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3 years ago

Assuming ideal solution behavior, what is the freezing point of a solution of 9.04 g of I2 in 75.5 g of benzene?

Chemistry

1 answer:

cluponka [151]3 years ago

3 0

Answer:

3.1°C

Explanation:

Using freezing point depression expression:

ΔT = Kf×m×i

<em>Where ΔT is change in freezing point, Kf is freezing point depression constant (5.12°c×m⁻¹), m is molality of the solution and i is Van't Hoff factor constant (1 For I₂ because doesn't dissociate in benzene).</em>

Molality of 9.04g I₂ (Molar mass: 253.8g/mol) in 75.5g of benzene (0.0755kg) is:

9.04g ₓ (1mol / 253.8g) = 0.0356mol I₂ / 0.0755kg = 0.472m

Replacing in freezing point depression formula:

ΔT = 5.12°cm⁻¹×0.472m×1

ΔT = 2.4°C

As freezing point of benzene is 5.5°C, the new freezing point of the solution is:

5.5°C - 2.4°C =

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A certain first order reaction has a half-life of 54. 3 s. How long will it take (in s) for the reactant concentration to decrea

Maksim231197 [3]

Answer:

82.4 s

Explanation:

Find the NUMBEr of half lives...then multiply by 54.3

2.27 = 6.5 (1/2)^n

log (2.27/6.5) / log (1/2) = n = 1.52 half lives

1.52 * 54.3 = 82.4 s

8 0

1 year ago

What is molarity of 47.0 g KCl dissolved in enough water to give 375 mL of solution?

natita [175]

This question provides us –

Weight of $\bf KCl$ is = 47 g
Volume, V = 375 mL

__________________________________________

Molar Mass of $\bf KCl$ –

$\qquad$ $\twoheadrightarrow\bf 39.0983 \times 35.453$

$\qquad$ $\twoheadrightarrow\bf 74.5513$

<u>Using formula</u> –

$\qquad$ $\purple{\twoheadrightarrow\bf Molarity _{(Solution)} = \dfrac{ W\times 1000}{MV}}$

$\qquad$ $\twoheadrightarrow\bf Molarity _{(Solution)} = \dfrac{ 47 \times 1000}{74.5513\times 375}$

$\qquad$ $\twoheadrightarrow\bf Molarity _{(Solution)} = \dfrac{47000}{27956.7375}$

$\qquad$ $\twoheadrightarrow\bf Molarity _{(Solution)} = \cancel{\dfrac{47000}{27956.7375}}$

$\qquad$ $\twoheadrightarrow\bf Molarity _{(Solution)} = 1.68117M$

$\qquad$ $\pink{\twoheadrightarrow\bf Molarity _{(Solution)} = 1.7M}$

Henceforth, Molarity of the solution is = 1.7M

___________________________________________

6 0

2 years ago

If a gas has a proportionality constant of 4.32 x 10-4 mol at room temperature for a particular solvent, what will the

inysia [295]

$0.0467 X 10^{-4} M/kPa$ is the solubility of the gas when it exerts a partial pressure of 92.4kPa.

<h3>What is Henry's law?</h3>

Mathematically, we can get this from Henry's law

From Henry law;

Concentration = Henry constant × partial pressure

Thus Henry constant = $\frac{Concentration}{partial \;pressure}$

Henry constant = $\frac{4.32 \;X \;10^{-4} mol}{92.4kPa}$

$= 0.0467 X 10^{-4} M/kPa$

Hence, $0.0467 X 10^{-4} M/kPa$ is the solubility of the gas when it exerts a partial pressure of 92.4kPa.

Learn more about the Henry's law here:

brainly.com/question/16222358

#SPJ1

6 0

2 years ago

Consider the balanced equation.

Zolol [24]

<span>The balanced reaction that describes the reaction between hydrogen and nitrogen to produce ammonia is expressed 3H2 + N2 = 2NH3. The yield of the reaction is equal to the actual amount of product divided to the theoretical amount of product multiplied by 100 percent. 26.3 grams of H2 theoretically produces149 grams. The yield is 79 divided by 149 equal to 53.02 percent. </span>

6 0

3 years ago

Read 2 more answers

Determine the pH of the resulting solution if 25 mL of 0.400 M strychnine (C21H22N2O2) is added to 50 mL of 0.200 M HCl? Assume

DIA [1.3K]

Answer:

pH = 4.56

Explanation:

The strychnine reacts with HCl as follows:

C₂₁H₂₂N₂O₂ + HCl ⇄ C₂₁H₂₂N₂O₂H⁺ + Cl⁻

<em />

For strychnine buffer:

pOH = 5.74 + log [C₂₁H₂₂N₂O₂H⁺] / [C₂₁H₂₂N₂O₂]

Initial moles of C₂₁H₂₂N₂O₂ are:

0.025L * (0.400 mol / L) = 0.01 moles C₂₁H₂₂N₂O₂

And of HCl are:

0.05L * (0.200 mol / L) = 0.01 moles HCl

That means after the reaction, you will have just 0.01 moles of C₂₁H₂₂N₂O₂H⁺ in 50mL + 25mL = 0.075L. And molarity is:

[C₂₁H₂₂N₂O₂H⁺] = 0.01 mol / 0.075L = 0.1333M

This conjugate acid, is in equilibrium with water as follows:

C₂₁H₂₂N₂O₂H⁺(aq) + H₂O(l) ⇄ C₂₁H₂₂N₂O₂ + H₃O⁺

<em />

<em>Where Ka = Kw / Kb = 1x10⁻¹⁴ / 1.8x10⁻⁶ = 5.556x10⁻⁹</em>

<em />

Ka is defined as:

Ka = 5.556x10⁻⁹ = [C₂₁H₂₂N₂O₂] [H₃O⁺] / [C₂₁H₂₂N₂O₂H⁺]

In equilibrium, concentrations are:

C₂₁H₂₂N₂O₂ = X

H₃O⁺ = X

C₂₁H₂₂N₂O₂H⁺ = 0.1333M - X

Replacing in Ka expression:

5.556x10⁻⁹ = [X] [X] / [0.1333M - X]

7.39x10⁻¹⁰ - 5.556x10⁻⁹X = X²

7.39x10⁻¹⁰ - 5.556x10⁻⁹X - X² = 0

Solving for X:

X = - 2.72x10⁻⁵M → False solution. There is no negative concentrations

X = 2.72x10⁻⁵M → Right solution.

As H₃O⁺ = X

H₃O⁺ = 2.72x10⁻⁵M

And pH = -log H₃O⁺

4 0

3 years ago