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GalinKa [24]
1 year ago
11

What is the electron configuration for vanadium (V)? The Periodic Table A. 1s22s22p63s23p64s24d3 B. 1s22s22p63s23p63d5 C. 1s22s2

2p63s23p64s24p3 D. 1s22s22p63s23p64s23d3
Chemistry
1 answer:
kramer1 year ago
5 0

The electronic configuration for vanadium (V) in the periodic table is as follows: 1s2 2s2 2p6 3s2 3p6 4s2 3d3 (option D).

<h3>What is electronic configuration?</h3>

Electronic configuration is the the arrangement of electrons in an atom, molecule, or other physical structure like a crystal.

Vanadium is the 23rd element on the periodic table and has chemical symbol V with atomic number 23. It is a transition metal, used in the production of special steels.

This suggests that the electronic configuration of Vanadium will be written as follows: 1s2 2s2 2p6 3s2 3p6 4s2 3d3

Therefore, the electronic configuration for vanadium (V) in the periodic table is as follows: 1s2 2s2 2p6 3s2 3p6 4s2 3d3.

Learn more about electronic configuration at: brainly.com/question/14283892

#SPJ1

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iren2701 [21]

Answer:  

1) Endothermic.  

2) Q_{rxn}=4435.04J  

3) \Delta _rH=15.8kJ/mol

Explanation:  

Hello there!  

1) In this case, for these calorimetry problems, we can realize that since the temperature decreases the reaction is endothermic because it is absorbing heat from the solution, that is why the temperature goes from 22.00 °C to 16.0°C.  

2) Now, for the total heat released by the reaction, we first need to assume that all of it is released by the solution since it is possible to assume that the calorimeter is perfectly isolated. In such a way, it is also valid to assume that the specific heat of the solution is 4.184 J/(g°C) as it is mostly water, therefore, the heat released by the reaction is:

Q_{rxn}=-(15.0g+250.0g)*4.184\frac{J}{g\°C}(16.0-20.0)\°C\\\\ Q_{rxn}=4435.04J    

3) Finally, since the enthalpy of reaction is calculated by dividing the heat released by the reaction over the moles of the solute, in this case NH4Cl, we proceed as follows:

\Delta _rH=\frac{ Q_{rxn}}{n}\\\\\Delta _rH= \frac{ 4435.04J}{15.0g*\frac{1mol}{53.49g} } *\frac{1kJ}{1000J} \\\\\Delta _rH=15.8kJ/mol

Best regards!  

Best regards!

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