Answer:
1,033.56 grams of carbon dioxide was emitted into the atmosphere.
Explanation:
Energy absorbed by pork,E =
(assuming)
Total energy produced by barbecue = Q
Percentage of energy absorbed by pork = 10%


Since, it is a energy produced in order to indicate the direction of heat produced we will use negative sign.
Q = 
Moles of propane burnt to produce Q energy =n


According to reaction , 1 mol of propane gives 3 moles of carbon dioxide. then 7.83 moles of will give:
carbon dioxide gas.
Mass of 23.49 moles of carbon dioxide gas:
23.49 mol × 44 g/mol =1,033.56 g
1,033.56 grams of carbon dioxide was emitted into the atmosphere.
-<u><em>Oxygen</em></u>
According to Google these are the percentages of the <em>Earths Atmosphere</em>
<em>1</em> 78% - Nitrogen
<u>2</u> 21% - Oxygen
<em>3</em> 0.9% - Argon
<em>4 </em>0.3 - Carbon Dioxide with very small percentage of other elements.
Answer:
Entropy increases
Explanation:
Entropy (S) is a measure of the degree of disorder. For a given substance - say water - across phases the following is true ...
S(ice) < S(water) << S(steam)
For a chemical process, entropy changes can be related to increasing or decreasing molar volumes of gas from reactant side of equation to product side of equation. That is ...
if molar volumes of gas increase, then entropy increases, and
if molar volumes of gas decrease, then entropy decreases.
For the reaction 2KClO₃(s) => 2KCl(s) + 3O₂(g)
molar volumes of gas => 0Vm* 0Vm 3Vm
*molar volumes (Vm) apply only to gas phase substances. Solids and liquids do not have molar volume.
Since the reaction produces 3 molar volumes of O₂(g) product vs 0 molar volumes of reactant, then the reaction is showing an increase in molar volumes of gas phase substances and its entropy is therefore increasing.