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frosja888 [35]
3 years ago
13

What does the electron sea model for metals suggest? A. valence electrons drift freely around the metal anions B. valence electr

ons are anchored to specific metal cations C. valence electrons drift freely around the metal cations D. electrons idle around the metal cations E. valence electrons are anchored to specific metal anions
Chemistry
2 answers:
Ainat [17]3 years ago
5 0

The electron sea model for metals suggest that the valence electrons drift freely around the metal cations.

Answer: B

Explanation

The sea model of electron is used for describing the mechanism of metallic bonding.

The metallic bonding generally occurs between 2 or more metals leading to the formation of alloys.

According to electron sea model, the electrons which contributes to the metallic bond are mostly the valence electrons of the atoms, these valence electrons get de-localized and can move freely around the nuclei of other atoms.

Overall, it seems like nuclei of positive charge is surrounded by sea of negative electrons.  

ratelena [41]3 years ago
3 0

Answer-   The correct option among all the options given is option C .

Explanation- Electron sea model is named because there are more than half of the periodic table covered by metals and the way they react among themselves is completely different from the way they react with other members of the group.

Valence electrons  that are present in metals in such cases such electrons are free to roam about and drift about cations.

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Volume of the gass will decrease by three times of the original volume

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How many moles of oxygen are formed when 58.6 g of KNO3 decomposes according to the following reaction? 4 KNO3(s) → 2 K2O(s) + 2
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0.725 mol

Explanation:

Moles are calculated as the given mass divided by the molecular mass.

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since,

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Therefore,

moles of KNO₃ = 58.6 g / 101 g / mol

moles of KNO₃ = 0.58 mol

From the balanced reaction ,

4 KNO₃ (s) ---> 2K₂O (s) + 2N₂ (g) + 5O₂ (g)

By the decomposition of 4 mol of KNO₃ , 5 mol of O₂ are formed ,

hence, unitary method is used as,

1  mol of KNO₃  gives 5 / 4 mol O₂

Therefore,

0.58 mol of KNO₃ , gives , 5 / 4  * 0.58 mol of O₂

Solving,

0.58 mol of KNO₃ , gives , 0.725 mol of O₂

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58.6g of KNO₃ gives 0.725 mol of O₂.

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Diethyl ether is a commonly used solvent for GC analyses because of its low boiling point. In this experiment, why was heptane u
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