Answer:
Explanation:
mass % of C = 0.27/0.45*100 = 60%
mass % of H = 0.02/0.45*100 = 4.4%
mass % of O = 0.16/0.45*100 = 35.6%
Total = 60%+4.4%+ 35.6% = 100%
The answer is electrical energy
The balanced equation for the combustion of octane is as follows
2C₈H₁₈ + 25O₂ ---> 16CO₂ + 18H₂O
stoichiometry of C₈H₁₈ to O₂ is 2:25
number of octane moles reacted - 17.0 g / 114.2 g/mol = 0.149 mol
according to molar ratio
if 2 mol of octane reacts with 25 mol of O₂
then 0.149 mol of octane reacts with - 25 /2 x 0.149 mol = 1.86 mol of O₂
mass of O₂ - 1.86 mol x 32 g/mol = 59.5 g
59.5 g of O₂ is required to react with
(1) CH4 is a nonpolar molecule, because the differences in electronegativity is so small, and also its molecular structure is causes for its dipole moments to cancel out. The rest have large differences in electronegativity and have structures that do not cancel out the dipole moments.
The reaction H2SO4 + 2 NaOH -> Na2SO4 + 2H2O applies to an acid-base titration.
moles NaOH = c · V = 0.2423 mmol/mL · 32.23 mL = 7.809329 mmol
moles H2SO4 = 7.809329 mmol NaOH · 1 mmol H2SO4 / 2 mmol NaOH = 3.9046645 mmol
Hence
[H2SO4]= n/V = 3.9046645 mmol / 37.21 mL = 0.1049 M
The answer to this question is [H2SO4] = 0.1049 M
