Answer:
The decomposition of ethane is 153.344 times much faster at 625°C than at 525°C.
Explanation:
According to the Arrhenius equation,
![\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7BK_2%7D%7BK_1%7D%29%3D%5Cfrac%7BEa%7D%7B2.303%5Ctimes%20R%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= rate of reaction at 
= rate of reaction at 
= activation energy of the reaction
R = gas constant = 8.314 J/K mol


![\log (\frac{K_2}{K_1})=\frac{300,000 J/mol}{2.303\times 8.314 J/K mol}[\frac{1}{798.15 K}-\frac{1}{898.15 K}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7BK_2%7D%7BK_1%7D%29%3D%5Cfrac%7B300%2C000%20J%2Fmol%7D%7B2.303%5Ctimes%208.314%20J%2FK%20mol%7D%5B%5Cfrac%7B1%7D%7B798.15%20K%7D-%5Cfrac%7B1%7D%7B898.15%20K%7D%5D)


The decomposition of ethane is 153.344 times much faster at 625°C than at 525°C.
The large piece of jewelry that has a mass of 132.6 g and when is submerged in a graduated cylinder that initially contains 48.6 ml water and the volume increases to 61.2 ml once the piece of jewelry is submerged, has a density of: 10.523 g/ml
To solve this problem the formulas and the procedures that we have to use are:
Where:
- d= density
- m= mass
- v= volume
- v(f) = final volume
- v(i) = initial volume
Information about the problem:
- m = 132.6 g
- v(i) = 48.6 ml
- v(f) = 61.2 ml
- v = ?
- d =?
Applying the volume formula we get:
v = v(f)-v(i)
v = 61.2 ml - 48.6 ml
v = 12.6 ml
Applying the density formula we get:
d = m/v
d = 132.6 g/12.6 ml
d = 10.523 g/ml
<h3>What is density?</h3>
It is a physical quantity that expresses the ratio of the body mass to the volume it occupies.
Learn more about density in: brainly.com/question/1354972
#SPJ4
Answer:
About 5 times faster.
Explanation:
Hello,
In this case, since the Arrhenius equation is considered for both the catalyzed reaction (1) and the uncatalized reaction (2), one determines the relationship between them as follows:

By replacing the corresponding values we obtain:

Such result means that the catalyzed reaction is about five times faster than the uncatalyzed reaction.
Best regards.
Velocity and mass are directly proportional to the quantity of momentum by:
p = mv. Therefore, and increase in either velocity or mass will lead to an increase in momentum and vice versa. Momentum during a reaction is always conserved, meaning that the mass and initial velocity before a reaction will always be equal to the change in mass and velocity produced after the reaction. Kinetic energy after a reaction, however, is not always conserved. For example if a fast moving vehicle collided with a stationary vehicle, and moved together, the overall kinetic energy would be after the reaction, as a heaver mass would be moved by the same velocity causing a decrease in kinetic energy.
I don't know if this is exactly what you are looking for, but in physics this is how it is understood.
<h3>
Answer:</h3>
8.01 mol MgO
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Atomic Structure</u>
<u>Stoichiometry</u>
- Using Dimensional Analysis
- Analyzing Reactions RxN
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[RxN - Unbalanced] Mg + O₂ → MgO
[RxN - Balanced] 2Mg + O₂ → 2MgO
[Given] 8.01 moles Mg
[Solve] moles MgO
<u>Step 2: Identify Conversions</u>
[RxN] 2 mol Mg → 2 mol MgO
<u>Step 3: Stoich</u>
- [DA] Set up:

- [DA] Multiply/Divide [Cancel out units]:
