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3 years ago

Which element has only two half filled orbitals in the 3D sub level

Chemistry

2 answers:

lions [1.4K]3 years ago

8 0

Which element has only two half filled orbitals in the 3D sub level ?

ANSWER:
Titanium

Kaylis [27]3 years ago

7 0

Ans: Titanium

The distribution or arrangement of electrons in elements is represented in terms of electron configuration which depicts the electron occupation in various orbitals.

the s orbital has only one level and holds 2 electrons

p-orbitals have 3 sublevels and holds a maximum of 6 electrons

d-orbitals have 5 sublevels and holds a maximum of 10 electrons.

The element with two half filled orbitals in the ed sublevel must have an outer electron configuration of 3d²

Therefore, the inner electrons must be filled as follows:

1s²2s²2p⁶3s²3p⁶4s²3d²

The total number of electrons is 22. The element with the atomic number 22 is titanium (Ti)

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What is nacl monosaccharide disaccharide organic or inorganic?

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3 years ago

The second electron shell of an atom can hold a maximum of ______ electron(s).

lawyer [7]

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5 0

2 years ago

Burka [1]

Answer:

$\large \boxed{\text{E) 721 K; B) 86.7 g}}$

Explanation:

Question 7.

We can use the Combined Gas Laws to solve this question.

a) Data

p₁ = 1.88 atm; p₂ = 2.50 atm

V₁ = 285 mL; V₂ = 435 mL

T₁ = 355 K; T₂ = ?

b) Calculation

$\begin{array}{rcl}\dfrac{p_{1}V_{1}}{T_{1}}& =&\dfrac{p_{2}V_{2}}{T_{2}}\\\\\dfrac{1.88\times285}{355} &= &\dfrac{2.50\times 435}{T_{2}}\\\\1.509& = &\dfrac{1088}{T_{2}}\\\\1.509T_{2} & = & 1088\\T_{2} & = & \dfrac{1088}{1.509}\\\\ & = & \textbf{721K}\\\end{array}\\\text{The gas must be heated to $\large \boxed{\textbf{721 K}}$}$

Question 8. I

We can use the Ideal Gas Law to solve this question.

pV = nRT

n = m/M

pV = (m/M)RT = mRT/M

a) Data:

p = 4.58 atm

V = 13.0 L

R = 0.082 06 L·atm·K⁻¹mol⁻¹

T = 385 K

M = 46.01 g/mol

(b) Calculation

$\begin{array}{rcl}pV & = & \dfrac{mRT}{M}\\\\4.58 \times 13.0 & = & \dfrac{m\times 0.08206\times 385}{46.01}\\\\59.54 & = & 0.6867m\\m & = & \dfrac{59.54}{0.6867 }\\\\ & = & \textbf{86.7 g}\\\end{array}\\\text{The mass of NO$_{2}$ is $\large \boxed{\textbf{86.7 g}}$}$

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3 years ago