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3 years ago

CHEMISTRY AGAIN PAPER 4

Chemistry

1 answer:

hichkok12 [17]3 years ago

4 0

Cant See it get closer to the paper

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Consider the following reaction at a high temperature. Br2(g) ⇆ 2Br(g) When 1.35 moles of Br2 are put in a 0.780−L flask, 3.60 p

UNO [17]

Answer : The equilibrium constant $K_c$ for the reaction is, 0.1133

Explanation :

First we have to calculate the concentration of $Br_2$ .

$\text{Concentration of }Br_2=\frac{\text{Moles of }Br_2}{\text{Volume of solution}}$

$\text{Concentration of }Br_2=\frac{1.35moles}{0.780L}=1.731M$

Now we have to calculate the dissociated concentration of $Br_2$ .

The balanced equilibrium reaction is,

$Br_2(g)\rightleftharpoons 2Br(aq)$

Initial conc. 1.731 M 0

At eqm. conc. (1.731-x) (2x) M

As we are given,

The percent of dissociation of $Br_2$ = $\alpha$ = 1.2 %

So, the dissociate concentration of $Br_2$ = $C\alpha=1.731M\times \frac{1.2}{100}=0.2077M$

The value of x = 0.2077 M

Now we have to calculate the concentration of $Br_2\text{ and }Br$ at equilibrium.

Concentration of $Br_2$ = 1.731 - x = 1.731 - 0.2077 = 1.5233 M

Concentration of $Br$ = 2x = 2 × 0.2077 = 0.4154 M

Now we have to calculate the equilibrium constant for the reaction.

The expression of equilibrium constant for the reaction will be :

$K_c=\frac{[Br]^2}{[Br_2]}$

Now put all the values in this expression, we get :

$K_c=\frac{(0.4154)^2}{1.5233}=0.1133$

Therefore, the equilibrium constant $K_c$ for the reaction is, 0.1133

7 0

3 years ago

At standard temperature and pressure. 0.500 mole of xenon gas occupies

ANEK [815]

Answer:

0.500 mole of Xe (g) occupies 11.2 L at STP.

General Formulas and Concepts:

Gas Laws

STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K

Stoichiometry

Mole ratio
Dimensional Analysis

Explanation:

Step 1: Define

Identify.

0.500 mole Xe (g)

Step 2: Convert

[DA] Set up: $\displaystyle 0.500 \ \text{mole Xe} \bigg( \frac{22.4 \ \text{L Xe}}{1 \ \text{mole Xe}} \bigg)$
[DA] Evaluate: $\displaystyle 0.500 \ \text{mole Xe} \bigg( \frac{22.4 \ \text{L Xe}}{1 \ \text{mole Xe}} \bigg) = 11.2 \ \text{L Xe}$

Topic: AP Chemistry

Unit: Stoichiometry

3 0

2 years ago

One phosphorus, three chlorine, one oxygen

4vir4ik [10]

Phosphoryl chloride :

POCl3

hope this helps!.

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3 years ago

Which situation describes a reaction with a percentage yield of 100%?

HACTEHA [7]

Answer:

Explanation:

8 0

3 years ago

In a solution of salt and water which compound is the solute?

ra1l [238]

Solute is what gets dissovled, in this case the salt, and the solVENT is whatever a material is dissolved into, in this case the water.

7 0

3 years ago