What principle chemistry

If a ballon at 3.4 atm at 301k, what is the pressure when cooled to 212k

stepladder [879]

Answer:

2.39 atm

Explanation:

- Use Gay-Lussac's law

- P2 = P1T2/T1

- Fill in with our values

- Hope this helped! Please let me know if you need further explanation.

8 0

2 years ago

The purpose of this guide is to help you write a clear and concise report that summarizes the lab I just completed. This lab sho

Masja [62]

Answer:

The purpose of this guide is to help you write a clear and concise report that summarizes the lab I just completed. This lab showed me how water carries rocks to a new location. For instance how the first model showed how a stream would act with light drizzle to downpour as I saw the outcomes were all if not completely then partially different

I read the information that was given to me before I started each lab lesson and went on from there.

Explanation:

:)

3 0

1 year ago

Read 2 more answers

Find the quantinum numbers n,m,l,s for the last of potassium layer pleasee help explain correctly all

Fantom [35]

Answer:

Quantum numbers of the outermost electron in potassium:

$n = 4$ .
$l = 1$ .
$m_l = 0$ .
Either $m_s = 1/2$ .

Explanation:

Refer to the electron configuration of a potassium atom. The outermost electron in a ground-state potassium atom is in the $4s$ orbital (fourth $s$ orbital.)

The quantum number $n$ (the principal quantum number) specifies the main energy shell of an electron. This electron is in the fourth main energy shell (as seen in the number four in the orbital.) Hence, $n = 4$ for this electron.

The quantum number $l$ (the angular momentum quantum number) specifies the shape ( $s$ , $p$ , $d$ , etc.) of an electron. $l = 1$ for $s\!$ orbitals (such as the one that contains this electron.

Quantum numbers $n$ and $l$ specify the shape of an orbital. On the other hand, the magnetic quantum number $m_l$ specifies the orientation of these orbitals in space.

However, $s$ orbitals are spherical. Regardless of the value of $n$ , the only possible $m_l$ value for electrons in $s\!$ orbitals is $m_l = 0$ .

The spin quantum number $m_s$ distinguishes between the two electrons in an orbital. The two possible values of $m_s \!$ are $(+1/2)$ and $(-1/2)$ . Typically, the first electron in an orbital is assigned an upward ( $\uparrow$ ) spin, which corresponds to $m_s = (+1/2)$ .