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3 years ago

An atom holds 7 electrons. use orbital notation to model the probable location of its electrons.

Chemistry

1 answer:

zheka24 [161]3 years ago

7 0

Answers:

(a) 1s² 2s²2p³; (b) 1s² 2s²2p⁶ 3s²3p⁶ 4s²3d²; (c) 1s² 2s²2p⁶ 3s²3p⁵

Step-by-step explanation:

One way to solve this problem is to add electrons to the orbitals one-by-one until you have added the required amount.

Fill the subshells in the order listed in the diagram below. Remember that an s subshell can hold two electrons, while a p subshell can hold six, and a d subshell can hold ten.

(a) Seven electrons

1s² 2s²2p³

There are two electrons in the 2s subshell and three in the 2p subshell. The remaining two electrons are in the inner 1s subshell.

(b) 22 electrons

1s² 2s²2p⁶ 3s²3p⁶ 4s²3d²

There are two electrons in the 4s subshell and two in the 2p subshell. The remaining 18 electrons are in the inner subshells.

(c) 17 electrons

1s² 2s²2p⁶ 3s²3p⁵

There are two electrons in the 3s subshell and five in the 2p subshell. The remaining 10 electrons are in the inner subshells.

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Answer:

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Mrrafil [7]

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2 years ago

Part a use these data to calculate the heat of hydrogenation of buta-1,3-diene to butane. c4h6(g)+2h2(g)→c4h10(g)

Reptile [31]

Answer: The heat of hydrogenation of the reaction is coming out to be 234.2 kJ.

Explanation:

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H_{rxn}=\sum [n\times \Delta H_{(product)}]-\sum [n\times \Delta H_{(reactant)}]$

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$C_4H_6(g)+2H_2(g)\rightarrow C_4H_{10}(g)$

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$\Delta H_{rxn}=[(1\times \Delta H_{(C_4H_{10})})]-[(1\times \Delta H_{(C_4H_6)})+(2\times \Delta H_{(H_2)})]$

We are given:

$\Delta H_{(C_4H_{10})}=-2877.6kJ/mol\\\Delta H_{(C_4H_6)}=-2540.2kJ/mol\\\Delta H_{(H_2)}=-285.8kJ/mol$

Putting values in above equation, we get:

$\Delta H_{rxn}=[(1\times (-2877.6))]-[(1\times (-2540.2))+(2\times (-285.8))]\\\\\Delta H_{rxn}=234.2J$

Hence, the heat of hydrogenation of the reaction is coming out to be 234.2 kJ.

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