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weqwewe [10]
3 years ago
11

Indicate the electron pair geometry and the molecular geometry for each of the six compounds. Compound Electron pair geometry Mo

lecular geometry CO 2 CO2 linear linear BF 3 BF3 SO 2 SO2 trigonal planar bent SiCl 4 SiCl4 tetrahedral tetrahedral PF 3 PF3 tetrahedral trigonal pyramidal OF 2 OF2 tetrahedral bent
Chemistry
1 answer:
marusya05 [52]3 years ago
5 0

Answer:

CO2

Electron pair geometry- Linear

Molecular geometry- Linear

BF3

Electron pair geometry - Trigonal planar

Molecular geometry- trigonal planar

SO2

Electron geometry - Trigonal planar

Molecular geometry - bent

SiCl4

Electron geometry- tetrahedral

Molecular geometry - tetrahedral

PF3

Electron pair geometry - tetrahedral

Molecular geometry- trigonal pyramidal

OF2

Electron pair geometry- tetrahedral

Molecular geometry- bent

Explanation:

Considering the molecule CO2, there are two regions of electron density in the molecule positioned at an angle of 180 degrees from each other hence the molecule is linear.

For BF3, the three bond pairs are arranged at the corners of a triangle to give a trigonal planar geometry at a bond angle of 120 degrees.

SO2 has two bonding groups and one lone pair giving three regions of electron density and a trigonal planar electron pair geometry. Due to the distortion to geometry caused by the presence of a lone pair, the molecule is bent.

For SiCl4, the four bonding groups are arranged at the corners of a regular tetrahedron hence it is tetrahedral both in electron pair geometry and in molecular geometry.

PF3 molecule has four regions of electron density corresponding to tetrahedral electron pair geometry. The presence of the lone pair leads to a trigonal pyramidal molecular geometry.

For OF2, there are four regions of electron density around the central oxygen atom. Two bond pairs and two lone pairs leads to a tetrahedral electron pair geometry but a bent molecular geometry is observed due to the two lone pairs.

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aliya0001 [1]

Answer:

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Explanation:

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3 years ago
Natural gas is stored in a spherical tank at a temperature of 13°C. At a given initial time, the pressure in the tank is 117 kPa
drek231 [11]

Answer:

1.  the absolute pressure in the tank before filling = 217 kPa

2. the absolute pressure in the tank after filling = 312 kPa

3. the ratio of the mass after filling M2 to that before filling M1 = 1.44

The correct relation is option c (\frac{M_{2} }{M_{1} } = \frac{P_{2} T_{1} }{P_{1} T_{2} })

Explanation:

To find  -

1. What is the absolute pressure in the tank before filling?

2. What is the absolute pressure in the tank after filling?

3. What is the ratio of the mass after filling M2 to that before filling M1 for this situation?

As we know that ,

Absolute pressure = Atmospheric pressure + Gage pressure

So,

Before filling the tank :

Given - Atmospheric pressure = 100 kPa ,  Gage pressure = 117 kPa

⇒Absolute pressure ( p1 )  = 100 + 117 = 217 kPa

Now,

After filling the tank :

Given - Atmospheric pressure = 100 kPa ,  Gage pressure = 212 kPa

⇒Absolute pressure (p2)  = 100 + 212= 312 kPa

Now,

As given, volume is the same before and after filling,

i.e. V_{1} = V_{2}

As we know that, P ∝ M

⇒ \frac{p_{1} }{p_{2} } = \frac{m_{1} }{m_{2} }

⇒\frac{m_{2} }{m_{1} } = \frac{p_{2} }{p_{1} }

⇒\frac{m_{2} }{m_{1} } = \frac{312 }{217 } = 1.4378 ≈ 1.44

Now, as we know that PV = nRT

As V is constant

⇒ P ∝ MT

⇒\frac{P}{T} ∝ M

⇒\frac{M_{2} }{M_{1} } = \frac{P_{2} T_{1} }{P_{1} T_{2} }

So, The correct relation is c option.

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For example, diamond is a covalent network solid in which carbon atoms are arranged in a continuous lattice like structure.

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