Hydrogen is composed of H atom and oxygen is composed of O atom. For water, it is composed by both H and O atom. If you burn hydrogen in oxygen, you can get water. And if you electrolysis water, you can get hydrogen and oxygen.
this equation is balanced.
Mass of methane takne = 1.5g
moles of methane used = masss / molar mass = 1.5 / 16 = 0.094 moles
mass of water = 1000 g
Initial temperature of water = 25 C
final temperature = 37 C
specific heat of water = 4.184 J /g C
1) Heat absorbed by water = q =m• C• ΔT = 1000 X 4.184 x (37-25) = 50208 Joules
2) Heat absorbed by calorimeter = Heat capacity X ΔT = 695 X (37-25) = 8340 J
3) Total heat of combustion = heat absorbed by water + calorimeter = 50208 + 8340 = 58548 Joules
This heat is released by 0.094 moles of methane
So heat released by one mole of methane =
- 622851.06 Joules = 622.85 kJ / mole
4) standard enthalpy of combustion = -882 kJ / mole
Error = (882-622.85) X 100 / 882 = 24.84 %
Answer:
(b) BeF2 > OF2 > CH3OH
Explanation:
The degree and type of intermolecular forces present in a substance influences its vapour pressure considerably. The greater the magnitude and strength of intermolecular forces in the substance, the lower the vapour pressure of the substance.
BeF2 molecules are held together by weak vanderwaals forces hence BeF2 will exhibit the least degree of intermolecular interaction and have the highest vapour pressure. OF2 molecules are bound together by dipole interactions hence it will exhibit a lower vapour pressure compared to BeF2. CH3OH molecules form hydrogen bonds with water molecules hence it will exhibit the least vapour pressure among the trio.
atom x is sodium and then the atom y is chlorine