1. A block of plastic occupies a volume of 25.0 mL and weighs 20.5 g. What is its density? Circle or highlight

A mysterious white powder could be powdered sugar (C12H22O11), cocaine (C17H21NO4), codeine (C18H21NO3), norfenefrine (C8H11NO2)

rodikova [14]

Norfenefrine (C₈H₁₁NO₂).

<h3>Further explanation</h3>

We will solve a case related to one of the colligative properties, namely freezing point depression.

The freezing point of the solution is the temperature at which the solution begins to freeze. The difference between the freezing point of the solvent and the freezing point of the solution is called freezing point depression.

$\boxed{ \ \Delta T_f = T_f(solvent) - T_f(solution) \ } \rightarrow \boxed{ \ \Delta T_f = K_f \times molality \ }$

<u>Given:</u>

A mysterious white powder could be,

powdered sugar (C₁₂H₂₂O₁₁) with a molar mass of 342.30 g/moles,
cocaine (C₁₇H₂₁NO₄) with a molar mass of 303.35 g/moles,
codeine (C₁₈H₂₁NO₃) with a molar mass of 299.36 g/moles,
norfenefrine (C₈H₁₁NO₂) with a molar mass of 153.18 g/moles, or
fructose (C₆H₁₂O₆) with a molar mass of 180.16 g/moles.

When 82 mg of the powder is dissolved in 1.50 mL of ethanol (density = 0.789 g/cm³, normal freezing point −114.6°C, Kf = 1.99°C/m), the freezing point is lowered to −115.5°C.

<u>Question: </u>What is the identity of the white powder?

<u>The Process:</u>

Let us identify the solute, the solvent, initial, and final temperatures.

The solute = the powder
The solvent = ethanol
The freezing point of the solvent = −114.6°C
The freezing point of the solution = −115.5°C

Prepare masses of solutes and solvents.

Mass of solute = 82 mg = 0.082 g
Mass of solvent = density x volume, i.e., $\boxed{ \ 0.789 \ \frac{g}{cm^3} \times 1.50 \ cm^3 = 1.1835 \ g = 0.00118 \ kg \ }$

We must prepare the solvent mass unit in kg because the unit of molality is the mole of the solute divided by the mass of the solvent in kg.

The molality formula is as follows:

$\boxed{ \ m = \frac{moles \ of \ solute}{kg \ of \ solvent} \ } \rightarrow \boxed{ \ m = \frac{mass \ of \ solute \ (g)}{molar \ mass \ of \ solute \times kg \ of \ solvent} \ }$

Now we combine it with the formula of freezing point depression.

$\boxed{ \ \Delta T_f = K_f \times \frac{mass \ of \ solute \ (g)}{molar \ mass \ of \ solute \times kg \ of \ solvent} \ }$

It is clear that we will determine the molar mass of the solute (denoted by Mr).

We enter all data into the formula.

$\boxed{ \ -114.6^0C - (-115.5^0C) = 1.99 \frac{^0C}{m} \times \frac{0.082 \ g}{Mr \times 0.00118 \ kg} \ }$

$\boxed{ \ 0.9 = \frac{1.99 \times 0.082}{Mr \times 0.00118} \ }$

$\boxed{ \ Mr = \frac{0.16318}{0.9 \times 0.00118} \ }$

We get $\boxed{ \ Mr = 153.65 \ }$

These results are very close to the molar mass of norfenefrine which is 153.18 g/mol. Thus the white powder is norfenefrine.

<h3>Learn more</h3>

The molality and mole fraction of water brainly.com/question/10861444
About the mass and density of ethylene glycol as an antifreeze brainly.com/question/4053884
About the solution as a homogeneous mixture brainly.com/question/637791

Keywords: a mysterious white powder, sugar, cocaine, codeine, norfenefrine, fructose, the solute, the solvent, dissolved, ethanol, normal freezing point, the freezing point depression, the identity

7 0

3 years ago

Read 2 more answers

The American Heart Association recommends to eat no more than 2,301mg of sodium per day. Convert the mass of

castortr0y [4]

The mass of 2,301 grams of sodium in ounces is 0.0811757609 ounces.

The amount of sodium recommended by American heart association is 2301 mg. This limit should not be crossed in a day.

We have to convert Mass of sodium from mg to ounces.

We know,

1 ounce = 28.3459 grams.

We also know,

1 gram = 1000 milligrams.

So,

28.3459 grams = 28.3459 x 1000 milligrams.

28.3459 grams = 28345.9 milligrams.

1 ounce = 28345.9 mg.

1 mg = 1/28345.9 ounces

Weight of sodium 2301mg in ounces,

2301mg = 2301/28345.9 ounces

Dividing till last significant figure,

2301 mg = 0.0811757609 ounces.

Mass of sodium in ounces is 0.0811757609.

To know more about Unit conversion, visit,

brainly.com/question/97386

#SPJ9

4 0

1 year ago

How much energy is required to raise the temperature of 3kg of iron from 20 degrees Celsius to 25 degrees Celsius

Yuri [45]

Answer:

6750 J

Explanation:

The specific heat capacity of iron is 450 J/kg/K

Quantity of heat = mcθ; where m is the mass and c is the specific heat capacity and θ is the temperature change.

Therefore;

m = 3 kg, c = 450 J/kg/K and θ = 5°C

Quantity of heat = 3 × 450 × 5

<u>= 6150 Joules</u>

5 0

3 years ago

One of the students in lab decided to use two fractionating columns (one on top of the other) instead of just one. How would thi

Andrews [41]

Answer:

See detailed explanation.

Explanation:

Hello there!

In this case, according to the given description, it turns out possible for us to infer that the second fractionating column on top of the first one will favor the light product, in this case hexane as it has the lowest boiling point and molar mass; in such a way, we can tell the following:

a) The separation between hexane and heptane will be increased as a purer hexane-rich product will be obtained on the top of the second column.

b) Will be increased as well, because the second column will remove more heptane.

c) Also, more pure heptane will be obtained on the bottom of the two columns, yet the most favored yield will be that of hexane.

All of the aforementioned is possible due to the fact that the second column will remove the amount of heptane that could not be removed on the top of the first column by taking the vapor-liquid equilibrium further from the first column's maximum separation, which is known as distillation sequences.

Regards!

8 0

3 years ago

Si dos elementos tienen igual valor del número cuántico n y diferente l, ml y ms entonces deben estar en : El mismo orbital, per

OlgaM077 [116]

Answer:

El mismo nivel, pero diferente subnivel y orbital

Explanation:

En un átomo, el número cuántico principal n muestra el nivel de energía del electrón.

Si dos electrones tienen el mismo valor del número cuántico principal n, simplemente significa que están en el mismo nivel de energía.

Sin embargo, cuando los números cuánticos l y ml difieren, los electrones deben estar en diferentes subniveles y orbitales.

8 0

3 years ago