Answer:- 335 kcal of heat energy is produced.
Solution:- The balanced equation for the combustion of glucose in presence of oxygen to give carbon dioxide and water is:
From given info, 2803 kJ of heat is released bu the combustion of 1 mol of glucose. We need to calculate the energy produced when 3.00 moles of oxygen react with excess of glucose.
We could solve this using dimensional analysis as:
= 1401.5 kJ
Now, let's convert kJ to kcal.
We know that, 1kcal = 4.184kJ
So, 
= 335 kcal
Hence, 335 kcal of heat energy is produced by the use of 3.00 moles of oxygen gas.
What part of it are you confused about
The answer is A, between 0 and 7.
In a pH scale from 0 to 14, we can groups these numbers into acidic, neutral, and alkaline. 7 is the neutral pH value, therefore, 0-7 is always acidic, and 7-14 is alkaline.
The smaller the number is, the more acidic the solution will be. This applies same in alkalis, the larger the pH value is, the more alkaline the solution is.
We can measure the pH of solution with many methods, the easiest way include using a pH paper, more advanced and accurate methods includes using a pH meter.
Answer:
C2H3Br + O2 → CO2 + H2O + HBr
Explanation:
The term balancing of chemical reaction equation has a unique meaning in chemistry. What it actually means is to ensure that the number of atoms of each element on the left hand side of reaction equation becomes equal to the number of atoms of the same element on the right hand side of the reaction equation.
When we look at the equation; C2H3Br + O2 → CO2 + H2O + HBr, the number of atoms of each element on the left and right hand sides of the given equation are not the same hence the equation is unbalanced.
If we look at the equation; 2C2H3Br + 5O2 → 4CO2 + 2H2O + 2HBr, the number of atoms of each element on both sides of the reaction equation are now equal, thus the later equation is the balanced version of the former.
Answer : The fuel value and the fuel density of pentane is, 49.09 kJ/g and 
respectively.
Explanation :
Fuel value : It is defined as the amount of energy released from the combustion of hydrocarbon fuels. The fuel value always in positive and in kilojoule per gram (kJ/g).
As we are given that:
Fuel value = 
Molar mass of pentane = 72 g/mol
Fuel value = 
Fuel value = 49.09 kJ/g
Now we have to calculate the fuel density of pentane.
Fuel density = Fuel value × Density
Fuel density = (49.09 kJ/g) × (0.626g/mL)
Fuel density = 30.73 kJ/mL =
Thus, the fuel density of pentane is
