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Alenkinab [10]
3 years ago
11

In the name, iron(III) oxide, the (III) represents

Chemistry
1 answer:
Veronika [31]3 years ago
7 0
<h2>In the name, iron(III) oxide, the (III) represents: D) the electrical charge of iron</h2><h2>Explanation:</h2>

To attain stability the chemical bond is formed .

Chemical bond

It is a kind of linkage that binds one atom with the other .

The atoms do so in order to attain stable noble gas configuration .

To form chemical bond they either:

Loose electrons : when atoms loose electrons they acquire positive charge which is equal to the number of electrons lost .

Gain electrons: After gaining electrons they acquire negative charge which is equal to the number of electrons gained by an atom.

share electrons : With sharing no charges are develop .

<em>In the above asked question when iron combines with oxygen it forms iron oxide : where iron looses 3 electrons and oxygen gains 2 electrons .That is the reason ,III here represents  the electrical charge of iron</em>

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The reaction is not at equilibrium and reaction must run in forward direction.

Explanation:

At the given interval, concentration of NO = \frac{4.64\times 10^{-2}}{1}M=4.64\times 10^{-2}M

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Reaction quotient,Q_{c} , for this reaction = \frac{[NOBr]^{2}}{[NO]^{2}[Br_{2}]}

species inside third bracket represents concentrations at the given interval.

So, Q_{c}=\frac{(0.102)^{2}}{(4.64\times 10^{-2})^{2}\times (4.56\times 10^{-2})}=106

So, the reaction is not at equilibrium.

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What is the work required for the separation of air (21-mol-% oxygen and 79-mol-% nitrogen) at 25°C and 1 bar in a steady-flow p
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In the Haber process for ammonia synthesis, K " 0.036 for N 2 (g) ! 3 H 2 (g) ∆ 2 NH 3 (g) at 500. K. If a 2.0-L reactor is char
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Answer : The partial pressure of N_2,H_2\text{ and }NH_3 at equilibrium are, 1.133, 2.009, 0.574 bar respectively. The total pressure at equilibrium is, 3.716 bar

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Initial pressure of N_2 = 1.42 bar

Initial pressure of H_2 = 2.87 bar

K_p = 0.036

The given equilibrium reaction is,

                              N_2(g)+H_2(g)\rightleftharpoons 2NH_3(g)

Initially                   1.42      2.87             0

At equilibrium    (1.42-x)  (2.87-3x)     2x

The expression of K_p will be,

K_p=\frac{(p_{NH_3})^2}{(p_{N_2})(p_{H_2})^3}

Now put all the values of partial pressure, we get

0.036=\frac{(2x)^2}{(1.42-x)\times (2.87-3x)^3}

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From the values of 'x' we conclude that, x = 3.889 can not more than initial partial pressures. So, the value of 'x' which is equal to 3.889 is not consider.

Thus, the partial pressure of NH_3 at equilibrium = 2x = 2 × 0.287 = 0.574 bar

The partial pressure of N_2 at equilibrium = (1.42-x) = (1.42-0.287) = 1.133 bar

The partial pressure of H_2 at equilibrium = (2.87-3x) = [2.87-3(0.287)] = 2.009 bar

The total pressure at equilibrium = Partial pressure of N_2 + Partial pressure of H_2 + Partial pressure of NH_3

The total pressure at equilibrium = 1.133 + 2.009 + 0.574 = 3.716 bar

6 0
3 years ago
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