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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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The area of the gold electrodes on the quartz crystal microbalance at the opening of Chapter 2 is 3.3 mm^2. One gold electrode i
aleksandrvk [35]

This question is incomplete, the complete question is;

The area of the gold electrodes on the quartz crystal microbalance at the opening of Chapter 2 is 3.3 mm^2. One gold electrode is covered with DNA at a surface density of 1.2 pmol/cm2.

(a) How much mass of the nucleotide cytosine (C) is bound to the surface of the electrode when each bound DNA is elongated by one unit of C. The mass formula mass of the bound nucleotide is cytosine + deoxyribose + phosphate = C9H10N3O6P = 287.2 g/mol

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Explanation:

Given that the area of gold electrodes = 3.3 mm^2

surface density of one gold electrode = 1.2 pmol/cm^2

that is to that in every 1 cm^2 of area, 1.2 pmol DNA is present

therefore

mass of nucleotide present in 3.3 mm^2 is;

= (1.2/100 * 3.3) pmol

= 0.0396 pmol

we were given that formula mass of the bound nucleotide = 287.2 g/mol

so

mass of the nucleotide (c) bound = ( 287.2 * 0.0396 )g

mass of the nucleotide (c) bound =  11.37 g

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a cylinder container is filled with air. its radius is 6cm and its height is 3.5cm. what is the volume of the air inside?
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Answer:

pH=4.56

Explanation:

Hello there!

In this case, given the Henderson-Hasselbach equation, it is possible for us to compute the pH by firstly computing the concentration of the acid and the conjugate base; for this purpose we assume that the volume of the total solution is 0.025 L and the molar mass of the sodium base is 234 - 1 + 23 = 256 g/mol as one H is replaced by the Na:

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[acid]=0.000855mol/0.025L=0.0342M

[base]=0.000781mol/0.025L=0.0312M

Then, considering that the Ka of this acid is 2.5x10⁻⁵, we obtain for the pH:

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