Answer:
Option C is correct
Explanation:
Entropy is defined as the randomness in a system. In nature, a system tends to have high entropy. The higher the entropy , the spontaneous the reactions would be. As a system loses energy, it becomes more ordered and less random. Hence, as the reaction proceeds in a system, the system has higher energy/entropy and less order.
Hence, option c is correct
47% yield.
First, let's determine how many moles of ethane was used and how many moles of CO2 produced. Start with the respective atomic weights.
Atomic weight carbon = 12.0107
Atomic weight hydrogen = 1.00794
Atomic weight oxygen = 15.999
Molar mass C2H6 = 2 * 12.0107 + 6 * 1.00794 = 30.06904 g/mol
Molar mass CO2 = 12.0107 + 2 * 15.999 = 44.0087 g/mol
Moles C2H6 = 8 g / 30.06904 g/mol = 0.266054387 mol
Moles CO2 = 11 g / 44.0087 g/mol = 0.249950578 mol
Looking at the balanced equation, for every 2 moles of C2H6 consumed, 4 moles of CO2 should be produced. So at 100% yield, we should have 0.266054387 / 2 * 4 = 0.532108774 moles of CO2. But we only have 0.249950578 moles, or 0.249950578 / 0.532108774 = 0.46973587 =
46.973587% of what was expected.
Rounding to 2 significant figures gives 47% yield.
Answer:
number 2 becaus 1 and 3 look simler.
Explanation:
Answer:
(a) rate = -(1/3) Δ[O₂]/Δt = +(1/2) Δ[O₃]/Δt
(b) Δ[O₃]/Δt = 1.07x10⁻⁵ mol/Ls
Explanation:
By definition, t<u>he reaction rate for a chemical reaction can be expressed by the decrease in the concentration of reactants or the increase in the concentration of products:</u>
aX → bY (1)
![rate= -\frac{1}{a} \frac{\Delta[X]}{ \Delta t} = +\frac{1}{b} \frac{\Delta[Y]}{ \Delta t}](https://tex.z-dn.net/?f=%20rate%3D%20-%5Cfrac%7B1%7D%7Ba%7D%20%5Cfrac%7B%5CDelta%5BX%5D%7D%7B%20%5CDelta%20t%7D%20%3D%20%2B%5Cfrac%7B1%7D%7Bb%7D%20%5Cfrac%7B%5CDelta%5BY%5D%7D%7B%20%5CDelta%20t%7D%20)
<em>where, a and b are the coefficients of de reactant X and product Y, respectively. </em>
(a) Based on the definition above, we can express the rate of reaction (2) as follows:
3O₂(g) → 2O₃(g) (2)
![rate = -\frac{1}{3} \frac{\Delta[O_{2}]}{\Delta t} = +\frac{1}{2} \frac{ \Delta[O_{3}]}{ \Delta t}](https://tex.z-dn.net/?f=%20rate%20%3D%20-%5Cfrac%7B1%7D%7B3%7D%20%5Cfrac%7B%5CDelta%5BO_%7B2%7D%5D%7D%7B%5CDelta%20t%7D%20%3D%20%2B%5Cfrac%7B1%7D%7B2%7D%20%5Cfrac%7B%20%5CDelta%5BO_%7B3%7D%5D%7D%7B%20%5CDelta%20t%7D%20)
(3)
(b) From the rate of disappearance of O₂ in equation (3), we can find the rate of appearance of O₃:
![rate = +\frac{1}{2} \frac{\Delta[O_{3}]}{ \Delta t} = -\frac{1}{3} \frac{\Delta[O_{2}]}{ \Delta t}](https://tex.z-dn.net/?f=%20rate%20%3D%20%2B%5Cfrac%7B1%7D%7B2%7D%20%5Cfrac%7B%5CDelta%5BO_%7B3%7D%5D%7D%7B%20%5CDelta%20t%7D%20%3D%20-%5Cfrac%7B1%7D%7B3%7D%20%5Cfrac%7B%5CDelta%5BO_%7B2%7D%5D%7D%7B%20%5CDelta%20t%7D%20)
![+\frac{\Delta[O_{3}]}{ \Delta t} = -\frac{2}{3} \frac{\Delta[-1.61 \cdot 10^{-5}]}{ \Delta t}](https://tex.z-dn.net/?f=%20%2B%5Cfrac%7B%5CDelta%5BO_%7B3%7D%5D%7D%7B%20%5CDelta%20t%7D%20%3D%20-%5Cfrac%7B2%7D%7B3%7D%20%5Cfrac%7B%5CDelta%5B-1.61%20%5Ccdot%2010%5E%7B-5%7D%5D%7D%7B%20%5CDelta%20t%7D%20)
![\frac{\Delta[O_{3}]}{ \Delta t} = 1.07 \cdot 10^{-5} \frac{mol}{Ls}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5CDelta%5BO_%7B3%7D%5D%7D%7B%20%5CDelta%20t%7D%20%3D%201.07%20%5Ccdot%2010%5E%7B-5%7D%20%5Cfrac%7Bmol%7D%7BLs%7D%20)
So the rate of appearance of O₃ is 1.07x10⁻⁵ mol/Ls.
Have a nice day!
<u><em>Answer:</em></u>
<u><em>49.98</em></u>
<u><em>Explanation:49.98 should be the correct answer but if you start with 49.98 g H2O2 and decompose all of it you do not get 34.00 g H2O and 15.98 g O2. </em></u>
