Identify the following emission.

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

A chemistry student needs to standardize a fresh solution of sodium hydroxide. She carefully weighs out 385.mg of oxalic acid H2

Vlada [557]

Answer:

The molarity of the sodium hydroxide solution is 0.0692 M

Explanation:

Step 1: Data given

Mass of H2C2O4 = 385 mg = 0.385 grams

volume = 250 mL = 0.250 L

Volume of NaOH = 123.7 mL = 0.1237 L

Molar mass H2C2O4 = 90.03 g/mol

Step 2: The balanced equation

2NaOH + H2C2O4 → Na2C2O4 + 2H2O

Step 3: Calculate moles H2C2O4

Moles H2C2O4 = mass H2C2O4 / molar mass H2C2O4

Moles H2C2O4 = 0.385 grams / 90.03 g/mol

Moles H2C2O4 = 0.00428 moles

Step 4: Calculate molarity of H2C2O4

Molarity H2C2O4 = moles / volume

Molarity H2C2O4 = 0.00428 moles / 0.250 L

Molarity H2C2O4 = 0.01712 M

Step 5: Calculate molarity of NaOH

2*Ca*Va = n*Cb*Vb

⇒ with Ca = Molarity of H2C2O4 = 0.01712 M

⇒ Va = volume of H2C2O4 = 0.250 L

⇒Cb = molarity of NaOH = TO BE DETERMINED

⇒ Vb = volume of NaOH = 0.1237 L

Cb = (2*0.01712*0.250)/0.1237

Cb = 0.0691 M

The molarity of the sodium hydroxide solution is 0.0692 M

4 0

3 years ago

(3 Points)

prisoha [69]

Your answer is B. radio waves have shorter wavelenghts than microwaves.
Have a great day

5 0

3 years ago

Read 2 more answers

A student neutralized 16.4 milliliters of HCl by adding 12.7 milliliters of 0.620 M KOH. What was the molarity of the HCl acid?

Katen [24]

V ( HCl ) = 16.4 mL / 1000 => 0.0164 L

M( HCl) = ?

V( KOH) = 12.7 mL / 1000 => 0.0127 L

M(KOH) = 0.620 M

Number of moles KOH:

n = M x V

n = 0.620 x 0.0127

n = 0.007874 moles of KOH

number of moles HCl :

HCl + KOH = H2O + KCl

1 mole HCl ------ 1 mole KOH
? mole HCl--------0.007874 moles KOH

moles HCl = 0.007874 * 1 / 1

= 0.007874 moles of HCl

M = n / V

M = 0.007874 / 0.0164

= 0.480 M

Answer (2)

hope this helps!

4 0

3 years ago

Compare which element would have larger first ionization energy: an alkali metal in Period 2 or an alkali metal in Period 4?

maria [59]

Answer:

An alkali metal present in period 2 have larger first ionization energy.

Explanation:

Ionization energy:

The amount of energy required to remove the electron from the atom is called ionization energy.

Trend along period:

As we move from left to right across the periodic table the number of valance electrons in an atom increase. The atomic size tend to decrease in same period of periodic table because the electrons are added with in the same shell. When the electron are added, at the same time protons are also added in the nucleus. The positive charge is going to increase and this charge is greater in effect than the charge of electrons. This effect lead to the greater nuclear attraction. The electrons are pull towards the nucleus and valance shell get closer to the nucleus. As a result of this greater nuclear attraction atomic radius decreases and ionization energy increases because it is very difficult to remove the electron from atom and more energy is required.

Trend along group:

As we move down the group atomic radii increased with increase of atomic number. The addition of electron in next level cause the atomic radii to increased. The hold of nucleus on valance shell become weaker because of shielding of electrons thus size of atom increased.

As the size of atom increases the ionization energy from top to bottom also decreases because it becomes easier to remove the electron because of less nuclear attraction and as more electrons are added the outer electrons becomes more shielded and away from nucleus. Thus alkali metal present in period 2 have larger ionization energy because of more nuclear attraction as compared to the alkali metal present in period 4.

6 0

3 years ago