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Reptile [31]
3 years ago
9

Identify the strongest attractive forces between the particles of each of the following A.CH3OH B.CO C.CF4 D.CH3-CH3

Chemistry
2 answers:
Bumek [7]3 years ago
5 0

Answer: A) hydrogen bonding, B) dipole-dipole forces, C) London dispersion forces and D) London dispersion forces.

Explanation: Methanol is a polar covalent molecule means it has dipole-dipole forces of attraction. Since, hydrogen is bonded to more electron negative oxygen atom, this molecule has hydrogen bonding. So, strongest attractive forces for this molecule are hydrogen bonding.

CO is a polar covalent molecule and so it has dipole-dipole forces of attraction.

CF_4 has four polar covalent bonds but the over all molecule is nonpolar as the dipole moment of one C-F bond is canceled by its opposite C-F bond as the molecule is tetrahedral. Being nonpolar, the molecule has london dispersion forces only.

Ethane is also a nonpolar molecule and so it has london dispersion forces.

mylen [45]3 years ago
4 0
The <span> strongest attractive forces between the particles of each of the following are as follows:

</span><span>A.CH3OH
hydrogen bonding

B.CO
covalent bond

C.CF4
covalent bond

D.CH3-CH3
dipole-dipole bonding

Hope this answers the question. Have a nice day.</span>
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The volume of the gas sample at standard pressure is <u>819.5ml</u>

Explanation:

Solution Given:

let volume be V and temperature be T and pressure be P.

V_1=250ml

V_2=?

P_{total}=735 mmhg

1 torr= 1 mmhg

42.2 torr=42.2 mmhg

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P_{water}=42.2mmhg

T_1=35°C=35+273=308 K

Now

firstly we need to find the pressure due to gas along by subtracting the vapor pressure of water.

P_{gas}=P_{total}-P_{water}

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Now

By using combined gas law equation:

\frac{P_1*V_1}{T_1} =\frac{P_2*V_2}{T_2}

V_2=\frac{P_1*}{P_2}*\frac{T_2}{T_1} *V_1

V_2=\frac{P_gas}{P_2}*\frac{T_2}{T_1} *V_1

Here P_2 \:and\: T_2 are standard pressure and temperature respectively.

we have

P_2=750mmhg \:and\: T_2=273K

Substituting value, we get

V_2=\frac{692.8}{750}*\frac{273}{308} *250

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