The most condensed state of matter is (4 points)

Multiply the following and express your answer in appropriate scientific notation, and a correct number of significant figures.

7nadin3 [17]

Answer:

2

Explanation:

8 0

2 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

Romashka [77]

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

8 0

3 years ago

Why must sample spots be above the solvent in paper chromatography

marishachu [46]

Because if they are submerged in the solvent, they would dissolve! This would prevent them from seperating and not allow you to actually record anything

6 0

3 years ago

What is the mass sample of 0.0500 moles of zinc chloride ?

vlabodo [156]

Answer:

6.82g

0.59moles

Explanation:

1. What is the mass sample of 0.0500 moles of zinc chloride ?

Given parameters:

Number of moles ZnCl₂ = 0.05moles

Unknown:

Mass of the sample = ?

Solution:

To find the mass of a substance using the number of moles, it would be pertinent to understand what mole is.

A mole is a substance that contains the avogadro's number of particles.

It relates to the mass using the expression below;

Mass of a substance = number of moles x molar mass

Molar mass of ZnCl₂;

Atomic mass of Zn = 65.4g/mol

Cl = 35.5g/mol

Molar mass = 65.4 + 2(35.5) = 136.4g/mole

Mass of a substance = 0.05 x 136.4 = 6.82g

2. How many moles of potassium sulfide are in a 65.50g sample?

Given parameters:

Mass of K₂S = 65.5g

Unknown:

Number of moles = ?

Solution:

The number of moles of any substance is related to mass using the expression below;

Number of moles = $\frac{mass}{molar mass}$

Molar mass of K₂S = 2(39) + 32 = 110g/mol

Number of moles = $\frac{65.5}{110}$ = 0.59moles

8 0

3 years ago

What is the molarity of a 250.0 milliliter aqueous solution of sodium hydroxide that contains 15.5 grams of solute

tensa zangetsu [6.8K]

There are a number of ways to express concentration of a solution. This includes molarity. Molarity is expressed as the number of moles of solute per volume of the solution. The concentration of the solution is calculated as follows:

Molarity = 15.5 g NaOH (1 mol NaOH / 40 g NaOH) / .250 L solution

Molarity = 1.55 M

5 0

3 years ago