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:
The air molecules that are surrounding the metal will speed up, and the molecules in the metal will slow down.
Explanation:
There will be a transfer of thermal energy (heat) from the hot metal plate to the surrounding air. This transfer of energy equates to a transfer of kinetic energy in the molecules. As the plate loses heat, the molecules in the plate will lose kinetic energy and slow down. As the surrounding air gains heat, the molecules will gain kinetic energy and speed up.
Answer:
11.6g of NH₃(g) have to react
Explanation:
For the reaction:
4 NH₃(g) + 5 O₂(g) → 4 NO(g) + 6 H₂O(g) ΔH = -905kJ
<em>4 moles of ammonia produce 905kJ</em>
Thus, if you want to produce 154kJ of energy you need:
154kJ × (4 mol NH₃ / 905kJ) = <em>0.681moles of NH₃. </em>In mass -Molar mass ammonia is 17.031g/mol-
0.681mol NH₃ × (17.031g / mol) = <em>11.6g of NH₃(g) have to react</em>
N=14
H= 1(5)= 5
C=12(2)=24
O=16(2)=32
= 75