A sample of gas has a volume of 6.20 L at 23 degrees celsius at a pressure of 1.90 atm. What is its volume at STP?

To find the volume of a gas at STP (Standard Temperature and Pressure), you need to know the initial volume of the gas and the conditions it is under.

STP is defined as a temperature of 0 degrees Celsius (273.15 Kelvin) and a pressure of 1 atmosphere (atm). In this case, you are given the initial volume of the gas and its conditions at 23 degrees Celsius and 1.90 atmospheres.

To find the final volume, you will use the combined gas law equation, which relates the initial and final volumes of a gas at different temperatures and pressures. The combined gas law equation is:

P₁V₁ / T₁ = P₂V₂ / T₂

Where:
P₁ and P₂ are the initial and final pressures of the gas (in atm)
V₁ and V₂ are the initial and final volumes of the gas (in L)
T₁ and T₂ are the initial and final temperatures of the gas (in Kelvin)

Now, let's substitute the values given in the question into the equation:

P₁ = 1.90 atm
V₁ = 6.20 L
T₁ = 23 degrees Celsius + 273.15 = 296.15 K
P₂ = 1 atm (STP conditions)
V₂ = ? (what we're trying to find)
T₂ = 0 degrees Celsius + 273.15 = 273.15 K (STP conditions)

Plugging these values into the combined gas law equation, we get:

(1.90 atm)(6.20 L) / (296.15 K) = (1 atm)(V₂) / (273.15 K)

Now, we rearrange the equation to solve for V₂:

V₂ = (1 atm)(6.20 L)(273.15 K) / (1.90 atm)(296.15 K)

V₂ = (6.20 L)(273.15 K) / (1.90)(296.15 K)

V₂ = 1707.33 L / 561.74

V₂ ≈ 3.04 L (rounded to two decimal places)

Therefore, the volume of the gas at STP is approximately 3.04 liters.