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stiv31 [10]
3 years ago
7

Sulfur has an atomic number of 16 what is the ground-state electron configuration of sulfur​

Chemistry
2 answers:
guapka [62]3 years ago
5 0

Answer:

1s2 2s2 2p6 3s2 3p4

Explanation:

For ground state atoms you need to fill the orbitals in order.

Orbital s has max 2 electrons

Orbital p has max 6 electrons

Level 1 only has s orbital

Level 2 has s and p orbitals

Level 3 has s, p and d orbitals

You fill them in order until you reach 16 electrons:

1s2 2s2 2p6 3s2 3p4

Sphinxa [80]3 years ago
3 0

Answer: 1s^22s^22p^63s^23p^4

Explanation:

Electronic configuration represents the total number of electrons that a neutral element contains. We add all the superscripts to know the number of electrons in an atom.

The electrons are filled according to Afbau's rule in order of increasing energies. Sulphur with atomic number of 16 will have 16 electrons and thus the ground state electronic configuration is:

S:16:1s^22s^22p^63s^23p^4

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If 20 mL of gas is subjected to a temperature change from 10°C to 100°C and a pressure change from 1 atm to 10 atm,
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Answer:

E. None of these

Explanation:

We know, By GAS laws,

PV = NRT, where p- pressure, v- volume, n- number of moles, R- gas constant ,and T- temperature

Now, In the question, the number of moles remains the same as the gas is the same. so n is constant so we can compare n before and after a temperature change.

\frac{P1V1}{RT1} = \frac{P2V2}{RT2}

where P1= 1 atm, P2 = 10 atm, V1= 20 mL, T1= 10°C and T2= 100°C

We don't have to worry about the standard units as they are present equally on both the sides and get cut, same goes for R( gas constant)

So putting values, we get

\frac{1*20}{R*10} = \frac{10*V2}{R*100}

Cutting, R on both sides and moving contents to the right so that only V2 is left on the left.

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3 0
1 year ago
A solution contains [Ba2+] = 5.0 × 10−5 M, [Zn2+] = 2.0 × 10−7 M, and [Ag+] = 3.0 × 10−5 M. Sodium oxalate (Na2C2O4) is slowly a
Jet001 [13]

Answer:

BaC₂O₄, then ZnC₂O₄, then Ag₂C₂O₄  

Explanation:

1. Calculate the equilibrium concentrations of oxalate ion

Let [C₂O₄²⁻] = c

(a) Barium oxalate

                 BaC₂O₄ ⇌   Ba²⁺   + C₂O₄²⁻

E/mol·L⁻¹:                   5.0 × 10⁻⁵     c

Ksp = [Ba²⁺][C₂O₄²⁻] = 5.0 × 10⁻⁵c = 1.5 × 10⁻⁸

c = (1.5 × 10⁻⁸)/(5.0 × 10⁻⁵) = 3.0 × 10⁻⁴ mol·L⁻¹

(b) Zinc oxalate

                ZnC₂O₄ ⇌   Zn²⁺   + C₂O₄²⁻

E/mol·L⁻¹:                 2.0 × 10⁻⁷      c

Ksp = [Zn²⁺][C₂O₄²⁻] = 2.0 × 10⁻⁷c = 1.35 × 10⁻⁹

c = (1.35 × 10⁻⁹)/(2.0 × 10⁻⁷) = 6.8 × 10⁻³ mol·L⁻¹

(c) Silver oxalate

                 Ag₂C₂O₄ ⇌   2Ag⁺   +   C₂O₄²⁻  

E/mol·L⁻¹:                      3.0 × 10⁻⁵       c

Ksp = [Ag⁺]²[C₂O₄²⁻] = (3.0× 10⁻⁵)²c = 9.0 × 10⁻¹⁰c = 1.1 × 10⁻¹¹

c = (1.1 × 10⁻¹¹)/(9.0 × 10⁻¹⁰) = 0.012 mol·L⁻¹

2. Decide the order of precipitation

BaC₂O₄ will precipitate when   c > 3.0 × 10⁻⁴ mol·L⁻¹

ZnC₂O₄ will precipitate when   c > 6.8 × 10⁻³ mol·L⁻¹

Ag₂C₂O₄ will precipitate when c > 0.028       mol·L⁻¹

This happens to be the order of increasing concentration of oxalate ion.

The order of precipitation is

BaC₂O₄, then ZnC₂O₄, then Ag₂C₂O₄

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