Answer:
1) Endothermic.
2) 
3) 
Explanation:
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1) In this case, for these calorimetry problems, we can realize that since the temperature decreases the reaction is endothermic because it is absorbing heat from the solution, that is why the temperature goes from 22.00 °C to 16.0°C.
2) Now, for the total heat released by the reaction, we first need to assume that all of it is released by the solution since it is possible to assume that the calorimeter is perfectly isolated. In such a way, it is also valid to assume that the specific heat of the solution is 4.184 J/(g°C) as it is mostly water, therefore, the heat released by the reaction is:
3) Finally, since the enthalpy of reaction is calculated by dividing the heat released by the reaction over the moles of the solute, in this case NH4Cl, we proceed as follows:
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Missing question:
A. [3.40 mol Fe2O3 (s) × 26.3 kJ/1 mol Fe2O3 (s)] / 2
<span>B. 3.40 mol Fe2O3 (s) × 26.3 kJ/1 mol Fe2O3 (s) </span>
<span>C. 26.3 kJ/1 mol Fe2O3 (s) / 3.40 mol Fe2O3 (s) </span>
<span>D. 26.3 kJ/1 mol Fe2O3 (s) – 3.40 mol Fe2O3 (s).
</span>Answer is: B.
Chemical reaction: F<span>e</span>₂O₃<span>(s) + 3CO(g) → 2Fe(s) + 3CO</span>₂<span>(g);</span>ΔH = <span>+ 26.3 kJ.
When one mole of iron(III) oxide reacts 26,3 kJ of energy is required and for 3,2 moles of iron(III) oxide 3,2 times more energy is required.</span>
Answer:
<em>Argon</em><em> </em><em>can</em><em> </em><em>exi</em><em>st</em><em> </em><em>freely</em><em> </em><em>in</em><em> </em><em>nature</em><em> </em><em>because</em><em> </em><em>it</em><em> </em><em>has</em><em> </em><em>a</em><em> </em><em>full</em><em> </em><em>octet</em><em> </em><em>of</em><em> </em><em>electron</em><em>s</em><em> </em><em>the</em><em> </em><em>way</em><em> </em><em>its</em><em> </em><em>found</em><em> </em><em>in</em><em> </em><em>the</em><em> </em><em>nature</em><em> </em><em>is</em><em> </em><em>the</em><em> </em><em>same</em><em> </em><em>way</em><em> </em><em>its</em><em> </em><em>found</em><em> </em><em>in</em><em> </em><em>periodic </em><em>table</em><em> </em><em>of</em><em> </em><em>element </em><em>in</em><em> </em><em>vast</em><em> </em><em>amouts</em><em> </em><em>of</em><em> </em><em>stabilization</em><em>.</em>
Answer:
Ag 0 is the reducing agent.
Explanation:
Reducing -> gaining electrons
Oxidizing -> losing electrons
Ag lost electrons (became more positive) since it went from a 0 charge to a +1 charge. Therefore it was oxidized. Ag+ is the oxidized product. Reactants that create an oxidized product are called reducing agents. This would make Ag 0 the reducing agent in this reaction.
