Two bulbs are connected by a stopcock. The large bulb, with a volume of 6.00 L, contains nitric oxide at a pressure of 0.750 atm, and the small bulb, with a volume of 1.50 L, contains oxygen at a pressure of 2.50 atm. The temperature at the beginning and the end of the experiment is 22 ∘ C . After the stopcock is opened, the gases mix and react. 2 NO ( g ) + O 2 ( g ) → 2 NO 2 ( g )

Determine which gases remain after the reaction goes to completion and calculate their partial pressures.
The temperature at the beginning and the end of the experiment is 22 degree celsius.

Which gases are present at the end of the experiment?

What are the partial pressures of the gases?

If the gas was consumed completely, put 0 for the answer.

Aballoon inflated with three breaths of air has a volume of 1.7 l. at the same temperature and pressure, what is the volume of the balloon if five more same-sized breaths are added to the balloon? a balloon inflated with three breaths of air has a volume of 1.7 l. at the same temperature and pressure, what is the volume of the balloon if five more same-sized breaths are added to the balloon?

Convert a speed of 201 cm/s to units of inches per minute. also, show the unit analysis by dragging components into the unit‑factor slots.

Challenge question: this question is worth 6 points. as you saw in problem 9 we can have species bound to a central metal ion. these species are called ligands. in the past we have assumed all the d orbitals in some species are degenerate; however, they often are not. sometimes the ligands bound to a central metal cation can split the d orbitals. that is, some of the d orbitals will be at a lower energy state than others. ligands that have the ability to cause this splitting are called strong field ligands, cnâ’ is an example of these. if this splitting in the d orbitals is great enough electrons will fill low lying orbitals, pairing with other electrons in a given orbital, before filling higher energy orbitals. in question 7 we had fe2+, furthermore we found that there were a certain number (non-zero) of unpaired electrons. consider now fe(cn)6 4â’: here we also have fe2+, but in this case all the electrons are paired, yielding a diamagnetic species. how can you explain this?