A spherical balloon can hold 1.2 liters of helium. The balloon has a mass of 26 milligrams when empty. Helium has a density of 0.1785 grams per liter. What is the mass of the balloon when it is filled with helium?

Now of course you wanted the number of atoms remaining in the balloon. But the use of a $"mole"$ specifies that number. One mole of stuff specifies $6.022xx10^23$ individual items...

Given $n="Mass"/"Molar mass"=(PV)/(RT)$. $"Mass"=(PV)/(RT)xx"molar mass"$, .........Agreed........? And, as is typical, we use the value of.......... $R=0.0821*L*atm*K^-1*mol^-1$ The which has units that chemists typically use. And so we plugs in the...

The molar volume of an ideal gas at $"STP"$ is $22.4$ $L$. We have a volume of $3.8*L$ helium gas under these conditions. Assuming, quite reasonably, that helium behaves ideally,...

$"Charles' Law"$ states that a given quantity of gas has a volume directly proportional to the absolute temperature: $VpropT$, else $V=kT$. Solving for $k$, $V_1/T_1=V_2/T_2$, and $T_2=V_2/V_1xxT_1$ $T_2=(4.5xx10^2*L)/(1.89xx10^2*L)xx294*K$ $=$ $??K$

$T$ is specified to be $"Absolute Temperature"$: $V_2=V_1/T_1xxT_2$ $= (5.00*L)/(298*K)xx273*K~=4.5*L$

Since, for a given quantity of gas, $VpropT$ if $P$ is constant, $V=kT$. Solving for $k$, $V_1/T_1=V_2/T_2$, and $V_2=(V_1xxT_2)/T_1$; $T$ is reported in $"degrees Kelvin"$, $V$ may be in whatever...

At first glance, this appears to be an problem. However, water is a liquid at 24 °C and 2.27 atm. For argument's sake, let's say that the temperature on...

The number of solute particles may be determined by calculation of the molar quantities of each reagent. $NaCl$ has a molar mass of $58.44$ $g*mol^-1$; $KCl$ has a molar mass...

Multiply the soluble mass per liter by the number of liters. $36("g")/(cancel"L")xx5cancel"L"="180 g"$

You are starting with 80.0 mg of $"F-21"$. After 1 half-life, you will have 40.0 mg of $"F-21"$. After 2 half-lives, you will have 20.0 mg of $"F-21"$. Hence, $"2...