Answer:
The answer to the question is
The source of the nitro- gen and oxygen atoms is from the atmosphere
Explanation:
When gasoline burns oxygen from the atmosphere supports the combustion reaction.
For example the internal combustion engine of a motor vehicle pulls in air from the surrounding atmosphere to make use of the oxygen in aiding the combustion of the gasoline to drive the pistons that ultimately moves the vehicle. The air from which the oxygen is absorbed also contains more amount of Nitrogen by percentage than oxygen. The nitrogen therefore enters into the combustion reaction that eventually produces nitrogen oxide and nitrogen dioxide
Answer:
A switch
Battery or cell
Resistor
Rheostat
Appliance or load such as bulb
connecting wires.
Jockey
Inductor
Capacitor
Meter bridge
Potentiometer
Voltimeter
Ammeter
Galvanometer
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:
41.67 mol
Explanation:
1 Litre of water = 1000g
Mole = mass / molar mass
Mass of 1 L of water = 1000 g
Molar mass of water (H2O) :
(H = 1, O = 16)
H2O = (1 * 2) + 16 = (2 + 16) = 18g/mol
Amount of water consumed = (3/4) of 1 litre
= (3/4) * 1000g
= 750g
Therefore mass of water consumed = 750g
Mole = 750g / 18g/mol
Mole of water consumed = 41.6666
= 41.67 mol