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3 years ago

At a certain temperature, 0.900 mol of SO3 is placed in a 2.00-L container.

Chemistry

1 answer:

Goryan [66]3 years ago

3 0

Answer:

Kc = 2.34 mol*L

Explanation:

The calculation of the Kc of a reaction is performed using the values of the concentrations of the participants in the equilibrium.

A + B ⇄ C + D

Kc = [C] * [D] / [A] * [B]

According to the reaction

Kc = [SO2]^2 * [O2]^2 / [SO3]^2

Knowing the 0.900 mol of SO3 is placed in a 2.00-L it means we have a 0.450 mol/L of SO3

0.450 --> 0 + 0 (Beginning of the reaction)

0.260 --> 0.260 + 0.130 (During the reaction)

0.190 --> 0.260 + 0.130 (Equilibrium of the reaction)

Kc = [0.260]^2 + [0.130]^2 / [0.190]^2

Kc = 2.34 mol*L

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The thumb alone is controlled by nine separate muscles; The hand has been used as a symbol of protection since ancient times; The human hand is different to the hands or paws of other animals, because it has fingers and a thumb that can work together. Each finger has 3 bones and the thumb has two bones

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3 years ago

Read 2 more answers

Which buffer would be better able to hold a steady pH on the addition of strong acid, buffer 1 or buffer 2? Explain. Buffer 1: a

Talja [164]

Answer:

Buffer 1.

Explanation:

Ammonia is a weak base. It acts like a Bronsted-Lowry Base when it reacts with hydrogen ions.

$\rm NH_3\; (aq) + H^{+}\; (aq) \to {NH_4}^{+}\; (aq)$ .

$\rm NH_3$ gains one hydrogen ion to produce the ammonium ion $\rm {NH_4}^{+}$ . In other words, $\rm {NH_4}^{+}$ is the conjugate acid of the weak base $\rm NH_3$ .

Both buffer 1 and 2 include

the weak base ammonia $\rm NH_3$ , and
the conjugate acid of the weak base $\rm {NH_4}^{+}$ .

The ammonia $\rm NH_3$ in the solution will react with hydrogen ions as they are added to the solution:

$\rm NH_3\; (aq) + H^{+}\; (aq) \to {NH_4}^{+}\; (aq)$ .

There are more $\rm NH_3$ in the buffer 1 than in buffer 2. It will take more strong acid to react with the majority of $\rm NH_3$ in the solution. Conversely, the pH of buffer 1 will be more steady than that in buffer 2 when the same amount of acid has been added.

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3 years ago

The vapor pressure of water at 65oC is 187.54 mmHg. What is the vapor pressure of a ethylene glycol (CH2(OH)CH2(OH)) solution ma

Pavlova-9 [17]

Answer:

173.83 mmHg is the vapor pressure of a ethylene glycol solution.

Explanation:

Vapor pressure of water at 65 °C= $p_o= 187.54 mmHg$

Vapor pressure of the solution at 65 °C= $p_s$

The relative lowering of vapor pressure of solution in which non volatile solute is dissolved is equal to mole fraction of solute in the solution.

Mass of ethylene glycol = 22.37 g

Mass of water in a solution = 82.21 g

Moles of water= $n_1=\frac{82.21 g}{18 g/mol}=4.5672 mol$

Moles of ethylene glycol= $n_2=\frac{22.37 g}{62.07 g/mol}=0.3603 mol$

$\frac{p_o-p_s}{p_o}=\frac{n_2}{n_1+n_2}$

$\frac{187.54 mmHg-p_s}{187.54 mmHg}=\frac{0.3603 mol}{0.3603 mol+4.5672 mol}$

$p_s=173.83 mmHg$

173.83 mmHg is the vapor pressure of a ethylene glycol solution.

6 0

3 years ago

Question 2

Alenkinab [10]

Answer:

1.53 atm

Explanation:

From the question given above, the following data were obtained:

Volume = constant

Initial pressure (P₁) = stp = 1 atm

Initial temperature (T₁) = 273 K

Final temperature (T₂) = 144 °C = 144 °C + 273 = 417 K

Final pressure (P₂) =?

Since the volume is constant, the final pressure can be obtained as follow:

P₁ / T₁ = P₂ / T₂

1 / 273 = P₂ / 417

Cross multiply

273 × P₂ = 417

Divide both side by 273

P₂ = 417 / 273

P₂ = 1.53 atm

Therefore, the final pressure (i.e the pressure inside the hot water bottle) is 1.53 atm.

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3 years ago

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RUDIKE [14]

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Explanation:

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