>q</u></em>
- Substitute in variables [Specific Heat Formula]: q = (79 g)(0.897 J/g °C)(28 °C)
- Multiply [Cancel out units]: q = (70.863 J/°C)(28 °C)
- Multiply [Cancel out units]: q = 1984.16 J
<u>Step 3: Check</u>
<em>Follow sig fig rules and round. We are given 2 sig figs as our lowest.</em>
1984.16 J ≈ 2000 J
