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3 years ago

What is the reduction half-reaction for 2Mg + O2 → 2Mgo?

Chemistry

1 answer:

Helen [10]3 years ago

4 0

Answer:

When magnesium burns, it combines with oxygen (O2) from the air to form magnesium oxide (MgO) according to the following equation: 2Mg(s) + O2(g) → 2MgO(s) ... Thus, a reduction half reaction can be written for the O2 as it gains 4 electrons: O2(g) + 4e− → 2O.

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Read 2 more answers

Nitrogen dioxide is one of the many oxides of nitrogen (often collectively called "NOx") that are of interest to atmospheric che

Helga [31]

Answer:

Partial pressure of dinitrogen tetroxide after equilibrium is reached the second time is 0.82 atm.

Explanation:

$2NO_2(g)\rightleftharpoons N_2O_4(g)$

Initially

3.0 atm 0

At equilibrium

(3.0-2p) p

Equilibrium partial pressure of $NO_2=2.1atm=3.0-2p$

p = 0.45 atm

The value of equilibrium constant wil be given by :

$K_p=\frac{p_{N_2O_4}}{(p_{NO_2})^2}=\frac{p}{(3.0-2p)^2}$

$K_p=\frac{0.45}{(2.1)^2}=0.10$

After addition of 1.5 atm of nitrogen dioxide gas equilibrium reestablishes it self :

$2NO_2(g)\rightleftharpoons N_2O_4(g)$

After adding 1.5 atm of $NO_2$ :

(2.1+1.5) atm 0.45 atm

At second equilibrium:'

(3.6-2P) (0.45+P)

The expression of equilibrium can be written as:

$K_p=\frac{p'_{N_2O_4}}{(p'_{NO_2})^2}$

$0.10=\frac{(0.45+P)}{(3.6-2P)^2}$

Solving for P:

P = 0.37 atm

Partial pressure of dinitrogen tetroxide after equilibrium is reached the second time:

= (0.45+P) atm = (0.45 + 0.37 )atm = 0.82 atm

Partial pressure of dinitrogen tetroxide after equilibrium is reached the second time is 0.82 atm.

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Answer:

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