The reaction of iron (III) oxide and aluminum is initiated by heat released from a small amount "starter mixture". This reaction is an oxidation-reduction reaction, a single replacement reaction, producing great quantities of heat (flame and sparks) and a stream of molten iron and aluminum oxide which pours out of a hole in the bottom of the pot into sand.
The balanced chemical equation for this reaction is:
2 Al(s) + Fe2O3(s) --> 2Fe(s) + Al2O3(s) + 850 kJ/mol
Curriculum Notes
This chemical reaction can be used to demonstrate an exothermic reaction, a single replacement or oxidation-reduction reaction, and the connection between ∆H calculated for this reaction using heats of formation and Hess' Law and calculating ∆H for this reaction using qrxn = mc∆T and the moles of limiting reactant. This reaction also illustrates the role of activation energy in a chemical reaction. The thermite mixture must be raised to a high temperature before it will react.
To determine how much thermal energy is released in this reaction, heats of formation values and Hess' Law can be used.
By definition, the deltaHfo of an element in its standard state is zero.
2 Al(s) + Fe2O3(s) --> 2Fe (s) + Al2O3 (s)
The deltaH for this reaction is the sum of the deltaHfo's of the products - the sum of the deltaHfo's of the reactants (multiplying each by their stoichiometric coefficient in the balanced reaction equation), i.e.:
deltaHorxn = (1 mol)(deltaHfoAl2O3) + (2 mol)(deltaHfoFe) - (1 mol)(deltaHfoFe2O3) - (2 mol)(deltaHfoAl)
deltaHorxn = (1 mol)(-1,669.8 kJ/mol) + (2 mol)(0) - (1 mol)(-822.2 kJ/mol) - (2mol)(0 kJ/mol)
deltaHorxn = -847.6 kJ
The melting point of iron is 1530°C (or 2790°F).
MARK ME BRAINLIEST
<u />C. Water is an inexhaustible energy resource among these options. Coal, oil, and natural gas we can run of, but for the foreseeable future, there will always be water.
Answer:
19 °C
Explanation:
Step 1: Given and required data
- Mass of granite (m): 20 g
- Heat absorbed (Q): 300. 2 J
- Specific heat capacity of granite (c): 0.790 J/g.°C
Step 2: Calculate the temperature change (ΔT)
We will use the following expression.
Q = c × m × ΔT
ΔT = Q/c × m
ΔT = 300.2 J/(0.790 J/g.°C) × 20 g = 19 °C
Answer:
Ay high concentration of reactants
Explanation:
To slow down a reaction, you need to do the opposite. Factors that can affect rates of reactions include surface area, temperature, concentration, and the presence of catalysts and inhibitors. ... Concentration - another way to increase the rate of a chemical reaction is to increase the concentration of the reactants.
Answer:
b) add 130 g of NaCH₃CO₂ to 100 mL of H₂O at 80 °C while stirring until all the solid dissolves, then let the solution cool to room temperature.
Explanation:
The solubility of NaCH₃CO₂ in water is ~1.23 g/mL. This means that at room temperature, we can dissolve 1.23 g of solute in 1 mL of water (solvent).
<em>What would be the best method for preparing a supersaturated NaCH₃CO₂ solution?</em>
<em>a) add 130 g of NaCH₃CO₂ to 100 mL of H₂O at room temperature while stirring until all the solid dissolves.</em> NO. At room temperature, in 100 mL of H₂O can only be dissolved 123 g of solute. If we add 130 g of solute, 123 g will dissolve and the rest (7 g) will precipitate. The resulting solution will be saturated.
<em>b) add 130 g of NaCH₃CO₂ to 100 mL of H₂O at 80 °C while stirring until all the solid dissolves, then let the solution cool to room temperature. </em>YES. The solubility of NaCH₃CO₂ at 80 °C is ~1.50g/mL. If we add 130 g of solute at 80 °C and let it slowly cool (and without any perturbation), the resulting solution at room temperature will be supersaturated.
<em>c) add 1.23 g of NaCH₃CO₂ to 200 mL of H₂O at 80 °C while stirring until all the solid dissolves, then let the solution cool to room temperature.</em> NO. If we add 1.23 g of solute to 200 mL of water, the resulting solution will have a concentration of 1.23 g/200 mL = 0.00615 g/mL, which represents an unsaturated solution.
