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9966 [12]
2 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]2 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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3,2,0,1/2

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4,2,1,1/2

2,1,-1,-1/2

THEY ARE VALID


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3 years ago
Cho 6,72 gam Fe tan hết trong dung dịch H2SO4 đặc nóng dư thu được V lít khí SO2. Giá trị của V là
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2 years ago
The average bond energy (enthalpy) for a C=C double bond is 614 kJ/mol and that of a C−C single bond is 348 kJ/mol. Estimate the
AysviL [449]

Answer:

4.42x10⁻¹⁹ J/molecule

Explanation:

At a double bond, there's sigma and a pi bond, and at a single bond, there's only a sigma bond. Thus, if the energy to break both sigma and pi is 614 kJ/mol, and the energy to break only the sigma bond is 348 kJ/mol, the energy to break only the pi bond is:

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3 years ago
_________ often has a unique characteristic that is being measured or compared with the base.
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If PbI2(s) is dissolved in 1.0MNaI(aq) , is the maximum possible concentration of Pb2+(aq) in the solution greater than, less th
fredd [130]

Answer:

\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}

Less than the concentration of Pb2+(aq) in the solution in part ( a )

Explanation:

From the question:

A)

We assume that s to be  the solubility of PbI₂.

The equation of the reaction is given as :

PbI₂(s) ⇌ Pb²⁺(aq) + 2I⁻(aq); Ksp = 7 × 10⁻⁹

 [Pb²⁺] =   s

Then [I⁻] = 2s

K_{sp} =\text{[Pb$^{2+}$][I$^{-}$]}^{2} = s\times (2s)^{2} =  4s^{3}\\s^{3} = \dfrac{K_{sp}}{4}\\\\s =\mathbf{ \sqrt [3]{\dfrac{K_{sp}}{4}}}\\\\\text{The mathematical expressionthat can be used to determine the value of  }\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}

B)

The Concentration of Pb²⁺  in water is calculated as :

\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}

\mathbf{s =\sqrt [3]{\dfrac{7*10^{-9}}{4}}}

\mathbf{s} =\sqrt[3]{1.75*10^{-9}}

\mathbf{s} =\mathbf{1.21*10^{-3}  \ mol/L }

The Concentration of Pb²⁺  in 1.0 mol·L⁻¹ NaI

\mathbf{PbCl{_2}}  \leftrightharpoons    \ \ \ \ \ \ \  \mathbf{Pb^{2+}}   \ \ \ \  \ +   \ \  \ \ \ \ \ \mathbf{2 I^-}

                             \ \ \ \ \ \ \  \ \   \ \  \ \ \ \ \ \ \  \mathbf0}   \ \ \ \  \ \ \ \ \ \   \ \ \ \ \ \mathbf{1.0}

                            \ \ \ \ \ \ \ \ \ \ \ \ \ \    \ \ \ \ \  \mathbf{+x}   \ \ \ \  \    \ \  \ \ \ \ \ \mathbf{+2x}

                            \ \ \ \ \ \ \ \ \ \ \ \ \ \    \ \ \ \ \  \mathbf{+x}   \ \ \ \  \    \ \  \ \ \ \ \ \mathbf{1.0+2x}

The equilibrium constant:

K_{sp} =[Pb^{2+}}][I^-]^2 \\ \\ K_{sp} = s*(1.0*2s)^2 =7*1.0^{-9} \\ \\ s = 7*10^{-9} \ \  m/L

It is now clear that maximum possible concentration of Pb²⁺ in the solution is less than that in the solution in part (A). This happens due to the  common ion effect. The added iodide ion forces the position of equilibrium to shift to the left, reducing the concentration of Pb²⁺.

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