The equilibrium constant for the reaction is:
K subscript eq equals StartFraction StartBracket upper H subscript 2 EndBracket superscript 2 StartBracket upper O subscript 2 EndBracket over StartBracket upper H subscript 2 upper O EndBracket superscript 2 EndFraction.
<h3>What is equilibrium constant? </h3>
The equilibrium constant (Kₑq) for a given reaction is simply defined as the ratio of the concentration of the products raised to their coefficient to the concentration of the reactants raised to their coefficient.
For example, the equilibrium constant Kₑq for the reaction below is given as
2A <=> B
Kₑq = [B]/[A]²
<h3>How to determine the equilibrium constant </h3>
- 2H₂O(g) <=> 2H₂(g) + O₂(g)
- Equilibrium constant (Kₑq) =?
Kₑq = [H₂]²[O₂] / [H₂O]²
K subscript eq equals StartFraction StartBracket upper H subscript 2 EndBracket superscript 2 StartBracket upper O subscript 2 EndBracket over StartBracket upper H subscript 2 upper O EndBracket superscript 2 EndFraction.
Learn more about equilibrium constant:
brainly.com/question/17960050
Answer:
<h2>8.89 g/mL</h2>
Explanation:
The density of a substance can be found by using the formula
From the question we have
We have the final answer as
<h3>8.89 g/mL</h3>
Hope this helps you
Hydrogenation is a process that involves adding hydrogen to the double bonds in unsaturated fatty acids to extend the shelf-life of food products.
<h3>What is
hydrogenation?</h3>
Hydrogenation can be defined as a chemical process that involves an addition of hydrogen to the double bonds in unsaturated fatty acids to extend the shelf-life of food products.
<h3>The purpose of
hydrogenation.</h3>
In Science, some of the purpose of hydrogenation include the following:
- Conversion of liquid oil into solid fat.
- To alter a fat's consistency level.
- To stabilize a fat or an oil.
Read more on unsaturated fatty acids here: brainly.com/question/1580524
Answer
is: activation energy of this reaction is 212,01975 kJ/mol.<span>
Arrhenius equation: ln(k</span>₁/k₂) =
Ea/R (1/T₂ - 1/T₁).<span>
k</span>₁
= 0,000643 1/s.<span>
k</span>₂
= 0,00828 1/s.
T₁ = 622 K.
T₂ = 666 K.
R = 8,3145 J/Kmol.
<span>
1/T</span>₁ =
1/622 K = 0,0016 1/K.<span>
1/T</span>₂ =
1/666 K = 0,0015 1/K.<span>
ln(0,000643/0,00828) = Ea/8,3145 J/Kmol ·
(-0,0001 1/K).
-2,55 = Ea/8,3145 J/Kmol </span>· (-0,0001 1/K).<span>
Ea = 212019,75 J/mol = 212,01975 kJ/mol.</span>
Answer:
0.48 V
Explanation:
Usually in the cell notation, the left side shows oxidation. So,
Oxidation half reaction:
Reduction half reaction:
