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:
4 moles of neon
Explanation:
Given data:
Number of moles of neon = ?
Number of atoms of neon = 2.4×10²⁴ atoms
Solution:
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.
The number 6.022 × 10²³ is called Avogadro number.
For example,
18 g of water = 1 mole = 6.022 × 10²³ molecules of water
1.008 g of hydrogen = 1 mole = 6.022 × 10²³ atoms of hydrogen
For given neon atoms:
1 mol = 6.022 × 10²³ atoms
2.4×10²⁴ atoms × 1 mol / 6.022 × 10²³ atoms
0.4×10¹ mol = 4 mol
Independent variable: the student chooses to miss class.
dependent variable: in result of the student missing class his grade goes down.
hyposthesis: if the student misses class multiple times, then his grades will go down.
Answer: so the answer is A
Explanation: The relationship between an object's mass (m), its acceleration (a), and the applied force (f) is F=ma. ... This law requires that the direction of the acceleration vector is in the same direction as the force vectors.