Ask question

Ask question

All categories

3 years ago

13

when 0.863 mol acetone is dissolved in 500 g benzene, the freezing point of benzene drops 8.84°C. what is the molal freezing poi

nt depression constant (kf) for benzene?

1 answer:

sveticcg [70]3 years ago

3 0

Answer:

<u>Kf = 5.12 °C/m</u>

Explanation:

<u>1) Data:</u>

a) Solute: acetone

b) Solvent: benzene

c) n (solute) = 0.863 mol

d) mass of solvent = 500 g = 0.500 Kg

e) ΔTf = 8.84 °C

f) Kf = ?

<u>2) Formulae:</u>

Depression of freezing point: ΔTf = m . Kf

Molality: m = n (solute) / kg solvent

<u>3) Solution:</u>

a) m = 0.863 mol / 0.500 kg = 1.726 m

b) Kf = ΔTf / m = 8.84°C / 1.726 m = 5.12 °C/m ← answer

You might be interested in

4HCl(g)+O2(g)⟶2Cl2(g)+2H2O(g) Calculate the number of grams of Cl2 formed when 0.385 mol HCl reacts with an excess of O2. mass:

FromTheMoon [43]

The number of grams of Cl2 formed when 0.385 mol HCl reacts with an excess of O2 is 13.6675 g.

<h3>What are moles?</h3>

A mole is defined as 6.02214076 × $10^{23}$ of some chemical unit, be it atoms, molecules, ions, or others. The mole is a convenient unit to use because of the great number of atoms, molecules, or others in any substance.

Given data:

Moles of hydrochloric acid = 0.385 mol

Mass of chlorine gas =?

Chemical equation:

4HCl + O₂ → 2Cl₂ + 2H₂O

Now we will compare the moles of Cl₂ with HCl.

HCl : Cl₂

4 : 2

0.385 : 2÷4× 0.385 = 0.1925 mol

Oxygen is present in excess that's why the mass of chlorine produced depends upon the available amount of HCl.

Mass of Cl₂ :

Mass of Cl₂ = moles × molar mass

Mass of Cl₂ =0.1925 mol × 71 g/mol

Mass of Cl₂ = 13.6675 g

Hence, the number of grams of Cl2 formed when 0.385 mol HCl reacts with an excess of O2 is 13.6675 g.

Learn more about moles here:

brainly.com/question/8455949

#SPJ1

6 0

2 years ago

im found only in plant cells im as green as can be i make food for the plant using the suns energy what am i?

Cloud [144]

im found only in plant cells im as green as can be i make food for the plant using the suns energy what am i?

ANS:-Chloroplast

7 0

3 years ago

Read 2 more answers

A certain radioactive isotope decays at a rate of 0.2% annually. Determine the half-life of this isotope, to the nearest year

pychu [463]

Answer:

The half-life of the radioactive isotope is 346 years.

Explanation:

The decay rate of the isotope is modelled after the following first-order linear ordinary differential equation:

$\frac{dm}{dt} = -\frac{m}{\tau}$

Where:

$m$ - Current isotope mass, measured in kilograms.

$t$ - Time, measured in years.

$\tau$ - Time constant, measured in years.

The solution of this differential equation is:

$m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }$

Where $m_{o}$ is the initial mass of the isotope. It is known that radioactive isotope decays at a yearly rate of 0.2 % annually, then, the following relationship is obtained:

$\%e = \frac{m(t)-m(t+1)}{m(t)}\times 100\,\% = 0.2\,\%$

$1 - \frac{m(t+1)}{m(t)} = 0.002$

$1 - \frac{m_{o}\cdot e^{-\frac{t+1}{\tau} }}{m_{o}\cdot e^{-\frac{t}{\tau} }}=0.002$

$1 - e^{-\frac{1}{\tau} } = 0.002$

$e^{-\frac{1}{\tau} } = 0.998$

$-\frac{1}{\tau} = \ln 0.998$

The time constant associated to the decay is:

$\tau = -\frac{1}{\ln 0.998}$

$\tau \approx 499.500\,years$

Finally, the half-life of the isotope as a function of time constant is given by the expression described below:

$t_{1/2} = \tau \cdot \ln 2$

If $\tau \approx 499.500\,years$ , the half-life of the isotope is:

$t_{1/2} = (499.500\,years)\cdot \ln 2$

$t_{1/2}\approx 346.227\,years$

The half-life of the radioactive isotope is 346 years.

6 0

3 years ago

Are the following solutions acidic, basic, ot neutral.

Slav-nsk [51]

Answer:

1) Basic

2) Basic

3) Acidic

Explanation:

pH of a solution is a measure of the H3O+ ions in the solution and hence reflects its acidity.

$pH = -log[H3O+]$

Solutions are classified as acidic, basic or neutral based on the pH range

-pH < 7, acidic

- pH = 7, neutral

- pH > 7, basic

1) [H3O+] = 2.5*10^-9M

$pH = -log[H3O+]=-log[2.5*10^{-9}]=8.60$

Since pH > 7, solution is basic

2)[OH-] = 1.6*10^-2M

$pOH = -log[OH-]=-log[1.6*10^{-2}]=1.80$

pH = 14 - pOH = 14 - 1.80 =12.2

Since pH > 7, solution is basic

3) [H3O+] = 7.9*10^-3M

$pH = -log[H3O+]=-log[7.9*10^{-3}]=2.10$

Since pH < 7, solution is acidic

6 0

3 years ago

Pls help with me who is good in science!!!

anzhelika [568]

Here are my answers since I examined the questions.

Question 1.

No! as a matter of fact, the Rock cycle has no Beginning and No End it's an infinite loop.

Question 2.

The Rock cycle can happen in both ways such as, Volcanic rock melts and cool downs happen quite quickly as in others happen slower.

Question 3.

The 4 Major Steps of the rate of change are, Weathering, Transportation, Deposition, And Compaction.

Question 4.

The main factor for this one would be: The mineral composition of the Parent Rock.

Question 5.

The Parent Rock and the Pressure/temperature conditions.

These all my answers for this project, Good Luck.

7 0

2 years ago