Separa los metales de los no metales. Agregame como amiga, saludos.
<h3>
Answer:</h3>
56.11 g/mol
<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>
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[Compound] KOH
<u>Step 2: Identify</u>
[PT] Molar Mass of K - 39.10 g/mol
[PT] Molar Mass of O - 16.00 g/mol
[PT] Molar Mass of H - 1.01 g/mol
<u>Step 3: Find</u>
39.10 + 16.00 + 1.01 = 56.11 g/mol
Answer: The activation energy Ea for this reaction is 22689.8 J/mol
Explanation:
According to Arrhenius equation with change in temperature, the formula is as follows.
![ln \frac{k_{2}}{k_{1}} = \frac{-E_{a}}{R}[\frac{1}{T_{2}} - \frac{1}{T_{1}}]](https://tex.z-dn.net/?f=ln%20%5Cfrac%7Bk_%7B2%7D%7D%7Bk_%7B1%7D%7D%20%3D%20%5Cfrac%7B-E_%7Ba%7D%7D%7BR%7D%5B%5Cfrac%7B1%7D%7BT_%7B2%7D%7D%20-%20%5Cfrac%7B1%7D%7BT_%7B1%7D%7D%5D)
= rate constant at temperature
= 
= rate constant at temperature
=
= activation energy = ?
R= gas constant = 8.314 J/kmol
= temperature = 
= temperature = 
Putting in the values ::
![ln \frac{4.8\times 10^8}{2.3\times 10^8} = \frac{-E_{a}}{8.314}[\frac{1}{649} - \frac{1}{553}]](https://tex.z-dn.net/?f=ln%20%5Cfrac%7B4.8%5Ctimes%2010%5E8%7D%7B2.3%5Ctimes%2010%5E8%7D%20%3D%20%5Cfrac%7B-E_%7Ba%7D%7D%7B8.314%7D%5B%5Cfrac%7B1%7D%7B649%7D%20-%20%5Cfrac%7B1%7D%7B553%7D%5D)

The activation energy Ea for this reaction is 22689.8 J/mol
Answer:
Final volumeof the gas = 2.84 L
Explanation:
The formular to be used here is the general gas equation. the formular is being used because it gives the relationship between the three gas parameters (volume, temperature and pressure) mentioned.
The general gas equation is given as;

where;
P1 = initial pressure
V1 = initial volume
T1 = initial temperature
P2 = Final pressure
V2 = Final volume
T2 = Final temperature
From the question,
P1 = 1.00 atm
P2 = 0.85atm
T1 = 25C + 273 = 298K (Converting to kelvin)
T2 = 15C + 273 = 288K (Converting to kelvin)
V1 = 2.5L
V2 = ?
from the equation, making V2 subject of formula we have;

V2 = (1*2.5*288)/(298*0.85) = 2.84 L.