Answer:
If nitrogen = 30.4% then oxygen = 100-30.4 = 69.6%
divide each % value by atomic mass
N = 30.4/14 = 2.17
O = 69.6/16 = 4.35
Divide each y smaller:
N = 2.17/2.17 = 1
O = 4.35/2.17 = 2
Explanation:
put some water out with less salt and then some more water out with more salt and look at the differences with the freezing.
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-
In a number with decimal, zeroes to the right of last non-zero digit are significant.
Hence there are 4 significant digits in 3.140.
Hope this helps, have a great day ahead!
