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9966 [12]
3 years ago
9

Using principles of atomic structure, explain why the atomic radius of Te is less than the ionic radius of Te2− . Photoelectron

spectroscopy data for the 1s sublevel of Te and the 1s sublevel of O are represented below. In terms of Coulomb's law and atomic structure, explain why the peak for O is positioned so far to the right of the peak for Te .
Chemistry
1 answer:
ValentinkaMS [17]3 years ago
3 0

Answer:

See explanation

Explanation:

Let us recall that a negative ion is formed by addition of electrons to an atom. When electrons are added to the atom, greater interelectronic repulsion increases the size of the Te^2− hence it is greater in size than Te atom. Therefore, the ionic radius of Te^2− is greater than the atomic radius of Te.

In the second question, oxygen is positioned so far to the right because it has a far smaller nuclear charge compared to Te. Hence in the PES spectrum, the 1s sublevel of oxygen lies far to the right of that of Te.

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2. Chromium+oxygen - chromium(III) oxide <br>balanced​
sesenic [268]

Answer:

4Cr + 3O2 —> 2Cr2O3

Explanation:

Information from the question include:

Chromium + oxygen -> chromium(III) oxide

From the word equation given above, the equation can be written as follow:

Cr + O2 —> Cr2O3

The equation can be balance by doing the following:

There are 2 atoms of O on the left side and 3 atoms on the right side. It can be balance by putting 2 in front of Cr2O3 and 3 in front of O2 as shown below:

Cr + 3O2 —> 2Cr2O3

Now, we have 4 atoms of Cr on the right side and 1 atom on the left. It can be balance by putting 4 in front of Cr as shown below:

4Cr + 3O2 —> 2Cr2O3

Now the equation is balanced

3 0
3 years ago
Write a balanced equation for the oxidation-reduction reaction that occurs when hydrogen peroxide reacts with ferrous ion
Alex17521 [72]
H₂O₂ + 2FeSO₄ + H₂SO₄ → Fe₂(SO₄)₃ + 2H₂O

H₂O₂ + 2H⁺ + 2e⁻ → 2H₂O  k=1
Fe²⁺ → Fe³⁺ + e⁻                 k=2

H₂O₂ + 2H⁺ + 2Fe²⁺ → 2H₂O + 2Fe³⁺


7 0
3 years ago
202H8(9) + 7 O2(g) &gt; 4CO2(g) + 6 H2O(9).
lilavasa [31]

since the concentration of Carbon Dioxide will increase, it would make Q > K, cause equilibrium to shift in the direction with less moles of gas to alleviate the extra pressure. In this case, the reaction will shift left because there are fewer moles of gas present.

4 0
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Compounds X and Y are both C6H13Cl compounds formed in the radical chlorination of 3-methylpentane. Both X and Y undergo base-pr
zubka84 [21]

Answer:

Y is a 3-chloro-3-methylpentane.

The structure is shown in the figure attached.

Explanation:

The radical chlorination of 3-methylpentane can lead to a tertiary substituted carbon (Y) and to a secondary one (X).

The E2 elimination mechanism, as shown in the figure, will happen with a simulyaneous attack from the base and elimination of the chlorine. This means that primary and secondary substracts undergo the E2 mechanism faster than tertiary substracts.

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What is the molar mass of nitrogen?
sukhopar [10]
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