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2 years ago

When four atomic orbitals are mixed to form hybrid orbitals, how many hybrid orbitals are formed?

Physics

2 answers:

lilavasa [31]2 years ago

6 0

When four atomic orbitals are mixed to form hybrid orbitals, 4 x sp3 hybrid orbitals are formed.

Carbon's 2s and all three of its 3p orbitals hybridize to form four sp3 orbitals. These orbitals then bond with four hydrogen atoms through sp3-s orbital overlap, creating methane.

Snezhnost [94]2 years ago

5 0

Answer:

Four

Explanation:

When atomic orbitals are combined, hybrid orbitals are formed. Hybrid orbitals exist only in covalent-bonded atoms and not in isolated atoms. The shapes and orientations of hybrids are different from that of atomic orbitals. The number of hybrid orbitals is same as the number of atomic orbitals that combined to form atomic orbitals. Thus, when 4 atomic orbitals are mixed then 4 hybrid orbitals are formed.

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3 years ago

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Answer:

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3 years ago

A container with volume 1.64 L is initially evacuated. Then it is filled with 0.226 g of N2N

vaieri [72.5K]

Answer:

0.015 atm

Explanation:

The pressure of the gas can be calculated using Ideal Gas Law:

$p = \frac{nRT}{V}$

Where:

n: is the number of moles of the gas

R: is the gas constant = 0.082 L*atm/(K*mol)

V: is the volume of the container = 1.64 L

T: is the temperature

We need to find the number of moles and the temperature. The number of moles is:

$n = \frac{m}{M}$

Where:

M: is the molar mass of the N₂ = 14.007 g/mol*2 = 28.014 g/mol

m: is the mass of the gas = 0.226 g

$n = \frac{0.226 g}{28.014 g/mol} = 8.07 \cdot 10^{-3} moles$

Now, the temperature can be found using the following equation:

$v_{rms} = \sqrt{\frac{3RT}{M}}$

Where:

R: is the gas constant = 0.082 L*atm/K*mol = 8.314 J/K*mol

$v_{rms}$ : is the root-mean-square speed of the gas = 182 m/s

By solving the above equation for T, we have:

$T = \frac{v_{rms}^{2}*M}{3R} = \frac{(182 m/s)^{2}*28.014 \cdot 10^{-3} Kg/mol}{3*8.314 J K^{-1}mol^{-1}} = 37.20 K$

Finally, we can find the pressure of the gas:

$p = \frac{nRT}{V} = \frac{8.07 \cdot 10^{-3} mol*0.082 L*atm* K^{-1}*mol^{-1}*37.20 K}{1.64 L} = 0.015 atm$

Therefore, the pressure of the gas is 0.015 atm.

I hope it helps you!

8 0

3 years ago