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Ierofanga [76]
3 years ago
9

16. A small piece of silver was placed in a solution containing both magnesium nitrate and copper(II) nitrate.

Chemistry
1 answer:
sashaice [31]3 years ago
3 0
I believe it's about the dislodging response as Ag structures solvent salt with nitrate. For dislodging response, more receptive metal will uproot less receptive metal from the arrangement. In your question, the reactivity of the metal are positioned as takes after: Mg > Cu > Ag. Take note of that more responsive metals, that are Mg and Cu, are in the arrangement.
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The mentioned molecule with formula, CH₃NO where no bond is found between N and O can be depicted as formamide is shown in the attachment below.  

The formal charges = Number of valence electrons for the atom (V) - the number of electrons in lone pairs (N) - 1/2 (number of electrons in bond pairs, B)

FC = V - N - B/2

Thus, there is a need to calculate valence electrons, electrons in lone pairs, and the number of electrons in bond pairs for each atom in the mentioned molecule.  

V or valence electrons on C = 4e, on H = 1e, on N = 5e, and on O = 6e.  

N or electrons in lone pairs on C = 0e, on H = 0e, on N = 2e, and on O = 4e.  

B or number of electrons in bond pairs for C = 8e, for H = 2e, for N = 6e, and for O = 4e.  

Thus, the formal charges for each will be,  

C = 4 - 0 - (8/2) = 0

H = 1 - 0 - (2/2) = 0

N = 5 - 2 - (6/2) = 0

O = 6 - 4 - (4/2) = 0

Lewis dot structure for the given molecule is given in the attachment below:


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3 years ago
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Answer: Explanation:

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