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3 years ago

Why is a chemical equilibrium described as dynamic?

2 answers:

8 0

Because when equilibrium is reached, the reaction is still occurring in both directions, it's just that rate(forward) =rate(reverse) so there is no net change in the concentrations of the reactants or products.

Sergeu [11.5K]3 years ago

6 0

<span>The equilibrium constant does not indicate whether a reaction is in static ordynamic equilibrium because it is simply the concentrations of the products divided by the concentrations of the reactants. A reaction is at dynamic equilibrium if the rate of the forward reaction is equal to the rate of the reverse reaction.</span>

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What happens to the event horizon when a larger star “dies”?

Neporo4naja [7]

Black hole is likely to spin

3 0

3 years ago

) Given the following balanced equation, determine the rate of reaction with respect to [O2]. If the rate of formation of O2is 7

Travka [436]

Answer:

Rate of the reaction is 0.2593 M/s

-0.5186 M/s is the rate of the loss of ozone.

Explanation:

The rate of the reaction is defined as change in any one of the concentration of reactant or product per unit time.

$2O_3\rightleftharpoons 3O_2$

Rate of formation of oxygen : $7.78\times 10^{-1} M/s$

Rate of the reaction(R) = $\frac{-1}{2}\frac{d[O_3]}{dt}=\frac{1}{3}\frac{d[O_2]}{dt}$

$R=\frac{1}{3}\frac{d[O_2]}{dt}$

Rate of formation of oxygen=3 × (R)

$7.78\times 10^{-1} M/s=3\times R$

Rate of the reaction(R): $0.2593 M/s$

Rate of the reaction is 0.2593 M/s

Rate of disappearance of the ozone:

$R=-\frac{1}{2}\frac{d[O_3]}{dt}$

$\frac{d[O_3]}{dt}=-2\times R=-2\times 0.2593\times M/s=-0.5186M/s$

-0.5186 M/s is the rate of the loss of ozone.

6 0

3 years ago

Property of a substance

Scrat [10]

There are many properties to substances.

I'll list some examples below:

Mass
Volume
Density
Conductivity
Malleability
Boiling point
Melting point
Heat capacity

Hope this helps! :3

8 0

3 years ago

What is the relationship between the number of protons and the number of electrons in an atom?

aleksley [76]

In a neutral atom they are both equal, and their even quantities makes the atom neutral...

6 0

3 years ago

An ideal gas in a cylindrical container of radius r and height h is kept at constant pressure p. The bottom of the container is

Juli2301 [7.4K]

Answer:

$m =\frac{p*(pi)*r^{2}*h*mw}{R*\frac{T_{1} + T_{O}}{2}}$

Explanation:

The gas ideal law is

PV= nRT (equation 1)

Where:

P = pressure

R = gas constant

T = temperature

n= moles of substance

V = volume

Working with equation 1 we can get

$n =\frac{PV}{RT}$

The number of moles is mass (m) / molecular weight (mw). Replacing this value in the equation we get.

$\frac{m}{mw} =\frac{PV}{RT}$ or

$m =\frac{P*V*mw}{R*T}$ (equation 2)

The cylindrical container has a constant pressure p

The volume is the volume of a cylinder this is

$V =(pi)*r^{2}*h$

Where:

r = radius

h = height

(pi) = number pi (3.1415)

This cylinder has a radius, r and height, h so the volume is $V =(pi)*r^{2}*h$

Since the temperatures has linear distribution, we can say that the temperature in the cylinder is the average between the temperature in the top and in the bottom of the cylinder. This is:

$T =\frac{T_{1} + T_{O}}{2}$

Replacing these values in the equation 2 we get:

$m =\frac{P*V*mw}{R*T}$ (equation 2)

$m =\frac{p*(pi)*r^{2}*h*mw}{R*\frac{T_{1} + T_{O}}{2}}$

8 0

3 years ago