A13.0 l sample of helium gas has a pressure of 29.0 atm. what volume would this gas occupy at 3.50 atm? assume ideal behavior.

From the equation;
P1V1=P2V2
V2=P1V1÷P2
V2=29.0×13.0÷3.50
V2=107.71L
therefore V2=107.7L

This uses an equation known as Boyle's law. Boyle's law states that at a constant temperature, the product of the pressure that a gas exerts and the volume it occupies is constant, meaning they are inversely proportional. This means the equation is:
P1V1 = P2V2
We have 3 knowns:
P1 = 29.0 atm
V1 = 13.0 L
P2 = 3.50 atm
V2 = unknown
Plugging into our equation:
(29.0 atm)*(13.0 L) = (3.50 atm)*(V2)
V2 = (29.0 atm * 13.0 L)/(3.50 atm)
V2 = 108 L = Volume occupied at 3.50 atm
This makes sense, as when the pressure decreased, Boyle's Law predicts that the volume will increase to compensate for the lost pressure, and it did.

There is one s orbital, and there are three p orbitals, five d orbitals, and seven f orbitals. thus, the answer is b. do not confuse the number of orbitals in a subshell with the number of electrons the subshell can hold. each orbital can hold two electrons, so the capacity of an nf subshell is 14 electrons.

16- counting by fours