Which of the following gas samples would contain the same amount of gas as 200 mL of helium, He(g), at 25° C and 1 atm?
1 answer:
Answer:
of neon
at
and
(the second choice) would contain an equal number of gas particles as
of
at
and
(assuming that all four gas samples behave like ideal gases.)
Explanation:
By Avogadro's Law, if the temperature and pressure of two ideal gases is the same, the number of gas particles in each gas would be proportional to the volume of that gas.
All four gas samples in this question share the same temperature and pressure. Hence, if all these gases are ideal gases, the number of gas particles in each sample would be proportional to the volume of that sample. Two of these samples would contain the same number of gas particles if and only if the volume of the two samples is equal to one another.
The second choice, of neon
at
and
, is the only choice where the volume of the sample is also
. Hence, that choice would be the only one with as many gas particles as
of
at
and
.
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Answer:
\text{0.30 cm}^{3} \times \left (\dfrac{10^{-2}\text{ m}}{\text{1 cm}}\right )^{3} = 3.0 \times 10^{-7} \text{ m}^{3}
Explanation:
0.030 cm³ × ? = x m³
You want to convert cubic centimetres to cubic metres, so you multiply the cubic centimetres by a conversion factor.
For example, you know that centi means "× 10⁻²", so
1 cm = 10⁻² m
If we divide each side by 1 cm, we get 1 = (10⁻² m/1 cm).
If we divide each side by 10⁻² m, we get (1 cm/10⁻² m) = 1.
So, we can use either (10⁻² m/1 cm) or (1 cm/10⁻² m) as a conversion factor, because each fraction equals one.
We choose the former because it has the desired units on top.
The "cm" is cubed, so we must cube the conversion factor.
The calculation becomes