Answer: This was because the experiment showed that a substance could emit radiation even while it was not exposed to light.
Answer is: -963,8 kJ.
Q₁ = m(Fe) · C · ΔT₁.
C - specific heat capacity of liquid iron, C(Fe) = 0,82 J/g°<span>C.
</span>m(Fe) = 575 g.
ΔT₁ = 1181 - 1825 = -644°C.
Q₁ = -859306,5 J = -859,3 kJ.
Q₂ = m(Fe) · C · ΔT₂.
ΔT₂ = 293 - 1181 = -888°C.
C - specific heat capacity, C(Fe) = 0,44 J/g°C.
Q₂ = -224664 J = -224,66 kJ.
Q₃ =- heat of fusion, ΔH = 209 J/g.
Q₃ = 120175 J = 120,17 kJ.
Q = Q₁ + Q₂ + Q₃ = -963,8 kJ.
1. O2 is not a compound because it only contains one or more type of the same element atom.
2. O2 is a molecule because a molecule is one or more of the same element atom.
3. The law of conversion is that the mass of the system will stay the same when transfer takes place. Like if you had an equation O+H2—> H2O the mass will remain the same.
4. It will be equal to 10 because of law of conservation of matter.
5. One observation can be that the compound, reaction you’re observing, has change states.
Answer:
½O 2 + 2e - + H 2O → 2OH.
Explanation:
Redox reactions - Higher
In terms of electrons:
oxidation is loss of electrons
reduction is gain of electrons
Rusting is a complex process. The example below show why both water and oxygen are needed for rusting to occur. They are interesting examples of oxidation, reduction and the use of half equations:
iron loses electrons and is oxidised to iron(II) ions: Fe → Fe2+ + 2e-
oxygen gains electrons in the presence of water and is reduced: ½O2 + 2e- + H2O → 2OH-
iron(II) ions lose electrons and are oxidised to iron(III) ions by oxygen: 2Fe2+ + ½O2 → 2Fe3+ + O2-
it will be hard, but you can do it. Just study given the materials for the course. Understand enthalpy and entropy, and various types of bonding and you'll be fine.
