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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 reform reaction between steam and gaseous methane (CH4) produces "synthesis gas," a mixture of carbon monoxide gas and dihyd
kenny6666 [7]

Answer:

The answer is "= 0.078 \ kg \ H_2".

Explanation:

calculating the moles in CH_4 =\frac{PV}{RT}

                                                =\frac{(0.58 \ atm) \times (923 \ L) }{ (0.0821 \frac{L \cdot atm}{K \cdot mol})(232^{\circ} C +273)}\\\\=\frac{(535.34 \ atm \cdot \ L) }{ (0.0821 \frac{L \cdot atm}{K \cdot mol})(505)K}\\\\=\frac{(535.34 \ atm \cdot \ L) }{ (41.4605 \frac{L \cdot atm}{mol})}\\\\= 12.9 \ mol

Eqution:

CH_4 +H_2O \to  3H_2+ CO \ (g)

Calculating the amount of H_2 produced:

= 12.9 \ mol CH_4 \times  \frac{3 \ mol \ H_2 }{1 \ mol \ CH_4}\times \frac{2.016 g H_2}{1 \ mol \ H_2}\\\\= 78 \ g \ H_2 \\\\= 0.078 \ kg \ H_2

So, the amount of dihydrogen produced = 0.078 \frac{kg}{s}

5 0
3 years ago
What is the concentration of a solution with a volume of 2.5 liters containing 600 grams of calcium phosphate?​
trapecia [35]

Answer:

1.12M

Explanation:

Given parameters:

Volume of solution  = 2.5L

Mass of Calcium phosphate  = 600g

Unknown:

Concentration  = ?

Solution:

Concentration is the number of moles of solute in a particular solution.

Now, we find the number of moles of the calcium phosphate from the given mass;

        Formula of calcium phosphate  = Ca₃PO₄

         molar mass = 3(40) + 31 + 4(16) = 215g/mol  

Number of moles of  Ca₃PO₄  = \frac{600}{215}   = 2.79moles

   Now;

  Concentration  = \frac{Number of moles }{volume }  

 Concentration  = \frac{2.79}{2.5}   = 1.12M

7 0
3 years ago
1. The solubility of lead(II) chloride at some high temperature is 3.1 x 10-2 M. Find the Ksp of PbCl2 at this temperature.
solniwko [45]

Answer:

1) The solubility product of the lead(II) chloride is 1.2\times 10^{-4}.

2) The solubility of the aluminium hydroxide is 1.6\times 10^{-10} M.

3)The given statement is false.

Explanation:

1)

Solubility of lead chloride = S=3.1\times 10^-2M

PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)

                            S     2S

The solubility product of the lead(II) chloride = K_{sp}

K_{sp}=[Pb^{2+}][Cl^-]^2

K_{sp}=S\times (2S)^2=4S^3=4\times (3.1\times 10^{-2})^3=1.2\times 10^{-4}

The solubility product of the lead(II) chloride is 1.2\times 10^{-4}.

2)

Concentration of aluminium nitrate = 0.000010 M

Concentration of aluminum ion =1\timed 0.000010 M=0.000010 M

Solubility of aluminium hydroxide in aluminum nitrate solution = S

Al(OH)_3(aq)\rightleftharpoons Al^{3+}(aq)+3OH^-(aq)

                            S     3S

The solubility product of the aluminium nitrate = K_{sp}=1.0\times 10^{-33}

K_{sp}=[Al^{3+}][OH^-]^3

1.0\times 10^{-33}=(0.000010+S)\times (3S)^3

S=1.6\times 10^{-10} M

The solubility of the aluminium hydroxide is 1.6\times 10^{-10} M.

3.

Molarity=\frac{Moles}{Volume (L)}

Mass of NaCl= 3.5 mg = 0.0035 g

1 mg = 0.001 g

Moles of NaCl = \frac{0.0035 g}{58.5 g/mol}=6.0\times 10^{-5} mol

Volume of the solution = 0.250 L

[NaCl]=\frac{6.0\times 10^{-5} mol}{0.250 L}=0.00024 M

1 mole of NaCl gives 1 mole of sodium ion and 1 mole of chloride ions.

[Cl^-]=[NaCl]=0.00024 M

Moles of lead (II) nitrate = n

Volume of the solution = 0.250 L

Molarity lead(II) nitrate = 0.12 M

n=0.12 M]\times 0.250 L=0.030 mol

1 mole of lead nitrate gives 1 mole of lead (II) ion and 2 moles of nitrate ions.

[Pb^{2+}]=[Pb(NO_2)_3]=0.030 M

PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)

Solubility of lead(II) chloride = K_{sp}=1.2\times 10^{-4}

Ionic product of the lead chloride in solution :

Q_i=[Pb^{2+}][Cl^-]^2=0.030 M\times (0.00024 M)^2=1.7\times 10^{-9}

Q_i ( no precipitation)

The given statement is false.

3 0
4 years ago
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