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Anna35 [415]
4 years ago
10

How are these equilibrium quantities affected by the initial amount of x(g) placed in the container? assume constant temperature

?
Chemistry
2 answers:
Damm [24]4 years ago
8 0
The equilibrium can be affected when there is a change of concentration, pressure or temperature of the system. Since it is mentioned that the temperature is constant, and the parameter being focused on is the initial amount, that should be the concentration factor. It depends where substance X is placed in the reaction. If it is a reactant (left side), then the equilibrium would shift and favor the forward reaction. If it is a product (right side), then the equilibrium would shift and favor the reverse reaction.
nikklg [1K]4 years ago
6 0

Equilibrium quantities of Y and Z are directly affected by initial amount of X placed in container.

Further Explanation:

Le Chatelier’s principle:

According to this principle, if changes are done in pressure, temperature or concentration in any equilibrium reaction, equilibrium shifts to that direction in which effect of altered value is reversed.

Given reaction occurs as follows:

 \text{2X(g)}\rightleftharpoons\text{Y(g)}+\text{Z(g)}

From this reaction, it is evident that two moles of X decompose to produce one mole of Y and one mole of Z.

If amount of X used is large, it results in more formation of Y and Z and vice-versa. This implies amount of product formation depends on initial amount of X taken in reaction.

According to Le Chatelier's principle, if amount of X is increased, equilibrium will tend to shift in that direction in which more consumption of X is observed. This is done only if more products are formed. Opposite effects are observed in case less amount of X is taken. Hence formation of Y and Z is dependent on initial amount of X.

Learn more:

1. Sort the solubility of gas will increase or decrease: brainly.com/question/2802008

2. Calculation of equilibrium constant of pure water at 25°C: brainly.com/question/3467841

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Chemical Equilibrium

Keywords: Le Chatelier’s principle, equilibrium, shift, direction, X, Y, Z, 2X, pressure, temperature, concentration, consumption, increase, decrease, two moles, one mole.

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the hcp ordered 0.2 mg of methylergonovine, and the vial contains 0.8 mg/ml. how many ml of methylergonovine should the nurse dr
olasank [31]

The volume (in mL) of methylergonovine the nurse should draw up in the syringeis is 0.25 mL

<h3>What is density? </h3>

The density of a substance is simply defined as the mass of the subtance per unit volume of the substance. Mathematically, it can be expressed as

Density = mass / volume

With the concept of density, we can obtain the volume of methylergonovine. Details below

<h3>How to determine the volume </h3>

The following data were obtained from the question:

  • Mass of methylergonovine = 0.2 mg
  • Density of methylergonovine = 0.8 mg/mL
  • Volume of methylergonovine =?

Density = mass / volume

Cross multiply

Density × volume = mass

Divide both sides by density

Volume = mass / density

Volume = 0.2 / 0.8

Volume of methylergonovine = 0.25 mL

Learn more about density:

brainly.com/question/952755

#SPJ1

8 0
2 years ago
What is The pH and pOH of a 1.00x10-3M solution of CH3COOH (Ka=1.75x10-5) ?
zaharov [31]

Answer:

<h2>pH = 3.9</h2><h2>pOH = 10.1</h2>

Explanation:

Since CH _ 3COOH is a weak acid to find the pH of CH _ 3COOH we use the formula

pH =  -  \frac{1}{2}   log(Ka)  -  \frac{1}{2}  log(c)

where

Ka is the acid dissociation constant

c is the concentration

From the question

Ka of CH _ 3COOH = 1.75 × 10^-5

c = 1.00 × 10-³M

Substitute the values into the above formula and solve for the pH

That's

pH =  \frac{1}{2} ( -  log(1.75 \times {10 }^{ - 5} -  log(1.00 \times  {10}^{ - 3} )  )  \\   =  \frac{1}{2} (4.757  + 3) \\  =  \frac{1}{2}  \times 7.757) \\  = 3.8785 \:  \:  \:  \:  \:  \:  \:  \:

We have the answer as

<h3>pH = 3.9</h3>

To find the pOH we use the formula

pH + pOH = 14

pOH = 14 - pH

pOH = 14 - 3.9

We have the answer as

<h3>pOH = 10.1</h3>

Hope this helps you

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