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Klio2033 [76]
3 years ago
7

What element has the noble gas electron configuration [Kr]5524d105p?

Chemistry
1 answer:
Digiron [165]3 years ago
3 0

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!).

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When you dissolve salt in hot water, the salt is referred to as the
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3 years ago
2C(s)+O2(g)2H2(g)+O2(g)H2O(l)→2CO(g)→2H2O(g)→H2O(g)ΔHΔHΔH=−222kJ=−484kJ=+44kJ Use the thermochemical data above to calculate the
sveticcg [70]

Answer : The change in enthalpy for the reaction is, 175 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The main reaction is:

H_2O(l)+C(s)\rightarrow CO(g)+H_2(g)    \Delta H=?

The intermediate balanced chemical reaction will be,

(1) 2C(s)+O_2(g)\rightarrow 2CO(g)     \Delta H_1=-222kJ

(2) 2H_2(g)+O_2(g)\rightarrow 2H_2O(g)    \Delta H_2=-484kJ

(3) H_2O(l)\rightarrow H_2O(g)    \Delta H_3=44kJ

Now we are dividing reaction 1 by 2, dividing reverse reaction 2 by 2 and then adding all the equations, we get :

(1) C(s)+\frac{1}{2}O_2(g)\rightarrow CO(g)     \Delta H_1=\frac{-222kJ}{2}=-111kJ

(2) H_2O(g)\rightarrow H_2(g)+\frac{1}{2}O_2(g)    \Delta H_2=\frac{484kJ}{2}=242kJ

(3) H_2O(l)\rightarrow H_2O(g)    \Delta H_3=44kJ

The expression for change in enthalpy of the given reaction is:

\Delta H=\Delta H_1+\Delta H_2+\Delta H_3

\Delta H=(-111)+(242)+(44)

\Delta H=175kJ

Therefore, the change in enthalpy for the reaction is, 175 kJ

6 0
4 years ago
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