Suppose a piston automatically adjusts to maintain a gas at a constant pressure of 13.00 atm . For the initial conditions, consider 0.05 mol of helium at a temperature of 220.00 K . This gas occupies a volume of 0.07 L under those conditions. What volume will the gas occupy if the number of moles is increased to 0.08 mol (n2) from the initial conditions

The final volume of the gas is 0.112 L

Explanation:

Step 1: Data given

Pressure is constant at 13.00 atm

The initial number of moles = 0.05 moles

The  temperature = 220.00 K

Initial volume = 0.07 L

The number of moles is increased to 0.08 moles

Step 2: Calculate the new volume

V1/ n1 = V2/n2

⇒with V1 = the initial volume of the gas = 0.07 L

⇒with n1 = the initial number of moles of gas = 0.05 moles

⇒with V2 = the final volume = TO BE DETERMINED

⇒with n2 = the final number of moles of gas = 0.08 moles

0.07 L / 0.05 moles = V2 / 0.08 moles

V2 = 0.07 L * 0.08 / 0.05 moles

V2 = 0.112 L

The final volume of the gas is 0.112 L

2h2o2--> 2h2o + o2

this reaction is of the spontaneous decomposition of hydrogen peroxide down into water and oxygen.

add 2 molecules of hydrogen peroxide and 2 molecules of water. since oxygen is naturally diatomic, the total number of atoms of each element is now the same on both sides of the equation so it is balanced.

this may not be entirely correct but let me know.