All categories

3 years ago

What is the boiling point of kerosene

1 answer:

6 0

Answer: The boiling point of kerosene ranges from around 150 to 300 degrees Celsius normally being closer to 300 degrees Celsius.

Explanation:

You might be interested in

Answer the following question:

zhenek [66]

poste en français s’il vous plaît

8 0

3 years ago

Read 2 more answers

How many atoms are present in a sample of Potassium (K) weighing 33.49g?

madam [21]

Answer:

5.158 × 10²³ atoms K

General Formulas and Concepts:

Chemistry - Atomic Structure

Reading a Periodic Table
Using Dimensional Analysis
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Explanation:

Step 1: Define

33.49 g K

Step 2: Identify Conversions

Avogadro's Number

Molar Mass of K - 39.10 g/mol

Step 3: Convert

$33.49 \ g \ K(\frac{1 \ mol \ K}{39.10 \ g \ K} )(\frac{6.022 \cdot 10^{23} \ atoms \ K}{1 \ mol \ K} )$ = 5.15797 × 10²³ atoms K

Step 4: Check

We are given 4 sig figs. Follow sig figs and round.

5.15797 × 10²³ atoms K ≈ 5.158 × 10²³ atoms K

3 0

3 years ago

Think about a hot summer afternoon at the beach

NeTakaya

A. Would be the correct answer :-)

3 0

3 years ago

Read 2 more answers

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

To what volume should 25ml of 15m nitric acid be diluted to prepare a 3m solution

Maurinko [17]

We can use the dilution formula to find the volume of the diluted solution to be prepared

c1v1 = c2v2

Where c1 is concentration and v1 is volume of the concentrated solution

And c2 is concentration and v2 is volume of the diluted solution to be prepared

Substituting the values in the equation

15 M x 25 mL = 3 M x v2

v2 = 125 mL

The 25 mL concentrated solution should be diluted with distilled water upto 125 mL to make a 3 M solution

6 0

3 years ago