Answer: The coefficient for
is 12.
Explanation:
According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

Thus in the reactants, there are 12 molecules of oxygen in balanced chemical equation. Thus the coefficient for
is 12.
Answer:
Theoretical yield for CO₂ is 5.10g
Explanation:
Reaction: 2C₆H₆(l) + 15O₂(g) → 12CO₂(g) + 6H₂O(g)
We convert the mass of oxygen to moles:
4.64 g /32 g/mol = 0.145 moles of O₂
Let's find out the 100% yield reaction of CO₂ (theoretical yield)
Ratio is 15:12. So let's make this rule of three:
15 moles of O₂ can produce 12 moles of CO₂
Therefore 0.145 moles of oxygen will produce (0.145 . 12) /15 = 0.116 moles
We convert the moles to mass: 0.116 mol . 44 g / 1mol = 5.10 g
Answer :
(A) The rate expression will be:
![Rate=-\frac{1}{2}\frac{d[HBr]}{dt}=+\frac{d[H_2]}{dt}=+\frac{d[Br_2]}{dt}](https://tex.z-dn.net/?f=Rate%3D-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BHBr%5D%7D%7Bdt%7D%3D%2B%5Cfrac%7Bd%5BH_2%5D%7D%7Bdt%7D%3D%2B%5Cfrac%7Bd%5BBr_2%5D%7D%7Bdt%7D)
(B) The average rate of the reaction during this time interval is, 0.00176 M/s
(C) The amount of Br₂ (in moles) formed is, 0.0396 mol
Explanation :
Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.
The given rate of reaction is,

The expression for rate of reaction :
![\text{Rate of disappearance of }HBr=-\frac{1}{2}\frac{d[HBr]}{dt}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20disappearance%20of%20%7DHBr%3D-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BHBr%5D%7D%7Bdt%7D)
![\text{Rate of disappearance of }H_2=+\frac{d[H_2]}{dt}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20disappearance%20of%20%7DH_2%3D%2B%5Cfrac%7Bd%5BH_2%5D%7D%7Bdt%7D)
![\text{Rate of formation of }Br_2=+\frac{d[Br_2]}{dt}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20formation%20of%20%7DBr_2%3D%2B%5Cfrac%7Bd%5BBr_2%5D%7D%7Bdt%7D)
<u>Part A:</u>
The rate expression will be:
![Rate=-\frac{1}{2}\frac{d[HBr]}{dt}=+\frac{d[H_2]}{dt}=+\frac{d[Br_2]}{dt}](https://tex.z-dn.net/?f=Rate%3D-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BHBr%5D%7D%7Bdt%7D%3D%2B%5Cfrac%7Bd%5BH_2%5D%7D%7Bdt%7D%3D%2B%5Cfrac%7Bd%5BBr_2%5D%7D%7Bdt%7D)
<u>Part B:</u>
![\text{Average rate}=-\frac{1}{2}\frac{d[HBr]}{dt}](https://tex.z-dn.net/?f=%5Ctext%7BAverage%20rate%7D%3D-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BHBr%5D%7D%7Bdt%7D)


The average rate of the reaction during this time interval is, 0.00176 M/s
<u>Part C:</u>
As we are given that the volume of the reaction vessel is 1.50 L.
![\frac{d[Br_2]}{dt}=0.00176M/s](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BBr_2%5D%7D%7Bdt%7D%3D0.00176M%2Fs)
![\frac{d[Br_2]}{15.0s}=0.00176M/s](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BBr_2%5D%7D%7B15.0s%7D%3D0.00176M%2Fs)
![[Br_2]=0.00176M/s\times 15.0s](https://tex.z-dn.net/?f=%5BBr_2%5D%3D0.00176M%2Fs%5Ctimes%2015.0s)
![[Br_2]=0.0264M](https://tex.z-dn.net/?f=%5BBr_2%5D%3D0.0264M)
Now we have to determine the amount of Br₂ (in moles).



The amount of Br₂ (in moles) formed is, 0.0396 mol
<em>Thermal energy</em> is the sum of the kinetic and potential energies of all the particles in an object.
Assume that you have 250 gL of water and 1 kg of water at the same temperature.
Then, each water molecule has the same kinetic energy.
The larger sample contains four times as many molecules, so it contains four times as much thermal energy.
Thus, thermal energy is directly proportional to mass.
In symbols, <em>KE </em>∝ <em>m</em> or <em>KE = km</em>.
The graph of a direct proportion is a <em>straight line passing trough the origin</em>.
It should look something like the graph below.
I believe the answer is "After adding two substances together a new order is detected" because nearly all of the other answers suggest a chemical change has taken place. The reason I say "almost" is because answer two and four results in bubbles forming so the answer cannot be both. The only one to not clearly show a chemical change has occurred is the last answer.