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
I am so sorry I do not know but I need to answer a question in order to ask one I hope u pass:))
Answer:
Below.
Explanation:
Coal gas is a mixture of a variety of gases: inflammable gases including, hydrogen, methane, ethylene, carbon monoxide and volatile hydrocarbons and small amounts of non flammable gases like nitrogen and carbon dioxide.
Water gas consists mainly of carbon monoxide and hydrogen.
Producer gas is similar to water gas and consists mainly of carbon monoxide and hydrogen together with nitrogen and carbon dioxide.
Natural gas occurs naturally and consists mainly of methane with small amounts of other hydrocarbon gases.
<h3>
Answer:</h3>
150000 J
<h3>
General Formulas and Concepts:</h3>
<u>Chemistry</u>
<u>Thermodynamics</u>
Specific Heat Formula: q = mcΔT
- <em>q</em> is heat (in J)
- <em>m</em> is mass (in g)
- <em>c</em> is specific heat (in J/g °C)
- ΔT is change in temperature (in °C or K)
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
<em>Identify variables</em>
[Given] <em>m</em> = 225 g
[Given] <em>c</em> = 4.184 J/g °C
[Given] ΔT = 133 °C - -26.8 °C = 159.8 °C
[Solve] <em>q</em>
<u>Step 2: Solve for </u><em><u>q</u></em>
- Substitute in variables [Specific Heat Formula]: q = (225 g)(4.184 J/g °C)(159.8 °C)
- Multiply: q = (941.4 J/°C)(159.8 °C)
- Multiply: q = 150436 J
<u>Step 3: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
150436 J ≈ 150000 J
Topic: AP Chemistry
Unit: Thermodynamics
Book: Pearson AP Chemistry
to neutralize 1 mole of H2 S o4 we need one mole of any if we are having 50 grams of H2 S o4 it means the mole of H2 S o4 in 50 gram will be 50×40)/98 hence
utilising 50 grams of H2 S o4 we need approximately 20. 5 gm of Naoh
