Answer:
Decreased because the method can not distinguish oxygen from the gas added
Explanation:
Answer:
This is one of the factors that affects chemical reactions
Temperature:This is because when the temperature is raised energy in form of heat is supplied to the reacting particles and so the rate of reaction is increased.
Answer:

Explanation:
Hello,
In this case, the enthalpy of combustion is understood as the energy released when one mole of fuel, in this case octene, is burned in the presence of oxygen and is computed with the enthalpies of formation of the fuel, carbon dioxide and water as shown below (oxygen is circumvented as it is a pure element):

Thus, since we already know the enthalpy of combustion of the fuel, for carbon and water we have -393.5 and -241.8 kJ/mol respectively, thereby, the enthalpy of combustion turns out:

Best regards.
Answer:
<h3>The answer is 8.29 %</h3>
Explanation:
The percentage error of a certain measurement can be found by using the formula

From the question
actual density = 19.30g/L
error = 20.9 - 19.3 = 1.6
We have

We have the final answer as
<h3>8.29 %</h3>
Hope this helps you