Question

Question #2e2b3

Answer 1

The number of moles is `~~1.10` and the mass is `~~4.36` `g`

That's a Chemistry question and you need to use the ideal gas law to find the number of moles `n`.

`pV = nRT`, where `p` is the pression in Pascal, `V` the volume in `m^3`, `R` is the ideal gas constant (`=8.134[J/(K*mol)]`), `T` the temperature in Kelvin.

Here, we have a ballon with a radius of `r = 18 cm = 0.18m`.
The volume is given by `V = 4/3 pi r^3~~0.0078pi` `m^3`

The pression `p = 1.05` `atm`. We know that `1` `atm = 101325` `Pa`.

Therefore, by the Rule of Three :

`p = (1.05*101325)/1 = 106391.25` `Pa`

For the temperature, `0°` is `273.15` `K`, so `20°` is `293.15` `K`.

`n = (pV)/(RT) = (106391.25*0.0078pi)/(8.134*293.15) ~~ 1.10` `mol`.

For the mass, `m` `[g] = n * M`, where M is the molar mass.

The molar mass of `He` is `4 [g/(mol)]`.

Therefore, `m ~~ 1.10*4~~4.36` `g`.

Answer 2

The number of moles of helium will be equal to 1.1 and the mass of helium will be 4.4 g.

SIDE NOTE Small correction - helium cannot exist as a diatomic molecule, so the correct notation is `"He"`, not `"He"_2`.

There's nothing wrong with the method used in the other answer, but I want to show you how you can get the number of moles of helium without doing that many conversions.

You that your balloon has a radius of 18 cm. You can determine its volume in `"cm"^3` by

`V_"balloon" = 4/3 * pi * "radius"^3`

`V_"balloon" = 4/3 * pi * ("18 cm")^3 = "24429 cm"^3`

Use the fact that `"1 L" = "1 dm"^3` to express the volume in liters

`24429cancel("cm"^3) * ("1 L")/(10^3cancel("cm"^3)) = "24.429 L"`

Now use the ideal gas law equation to solve for `n` by using `R` expressed in L atm/mol K

`PV = nRT => n = (PV)/(RT)`

`n = (1.05cancel("atm") * 24.429cancel("L"))/(0.082(cancel("atm") * cancel("L"))/("mol" * cancel("K")) * (273.15 + 20)cancel("K")) = color(green)("1.1 moles")`

To get the mass of helium, use its molar mass, which expresses the mass of 1 mole of helium

`1.1cancel("moles He") * "4 g"/(1cancel("mole He")) = color(green)("4.4 g")`