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Natali5045456 [20]
3 years ago
10

Atoms of elements at the top of a group on the periodic table are smaller than the atoms of elements at the bottom of the group.

How does this help explain the difference in the reactivity of metals within a group? The valence electrons of the larger atoms are farther from the nucleus and are easier to remove, so the metals near the top are more reactive than those at the bottom. The valence electrons of the larger atoms are farther from the nucleus and are harder to remove, so the metals near the bottom are more reactive than those at the top. The valence electrons of the larger atoms are farther from the nucleus and are harder to remove, so the metals near the top are more reactive than those at the bottom. The valence electrons of the larger atoms are farther from the nucleus and are easier to remove, so the metals near the bottom are more reactive than those at the top.
Chemistry
2 answers:
krok68 [10]3 years ago
8 0

Answer:

the answer is D.The valence electrons of the larger atoms are farther from the nucleus and are easier to remove, so the metals near the bottom are more reactive than those at the top.

Explanation:

Serjik [45]3 years ago
7 0

I think the correct answer would be the last one.

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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

Solution :  Given,

Initial pressure of N_2 = 1.42 bar

Initial pressure of H_2 = 2.87 bar

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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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