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3 years ago

HELP? What is the major difference between an ion and a dipole?

Chemistry

1 answer:

Anastasy [175]3 years ago

8 0

Answer:

Key differences between ion/ion and ion/dipole interactions

Ion-ion interactions fall off slower than ion-dipole. Tripling the distance between two ions reduces the energy by 1/3, while tripling the distance between the ion and a dipole reduces it by 1/9

Letter answer: B

Explanation:

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2 years ago

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5 0

3 years ago

Read 2 more answers

Element

Svetllana [295]

Answer:

Number of moles = 10.6 mol

Explanation:

Given data:

Molar mass of H = 1.008 g/mol

Molar mass of C = 12.01 g/mol

Molar mass of O = 16.00 g/mol

Mass of citric acid = 2.03 kg (2.03×1000 = 2030 g)

Number of moles of citric acid = ?

Solution:

Formula:

Number of moles = mass/molar mass

Now we will calculate the molar mass of citric acid:

C₆H₈O₇ = (12.01× 6) + (1.008×8) + (16.00×7)

C₆H₈O₇ = 72.06 + 8.064+112

C₆H₈O₇ = 192.124g/mol

Number of moles = 2030 g/ 192.124g/mol

Number of moles = 10.6 mol

7 0

3 years ago

A solution was prepared by dissolving 0.800 g of sulfur S8, in 100.0 g of acetic acid, HC2H3O2. Calculate the freezing point and

sammy [17]

Answer: The freezing point of solution is 16.5°C and the boiling point of solution is 118.2°C

Explanation:

To calculate the molality of solution, we use the equation:

$Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

Where,

$m_{solute}$ = Given mass of solute $(S_8)$ = 0.800 g

$M_{solute}$ = Molar mass of solute $(S-8)$ = 256.52 g/mol

$W_{solvent}$ = Mass of solvent (acetic acid) = 100.0 g

Putting values in above equation, we get:

$\text{Molality of solution}=\frac{0.800\times 1000}{256.52\times 100.0}\\\\\text{Molality of solution}=0.0312m$

Calculation for freezing point of solution:

Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.

$\Delta T_f=\text{freezing point of acetic acid}-\text{Freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

or,

$\text{Freezing point of acetic acid}-\text{Freezing point of solution}=iK_fm$

where,

Freezing point of acetic acid = 16.6°C

i = Vant hoff factor = 1 (for non-electrolyte)

$K_f$ = molal freezing point depression constant = 3.59°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

$16.6^oC-\text{freezing point of solution}=1\times 3.59^oC/m\times 0.0312m\\\\\text{Freezing point of solution}=16.5^oC$

Hence, the freezing point of solution is 16.5°C

Calculation for boiling point of solution:

Elevation in boiling point is defined as the difference in the boiling point of solution and freezing point of pure solution.

The equation used to calculate elevation in boiling point follows:

$\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of acetic acid}$

To calculate the elevation in boiling point, we use the equation:

$\Delta T_b=iK_bm$

or,

$\text{Boiling point of solution}-\text{Boiling point of acetic acid}=iK_fm$

where,

Boiling point of acetic acid = 118.1°C

i = Vant hoff factor = 1 (for non-electrolyte)

$K_f$ = molal boiling point elevation constant = 3.08°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

$\text{Boiling point of solution}-118.1^oC=1\times 3.08^oC/m\times 0.0312m\\\\\text{Boiling point of solution}=118.2^oC$

Hence, the boiling point of solution is 118.2°C

5 0

3 years ago