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3 years ago

In the equilibrium constant expression for the reaction below what is the correct exponent for N2O4?

1 answer:

7 0

As we have the balanced reaction equation is:

N2O4 (g) ↔ 2NO2(g)

from this balanced equation, we can get the equilibrium constant expression

KC = [NO2]^2[N2O4]^1

from this expression, we can see that [NO2 ] is with 2 exponent of the stoichiometric and we can see that from the balanced equation as NO2
is 2NO2 in the balanced equation.

and [N2O4] is with 1 exponent of the stoichiometric and we can see that from the balanced equation as N2O4 is 1 N2O4 in the balanced equation.

∴ the correct exponent for N2O4 in the equilibrium constant expression is 1

You might be interested in

600 s after initiation of a first order reaction 48.5% of the initial reactant concentration remains present. What is the rate c

Ludmilka [50]

Answer:

$k=1.20x10^{-3} s^{-1}$

Explanation:

For a first order reaction the rate law is:

$v=\frac{-d[A]}{[A]}=k[A]$

Integranting both sides of the equation we get:

$\int\limits^a_b {\frac{d[A]}{[A]}} \, dx =-k\int\limits^t_0 {} \, dt$

where "a" stands for [A] (molar concentration of a given reagent) and "b" is {A]0 (initial molar concentration of a given reagent), "t" is the time in seconds.

From that integral we get the integrated rate law:

$ln\frac{[A]}{[A]_{0} } =-kt$

$[A]=[A]_{0}e^{-kt}$

$ln[A]=ln[A]_{0} -kt$

$k=\frac{ln[A]_{0}-ln[A]}{t}$

therefore k is

$k=\frac{ln1-ln0,485}{600}=1,20x10^{-3}$

8 0

3 years ago

You have a mass of 54 kg and are sitting on a stool. The normal force the stool applies on you is ...

Karo-lina-s [1.5K]

Answer:

529.2 N

Explanation:

As we have studied the first law of motion, which states that every action has some reaction, equal in magnitude but having an opposite direction.

The force that is acting on the student will be due to gravitational force, that is equal to his weight.

F=mg: 54kg x 9.8m/s^2 =529.2 N

So the weight of student is exerting downwards towards the stool and land. The stool will also exert a force on the student that will be equal in magnitude but opposite in direction, then it will be 529.2 N.

This is because the student is sitting in a constant state and all the weight is exerted on the stool.

Note: This answer is very generic supposing that all the weight of the student is on stool. But, if we suppose that student's legs are on floor so it means the force of gravity acting on the stool has become less because student's mass on stool is less. So the answer would be a force somehow less than 529.2 N. However, since the question asked normal force, it would be weight of student in general terms.

Hope it helps!

7 0

2 years ago

What is the molar mass of 81.50 g of gas

garik1379 [7]

The molar mass of gas = 238.29 g/mol

<h3>Further explanation</h3>

Given

mass = 81.5 g

P=1.75 atm

V=4.92 L

T=307 K