2(3a-3)-2(a-2)=3<br>thank you

Without consulting Figure 28 or Table 22, determine whether each of the following electron configurations is an inert gas,a halo

Anna [14]

This is an incomplete question, here is a complete question.

Determine whether each of the following electron configurations is an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a transition metal. Justify your choices.

(a) $1s^22s^22p^63s^23p^5$

(b) $1s^22s^22p^63s^23p^63d^74s^2$

(c) $1s^22s^22p^63s^23p^63d^{10}4s^24p^6$

(d) $1s^22s^22p^63s^23p^64s^1$

(e) $1s^22s^22p^63s^23p^63d^24s^2$

(f) $1s^22s^22p^63s^2$

Answer :

(a) $1s^22s^22p^63s^23p^5$ → Halogen

(b) $1s^22s^22p^63s^23p^63d^74s^2$ → Transition metal

(c) $1s^22s^22p^63s^23p^63d^{10}4s^24p^6$ → Transition metal

(d) $1s^22s^22p^63s^23p^64s^1$ → Halogen metal

(e) $1s^22s^22p^63s^23p^63d^24s^2$ → Transition metal

(f) $1s^22s^22p^63s^2$ → Alkaline earth metal

Explanation :

Inert gas : These are the gases which lie in group 18.

Their general electronic configuration is: $ns^2np^6$ where n is the outermost shell.

Halogen : These are the elements which lie in group 17.

Their general electronic configuration is: $ns^2np^5$ where n is the outermost shell.

An alkali metal : These are the elements which lie in group 1.

Their general electronic configuration is: $ns^1$ where n is the outermost shell.

An alkaline earth metal : These are the elements which lie in group 2.

Their general electronic configuration is: $ns^2$ where n is the outermost shell.

Transition elements : They are the elements which lie between 's' and 'p' block elements. These are the elements which lie in group 3 to 12. The valence electrons of these elements enter d-orbital.

Their general electronic configuration is: where n is the outermost shell.

(a) $1s^22s^22p^63s^23p^5$

The element having this electronic configuration belongs to the halogen family.

(b) $1s^22s^22p^63s^23p^63d^74s^2$

The element having this electronic configuration belongs to the transition family.

(c) $1s^22s^22p^63s^23p^63d^{10}4s^24p^6$

The element having this electronic configuration belongs to the transition family.

(d) $1s^22s^22p^63s^23p^64s^1$

The element having this electronic configuration belongs to the halogen family.

(e) $1s^22s^22p^63s^23p^63d^24s^2$

The element having this electronic configuration belongs to the transition family.

(f) $1s^22s^22p^63s^2$

The element having this electronic configuration belongs to the an alkaline earth metal.

7 0

3 years ago

Calculate the equilibrium concentration of c2o42− in a 0.20 m solution of oxalic acid.

spin [16.1K]

Answer:

[C2O2−4] =1.5⋅10−4⋅mol⋅dm−3

Explanation:

For the datasheet found at Chemistry Libretext,

Ka1=5.6⋅10−2 and Ka2=1.5⋅10−4 [1]

for the separation of the primary and second nucleon once oxalic corrosive C2H2O4 breaks up in water at 25oC (298⋅K).

Build the RICE table (in moles per l, mol⋅dm−3, or identically M) for the separation of the primary oxalic nucleon. offer the growth access H+(aq) fixation be x⋅mol⋅dm−3.

R C2H2O4(aq)⇌C2HO−4(aq)+H+(aq)

I 0.20

C −x +x

E 0.20−x x

By definition,

Ka1=[C2HO−4(aq)][H+(aq)][C2H2O4(aq)]=5.6⋅10−2

Improving the articulation can provides a quadratic condition regarding x, sinking for x provides [C2HO−4]=x=8.13⋅10−2⋅mol⋅dm−3

(dispose of the negative arrangement since fixations can faithfully be additional noteworthy or appreciate zero).

Presently build a second RICE table, for the separation of the second oxalic nucleon from the amphoteric C2HO−4(aq) particle. offer the modification access C2O2−4(aq) be +y⋅mol⋅dm−3. None of those species was out there within the underlying arrangement. (Kw is neglectable) therefore the underlying centralization of each C2HO−4 and H+ are going to be appreciate that at the harmony position of the first ionization response.

R C2HO−4(aq)⇌C2O2−4(aq)+H+(aq)

I 8.13⋅10−2 zero eight.13⋅10−2

C −y +y

E 8.13⋅10−2−y y eight.13⋅10−2+y

It is smart to just accept that

a. 8.13⋅10−2−y≈8.13⋅10−2,

b. 8.13⋅10−2+y≈8.13⋅10−2 , and

c. The separation of C2HO−4(aq)

Accordingly

Ka2=[C2O2−4(aq)][H+(aq)][C2HO−4(aq)]=1.5⋅10−4≈(8.13⋅10−2 )⋅y8.13⋅10−2

Consequently [C2O2−4(aq)]=y≈Ka2=1.5⋅10−4]⋅mol⋅dm−3

3 0

4 years ago

Not yet answered Points possible: 1.00 The equivalence point of any acid-base titration can be determined visually from a titrat

pshichka [43]

The volume of the base should be gradually increased, stopping once the equivalence point is achieved. At the volume halfway between the equivalence point and the acid, the acid's pKa is equal to the pH.

<h3>What is titration, for instance?</h3>

Titration, sometimes referred to as titrimetry, is a method for calculating the concentration of a specific analyte in a mixture that is used in chemical qualitative analysis. Titration, which is also sometimes referred to as volumetric analysis, is a crucial analytical chemistry method.

<h3>What governs titration in its basic form?</h3>

The following is the fundamental titration principle: The sample being studied is given a solution, referred to as a titrant or standard solution. A chemical is present in the titrant in a known concentration.

To know more about Titration visit:

brainly.com/question/29590776

#SPJ4

7 0

1 year ago

Where would you expect snow or rain to fall on a mountain

patriot [66]

You would expect snow to fail at the peak or the top because the weather is coldest there.

7 0

4 years ago

Which type of mixture is formed when one substance dissolves into another and fills the spaces between other kinds of molecules?

Rashid [163]

The type of mixture that is formed when one substance dissolves into another and fills the spaces between other kinds of molecules is a colloidal mixture. It is a type of mixture where a solute is dispersed evenly in a solvent. The particles' sizes range from 1 to 1000 nm.

8 0

4 years ago

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2(3a-3)-2(a-2)=3thank you ​

2(3a-3)-2(a-2)=3thank you