Answer:
What is a noble gas electron configuration?
A noble gas electron configuration is a configuration that completes the Octet Rule of achieving 8 valence electrons.
Atoms always behave in ways to achieve stability and as you probably know, Noble Gases are the most stable. Their configuration, with a full valence electron shell (8 electrons, when you add both the S & P sublevels together, this is why it’s called the Octet), is therefore desirable. This means metals on the far left of the table will lose electrons to achieve this noble gas configuration and nonmetals on the right will gain electrons (generally speaking).
For example; take Argon. Its electron configuration is 1s^2 2s^2 2p^6 3s^2 3p^6, meaning it has 8 valence electrons. Then, take Chlorine. It has the electron configuration 1s^2 2s^2 2p^6 3s^2 3p^5, meaning it has 7 valence electrons so it’s a very unhappy camper. It typically gains an electron to achieve the 8 valence electrons Argon has (even using the same configuration of 1s^2 2s^2 2p^6 3s^2 3p^6) because it’s Mega jealous of Argon’ s stability. Side note: this is why chlorine typically has a -1 charge!
In summary, an atom achieving a “Noble Gas configuration” is the same as saying an atom fulfilling the Octet Rule. Both mean that there are 8 valence electrons (electrons in shell furthest from nucleus). This is a stable form many atoms seek to achieve (of course, what’s a good rule in chemistry if there aren’t exceptions!).
Answer:
Nitrogen and Argon
Explanation:
These gases are part of a class of "noble" gases because they don't react with anything.
Answer:
It is considered an ionic compound since it is composed of the magnesium cation (Mg2+) and two hydroxide anions (OH-). There are no covalently shared electrons here. ... The reaction of the hydroxide anions with acid causes more of the magnesium hydroxide to dissolve until all of the acid has been neutralized.
Protons are positively charged and are found inside the nucleus of an atom
<u>Answer:</u> The vapor pressure of mercury at 322°C is 0.521 atm
<u>Explanation:</u>
To calculate the final pressure, we use the Clausius-Clayperon equation, which is:
where,
= initial pressure which is the pressure at normal boiling point = 1 atm
= final pressure which is vapor pressure of mercury = ?
= Enthalpy of vaporization = 58.51 kJ/mol = 58510 J/mol (Conversion factor: 1 kJ = 1000 J)
R = Gas constant = 8.314 J/mol K
= initial temperature which is normal boiling point =
= final temperature =
Putting values in above equation, we get:
Hence, the vapor pressure of mercury at 322°C is 0.521 atm