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The volume occupied by 5g NEON is ______ than the volume occupied by 2g HYDROGEN.

By the way, I want someone who actually knows this, not someone who just assumes that as 5g is a bigger amount it must be that one, k?

2007-11-01 08:43:58 · 5 answers · asked by Anonymous in Science & Mathematics Chemistry

5 answers

Moles Ne = 5 g / 20.183 g/mol = 0.248
Moles H2 = 2 g / 2.02 g/mol = 0.990
If p and T are the same
V of Ne is less than the volume occupied by 2 g of H2

V = nRT /p
R , T and p are constants
V = nK
Since the number of moles of Ne is less than the number of moles of He the volume of Ne is less than the volume of H2

2007-11-01 08:51:03 · answer #1 · answered by Dr.A 7 · 0 0

You need to find out how many moles of each molecule there is.

so, for Neon, 5g/atomic weight and for Hydrogen 2g/molecular weight (which for hydrogen is 2, because Hydrogen is present as H2).

Offhand, I'm not certain of the atomic weight of Neon, you can look that up.


The volume of gases is found by V=nRT/P, thus it is determined by n (number of moles), R (constant), T (temperature) and P (pressure). Since all of those values are the same for each gas, the volume will be proportional to the number of moles of each.

So look up the atomic masses of each element and figure it out from there.

2007-11-01 08:57:33 · answer #2 · answered by Anonymous · 0 0

volume Ne= 5 /20 X 22.4 L
= 5.6 L
volume H2= 2 /2 X 22.4
= 22.4 L
so the volume of Neon Ne is smaller than
of hydrogen H2

2007-11-01 08:57:48 · answer #3 · answered by sami_dodeen 3 · 0 0

Atomic weights: Ne=20 H=1 H2=2

5gNe x 1molNe/20gNe x 22.4LNe/1molNe = 5.6L Ne

2gH2 x 1molH2/2gH2 x 22.4LH2/1molH2 = 22.4L H2

So the answer is "one fourth."

2007-11-01 08:56:45 · answer #4 · answered by steve_geo1 7 · 0 0

density of neon = 0.839kg/m3
given mass=5 g density=mass/vol.
vol.=den./mass =0.839/5=0.1678m3....

density of H =0.08987kg/m3
given mass=2 g
vol.=den./mass =0.08987/2=0.044935m3


got it now ? say now which is lighter??

2007-11-01 08:57:20 · answer #5 · answered by Anonymous · 0 0

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