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3 years ago

What is the freezing point of an aqueous solution that boils at 105.0 ∘C? Express your answer using two significant figures.

Chemistry

1 answer:

konstantin123 [22]3 years ago

5 0

Answer:

T°fussion of solution is -18°C

Explanation:

We have to involve two colligative properties to solve this. Let's imagine that the solute is non electrolytic, so i = 1

First of all, we apply boiling point elevation

ΔT = Kb . m . i

ΔT = Boiling T° of solution - Boiling T° of pure solvent

Kb = ebuliloscopic constant

105°C - 100° = 0.512 °C kg/mol . m . 1

5°C / 0.512 °C mol/kg = m

9.7 mol/kg = m

Now that we have the molality we can apply, the Freezing point depression.

ΔT = Kf . m . i

Kf = cryoscopic constant

0° - (T°fussion of solution) = 1.86 °C/m . 9.76 m . 1

- (1.86°C /m . 9.7 m) = T°fussion of solution

- 18°C = T°fussion of solution

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What is the pressure of 4 moles of helium in a 50 L tank at 308k using PV=nRT

Ilya [14]

Hey there!:

p = ??

n = 4 moles

V = 50 L

T = 308 K

R = 0.082 atm

p*V = n * R * T

p * 50 = 4 * 0.082 * 308

p*50 = 101.024

p = 101.024 / 50

p = 2.020 atm

Hope that helps!

8 0

3 years ago

Read 2 more answers

Technetium (Tc; Z = 43) is a synthetic element used as a radioactive tracer in medical studies. A Tc atom emits a beta particle

Sedaia [141]

Question in incomplete, complete question is:

Technetium (Tc; Z = 43) is a synthetic element used as a radioactive tracer in medical studies. A Tc atom emits a beta particle (electron) with a kinetic energy (Ek) of $4.71\times 10^{-15}J$ . What is the de Broglie wavelength of this electron (Ek = ½mv²)?

Answer:

$6.762\times 10^{-12} m$ is the de Broglie wavelength of this electron.

Explanation:

To calculate the wavelength of a particle, we use the equation given by De-Broglie's wavelength, which is:

$\lambda=\frac{h}{\sqrt{2mE_k}}$

where,

= De-Broglie's wavelength = ?

h = Planck's constant = $6.624\times 10^{-34}Js$

m = mass of beta particle = $9.1094\times 10^{-31} kg$

$E_k$ = kinetic energy of the particle = $4.71\times 10^{-15}J$

Putting values in above equation, we get:

$\lambda =\frac{6.624\times 10^{-34}Js}{\sqrt{2\times 9.1094\times 10^{-31} kg\times 4.71\times 10^{-15}J}}$

$\lambda = 6.762\times 10^{-12} m$

$6.762\times 10^{-12} m$ is the de Broglie wavelength of this electron.

3 0

3 years ago

In a sample containing a mixture of only these gases at exactly one atmosphere pressure, the partial pressures of carbon dioxide

Black_prince [1.1K]

Answer:

Explanation:

The pressure of a gaseous mixture is equal to the sum of the partial pressures of the individual gases:

Σ $P_g_a_s$ $= P_1+P_2+P_3+...+P_n$

The prompt is trying to confuse you, but it actually tells us the pressure of the mixture to be 1 atm, but this can be converted to torr. Furthermore, we are informed only three gases are in the mixture: diatomic nitrogen, diatomic oxygen, and carbon dioxide:

$P_g_a_s=1 \ atm = 760 \ torr= P_N_2+P_O_2+P_C_O_2\\760 \ torr = 582.008 \ torr + P_O_2 \ + 0.285 \ torr$