Let us assume propane was the fuel
C3H8(g) + 5O2(g) ---> 3CO2(g) + 4H2O(g) = 2217kJ
1 mole ofpropane produces 3 moles of CO2
heat absorbed by pork = 0.11 x 2217
= 243.87 kJ/mol
number of moles of propane = 1700kJ / 243.87 kJ/mol
= 6.971 moles
1 mole of C3H8 = 3 moles ofCO2
6.971 moles of C3H8 = ?
3 x 6.971 = 20.913 moles of CO2
Convert to grams
mass = MW x mole
= 44 x 20.913
= 920.172g of CO2 emitted
Answer:The volume increases by 11%
Explanation:
When the pressure is decreased, the volume must definitely increase, to maintain the same pressure, the temperature must then also be increased as we see in the question. The volume was increased just as the pressure decreased and the temperature had to be increased to maintain a constant pressure. We have applied the general gas equation in solving the problem.
The decay mode of cesium-137 is beta decay. This means that the cesium-137 decays into a beta particle and a nuclide with the same mass number, but with a charge number that is 1 more than that of cesium.
Therefore, this means Cs-137 decays into an electron and Barium-137, meaning the answer is choice 1.
is a reducing agent,
is an oxidizing agent.
<h3>What is an oxidizing agent?</h3>
An oxidizing agent is a substance that causes oxidation by accepting electrons.
Al is a reducing agent,
is an oxidizing agent.
Hence, option C is correct.
Learn more about the oxidizing agent here:
brainly.com/question/10547418
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Answer:
it would be 53
Explanation:
because 4.00 L at 158kPa and 28c would be 53 grams of oxygine