Answer this please thank you

Part B Identify the sets of quantum numbers that describe all the electrons in the ground state of a neutral beryllium atom, Be.

dmitriy555 [2]

Answer: The set of quantum numbers for the electrons in Be atom are (2, 1, -1, 1/2), (1, 0, 0, 1/2), (1, 0, 0, -1/2) and (2, 0, 0, -1/2)

Explanation:

There are 4 quantum numbers:

Principal Quantum number (n) specifies the energy of the electron in a shell.
Azimuthal Quantum number (l) specifies the shape of an orbital. The value of it lies in the range of 0 to (n-1)
Magnetic Quantum number (m) specifies the orientation of the orbital in space. The value of it lies in the range of -l to +l
Spin Quantum number (s) specifies the spin of an electron in an orbital. It can either have a value of $+\frac{1}{2}$ or $-\frac{1}{2}$

Berylium (Be) is the 4th element of periodic table having electronic configuration of $1s^22s^2$

For electrons in 1s-orbital, the quantum numbers can be:

For first electron:

$n=1\\l=0\text{ (for s-subshell)}\\m=0\\s=+\frac{1}{2}$

For second electron:

$n=1\\l=0\\m=0\\s=-\frac{1}{2}$

For electrons in 2s-orbital, the quantum numbers can be:

For first electron:

$n=2\\l=0\text{ (for s-subshell)}\\m=0\\s=+\frac{1}{2}$

For second electron:

$n=2\\l=0\text{ (for s-subshell)}\\m=0\\s=-\frac{1}{2}$

Hence, the set of quantum numbers for the electrons in Be atom are (2, 1, -1, 1/2), (1, 0, 0, 1/2), (1, 0, 0, -1/2) and (2, 0, 0, -1/2)

4 0

3 years ago

Which substance can be decomposed by a chemical change? (1) Co (3) Cr(2) CO (4) Cu

kotykmax [81]

Answer: (2) CO

Co, Cr, and Cu are elements. Elements cannot be broken down by a chemical change, but compounds can. Since CO is a compound, it can be decomposed by a chemical change.

7 0

3 years ago

How many atoms are in a single molecule of FeCl2

Nady [450]

3
One from fe
And two from cl2

7 0

3 years ago

Nitric oxide reacts with chlorine to form nocl. the data refer to 298 k. 2no(g) + cl2(g) → 2nocl(g) substance: no(g) cl2(g) nocl

tigry1 [53]

Answer:

- 10.555 kJ/mol.

Explanation:

∵ ∆G°rxn = ∆H°rxn - T∆S°rxn.

Where, ∆G°rxn is the standard free energy change of the reaction (J/mol).

∆H°rxn is the standard enthalpy change of the reaction (J/mol).

T is the temperature of the reaction (K).

∆S°rxn is the standard entorpy change of the reaction (J/mol.K).

Calculating ∆H°rxn:

∵ ∆H°rxn = ∑∆H°products - ∑∆H°reactants

∴ ∆H°rxn = (2 x ∆H°f NOCl) - (1 x ∆H°f Cl₂) - (2 x ∆H°f NO) = (2 x 51.71 kJ/mol) - (1 x 0) - (2 x 90.29 kJ/mol) = - 77.16 kJ/mol.

Calculating ∆S°rxn:

∵ ∆S°rxn = ∑∆S°products - ∑∆S°reactants

∴ ∆S°rxn = (2 x ∆S° NOCl) - (1 x ∆S° Cl₂) - (2 x ∆S° NO) = (2 x 261.6 J/mol.K) - (1 x 223.0 J/mol.K) - (2 x 210.65 J/mol.K) = - 121.1 J/mol.K. = - 0.1211 kJ/mol.K.

Calculating ∆G°rxn:

∵ ∆G°rxn = ∆H°rxn - T∆S°rxn.

∴ ∆G°rxn = ∆H°rxn - T∆S°rxn = (- 77.16 kJ/mol) - (550 K)(- 0.1211 kJ/mol.K) = - 10.555 kJ/mol.

4 0

3 years ago

Suppose that 25,0 mL of a gas at 725 mmHg and 298K is converted to

posledela

The new volume : 21.85 ml

<h3>Further explanation</h3>

Given

V1=25,0 ml

P1=725 mmHg

T1=298K is converted to

T2=273'K

P2=760 mmHg atm

Required

V2

Solution

Combined gas law :

$\tt \dfrac{P_1.V_1}{T_1}=\dfrac{P_2.V_2}{T_2}$

Input the value :

V2=(P1.V1.T2)/(P2.T1)

V2=(725 x 25 ml x 273)/(760 x 298)

V2=21.85 ml

5 0

3 years ago