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Komok [63]
3 years ago
13

Is the electron configuration given below for an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a transition

metal.
(a) 1s2 2s2 2p6 3s2 3p5
(b) 1s2 2s2 2p6 3s2 3p6 3d7 4s2
(c) 1s2 2s2 2p6 3s2 3p6 3d7 4s2 4p6
(d) 1s2 2s2 2p6 3s2 3p6 3d7 4s1
Chemistry
1 answer:
OleMash [197]3 years ago
7 0

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^63d^4s^1   → Transition 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: (n-1)d^{1-10}ns^{0-2} 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^63d^4s^1

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

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The Density of pure carbon in Diamond form is 3.52 g/cm^3. How many cubic inches would 23.7 moles of pure diamond occupy?
Vedmedyk [2.9K]

Answer : The volume of pure diamond is 0.493inch^3

Explanation : Given,

Density of pure carbon in diamond = 3.52g/cm^3

Moles of pure diamond = 23.7 moles

Molar mass of carbon = 12 g/mol

First we have to calculate the mass of carbon or pure diamond.

\text{ Mass of carbon}=\text{ Moles of carbon}\times \text{ Molar mass of carbon}

Molar mass of carbon = 12 g/mol

\text{ Mass of carbon}=(23.7moles)\times (12g/mole)=284.4g

Now we have to calculate the volume of carbon or pure diamond.

Formula used:

Density=\frac{Mass}{Volume}

Now putting all the given values in this formula, we get:

3.52g/cm^3=\frac{284.4g}{Volume}

Volume = 80.8cm^3

As we know that:

1cm^3=0.061inch^3

So,

Volume = 0.061\times 80.8inch^3

Volume = 0.493inch^3

Therefore, the volume of pure diamond is 0.493inch^3

5 0
3 years ago
How many mL of 2.0 M KOH are necessary to neutralize 50 mL of 1 M HCl?
worty [1.4K]

Answer:

25mL

Explanation:

Please see the step-by-step solution in the picture attached below.

Hope this can help you. Have a nice day!

6 0
3 years ago
Ammonia, NH3 is a common base with Kb of 1.8 X 10-5. For a solution of 0.150 M NH3:
Vesnalui [34]

The concentrations : 0.15 M

pH=11.21

<h3>Further explanation</h3>

The ionization of ammonia in water :

NH₃+H₂O⇒NH₄OH

NH₃+H₂O⇒NH₄⁺ + OH⁻

The concentrations of all species present in the solution = 0.15 M

Kb=1.8 x 10⁻⁵

M=0.15

\tt [OH^-]=\sqrt{Kb.M}\\\\(OH^-]=\sqrt{1.8\times 10^{-5}\times 0.15}\\\\(OH^-]=\sqrt{2.7\times 10^{-6}}=1.64\times 10^{-3}

\tt pOH=-log[OH^-]\\\\pOH=3-log~1.64=2.79\\\\pH=14-2.79=11.21

8 0
2 years ago
Is a molecule positive or negative?
elixir [45]

It can be either they can have a negative or positive charge more specifically negatively charged molecule would called an anion and a positive one would be cation.

3 0
3 years ago
A student sees an absorbance a=1.140 for his solution that has a concentration of c=1.50*10-4 m using 0.50 cm cuvette. what is t
777dan777 [17]

The molar extinction coefficient is 15,200 M^{-1} cm^{-1}.

The formula to be used to calculate molar extinction coefficient is -

A = ξcl, where A represents absorption, ξ refers molar extinction coefficient, c refers to concentration and l represents length.

The given values are in required units, hence, there is no need to convert them. Directly keeping the values in formula to find the value of molar extinction coefficient.

Rewriting the formula as per molar extinction coefficient -

ξ = \frac{A}{cl}

ξ = \frac{1.140}{1.5*10^{-4}*0.5 }

Performing multiplication in denominator to find the value of molar extinction coefficient

ξ = \frac{1.140}{0.000075}  

Performing division to find the value of molar extinction coefficient

ξ = 15,200 M^{-1} cm^{-1}

Hence, the molar extinction coefficient is  15,200 M^{-1} cm^{-1}.

Learn more about molar extinction coefficient -

brainly.com/question/14744039

#SPJ4

6 0
1 year ago
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