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.
Answer:
Palms and other small trees
Molality is one way of expressing concentration of a solute in a solution. It is expressed as the mole of solute per kilogram of the solvent. To calculate for the molality of the given solution, we need to convert the mass of solute into moles and divide it to the mass of the solvent.
Molality = 29.5 g glucose (1 mol / 180.16 g ) / .950 kg water
Molality = 0.1724 mol / kg
Answer:
[NaCH₃COO] = 2.26M
Explanation:
17% by mass is a sort of concentration. Gives the information about grams of solute in 100 g of solution. (In this case, 17 g of NaCH₃COO)
Let's determine the volume of solution, by density
Mass of solution / Volume of solution = Solution density
100 g / Volume of solution = 1.09 g/mL
100 g / 1.09 g/mL = 91.7 mL
17 grams of solute is contained in 91.7 mL
Molarity (M) = Mol of solute /L of solution
91.7 mL / 1000 = 0.0917L
17 g / 82 g/m = 0.207 moles
Molariy = 0.207 moles / 0.0917L → 2.26M