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ollegr [7]
3 years ago
11

Find the quantinum numbers n,m,l,s for the last of potassium layer pleasee help explain correctly all

Chemistry
1 answer:
Fantom [35]3 years ago
5 0

Answer:

Quantum numbers of the outermost electron in potassium:

  • n = 4.
  • l  = 1.
  • m_l = 0.
  • Either m_s = 1/2.

Explanation:

Refer to the electron configuration of a potassium atom. The outermost electron in a ground-state potassium atom is in the 4s orbital (fourth s orbital.)

The quantum number n (the principal quantum number) specifies the main energy shell of an electron. This electron is in the fourth main energy shell (as seen in the number four in the orbital.) Hence, n = 4 for this electron.

The quantum number l (the angular momentum quantum number) specifies the shape (s, p, d, etc.) of an electron. l = 1 for s\! orbitals (such as the one that contains this electron.

Quantum numbers n and l specify the shape of an orbital. On the other hand, the magnetic quantum number m_l specifies the orientation of these orbitals in space.

However, s orbitals are spherical. Regardless of the value of n, the only possible m_l value for electrons in s\! orbitals is m_l = 0.

The spin quantum number m_s distinguishes between the two electrons in an orbital. The two possible values of m_s \! are (+1/2) and (-1/2). Typically, the first electron in an orbital is assigned an upward (\uparrow) spin, which corresponds to m_s = (+1/2).

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Pu is a nuclear waste byproduct with a half-life of 24,000 y. What fraction of the 239Pu present today will be present in 1000 y
olya-2409 [2.1K]

Answer:

0.9715 Fraction of Pu-239 will be remain after 1000 years.

Explanation:

\lambda =\frac{0.693}{t_{\frac{1}{2}}}

A=A_o\times e^{-\lambda t}

Where:

\lambda= decay constant

A_o =concentration left after time t

t_{\frac{1}{2}} = Half life of the sample

Half life of Pu-239 = t_{\frac{1}{2}}=24,000 y[

\lambda =\frac{0.693}{24,000 y}=2.8875\times 10^{-5} y^{-1]

Let us say amount present of  Pu-239 today = A_o=x

A = ?

A=x\times e^{-2.8875\times 10^{-5} y^{-1]\times 1000 y}

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\frac{A}{A_0}=\frac{A}{x}=0.9715

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7 0
3 years ago
Suppose of nickel(II) chloride is dissolved in of a aqueous solution of potassium carbonate. Calculate the final molarity of chl
stich3 [128]

Answer: Molarity of chloride anion = 0.32 M

<em>Note: the question is missing some values. The full question is given below;</em>

<em>Suppose 7.26 g of nickel(II) chloride is dissolved in 350 mL of a 0.50 M aqueous solution of potassium carbonate. Calculate the final molarity of chloride anion in the solution. You can assume the volume of the solution doesn't change when the nickel(II) chloride is dissolved in it. Be sure your answer has the correct number of significant digits.</em>

Explanation:

Molarity or molar concentration is the number of moles (mol) of component per volume (liters) concentration of solution in mol/L or M

The mass of nickel (II) chloride is 7.26 g.

The volume of potassium carbonate is 350 mL = 0.35 L

The molarity of potassium carbonate solution is 0.50 M

The reaction of nickel (II) chloride and potassium carbonate is given below.

NiCl₂(aq) + KCO₃(aq) --------> KCl(aq) +NiCO₃(s)

The dissociation of nickel (II) chloride is given below.

NiCl₂   -----> Ni²⁺ + 2Cl⁻

The molar mass of nickel (II) chloride is  129.6 g/mol

The moles of nickel (II) chloride can be calculated by the formula given below;

No of moles  = mass(g) / molar mass (g/mol)

No of moles = 7.26 / 129.6 = 0.056 moles

Therefore, molarity of NiCl₂ = 0.056 moles/ 0.35 L = 0.16 M

The molarity of 1 mole nickel (ii) chloride is 0.16 m and according to dissociation of nickel (II) chloride, 1 mole of nickel (II) chloride gives 2 moles of chloride anion.

Therefore, the molarity of chloride anion = 0.16 * 2 = 0.32 M

3 0
3 years ago
Someone help me with question pls
gizmo_the_mogwai [7]

Explanation:

the conductors are the three u have checked

8 0
2 years ago
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