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3 years ago

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V _ionic or compound

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3241004551 [841]3 years ago

8 0

Ionic compounds are formed by the bonding of non-metals to metals. Because inonic bonds involve electon transfer, the result is a mixture of ions; positive and negative, following the laws of electrostatics, like charges attract. These ions then form a giant structure (lattice) of ionic bonds and it is because of this that ionic compounds are solids (unless dissolved in water) and have extremely high melting and boling temparatures. In additions, beucase of the fixed placement of ions (charged particles) in a lattice, in the solid state ionic compounds don't conduct electricity very well but if dissolved in water (aqueous solution) the ions are free to move and carry a current.

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Read 2 more answers

calculate the freezing point of 3.46 gram of a compound X in 160 gram of benzene when a separate sample of X was vaporized it's

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Answer: The freezing point of 3.46 gram of a compound X in 160 gram of benzene is $4.38^0C$

Explanation:

The relation of density and molar mass is:

$d=\frac{PM}{RT}$

where

d = density = 3.27 g/ L

P = pressure of the gas = 773 torr = 1.02 atm (760 torr = 1atm)

M = molar mass of the gas = ?

T = temperature of the gas = $116^0C=(116+273)K=389K$

R = gas constant = $0.0821Latm/Kmol$

$M=\frac{dRT}{P}=\frac{3.27g/L\times 0.0821Latm/Kmol\times 389K}{1.02atm}=102.3g/mol$

The relation of depression in freezing point with molality:

$\Delta T_f=k_f\times m$

$\Delta T_f$ = depression in freezing point = $T_f^0-T_f$ = $5.45-T_f$

$k_f$ = freezing point constant = 5.1

m = molality = $\frac{\text {moles of X}}{\text {weight of solvent in kg}}=\frac{3.46\times 1000}{102.3\times 160}=0.21$

$5.45-T_f=5.1\times 0.21$

$T_f=4.38^0C$

Thus the freezing point of 3.46 gram of a compound X in 160 gram of benzene is $4.38^0C$

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I hope this helps. <u>Please mark brainliest!</u> <em>:)</em>

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