Density/Earth’s gravitational pull.
Answer:
the compound contains C, H, and some other element of unknownidentity, so we can’t calculate the empirical formula
Explanation:
Mass of CO2 obtained = 3.14 g
Hence number of moles of CO2 = 3.14g/44.0 g = 0.0714 mol
The mass of the carbon in the sample = 0.0714 mol × 12.0g/mol = 0.857 g
Mass of H2O obtained = 1.29 g
Hence number of moles of H2O = 1.29g/18.0 g = 0.0717 mol
The mass of the carbon in the sample = 0.0717 mol × 1g/mol = 0.0717 g
% by mass of carbon = 0.857/1 ×100 = 85.7 %
% by mass of hydrogen = 0.0717/1 × 100 = 7.17%
Mass of carbon and hydrogen = 85.7 + 7.17 = 92.87 %
Hence, there must be an unidentified element that accounts for (100 - 92.87) = 7.13% of the compound.
Answer:
1 gramo de metano aporta 50.125 kilojoules.
1 gramo de metano aporta 48.246 kilojoules.
Explanation:
La cantidad de energía liberada por la combustión de una unidad de masa del hidrocarburo (
), en kilojoules por mol, es igual a la cantidad de energía liberada por mol de compuesto (
), en kilojoules por mol, dividido por su masa molar (
), en gramos por mol:
(1)
A continuación, analizamos cada caso:
Metano


1 gramo de metano aporta 50.125 kilojoules.
Octano


1 gramo de metano aporta 48.246 kilojoules.
The elements present in Ammonium Nitrate are Hydrogen, Nitrogen, and Oxygen at a ratio of 4:2:3, respectively. Hydrogen weighs in at 1.008 amu, Nitrogen at 14.007, and Oxygen at 15.999. This means that the molar mass would be:
Hydrogen
4 x 1.008 = 4.032 amu
Nitrogen
2 x 14.007 = 28.014 amu
Oxygen
3 x 15.999 = 47.997 amu
Total
4.032 + 28.014 + 47.997 = 80.043 amu
The molar mass of Ammonium Nitrate is 80.043 grams per mole.