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nikdorinn [45]
2 years ago
12

Can someone help me PLEASE

Chemistry
1 answer:
AVprozaik [17]2 years ago
3 0
Lattice energy k bye
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Give the chemical symbol for the element with the ground‑state electron configuration [ Ar ] 4 s 2 3 d 1 . symbol: Determine the
Rus_ich [418]

Answer:

Sc (Scandium) has the given electronic configuration.

Explanation:

The given electronic configuration is [Ar]4s^{2}3d^{1}.

The last electron enters the d-subshell and hence is a d-block element known as Scandium with chemical symbol Sc.

For 4s subshell

n=4,l=0 and m ranges from -l to +l so m=0.

For 3d subshell

n=3,l=2 and m ranges from -l to +l so m can take values -2,-1,0,+1,+2

Note:

l values for subshells:

s : 0

p : 1

d : 2

f : 3 and so on.

5 0
3 years ago
Н - О - Н<br> I have to find the Molecular formula
Vladimir [108]
Answer
H2O is the answer
5 0
3 years ago
Calculate the energy (in kj/mol) required to remove the electron in the ground state for each of the following one-electron spec
Bess [88]

Explanation:

E_n=-13.6\times \frac{Z^2}{n^2}ev

where,

E_n = energy of n^{th} orbit

n = number of orbit

Z = atomic number

a) Energy change due to transition from n = 1 to n = ∞ ,hydrogen atom .

Z = 1

Energy of n = 1 in an hydrogen like atom:

E_1=-13.6\times \frac{1^2}{1^2}eV=-13.6 eV

Energy of n = ∞ in an hydrogen like atom:

E_{\infty}=-13.6\times \frac{1^2}{(\infty)^2}eV=0

Let energy change be E for 1 atom.

E=E_{\infty}-E_1=0-(-13.6  eV)=13.6 eV

1 mole = 6.022\times 10^{-23}

Energy for 1 mole = E'

E'=6.022\times 10^{-23} mol^{-1}\times 13.6 eV

1 eV=1.60218\times 10^{-22} kJ

E'=6.022\times 10^{23}\times 13.6 \times 1.60218\times 10^{-22} kJ/mol

E'=1,312.17 kJ/mol

The energy  required to remove the electron in the ground state is 1,312.17 kJ/mol.

b) Energy change due to transition from n = 1 to n = ∞ ,B^{4+} atom .

Z = 5

Energy of n = 1 in an hydrogen like atom:

E_1=-13.6\times \frac{5^2}{1^2}eV=-340 eV

Energy of n = ∞ in an hydrogen like atom:

E_{\infty}=-13.6\times \frac{5^2}{(\infty)^2}eV=0

Let energy change be E.

E=E_{\infty}-E_1=0-(-340eV)=340 eV

1 mole = 6.022\times 10^{-23}

Energy for 1 mole = E'

E'=6.022\times 10^{-23} mol^{-1}\times 340eV

1 eV=1.60218\times 10^{-22} kJ

E'=6.022\times 10^{23}\times 340\times 1.60218\times 10^{-22} kJ/mol

E'=32,804.31 kJ/mol

The energy  required to remove the electron in the ground state is 32,804.31 kJ/mol.

c) Energy change due to transition from n = 1 to n = ∞ ,Li^{2+}atom .

Z = 3

Energy of n = 1 in an hydrogen like atom:

E_1=-13.6\times \frac{3^2}{1^2}eV=-122.4 eV

Energy of n = ∞ in an hydrogen like atom:

E_{\infty}=-13.6\times \frac{3^2}{(\infty)^2}eV=0

Let energy change be E.

E=E_{\infty}-E_1=0-(-122.4 eV)=122.4 eV

1 mole = 6.022\times 10^{-23}

Energy for 1 mole = E'

E'=6.022\times 10^{-23} mol^{-1}\times 122.4 eV

1 eV=1.60218\times 10^{-22} kJ

E'=6.022\times 10^{23}\times 122.4\times 1.60218\times 10^{-22} kJ/mol

E'=11,809.55 kJ/mol

The energy  required to remove the electron in the ground state is 11,809.55 kJ/mol.

d) Energy change due to transition from n = 1 to n = ∞ ,Mn^{24+}atom .

Z = 25

Energy of n = 1 in an hydrogen like atom:

E_1=-13.6\times \frac{25^2}{1^2}eV=-8,500 eV

Energy of n = ∞ in an hydrogen like atom:

E_{\infty}=-13.6\times \frac{25^2}{(\infty)^2}eV=0

Let energy change be E.

E=E_{\infty}-E_1=0-(-8,500 eV)=8,500 eV

1 mole = 6.022\times 10^{-23}

Energy for 1 mole = E'

E'=6.022\times 10^{-23} mol^{-1}\times 8,500eV

1 eV=1.60218\times 10^{-22} kJ

E'=6.022\times 10^{23}\times 8,500 \times 1.60218\times 10^{-22} kJ/mol

E'=820,107.88 kJ/mol

The energy  required to remove the electron in the ground state is 820,107.88 kJ/mol.

4 0
3 years ago
Magnesium hydroxide, the active ingredient in milk of magnesia, neutralizes stomach acid, primarily hcl, according to the reacti
Tema [17]
For every 1 molecule of Magnesium hydroxide or Mg(OH)2 there will be 2 molecules of HCl neutralized.
If molar mass of magnesium hydroxide is 58.3197g/mol, the amount of mol in 5.50 g magnesium hydroxide should be: 5.50g/ (<span>58.3197g/mol)= 0.0943mol.
Then, the amount of HCl molecule neutralized would be: 2* </span>0.0943mol= 0.18861 mol

If molar mass of HCl is 36.46094 g/mol, the mass of the molecule would be: 0.18861 mol* 36.46094g/mol = 6.88grams
5 0
3 years ago
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What kind of bond is formed when electrons are gained or lost ?
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C.) Ionic bond is formed between that...
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