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
I think its D(a battery,a length of wire and so on...)
Answer:
Option C. 52.057
Explanation:
The following data were obtained from the question:
Isotope >> Mass number > Percentage
A (Cr-50) > 50 >>>>>>>>>> 4.3
B (Cr-52) > 52 >>>>>>>>>> 83.8
C (Cr-53) > 53 >>>>>>>>>> 9.5
D (Cr-54) > 54 >>>>>>>>>> 2.4
Average atomic mass =?
The average atomic mass of chromium, Cr can be obtained as follow:
Average atomic mass = [(Mass of A × A%) /100] + [(Mass of B × B%) /100] + [(Mass of C × C%) /100] + [(Mass of D × D%) /100]
Atomic mass of Cr = [50×4.3)/100] + [52×83.8)/100] + [53×9.5)/100] + [54×2.4)/100]
= 2.15 + 43.576 + 5.035 + 1.296
Atomic mass of Cr = 52.057
Therefore, the atomic mass of chromium, Cr is 52.057
Answer:
Explanation:
The movement of the electrons is illustrated in the picture attached to this answer. It is a four-step reaction mechanism.
First STEP: The first step involves the transfer of an electron from sodium to form a radical anion.
Second STEP: This radical anion then removes a proton/hydrogen from ammonia in a bid to neutralize itself (hence the hydrogen becomes bonded to the anion).
Third STEP: The sodium (from NaNH₂ formed) transfers an electron again to produce a vinyl carbanion.
Fourth STEP: The carbanion then removes a proton/hydrogen from ammonia (like in the second step) to form a neutral trans-alkene.
NOTE: The circled numbers denote each step while the mechanism on the left represents the use of any alkyl group (R and R') while the mechanism on the right assumes both alkyl groups are methyl. Hence, 2-butyne started the reaction and the final product was trans-2-butene.
