Answer:
molecular so number 3. ...
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
Newton's first law of motion states that an object at rest will remain at rest unless an unbalanced force acts on it. If you apply balanced forces on the object there would be no net force. The body does not accelerate but instead stays at rest.
Another way to look at this problem is to use Newton's second law of motion. The first law states that
, where
is the acceleration
is the net force and
is the mass of the object.
When F is zero, the acceleration of the object is zero. This means that if the object had a velocity of zero before the balanced forces started acting, the velocity will stay at zero after the balanced forces begin to act. If the object was moving at a constant velocity before the balanced forces started acting on it, it would continue at that constant velocity after the balanced forces begin to act.
Ammonia undergoes combustion with oxygen to produce nitric oxide and water. The volume of the oxygen required to react with 720 ml of ammonia is 900 ml.
<h3>What is volume?</h3>
Volume is the area occupied by the substance and is the ratio of the mass to the density.
At STP, 1 mole of gas occupies 22.4 L of volume
Given,
Volume of ammonia reacted = 0.720 L
The combustion reaction is shown as,

From the stoichiometry of the reaction, it can be said that,
L of ammonia reacts with
L of oxygen gas.
So, 0.720 L of ammonia will react with:

Therefore, the volume of oxygen required is 900 mL.
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The IUPAC rules are
a) Find out the longest chain of carbon in the given organic compound
b) We will name the longest chain.
c) We will identify the main functional group and will assign a suffix to the compound.
d) We will number the carbons in the longest chain selected so that the attached groups attain lowest numeral as substituent
e) We will name the side groups or chains.