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Mkey [24]
1 year ago
5

4.a) Draw two Lewis structures for HCP; one where C is central and one where P is central: b) Calculate formal charge for each a

tom in each structure: Which structure is favored? Explain your answer: 5. a) Draw all possible resonance structures for HCOSel (C is central) b) Calculate formal charge for each atom in each structure: Which structure is favored? Explain your answer:
Chemistry
1 answer:
love history [14]1 year ago
8 0

One Hydrogen atom (H) and one Oxygen atom (O) surround the central Carbon atom (C) in the HCP Lewis structure (O). Carbon (C) and Phosphorus (P) have a triple bond, and Carbon (C) and Hydrogen (H) have a single bond.

<h3>How can you choose the ideal format for a formal charge?</h3>

The Lewis structure with the negative formal charges on the most electronegative atoms is the one to choose from when faced with a choice between numerous Lewis structures with similar formal charge distributions.

<h3>How do you determine the preferred resonance structure?</h3>

The resonance forms with the fewest non-zero formal charge atoms are selected. Resonance develops atoms that have a negative formal charge or are the most electronegative are preferred.

To know more about Lewis structure visit:-

brainly.com/question/20300458

#SPJ4

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Nat2105 [25]
To answer the problem above first we need to find the difference of molar mass of NI3 from I2, 394.71 g/mol - 253.80 g/mol = 140.91 g/mol. Knowing the molar mass of the difference of NI3 from I2, in equation mass (g) / moles (mol) = molar mass, then we substitute. 3.58g / moles = 140.91 g/mol.
moles = 3.58 / 140.91 = 0.025 moles.
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How is primary selection different from secondary succession?
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Answer:

Explanation:

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A balloon is filled to a volume of 1.50 L with 3.00 mol of gas at 25 C. with pressure and temperature held constant, what will b
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Answer:

3.2L

Explanation:

PV=nRT

since pressure and temperature are held constant we have V=nR

R is a constant also,

Thus; \frac{v1}{n1}=\frac{v2}{n2}

v1=1.5L  , n1=3mol, n2=1.4mol

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3 years ago
The decomposition of a compound at 400⁰C is first order with half-life of 1570 seconds. what fraction of an initial amount of th
kirill115 [55]

Answer: After 4710 seconds, 1/8 of the compound will be left

Explanation:

Using the formulae

Nt/No = (1/2)^t/t1/2

Where

N= amount of the compound  present at time t

No= amount of compound present at time t=0

t= time taken for N molecules of the compound to remain = 4710 seconds

t1/2 = half-life of compound  = 1570 seconds

Plugging in the values, we have  

Nt/No = (1/2)^(4710s/1570s)

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Nt/No= 1/8

Therefore after 4710 seconds, 1/8 molecules of the compound will be left

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