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serious [3.7K]
2 years ago
6

Select correct examples of linear molecules for five electron groups. Select correct examples of linear molecules for five elect

ron groups.
a. BeCl2

b. B e C l 2

c. H2O

d. XeF2 X e F 2

e. H2S H 2 S
Chemistry
1 answer:
Shalnov [3]2 years ago
7 0

Explanation:

Formula to calculate hybridization is as follows.

                Hybridization = \frac{1}{2}[V+N-C+A]

where,

V = number of valence electrons present in central atom

N = number of monovalent atoms bonded to central atom

C = charge of cation

A = charge of anion

So, hybridization of BeCl_{2} is as follows.

              Hybridization = \frac{1}{2}[V + N - C + A]

                                    = \frac{1}{2}[2 + 2]

                                    = 2

Hybridization of BeCl_{2} is sp. Therefore, BeCl_{2} is a linear molecule. There will be only two electron groups through which Be is attached.

Similarly, hybridization of XeF_{2} is calculated as follows.

         Hybridization = \frac{1}{2}[V + N - C + A]

                                    = \frac{1}{2}[8 + 2]

                                    = 5

Therefore, hybridization of XeF_{2} is sp^{3}d. Therefore, [tex]XeF_{2} is also a linear molecule. Though there are three lone pair of electrons present on a xenon atom and it is further attached with fluorine atoms through two electron pairs. Hence, there are in total five electron groups.

Thus, we can conclude that out of the given options XeF_{2} is the correct examples of linear molecules for five electron groups.

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8 0
2 years ago
15A.4(a) What is the temperature of a two-level system of energy separation equivalent to 400 cm−1 when the population of the up
maksim [4K]

Answer:

T = 525K    

Explanation:

The temperature of the two-level system can be calculated using the equation of Boltzmann distribution:

\frac{N_{i}}{N} = e^{-\Delta E/kT}  (1)

<em>where Ni: is the number of particles in the state i, N: is the total number of particles, ΔE: is the energy separation between the two levels, k: is the Boltzmann constant, and T: is the temperature of the system </em>         

The energy between the two levels (ΔE) is:

\Delta E = hck    

<em>where h: is the Planck constant, c: is the speed of light and k: is the wavenumber</em>      

\Delta E = 6.63\cdot 10^{-34} J.s \cdot 3\cdot 10^{8}m/s \cdot 4 \cdot 10^{4}m^{-1} = 7.96 \cdot 10^{-21}J  

Solving the equation (1) for T:

T = \frac{-\Delta E}{k \cdot Ln(N_{i}/N)}  

<em>With Ni = N/3 and k = 1.38x10⁻²³ J/K, </em><em>the temperature of the two-level system is:</em><em> </em>

T = \frac{-7.96 \cdot 10^{-21}J}{1.38 \cdot 10^{-23} J/K \cdot Ln(N/3N)} = 525K                                  

I hope it helps you!

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Answer:

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Hope this helps!

Explanation:

7 0
2 years ago
A laboratory experiment requires 250 millimeters of water for boiling. It also requires 100 millimeters of water for a cooling p
Wewaii [24]

Answer:

1.05 L

Explanation:

There is some info missing I think this is the original question.

<em>A laboratory experiment requires 250 millimeters of water boiling. it also requires 100 mills of water for a cooling process. If a student performs the experiment three times, how much total water will the student need? Give your answer in liters.</em>

<em />

Step 1: Calculate the volume required for each experiment.

The volume required is the sum of the volumes used: 250 mL + 100 mL = 350 mL

Step 2: Calculate the volume required for the 3 experiments

We have to multiply the volume required for each experiment by 3.

3 × 350 mL = 1050 mL

Step 3: Convert the volume to liters

We use the relation 1 L = 1000 mL.

1050 mL × (1 L/1000 mL) = 1.05 L

7 0
3 years ago
Calculate the value of ΔG°, ΔH°, &amp; ΔS° for the oxidation of solid elemental sulfur to gaseous sulfur trioxide at at 25°C.​2S
Liula [17]

Answer:

ΔG° = -750324.96 J/mol = -750.324 Kj/mol

ΔH° = - 790.4 kJ/mol = -790400 j/mol

ΔS° =  = - 134.48 J K-1mol-1

Explanation:

(S, rhombic) + 3O2( g) → 2SO3 (g)

Temperature = 25°C + 273 = 298K (Converting to kelvin temperature.)

The formulae relating all four paramenters (ΔG°, T, ΔH°, & ΔS°) is given as;

ΔG° = ΔH° - TΔS°

All H and S values used are measurements at 25°C

Calculating ΔH

You started with 1 mole of Sulphur and 3 moles of oxygen.

Total starting enthalpy = 0+ 3(0) = 0 kJ/mol

You ended up with 2 moles of SO3.

Total enthalpy at the end = 2(-395.2) = - 790.4 kJ/mol

enthalpy change = what you end up with - what you started with.

ΔH = enthalpy change =  - 790.4 kJ/mol - 0 kJ/mol = - 790.4 kJ/mol = -790400 j/mol

Calculating ΔS

You started with 1 mole of Sulphur and 3 moles of oxygen.

Total starting entropy =31.88 + 3(205) = 646.88 J K-1mol-1

You ended up with 2 moles of SO3.

Total entropy  at the end = 2(256.2) = 512.4 J K-1mol-1

entropy change = what you end up with - what you started with.

ΔS = entropy change =  512.4 J K-1mol-1 - 646.88 J K-1mol-1

= - 134.48 J K-1mol-1

Inputing the values into the formular, we have;

ΔG° = ΔH° - TΔS°

ΔG° = - 790400 - 298 (- 134.48)

ΔG° =  - 790400 + 40075.04

ΔG° = -750324.96 J/mol = -750.324 Kj/mol

4 0
2 years ago
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