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3 years ago

Given that π = n M R T, rearrange the equation to solve for V

Chemistry

1 answer:

Pachacha [2.7K]3 years ago

4 0

Answer:

V= n/M

Explanation:

From;

π = nRT/V = MRT

Where;

n= number of moles

R= gas constant

T= absolute temperature

M= molar mass

V= volume of the solution

π= osmotic pressure

Thus;

nRT/V = MRT

nRT = VMRT

V= nRT/MRT

V= n/M

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Which organism on the food chain is a carnivore? fungus grass grasshopper spider

Lady_Fox [76]

Answer:

I think it would it spider

Explanation:

Spider catch there meals in there web then suck the blood out after it wraps them up in webing

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4 years ago

What determines the strength of the attraction between molecules?

alexandr402 [8]

To determine strength of attractive forces between the molecules the size of the molecules, their polarity (dipole moment), and their shape. ... If two molecules have about the same size and similar shape, the dipole-dipole intermolecular attractive force increases with increasing polarity.

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3 years ago

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You have a solution of 600 mg of caffeine dissolved in 100 mL of water. The partition coefficient for aqueous caffeine extracted

klio [65]

Answer:

159 mg caffeine is being extracted in 60 mL dichloromethane

Explanation:

Given that:

mass of caffeine in 100 mL of water = 600 mg

Volume of the water = 100 mL

Partition co-efficient (K) = 4.6

mass of caffeine extracted = ??? (unknown)

The portion of the DCM = 60 mL

Partial co-efficient (K) = $\frac{C_1}{C_2}$

where; $C_1=$ solubility of compound in the organic solvent and $C_2$ = solubility in aqueous water.

So; we can represent our data as:

$K=(\frac{A_{(g)}}{60mL} )$ ÷ $(\frac{B_{(mg)}}{100mL} )$

Since one part of the portion is A and the other part is B

A+B = 60 mL

A+B = 0.60

A= 0.60 - B

4.6= $(\frac{0.6-B(mg)}{60mL} )$ ÷ $(\frac{B_{(mg)}}{100mL})$

4.6 = $\frac{(\frac{0.6-B(mg)}{60mL} )}{(\frac{B_{(mg)}}{100mL})}$

4.6 × $(\frac{B_{(mg)}}{100mL})$ = $(\frac{0.6-B(mg)}{60mL} )$

4.6 B $*\frac{60}{100}$ = 0.6 - B

2.76 B = 0.6 - B

2.76 + B = 0.6

3.76 B = 0.6

B = $\frac{0.6}{3.76}$

B = 0.159 g

B = 159 mg

∴ 159 mg caffeine is being extracted from the 100 mL of water containing 600 mg of caffeine with one portion of in 60 mL dichloromethane.

4 0

3 years ago

Read 2 more answers

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

g Given that 50.0 mL of 0.100 M magnesium bromide reacts with 13.9 mL of silver nitrate solution according to the unbalanced equ

ipn [44]

Answer:

0.719M AgNO₃

Explanation:

Based on the reaction:

MgBr₂ + 2AgNO₃ ⇄ 2AgBr + Mg(NO₃)₂

<em>1 mole of magnesium bromide reacts completely with 2 moles of AgNO₃</em>

<em />

To find molarity of AgNO₃ solution we need to determine moles of AgNO₃ and, as molarity is the ratio of moles over liter (13.9mL = 0.0139L). Now, to determine moles of AgNO₃ we need to use the reaction, thus:

<em>Moles AgNO₃:</em>

<em />

Moles of MgBr₂ are:

50.0mL = 0.050L * (0.100mol / L) = 0.00500 moles of MgBr₂.

As the silver nitrate reacts completely and 2 moles of AgNO₃ reacts per mole of MgBr₂:

0.00500 moles MgBr₂ * (2 moles AgNO₃ / 1 mole MgBr₂) =

0.0100 moles of AgNO₃ are in the solution.

And molarity is:

0.0100 moles AgNO₃ / 0.0139L =

3 0

3 years ago