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Tom [10]
3 years ago
12

Which of the following is true regarding the inner transition elements? A. These include all elements in groups 3–12. B. They oc

cupy the d block of the periodic table. C. These include the lanthanides and actinides and do not have f sublevels. D. Their valence electrons can be located in both s and f sublevels.
Chemistry
2 answers:
sesenic [268]3 years ago
7 0

Answer:

D. Their valence electrons can be in both s and f sublevels.

Explanation:

The inner transition elements are those in the two long rows at the bottom of the Periodic Table.

The <em>lanthanide series </em>starts after Ba in Period 6, and the <em>actinide series</em> starts after Ra in Period 7.

Thus, we would predict their electron configurations to be of the form

n\text{s}^{2}(n-2)\text{f}^{n}

However, the energy levels of the <em>n</em>s, (<em>n</em>-1)d, and (<em>n</em>-2)f orbitals are so close in energy that there are many exceptions to our predictions

For example, here are some electron configurations.

La = [Xe]6s²5d (not [Xe]6s²4f)

Ce = [Xe]6s²4f5d (not [Xe]6s²4f²)

Pr = [Xe]6s²4f³ (as predicted)

Thus, their valence electrons can be in both s and f (and sometimes d) sublevels.

A. <em>Wrong</em>. The inner transition elements do not include the elements in Groups 3 to 12. They are the elements between Groups 2 and 3.

B. <em>Wrong.</em> They do not occupy the d block (those are the transition metals). They occupy the f block.

C. <em>Wrong</em>. They include the lanthanides and actinides, but most of them have at least one electron in an f sublevel.

Triss [41]3 years ago
5 0

I think B is your answer

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Suppose a 250. mL flask is filled with 0.30 mol of N_2 and 0.70 mol of NO. The following reaction becomes possible:N_2(g) +O2 →
Inessa [10]

Answer:

0.4 M

Explanation:

Equilibrium occurs when the velocity of the formation of the products is equal to the velocity of the formation of the reactants. It can be described by the equilibrium constant, which is the multiplication of the concentration of the products elevated by their coefficients divided by the multiplication of the concentration of the reactants elevated by their coefficients. So, let's do an equilibrium chart for the reaction.

Because there's no O₂ in the beginning, the NO will decompose:

N₂(g) + O₂(g) ⇄ 2NO(g)

0.30 0 0.70 Initial

+x +x -2x Reacts (the stoichiometry is 1:1:2)

0.30+x x 0.70-2x Equilibrium

The equilibrium concentrations are the number of moles divided by the volume (0.250 L):

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[O₂] = x/0.25

[NO] = (0.70 - 2x)/0.250

K = [NO]²/([N₂]*[O₂])

K = \frac{(\frac{0.70 -2x}{0.250})^2 }{\frac{0.30+x}{0.250}*\frac{x}{0.250} }

7.70 = (0.70-2x)²/[(0.30+x)*x]

7.70 = (0.49 - 2.80x + 4x²)/(0.30x + x²)

4x² - 2.80x + 0.49 = 2.31x + 7.70x²

3.7x² + 5.11x - 0.49 = 0

Solving in a graphical calculator (or by Bhaskara's equation), x>0 and x<0.70

x = 0.09 mol

Thus,

[O₂] = 0.09/0.250 = 0.36 M ≅ 0.4 M

3 0
3 years ago
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