Give the half equation to show what happens to oxygen when it is rusting

Chemistry

1 answer:

ycow [4]2 years ago

3 0

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-

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Two billiard balls having masses of 0.2 kg and 0.15kg approach each other. The first ball having a velocity of 2m/s hit the seco

Gekata [30.6K]

Answer: The velocity of the first ball is 0.875 m/s if the second ball travels at 1.5 m/s after collision.

Explanation:

Given: $m_{1}$ = 0.2 kg, $m_{2}$ = 0.15 kg

$v_{1}$ = 2 m/s, $v_{2}$ = 0 m/s, $v'_{1}$ = ?, $v'_{2}$ = 1.5 m/s

Formula used is as follows.

$m_{1}v_{1} + m_{2}v_{2} = m_{1}v'_{1} + m_{2}v'_{2}$

where,

v = velocity before collision

v' = velocity after collision

Substitute the values into above formula as follows.

$m_{1}v_{1} + m_{2}v_{2} = m_{1}v'_{1} + m_{2}v'_{2}\\0.2 kg \times 2 m/s + 0.15 kg \times 0 m/s = 0.2 kg \times v'_{1} + 0.15 kg \times 1.5 m/s\\0.4 kg m/s + 0 = 0.2v'_{1} + 0.225 kg m/s\\0.2v'_{1} kg = 0.175 kg m/s\\v'_{1} = \frac{0.175 kg m/s}{0.2 kg}\\= 0.875 m/s$