Answer:
Explanation:
Oxidation:
Oxidation involve the removal of electrons and oxidation state of atom of an element is increased.
Reduction:
Reduction involve the gain of electron and oxidation number is decreased.
Oxidizing agents:
Oxidizing agents oxidize the other elements and itself gets reduced.
Reducing agents:
Reducing agents reduced the other element are it self gets oxidized.
Consider the following reaction:
2AgCl + Zn → 2Ag + ZnCl₂
In this reaction oxidation state of Zn on left side is 0 while on right side +2 so it gets oxidized and oxidation state of Ag on left side is +1 and on right side 0 so it get reduced.
4NH₃ + 3O₂ → 2N₂ + 6H₂O
In this reaction oxidation state of nitrogen on left side is -3 while on right side 0 so it gets oxidized and oxidation state of oxygen on left side is 0 and on right side -2 so it get reduced.
Fe₂O₃ + 2Al → Al₂O₃ + 2Fe
In this reaction oxidation state of iron on left side is +3 while on right side 0 so it gets reduced and oxidation state of Al on left side is 0 and on right side +3 so it get oxidized.
<span>Answer: 56.6 moles
Explanation:
28.3 moles of Pb would produce twice as much moles as Ag.
28.3 X (2moles Ag/ 1 mol Pb) = 56.6 moles of Ag.</span>
Answer:
if balanced reaction then 2 NaOH + MgSO4 → Mg(OH)2 + Na2SO4
Explanation:
London dispersion
dipole-dipole
Answer:
ΔH° = -186.2 kJ
Explanation:
Hello,
This case in which the Hess method is applied to compute the required chemical reaction. Thus, we should arrange the given first two reactions as:
(1) it is changed as:
SnCl2(s) --> Sn(s) + Cl2(g)...... ΔH° = 325.1 kJ
That is why the enthalpy of reaction sign is inverted.
(2) remains the same:
Sn(s) + 2Cl2(g) --> SnCl4(l)......ΔH° = -511.3 kJ
Therefore, by adding them, we obtain the requested chemical reaction:
(3) SnCl2(s) + Cl2(g) --> SnCl4(l)
For which the enthalpy change is:
ΔH° = 325.1 kJ - 511.3 kJ
ΔH° = -186.2 kJ
Best regards.
