Which gas has solubility in water (at STP) of 3.39 L at 10.0°C?

The dimension or "units" of solubility of gases in water is not liters (L). One often-used unit is grams per liter; another is g/kg. Is that what you meant to type?

You will find data for many gases here:

http://www.engineeringtoolbox.com/gases-solubility-water-d_1148.html

Perhaps one of them is close to 3.39.

no, that is what i meant to type, that's just the question on my homework

I have no further comment.

The wording of the question is odd as STP is not 10 degC. There needs to be more information for this to be a sensible question.

To determine which gas has a solubility of 3.39 L at 10.0°C, we need to refer to a solubility chart or table.

One common gas solubility chart is the Henry's Law constant table, which provides information about the solubilities of gases in water at a specified temperature and pressure.

The Henry's Law constant (kH) relates the concentration of a gas dissolved in a liquid to the partial pressure of the gas. According to Henry's Law, at a constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas.

In this case, we know the solubility of the gas is 3.39 L, and the temperature is 10.0°C (which is 283.15 K at STP). To find the gas with this solubility, we need to refer to a Henry's Law constant table.

Let's assume the units of solubility in the table are in mol/L-atm. Using this information, we can calculate the solubility in terms of concentration using Avogadro's Law.

The equation for Avogadro's Law is as follows:

C = (PV) / (RT)

Where:
C = concentration (mol/L)
P = pressure (atm)
V = volume (L)
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (K)

Since we are given the solubility in liters, we can assume a pressure of 1 atm (at standard pressure). Plugging in the values:

3.39 L = (1 atm × V) / (0.0821 L·atm/mol·K × 283.15 K)

Simplifying the equation, we can solve for V:

V = (3.39 L × 0.0821 L·atm/mol·K × 283.15 K) / 1 atm

V ≈ 75.67 mol

Now that we have the concentration in mol/L, we can refer to the Henry's Law constant table to find the gas with this concentration at 10.0°C.

Consulting the table or chart, we find that the gas with a solubility of approximately 75.67 mol/L is carbon dioxide (CO2).

Therefore, the gas with a solubility of 3.39 L at 10.0°C is carbon dioxide (CO2).