Option C is the correct set of the problem for mass of water produced by 3.2 moles of oxygen and an excess ethene.
<h3>
Reaction between oxygen and ethene</h3>
Ethene (C2H4) burns in the presence of oxygen (O2) to form carbon dioxide (CO2) and water (H2O) along with the evolution of heat and light.
C₂H₄ + 3O₂ ----- > 2CO₂ + 2H₂O
from the equation above;
3 moles of O₂ ---------> 2(18 g) of water
3.5 moles of O₂ ----------> x
![x = 3.2 \times [\frac{2 \ moles \ H_2O}{3 \ moles \ O_2} ] \times[ \frac{18.02 \ g \ H_2O}{1 \ mole \ H_2O} ]](https://tex.z-dn.net/?f=x%20%3D%203.2%20%5Ctimes%20%5B%5Cfrac%7B2%20%5C%20moles%20%5C%20H_2O%7D%7B3%20%5C%20moles%20%5C%20O_2%7D%20%20%5D%20%5Ctimes%5B%20%5Cfrac%7B18.02%20%5C%20g%20%5C%20H_2O%7D%7B1%20%5C%20mole%20%5C%20H_2O%7D%20%5D)
Thus, option C is the correct set of the problem for mass of water produced by 3.2 moles of oxygen and an excess ethene.
Learn more about reaction of ethene here: brainly.com/question/4282233
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Answer:
e. 3
Explanation:
In order to solve this problem we need to keep in mind the definition of pH:
As stated by the problem, the hydrogen ion concentration, [H⁺], is 1x10⁻³ M.
As all required information is available, we now can <u>calculate the pH</u>:
The correct option is thus e.
The
equation for the photosynthesis reaction in which carbon dioxide and water
react to form glucose is .
The hear reaction is the difference between the bond dissociation energies in
the products and the bond dissociation energies of the reactants
The
reactant molecules have 12 C = O, 12 H - O bonds while the product molecules
have 5 C - C, 7 C – O, 5 H – O, and 6 O = O bonds. The average bond
dissociation energies for the bonds involved in the reaction are 191 for C = O,
112 for H – O, 83 C –C, 99 C – H, 86 C – O, 119 O = O.
Substitute
the average bond dissociation energies in the equation for and
calculate as follows
=
[12 (C=O) + 12 (H-O)] – [5(C-C) + 7(C-H) + 7 (C-O) + 5(H-O) + 6(O=O)]
=
[12x191 kcal/mol + 12x112 kcal//mol] – [5x83 kcal/mol + 7x99 kcal/mol + 7x86
kcal/mol + 5x112 kcal/mol + 6x119 kcal/mol]
=
3636 kcal/mol – 2984 kcal/mol = 652 kcal/mol x 4.184 Kj/1kcal = 2.73x10^3 kJ/mol
So,
enthalpy change for the reaction is 652 kcal/mol or 2.73x10^3 kJ/mol
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