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Amanda [17]
1 year ago
7

Determine the electron-group arrangement, molecular shape, and ideal bond angle(s) for each of the following:

Chemistry
1 answer:
Bad White [126]1 year ago
5 0

The electron group arrangement of NO²⁻is trigonal planar. The molecular shape is bent, and the bond angle is 120°.

<h3>What is the molecular shape of a compound?</h3>

The molecular geometry of the compound shows the position of nuclei and the electron of the compound. It shows how the joining of electrons and nuclei makes the shape of the compound.

Like here, the shape of nitrite is bent with lone pair which is shown by Lewis's structure The bond angle will be the distance between the nuclei of the neighbor atoms.

Thus, the electron geometry arrangement of nitrite is trigonal planer with a bent shape and the bond angle will be 120°.

To learn more about molecular geometry, refer to the link:

brainly.com/question/7558603

#SPJ4

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Calculate the ratio of effusion rates of cl2 to f2 .
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3 years ago
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
lisabon 2012 [21]

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

Solution :  Given,

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}

By solving the term x, we get

x=0.287\text{ and }3.889

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

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8 0
3 years ago
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