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3 years ago

Difference b\w bonding nd Antibonding molecular orbitals

1 answer:

7 0

Answer: Electrons in bonding orbitals stabilize the molecule because they are between the nuclei. They also have lower energies because they are closer to the nuclei. Antibonding orbitals place less electron density between the nuclei. The nuclear repulsions are greater, so the energy of the molecule increases.

Explanation:

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100 points !!!! Write your answer beside each number. You should have 13 items labeled in all.

Arte-miy333 [17]

Answer:

Explanation:

1) Valence electrons

2) Electrons

3) 2 electrons on K-shell

4) 8 electrons on L-Shell

5) 18 electrons on M-shell

6) nucleus

7) Proton

8) neutron

9) principal energy level

10) atomic number

11) symbol of the atom

12) name of the atom

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4 years ago

Which of the following represents the greater concentration?<br> 1. 5 ppm<br> 2. 5%

Elanso [62]

Answer:

Explanation:

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3 years ago

It has been suggested that the surface melting of ice plays a role in enabling speed skaters to achieve peak performance. Carry

vesna_86 [32]

Explanation:

Relation between pressure, latent heat of fusion, and change in volume is as follows.

$\frac{dP}{dT} = \frac{L}{T \times \Delta V}$

Also, $\frac{L}{T} = \Delta S^{fusion}_{m}$

where, $\Delta V^{fusion}_{m}$ is the difference in specific volumes.

Hence, $\frac{dP}{dT} = \frac{\Delta S^{fusion}_{m}}{\Delta V^{fusion}_{m}}$

As, $\Delta S^{fusion}_{m} = \frac{L}{T} = \frac{6010}{273.15}$ = 22.0 J/mol K

And, $\Delta V^{fusion}_{m} = \frac{M}{d_{H_{2}O}} - \frac{M}{d_{ice}}$ ...... (1)

where, $d_{H_{2}O}$ = density of water

$d_{ice}$ = density of ice

M = molar mass of water = $18.02 \times 10^{-3} kg$

Therefore, using formula in equation (1) we will calculate the volume of fusion as follows.

$\Delta V^{fusion}_{m} = \frac{M}{d_{H_{2}O}} - \frac{M}{d_{ice}}$

= $\frac{18.02 \times 10^{-3}}{997} - \frac{18.02 \times 10^{-3}}{920}$

= $-1.51 \times 10^{-6}$

Therefore, calculate the required pressure as follows.

$\frac{dP}{dT} = \frac{22}{-1.51 \times 10^{-6}}$

= $1.45 \times 10^{7} Pa/K$

or, = 145 bar/K

Hence, for change of 1 degree pressure the decrease is 145 bar and for 4.7 degree change dP = $145 \times 4.7 bar$

= 681.5 bar

Thus, we can conclude that pressure should be increased by 681.5 bar to cause 4.7 degree change in melting point.

5 0

3 years ago

A 4.27 g sample of a lab solution contains 1.81g of acid. What is the concentration of the solution as a mass percentage?

vaieri [72.5K]

Answer:

42.38875878%

Explanation:

i divided 4.27 g from 1.81g using a percentage calculator but im not sure if its correct

5 0

2 years ago

Arizona was the site of a 400,000-acre wildfire in June 2002. How much carbon dioxide (CO2) was produced into the atmosphere by

kodGreya [7K]

Answer:

2.97 × 10¹³ g

Explanation:

First, we have to calculate the biomass the is burned. We can establish the following relations:

2.47 acre = 10,000 m²
10 kg of C occupy an area of 1 m²
50% of the biomass is burned

The biomass burned in the site of 400,000 acre is:

$400,000acre\times\frac{10,000m^{2} }{2.47acre} \times \frac{10kgC}{m^{2} } \times 50\% = 8.10 \times 10^{9} kgC$

Let's consider the combustion of carbon.

C(s) + O₂(g) ⇒ CO₂(g)

We can establish the following relations:

The molar mass of C is 12.01 g/mol
1 mole of C produces 1 mole of CO₂
The molar mass of CO₂ is 44.01 g/mol

The mass of produced is CO₂:

$8.10 \times 10^{12}gC \times \frac{1molC}{12.01gC} \times \frac{1molCO_{2}}{1molC} \times \frac{44.01gCO_{2}}{1molCO_{2}} =2.97 \times 10^{13} gCO_{2}$

4 0

3 years ago

Difference b\w bonding nd Antibonding molecular orbitals​

Difference b\w bonding nd Antibonding molecular orbitals