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The addition of 9.54 kJ of heat is required to raise the temperature of 225.0 g of a liquid hydrocarbon from 20.5oC to 45.0oC. What is the heat capacity of this hydrocarbon
q= mc∆t9540 = 225.0 × s × (45 - 20.5)c = 9540/ 5512.5c = 1.73 J /gC
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what product of the combustion of a hydrocarbon occurs only when the supply of oxygen is limited? a. sulfur dioxide b. nitrogen dioxide c. carbon dioxide d. carbon monoxide
c. carbon dioxide
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Combustion of hydrocarbon in a limited supply of air produces CO in addition to CO2 and h2o when 1.31 g of particular hydrocarbon was burned in air ,1.26 g CO ,1.98 g CO2
Molar mass of CO = 28.011.26 = 0.045mMolar mass of CO2= 44.011.98/44.01 = 0.045mMolar mass of hydrocarbon = 28.06 mol1.31/28.06 = 0.047m
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How many elements have discovered in nature? not only on earth crust but the whole nature.
As of November 2011, 118 elements have been identified, the latest being ununseptium in 2010. Of the 118 known elements, only the first 98 are known to occur naturally on...
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Explain the meaning of the following words :
volatility
reducing nature
acidic nature
thermal stability
Volatility is a property of substance to vaporize. Since high volatile compounds have weak intermolecular interaction, they exhibit high vapor pressure and low boiling point. Reducing nature is a characteristic...
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Complete combustion of a 0.70-mol sample of a hydrocarbon, CxHy, gives 4.20 mol of CO2 and 3.50 mol of H20. The molecular formula of the original hydrocarbon is
0.7 CxHy + xO2 =4.2CO2 + 3.5H2O /0.7 CxHy + x/0.7O2 =6CO2 + 5H2O C6H10
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128 grams of oxygen reacted. After the combustion of an unknown hydrocarbon, the total pressure of a 18.91-L container is 7.3347 atm at 22.7 degrees celsius. What is the hydrocarbon?
PV=nRT n1=PV/RT n=7.3347*101.3*18.91/8.31*295.7=5.718 moles n2=128/32=4 CO2 is gas at this condition CxHy + (x+y/4)O2=xCO2 +y/2H2O as we see,coefficient near O2 (x+y/4) must be biger that near CO2 x.Coefficients are proportional...
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A compound containing only carbon and hydrogen is found to contain 4.1315 g of carbon and 0.8685 g of hydrogen. Determine the empirical formula of hydrocarbon
CxHy w(C) = 4.1315 g / (4.1315 g + 0.8685 g) = 0.8263 w(H) = 0.8685 g / (4.1315 g + 0.8685 g) = 0.1737 x : y = 0.8263/12,0107...
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Complete combustion of 8.10 g of a hydrocarbon produced 26.0 g of CO2 and 8.88 g of H2O. What is the empirical formula for the hydrocarbon? Insert subscripts as necessary.
CxHy + O2= xCO2 + y/2H2O8.1. 26. 8.8812 x 26/44 = 7.2 g (C)2 x 8.88/18= 0.9 g (H)7.2+0.9= 8.17.2/12 : 0.9:1= 0.6:0.9= 1:1.5 = 4:6C4H6.
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A certain hydrocarbon on complete combustion at s.t.p produced 89.6dm3 of CO2 and 54g of water. The hydrocarbon should be
A. C6H6
B. C4H10
C. C5H10
D. C4H6
Please show working
CxHy + O2 = xCO2 + y/2 H2OIf we burn 1 mol of certain hydrocarbon, we would get x mol of CO2 and y/2 mol of H2O.During combustion, we received...
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