Answer:
32.8 C
Explanation:
- Use combined gas law formula and rearrange.
- Hope that helped! Please let me know if you need further explanation.
1) 1.8 micrograms(least)
2) 1.8 grams
3) 1.8 kilograms(greatest)
Answer:
2.4
⋅
10
24
Explanation:
All that you need to know here is that in order for a given sample of water to contain exactly 1 mole of water, it must contain 6.022
⋅
10
23 molecules of water.
This is known as Avogadro's constant and essentially acts as the definition of a mole. If you have 6.022
⋅
10
23 molecules of water, then you can say for a fact that you have 1 mole of water.
Answer : The excess reactant in the combustion of methane in opem atmosphere is
molecule.
Solution : Given,
Mass of methane = 23 g
Molar mass of methane = 16.04 g/mole
The Net balanced chemical reaction for combustion of methane is,

First we have to calculate the moles of methane.
=
= 1.434 moles
From the above chemical reaction, we conclude that
1 mole of methane react with the 2 moles of oxygen
and 1.434 moles of methane react to give
moles of oxygen
The Moles of oxygen = 2.868 moles
Now we conclude that the moles of oxygen are more than the moles of methane.
Therefore, the excess reactant in the combustion of methane in open atmosphere is
molecule.