This is thermal conduction.
In conduction area with higher kinetic energy (in this case how water) transfer thermal energy to an area with lower kinetic energy (in this case frozen vegetables). Thermal c<span>onduction takes place in all phases of </span>matter (solids, liquids, gases). Rate of conduction is proportional with <span>temperature difference between substances.</span>
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
We should describe a little bit the legend.
A - Element - we should have circles with same color and not bonded together (argon gas).
B - Compound - here we may have circles with same or different color bonded together (water or oxygen which is a diatomic molecule).
C - Mixture of elements - circles with different colors not not bonded together (mixture of noble gases).
D - Mixture of compounds - circles with same or different color bonded together but we should see two or more types of connectivity between circles (mixture of water and ethanol).
E - Mixture of elements and compounds - circles with same or different color bonded together mixed with circles with same color and not bonded together (a mixture between oxygen which is a diatomic molecule and noble gas like argon).
Now we may answer the question:
1) B
2) C
3) D
4) D
5) A
6) B
7) B
8) E
9) E
10) D
11) B
12) D
13) D
14) D
15) D
For this problem we can use half-life formula and radioactive decay formula.
Half-life formula,
t1/2 = ln 2 / λ
where, t1/2 is half-life and λ is radioactive decay constant.
t1/2 = 8.04 days
Hence,
8.04 days = ln 2 / λ
λ = ln 2 / 8.04 days
Radioactive decay law,
Nt = No e∧(-λt)
where, Nt is amount of compound at t time, No is amount of compound at t = 0 time, t is time taken to decay and λ is radioactive decay constant.
Nt = ?
No = 1.53 mg
λ = ln 2 / 8.04 days = 0.693 / 8.04 days
t = 13.0 days
By substituting,
Nt = 1.53 mg e∧((-0.693/8.04 days) x 13.0 days))
Nt = 0.4989 mg = 0.0.499 mg
Hence, mass of remaining sample after 13.0 days = 0.499 mg
The answer is "e"
