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3 years ago

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 strongest attractive forces between the particles of each of the following are as follows:

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.

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When 74.8g of alanine C3H7NO2 are dissolved in 1450.g of a certain mystery liquid X, the freezing point of the solution is 8.30°

dlinn [17]

Answer: The mass of potassium bromide that must be dissolved in the same mass of X to produce the same depression in freezing point is 58.2 grams

Explanation:

Depression in freezing point is given by:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f=T_f^0-T_f=8.30^0C$ = Depression in freezing point

i= vant hoff factor = 1 (for non electrolyte)

$K_f$ = freezing point constant =

m= molality = $\frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}=\frac{74.8g\times 1000}{1450g\times 89.09g/mol}=0.579$

$8.30^0C=1\times K_f\times 0.579$

$K_f=14.3^0C/m$

Let Mass of solute (KBr) = x g

$8.3^0C=1.72\times 14.3\times \frac{xg\times 1000}{119g/mol\times 1450g}$

$x=58.2g$

Thus the mass of potassium bromide that must be dissolved in the same mass of X to produce the same depression in freezing point is 58.2 grams

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