Answer:
110L
Explanation:
Boyle's Law states that P1×V1=P2×V2
Volume is indirectly proportional to Pressure so P×V is constant
P1=55atm
V1=6L
P2=3atm
V2 is to be found
P1×V1=P2×V2
6×55=3×V2
330=3×V2
Answer: V2=110L
Evaporation, Condensation, precipitation and collection would be stages
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).
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Well, if you look at group 1 of the periodic table, you will notice a thrend. All elements in group 1 have 1 valence / outer electron. Then you look at period 2, 3, 4 and so on, you will see that the group number corresponds the number of valence/ outershell electrons. Hence, the group determines the electron(s) on the outershell.
