A sample containing 7.60 gas has a volume of 31.0 . Pressure and temperature remain constant.What is the volume after 3.00 is added to the gas already in the container
are 7.60 and 3.00 in moles? if it is,
recall the formula relating moles (n) and volume (V), at a constant P and T:
n1 / V1 = n2 / V2
plug in the given values to the equation:
7.60 mol/ 31.0 = (7.60 + 3) mol / V2
then solve for V2.
hope this helps. :)
With no units this is not a problem but just gibberish.
To find the new volume after adding additional gas to the container, we can use the combined gas law, which states that the ratio of the initial and final volumes of a gas is equal to the ratio of its initial and final quantities. The formula can be written as:
(V1 / V2) = (n1 / n2)
Where:
V1 is the initial volume of the gas in the container
V2 is the final volume after adding the additional gas
n1 is the initial quantity (number of moles) of the gas
n2 is the final quantity after adding the additional gas
Given:
V1 = 31.0 L
n1 = 7.60 mol
Let's calculate the final quantity (n2) by adding the amount of gas added:
n2 = n1 + 3.00 mol
= 7.60 mol + 3.00 mol
= 10.60 mol
Now we can substitute the values of V1, V2, n1, and n2 into the combined gas law equation to find V2:
(31.0 L) / V2 = (7.60 mol) / (10.60 mol)
To find V2, we can cross-multiply and solve for V2:
V2 = (31.0 L) x (10.60 mol) / (7.60 mol)
= 43.19 L
Therefore, the volume after adding 3.00 mol of gas to the initial sample is approximately 43.19 L.