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How is the reaction quotient used to determine whether a system is at equilibrium?

Juli2301 [7.4K]

Answer:

The answer to the question above is explained below

Explanation:

The reaction quotient, Q, is a measure of the relative amounts of reactants and products during a chemical reaction as it can be used to determine in which direction a reaction will proceed at a given point in time. Equilibrium constant is the numerical value of reaction quotient at the end of the reaction, when equilibrium is reached.

If Q = K then the system is already at equilibrium. If Q < Keq, the reaction will move toward the products to reach equilibrium. If Q > Keq, the reaction will move toward the reactants in order to reach equilibrium. Therefore, by comparing Q and K, we can determine the direction of a reaction.

Where Q= reaction quotient and Keq= equilibrium constant for the reaction.

The larger the equilibrium constant, the further the equilibrium lies toward the products. Reaction quotient is a quantity that changes as a reaction system approaches equilibrium.

We can determine the equilibrium constant based on equilibrium concentrations. K is the constant of a certain reaction when it is in equilibrium. Equilibrium occurs when there is a constant ratio between the concentration of the reactants and the products.

4 0

4 years ago

Shelby the Skater's rocket provides a forward force of

DochEvi [55]

Answer:

$a=3.17\ m/s^2$

Explanation:

Given that,

Force force acting on the Skater's rocket is 220 N

Resistive force acting on it, F' = 30 N

Mass of Shelby, m = 60 kg

We need to find the acceleration of Shelby. Net force acting on Shelby is given by :

Net force = 220-30

= 190 N

The formula of net force is :

F = ma

a is acceleration of Shelby

$a=\dfrac{F}{m}\\\\a=\dfrac{190}{60}\\\\a=3.17\ m/s^2$

So, the acceleration of Shelby is $3.17\ m/s^2$

4 0

3 years ago

A golfer imparts a speed of 29.0 m/s to a ball, and it travels the maximum possible distance before landing on the green. the te

Roman55 [17]

Explanation:

It is given that,

Initial speed of a golfer, u = 29 m/s

If it travels the maximum possible distance before landing. It means that it is projected at an angle of 45 degrees.

(a) We need to find the time spent by the ball in the air. It can be calculated by using second equation of motion.

$s=ut+\dfrac{1}{2}at^2$

Here,

a = -g

s = 0 (it is displacement and it is equal to 0 as the ball lands on the green).

So,

$0=29\sin(45)t-\dfrac{1}{2}\times 9.8t^2\ (\text{Initial vertical component of velocity is taken})\\\\-4.9t^2+29\times \dfrac{1}{\sqrt2}t=0\\\\-4.9t^2+20.5t=0\\\\t=0,4.184\ s$

So, it will take 4.184 seconds in the air.

(b) let x is the longest hole in one that the golfer can make if the ball does not roll when it hits the green. It can be given by :

$x=vt\cos\theta\\\\x=29\times 4.184\times \cos(45)\\\\x=85.79\ m$

Hence, this is the required solution.

8 0

4 years ago

PLZ HELP ASAP WILL GIVE BRAINLIEST!!!

igomit [66]

the answer is the forth one treatment of cancer

3 0

4 years ago

Read 2 more answers

An electron is constrained to the central perpendicular axis of a ring of charge of radius 2.2 m and charge 0.021 mC. Suppose th

igomit [66]

Answer:

T = 1.12 10⁻⁷ s

Explanation:

This exercise must be solved in parts. Let's start looking for the electric field in the axis of the ring.

All the charge dq is at a distance r

dE = k dq / r²

Due to the symmetry of the ring, the field perpendicular to the axis is canceled, leaving only the field in the direction of the axis, if we use trigonometry

cos θ = $\frac{dE_x}{dE}$