Answer:
The answer to your question is m = 4.7 kg
Explanation:
Data
Ice Water
mass = ? mass = 711 g
T₁ = -13°C T₁ = 87°C
T₂ = 10°C T₂ = 10°C
Ch = 2090 J/kg°K Cw = 4180 J/kg°K
Process
1.- Convert temperature to kelvin
T₁ = 273 + (-13) = 260°K
T₁ water = 87 + 273 = 360 °K
T₂ = 10 + 273 = 283°K
2.- Write the equation of interchange of heat
- Heat lost = Heat absorbed
- mwCw(T₂ - T₁) = miCi(T₂ - T₁)
-Substitution
- 0.711(4180)(10 - 87) = m(2090)(10 - (-13))
- Simplification
228842.46 = 48070m
m = 228842.46/48070
-Result
m = 4.7 kg
<span>D) recycling ;)
Waste Management's Aerobic-Anaerobic Bioreactor* is designed to accelerate waste degradation by combining attributes of the aerobic and anaerobic bioreactors. The objective of the sequential aerobic-anaerobic treatment is to cause the rapid biodegradation of food and other easily degradable waste in the aerobic stage in order to reduce the production of organic acids in the anaerobic stage resulting in the earlier onset of methanogenesis. In this system the uppermost lift or layer of waste is aerated, while the lift immediately below it receives liquids. Landfill gas is extracted from each lift below the lift receiving liquids. Horizontal wells that are installed in each lift during landfill construction are used convey the air, liquids, and landfill gas. The principle advantage of the hybrid approach is that it combines the operational simplicity of the anaerobic process with the treatment efficiency of the aerobic process. Added benefits include an expanded potential for destruction of volatile organic compounds in the waste mass. (*US Patent 6,283,676 B1)</span>
Answer:- As per the question is asked, 35.0 moles of acetylene gives 70 moles of carbon dioxide but if we solve the problem using the limiting reactant which is oxygen then 67.2 moles of carbon dioxide will form.
Solution:- The balanced equation for the combustion of acetylene is:
From the balanced equation, two moles of acetylene gives four moles of carbon dioxide. Using dimensional analysis we could show the calculations for the formation of carbon dioxide by the combustion of 35.0 moles of acetylene.
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The next part is, how we choose 35.0 moles of acetylene and not 84.0 moles of oxygen.
From balanced equation, there is 2:5 mol ratio between acetylene and oxygen. Let's calculate the moles of oxygen required to react completely with 35.0 moles of acetylene.
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Calculations shows that 87.5 moles of oxygen are required to react completely with 35.0 moles of acetylene. Since only 84.0 moles of oxygen are available, the limiting reactant is oxygen, so 35.0 moles of acetylene will not react completely as it is excess reactant.
So, the theoretical yield should be calculated using 84.0 moles of oxygen as:
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The answer to your question is the first one!
