As the volume of the container increases, the gas pressure decreases. They are directly proportional. This is supported by Boyle's Law.
When the balanced reaction equation of methane combustion is:
CH4 + 2O2 →CO2 + 2H2O
so, we can see that each 1 mole of methane combusted will give 2 moles of water as a product.
so first, we need to get the moles of methane =
= mass of methane /molar mass of methane
= 52.6 g / 16.04 g/mol
= 3.28 moles
when 1 mol of methane produces→ 2 moles of water
∴ 3.28 moles methane produces → X moles of water
∴ moles of water = 3.28 * 2
= 6.56 moles
when each 1 mole of water has 6.02 x 10^23 (Avogadro's number ) individual molecules:
∴number of molecules of water = 6.56 * 6.02 x 10^23
= 3.9 x 10^24 molecules
you need to <u>lose one electron </u>
Mixtures and solutions... what?
If 1mol ------- is ----------- 6,02×10²³
so x ------- is ----------- 1,39×10²⁴
