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kirill115 [55]
3 years ago
10

Consider the reaction. Upper H subscript 2 upper o (g) plus upper C l subscript 2 upper O (g) double-headed arrow 2 upper H uppe

r C l upper O (g). At equilibrium, the concentrations of the different species are as follows. [H2O] = 0.077 M [Cl2O] = 0.077 M [HClO] = 0.023 M What is the equilibrium constant for the reaction at this temperature?
Chemistry
2 answers:
Vlad [161]3 years ago
6 0

Answer:

0.089

Explanation:

Step 1:

The balanced equation for the reaction is given below:

H2O + Cl2O <=> 2HClO

Step 2:

Data obtained from the question. This includes:

Concentration of H2O, [H2O] = 0.077 M

Concentration of Cl2O, [Cl2O] = 0.077 M

Concentration of HClO, [HClO] = 0.023 M

Equilibrium constant, K =?

Step 3:

Determination of the equilibrium constant. This is illustrated below:

The equilibrium constant for the above reaction is given below:

K = [HClO]^2 / [H2O] [Cl2O]

K = (0.023)^2 / (0.077 x 0.077)

K = 0.089

Therefore, the equilibrium constant for the above reaction is 0.089

dedylja [7]3 years ago
5 0

Answer:

Equilibrium constant for the reaction is 0.0892

Explanation:

This is the reaction of equilibrium

         H₂O      +     Cl₂O     ⇄    2HClO

Eq   0.077M       0.077M           0.023M

Let's make the expression for Kc

Kc = [HClO]² / [Cl₂O] . [H₂O]

Kc= 0.023² / 0.077 . 0.077 = 0.0892

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Shkiper50 [21]

Answer:

0.87 pg

Explanation:

<em>GenAlex Medical, a leading manufacturer of medical laboratory equipment, is designing a new automated system that can detect normal levels of dissolved triiodothyronine (230. to 660 pg/dL), using a blood sample that is as small as 380 μL. Calculate the minimum mass in picograms of triiodothyronine that the new system must be able to detect. Be sure your answer has the correct number of significant digits.</em>

Step 1: Convert 380 μL to deciliters

We will use the following conversion factors.

  • 1 L = 10⁶ μL
  • 1 L = 10 dL

380 μL × 1 L/10⁶ μL × 10 dL/1 L = 3.8 × 10⁻³ L

Step 2: Calculate the minimum mass of triiodothyronine that can be found in a 3.8 × 10⁻³ L blood sample

Since we are looking for the minimum mass, we will use the lower limit of the concentration interval (230. pg/dL).

3.8 × 10⁻³ L × 230. pg/dL = 0.87 pg

7 0
2 years ago
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MAXImum [283]

Answer:

the vapor pressure of this solution would increase if some of the water were allowed to evaporate

4 0
3 years ago
The chemical hazard label indicates the class of hazard. It uses three major color-coded categories: Health (yellow), Flammabili
Rasek [7]

Answer:

False.

Explanation:

The chemical hazard label with colors indicates the specific class of hazard. Hazardous Materials Identification System is defined as the numerical hazard rating then incorporate the use of labels with colors.

Blue color: This sign conveys the health hazards of the material, means that long-term exposure to the material can cause the problems, for example kidney damage, and emphysema.

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5 0
3 years ago
En la reacción I2(g) + Br2(g) « 2 IBr(g), Keq = 280 a 150°C. Suponga que se permite que 0.500 mol de IBr en un matraz de 1.00 L
love history [14]

Answer:

P IBr: 15.454atm

I₂: 0.923 atm

P Br₂: 0.923atm

Explanation:

Basados en la reacción:

I₂(g) + Br₂(g) ⇄ 2 IBr(g)

La constante de equilibrio, Keq, es definida como:

Keq = \frac{P_{IBr}^2}{P_{I_2}P_{Br_2}}

<em>Se cumple la relación de Keq = 280 cuando las presiones están en equilibrio</em>

Usando PV = nRT, la presión inicial de IBr es:

P = nRT / V; 0.500mol*0.082atmL/molK*423.15K / 1.00L = <em>17.3 atm</em>

<em />

Siendo las presiones en equilibrio:

P IBr: 17.3 - 2X

P I₂: X

P Br₂: X

<em>Donde X representa el avance de reacción.</em>

Remplazando en Keq:

280 = (17.3 - 2X)² / X²

280X² = 4X² - 69.2X + 299.29

0 = -276X² - 69.2X + 299.29

<em>Resolviendo para X:</em>

X = -1.174 → Solución falsa. No existen presiones negativas

X = 0.923 → Solución real

Así, las presiones parciales en equilibrio de cada compuesto son:

P IBr: 17.3 - 2X = <em>15.454atm</em>

P I₂: X =<em> 0.923atm</em>

P Br₂: X = <em>0.923atm</em>

3 0
3 years ago
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sdas [7]

Answer: Ca^2+ and P^3-

Explanation:

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