Liquid octane CH3CH26CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. Suppose
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Answer:
According to the following position vs. Time graph the bicyclist was : D
D. Not moving for the first two seconds, then begin moving at the constant speed.
Explanation:
Along Y-axis = Distance in meters
Along X-axis = Time taken in second
- For first 2 seconds the The object has not changed its position.There is Zero distance covered.So , Speed = 0
- After 2 second,The Object changed the distance by equal amount in equal interval of time.(Uniform Speed).Hence at each point after 2 second, there is same value speed
The volume of oxygen at STP required would be 252.0 mL.
<h3>Stoichiometic problem</h3>The equation for the complete combustion of C2H2 is as below:
The mole ratio of C2H2 to O2 is 2:5.
1 mole of a gas at STP is 22.4 L.
At STP, 100.50 mL of C2H2 will be:
100.50 x 1/22400 = 0.0045 mole
Equivalent mole of O2 according to the balanced equation = 5/2 x 0.0045 = 0.01125 moles
0.01125 moles of O2 at STP = 0.01125 x 22400 = 252.0 mL
Thus, 252.0 mL of O2 gas will be required at STP.
More on stoichiometric problems can be found here: brainly.com/question/14465605
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