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SCORPION-xisa
2 months ago
10

Be sure to answer all parts. The sulfate ion can be represented with four S―O bonds or with two S―O and two S═O bonds. (a) Which

representation is better from the standpoint of formal charges? four S―O bonds two S―O bonds and two S═O bonds (b) What shape is the sulfate ion, and what hybrid orbitals of S are postulated for the σ bonding? bent tetrahedral trigonal planar trigonal pyramidal spmdn where m = and n = (c) In view of the answer to part (b), what orbitals of S must be used for the π bonds? What orbitals of O? sulfur: 3d 3p 2p 3s oxygen: 3p 2p 2s 3s

Chemistry
1 answer:
Anarel [2.9K]2 months ago
5 0

Response:

a) Representation - (in attachment)

b) The geometry is tetrahedral

c) sp^{3} hybrid orbitals involve sigma bonding.

\pi orbitals arise from the overlap of sulfur's d-orbitals with the p-orbitals of oxygen.

Clarification:

a)

Representation in attachment.

The total charge in both structures is -2. Structure (b) is preferred according to the resonance forms since the formal charges within the species are preserved.

Thus, structure (b) represents the sulfate ion more accurately.

b) In the sulfate ion, the sulfur atom forms connections with four distinct oxygen atoms. Using VSEPR theory, the sulfate ion is tetrahedral in shape.

There are four sigma bonds and no lone pairs surrounding the central atom.

Consequently, the sulfur atom's hybridization is sp^{3}

c)

The s and p orbitals of sulfur undergo hybridization to form sigma bonds. The orbitals involved in \pi bonding are

Therefore, \pi bonds result from the overlapping of sulfur's d-orbitals with the p-orbitals of oxygen.

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A flexible container has 5.00 L of nitrogen gas at 298 K. If the temperature is increased to 333K, what will the new volume of t
lorasvet [2795]

Givens

  • V1 = 5.00 L
  • V2 =?
  • T1 = 298 K
  • T2 = 333 K

Formula

V1/T1 = V2/T2

Note: This will be applicable only if the pressure remains constant.

Solution

5.00L / 298 K = x / 333 K. Multiply both sides by 333 K.

5.00 * 333 / 298 = x 333/ 333

V2 = 5.59 L

 

3 0
2 months ago
Acrylonitrile () is the starting material for many synthetic carpets and fabrics. It is produced by the following reaction. If 1
alisha [2963]

2C3H6 (g) + 2NH3 (g) + 3O2 (G) -> 2C3H3N (g) + 6H2O (g)

To begin, this isn't really a chemistry forum, but anyway.

This represents a limiting reagent scenario.

Set it up as a Dimensional Analysis issue.

Begin with your desired outcome.

Your goal is to find the mass of acrylonitrile (C3H3N)

so you should initiate with that (I'll abbreviate Acrylonitrile as ACL for convenience)

(g) = (53 g of ACL/1mol ACL) (2 mols ACL/2 mol C3H6)/ (1mol C3H6/42 grams) (15.0 grams)

If you calculate that, you will find that 15 grams of C3H6 yields 18.9 grams of acrylonitrile produced.

Utilize the same approach for the remaining two reactants.

So, I figured it out, and for

oxygen, I calculated 11.04 grams

and for ammonia, I calculated 15.29 grams

This means that the maximum possible production is 11.04 grams, since to create any additional amount, more O2 would be necessary, but with only 10 grams available, that's the upper limit for this reaction.

The other two reactants are in excess.

Please rate as brainliest!

3 0
2 months ago
A patient needs 40.0 mg of antibiotic per kilogram of body weight each day. If the patient weighs 55 kilograms.
VMariaS [2998]

Response:

2200 mg of antibiotic

Explanation:

The prescribed antibiotic dosage is 40 mg/kg of body weight.

For a patient weighing 55 kg, we calculate the dose of antibiotic as follows:

If we analyze 40/1000000, we can determine antibiotic allocation in kg per kg of weight

= 0.00004 kg of antibiotic for each kilogram

0.00004 multiplied by 55 (to find out the required amount for a 55 kg individual)

= 0.0022 kg

This 0.0022 figure converts to milligrams as follows

0.0022*10^6

= 2200 mg of antibiotic is indicated for a patient weighing 55 kg.

4 0
1 month ago
The nitrogen atom of NH2 would have The nitrogen atom of {\rm NH_2} would have blank electrons around the central nitrogen atom.
Alekssandra [3086]

Answer:

(a) The nitrogen atom in contains NH_2^-8 electrons surrounding the central nitrogen atom.

(b) The nitrogen atom in has NH_4^+8 electrons around the central nitrogen atom.

(c) The nitrogen atom in has NH_38 electrons surrounding the central nitrogen atom.

Explanation:

The Lewis dot structure illustrates the connections between atoms within a molecule, additionally showing unpaired electrons present in the molecule.In this structure, valence electrons are represented as 'dots'.

Now we will determine the number of electrons linked to the central nitrogen atom in the specified molecule.

(a) The identified molecule is,

NH_2^-Recognizing that nitrogen has '5' valence electrons while hydrogen has '1', we ascertain the total valence electrons in

= 5 + 2(1) + 1 = 8

According to the Lewis dot structure, this reveals 4 bonding and 4 non-bonding electrons.

NH_2^-From the Lewis structure, we conclude that the nitrogen atom in

has 8 electrons surrounding it.

(b) The identified molecule is, NH_2^-

Knowing that nitrogen has '5' valence electrons and hydrogen has '1' valence electron, the total number of valence electrons for NH_4^+ = 5 + 4(1) - 1 = 8

In the Lewis dot structure, there are 8 bonding electrons and 0 non-bonding electrons.

This leads us to conclude that the nitrogen atom in

contains 8 electrons surrounding it.NH_4^+

(c) The identified molecule is,

NH_4^+

Recognizing nitrogen’s '5' valence electrons and hydrogen’s '1' brings the total valence electrons for

= 5 + 3(1) = 8NH_3Utilizing the Lewis dot structure indicates there are 6 bonding and 2 non-bonding electrons.

According to the Lewis dot structure, we conclude that the nitrogen atom in

has 8 electrons surrounding it.

NH_2^-

3 0
1 month ago
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