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Oliga [24]
4 years ago
5

Write out the molecular orbital valence electron configurations for the ground state and first excited state for n2.

Chemistry
1 answer:
Bumek [7]4 years ago
7 0

Answer:

(a) (σ2s)² (σ2*s)² (π2p)⁴ (σ2p)²

(b) (σ2s)² (σ2*s)² (π2p)⁴ (σ2p) (π*2p)

Explanation:

An N atom has five valence electrons, so a molecule of N₂ has 10 valence electrons.

Use the Aufbau Principle to place 10 electrons in the MO diagram for B, C, and N.

1.Ground state

The valence electron configuration of N₂ is

(σ2s)² (σ2*s)² (π2p)⁴ (σ2p)²  

2. First excited state

One of the σ2p electrons is promoted to the next higher level,  π*2p.

The valence electron configuration becomes

(σ2s)² (σ2*s)² (π2p)⁴ (σ2p) (π*2p)

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A certain liquid X has a normal boiling point of 111.20 celsius and a boiling point elevation constant Kb=0.95. calculate the bo
Goshia [24]

Answer: the boiling point is = 137.325°C

Explanation:

From the formula: ∆Tb= Kb*m

From the question, Kb= 0.95, m= 27.5, T1= 111.2°C

Substitute into ∆Tb= Kb*m

∆Tb= 0.95*27.5= 26.125

∆Tb= T2-T1

Hence

T2- 111.2=26.125

T2= 26.125+ 111.2= 137.325°C

8 0
3 years ago
A point has no demension.​
Natalka [10]

Answer:

A point is a 0-D object, infinitely small.

Explanation:

6 0
3 years ago
if excess nitrogen gas reacts with 600 cm³ of hydrogen gas at room conditions , calculate the maximum volume of ammonia produced
spin [16.1K]

Answer:

400 cm³ of ammonia, NH₃.

Explanation:

The balanced equation for the reaction is given below:

N₂ + 3H₂ —> 2NH₃

From the balanced equation above,

3 cm³ of H₂ reacted to produce 2 cm³ of NH₃.

Finally, we shall determine the maximum volume of ammonia, NH₃ produced from the reaction. This can be obtained as illustrated below:

From the balanced equation above,

3 cm³ of H₂ reacted to produce 2 cm³ of NH₃.

Therefore, 600 cm³ of H₂ will react to produce = (600 × 2)/3 = 400 cm³ of NH₃.

Thus, 400 cm³ of ammonia, NH₃ were obtained from the reaction.

5 0
3 years ago
Which of the followings are true about D-glucose and L-glucose? A. They are furanose B. They are stereoisomers C. They are enant
zaharov [31]

Answer:

B. They are stereoisomers

C. They are enantiomers

Explanation:

Let us consider all the options

A. D and L-glucose are not necessarily furanose, they can also be in free form (open chain) or as a six-membered ring (pyranose)

B. These sugars are stereoisomers as they have the same molecular formula, same bonds but with the different spacial arrangement.

C. Two structures are called enantiomers, if they are stereoisomers and are mirror images of each other and are not-superimposable. The given pair of structures satisfy these conditions

D. Epimers are diastereoisomers (same molecular formula and connectivity having a different spacial arrangement but are not mirror images and non-superimposable) with only one different stereocenter (if there are more than one). This is not the case

E. All monosaccharides (any sugar that cannot be hydrolysed to a simpler sugar) are reducing sugars. So, this option is invalid

4 0
3 years ago
5. calculate the percentage of water of crystallization in blue vitriol (copper sulphate pentahydrate), cuso4.5h2o
ki77a [65]

The percentage of water of crystallization in blue vitriol is 36.07%.

M(H₂O) = 2Ar(H) + Ar(O) x g/mol

M(H₂O) = 2 + 16 x g/mol

M(H₂O) = 18 g/mol; molar mass of water

M(CuSO₄·5H₂O) = Ar(Cu) + Ar(S) + 4Ar(O) + 5Mr(H₂O) x g/mol

M(CuSO₄·5H₂O) = 63.5 + 32 + 64 + 90 x g/mol

M(CuSO₄·5H₂O) = 249.5 g/mol; molar mass of copper sulphate pentahydrate

The percentage of water: 5M(H₂O) / M(CuSO₄·5H₂O) x 100%

The percentage of water: 90 g/mol / 249.5 g/mol x 100%

The percentage of water = 36.07%

More about blue vitriol: brainly.com/question/8895853

#SPJ4

5 0
2 years ago
Read 2 more answers
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