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-
Answer:
28.75211 kj
Explanation:
Given data:
Mass of iron bar = 841 g
Initial temperature = 84°C
Final temperature = 7°C
Heat released = ?
Solution:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
specific heat capacity of iron is 0.444 j/g.°C
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 7°C - 84°C
ΔT = -77°C
By putting values,
Q = 841 g × 0.444 j/g.°C × -77°C
Q = 28752.11 j
In Kj:
28752.11 j × 1 kJ / 1000 J
28.75211 kj
Answer:
5- number of electrons=11
Explanation:
in a neutral atom number of protons=number of electrons which in this case=11
The higher the degree of ionization, the stronger the conductivity
Answer:
0.267 mol
Explanation:
The <em>unbalanced equation</em> for the reaction that takes place is:
Once we balance it we're left with:
- 2C₈H₁₈ + 25O₂ → 18H₂O + 16CO₂
Using the <em>stoichiometric ratio </em>of water and octane from the balanced reaction, we can <u>convert mol of water into mol of octane</u>:
- 2.40 mol H₂O *
= 0.267 mol C₈H₁₈
Thus 0.267 moles of octane are needed to produce 2.40 mol of water.
