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lyudmila [28]
3 years ago
14

What is penetration? How is it related to shielding? Use the penetration effect to explain the difference in relative orbital en

ergies of a 3p and a 3d electron in the same atom.
Chemistry
1 answer:
Whitepunk [10]3 years ago
5 0

Answer: penetration is the ability of an electron in a given orbital to approach the nucleus closely. Shielding refers to the fact that core electrons reduce the degree of nuclear attraction felt by the orbital electrons. Shielding is the opposite of penetration. The most penetrating orbital is the least screening orbital. The order of increasing shielding effect/decreasing penetration is s<p<d<f.

Explanation:

The order of penetrating power is 1s>2s>2p>3s>3p>4s>3d>4p>5s>4d>5p>6s>4f....

Since the 3p orbital is more penetrating than the 3d orbital, it will lie nearer to the nucleus and thus possess lower energy.

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In the word, thermohaline, thermo means _________ and haline means_____________
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Thermo means heat/related to heat and haline means salty/of salt

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4 years ago
Does a toxic isomer form by mistake &amp; what problems does it cause in the industrial process ?
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7 0
3 years ago
The chemistry of nitrogen oxides is very versatile. Given the following reactions and their standard enthalpy changes,
jok3333 [9.3K]

Answer:

The heat of the given reaction is -23.0 kJ.

Explanation:

We are given with ;

NO(g) + NO_2(g)\rightarrow N_2O_3(g) \Delta H_{1,rxn}=-39.kJ..[1]

NO(g) + NO_2(g) + O_2(g)\rightarrow N_2O_5(g), \Delta H_{2,rxn} = -112.5 kJ..[2]

2NO_2(g) \rightarrow N_2O_4(g) ,\Delta H_{3,rxn} = -57.2 kJ..[3]

2NO(g) + O_2(g)\rightarrow 2NO_2(g), \Delta H_{4,rxn} = -114.2 kJ..[4]

N_2O_5(s)\rightarrow N_2O_5(g) ,\Delta H_{5,sub}= 54.1 kJ..[5]

To find heat of reaction:

N_2O_3(g) + N_2O_5(s)\rightarrow 2N_2O_4(g),\Delta H_{6,rxn} = ?..[6]

Using Hess's law:

[5] +2 × [3] + [4] - [2] - [1] = [6]

\Delta H_{6,rxn}=\Delta H_{5,sub}+2\times \Delta H_{3,rxn}+\Delta H_{4,rxn}-\Delta H_{2,rxn}-\Delta H_{1,rxn}

\Delta H_{6,rxn}=54.1 kJ+(2\times (-57.2 kJ))+(-114.2 kJ)-(-112.5 kJ)-(-39.0 kJ)

\Delta H_{6,rxn}=-23.0 kJ

The heat of the given reaction is -23.0 kJ.

4 0
3 years ago
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