Answer:
0.162 moles of CO₂ are produced by this reaction
Explanation:
The reaction is:
C₃H₈(g) + 5O₂(g) → 3CO₂(g) +4H₂O(g)
As we have the volume of propane, we need to know the mass that has reacted, so we apply density's concept.
Density = Mass / Volume → Density . Volume = Mass
0.00183 g/mL . 1300 mL = Mass → 2.379 g
We determine the moles → 2.379 g . 1mol / 44 g = 0.054 moles
Ratio is 1:3. 1 mol of propane can produce 3 moles of dioxide
Then, 0.054 moles may produce (0.054 .3)/1 = 0.162 moles
Answer:
Explanation:
For the reaction
C2H5OH (l) + 3 O2(g) = 2CO2(g) + 3 H2O
We can calculate the standard molar enthalpy of combustion using the standard enthalpies of formation of the species involved in the reaction according to Hess law:
ΔHºc = 2ΔHºf CO2 (g) + 3ΔHºfH2O(l) - ( ΔHºf C2H5OH (l) - 3ΔHºfO2 (g) )
( we were not give the water state but we know we are at standard conditions so it is in its liquid state )
The ΔHºfs can be found in appropiate reference or texts.
ΔHºc = 2ΔHºf CO2 (g)+ 3ΔHºfH2O(l) - ( ΔHºf C2H5OH (l) -+3ΔHºfO2 (g) )
= [ 2 ( -393.52 ) + 3 ( -285.83 ) ] - [( -276.2 + 0 ) ] kJ
ΔHºc = -1368.33 kJ
Hydrogen mass is 1, the atom with mass number 7 is Lithium
Its <span>c.chromatography is the process of separating solutions on the basis of their boiling points </span>
