How many electrons are in the lowest occupied energy level in a Group 6A element?

Chemistry

2 answers:

zavuch27 [327]3 years ago

5 0

Answer:

Option a. 2

Explanation:

Thinking process:

The electron configuration of an atom always ensures that the lowest level orbital is filled first. The orbitals come in the different sizes and shapes. The first energy levels are made of the s orbitals. These hold two electrons. Next, the p orbitals hold 8 electrons. These are made of the s and p orbital. Then, the next level holds 18 electrons made up of the s, p, and d orbitals. Finally, the next level hold 32 electrons made up of s, p, d, and f orbitals.

kolbaska11 [484]3 years ago

4 0

Answer:

Explanation:

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A gas stream containing n-hexane in nitrogen with a relative saturation of 0.58 (as a fraction, multiply by 100% if you prefer %

kondor19780726 [428]

This problem is describing a gas mixture whose mole fraction of hexane in nitrogen is 0.58 and which is being fed to a condenser at 75 °C and 3.0 atm, obtaining a product at 3.0 atm and 20 °C, so that the removed heat from the system is required.

In this case, it is recommended to write the enthalpy for each substance as follows:

$H_{C-6}=y_{C-6}C_v(T_b-Ti)+\Delta _vH+C_v(T_f-Tb)\\\\H_{N_2}=y_{N_2}C_v(T_f-Ti)$

Whereas the specific heat of liquid and gaseous n-hexane are about 200 J/(mol*K) and 160 J/(mol*K) respectively, its condensation enthalpy is 31.5 kJ/mol, boiling point is 69 °C and the specific heat of gaseous nitrogen is about 29.1 J/(mol*K) according to the NIST data tables and $y_{C-6}$ and $y_{N_2}$ are the mole fractions in the gaseous mixture. Next, we proceed to the calculation of both heat terms as shown below:

$H_{C-6}=0.58*200(69-75)+(-31500)+160(20-69)=-40036J/mol\\\\H_{N_2}=0.42*29.1(20-75)=-672.21J/mol$

It is seen that the heat released by the nitrogen is neglectable in comparison to n-hexanes, however, a rigorous calculation is being presented. Then, we add the previously calculated enthalpies to compute the amount of heat that is removed by the condenser: