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Lina20 [59]
3 years ago
6

Write the electron configuration, the orbital notation, and the noble gas notation for the following elements:

Chemistry
1 answer:
V125BC [204]3 years ago
7 0

Answer:

See explanation

Explanation:

The electron configuration of an atom means a detailed arrangement of the electrons in the atom in orbitals. It normally begins from the least energetic orbitals to the most energetic orbital.

For each of the elements, their electronic configuration in terms of the nearest noble gas is shown below;

P - [Ne] 3s2 3p3

I- [Kr] 4d10 5s2 5p5

Pb- [Xe] 4f14 5d10 6s2 6p2

F- [He] 2s2 2p5

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Alex17521 [72]

Answer: If you think about it, B. would be the most reasonable answer with the given factors.

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3 years ago
If a stone with an original velocity of 0 is falling from a ledge and takes 8 seconds to hit the ground what is the final veloci
ziro4ka [17]
Also 0 as it hits the ground and stops
5 0
3 years ago
The chemical equation below is correctly balanced.
DIA [1.3K]

Answer:

44.8 L of O2 will react (option D)

Explanation:

Step 1: Data given

Number of moles of SO2 = 4.00 moles

STP = Pressure = 1 atm  and temperature = 273 K

Step 2: The balanced equation

2 SO2(g) + O2(g) → 2 SO3(g)

Step 3: Calculate moles of O2

For 2 moles SO2, we need 1 mol O2 to produce 2 moles SO3

For 4.00 moles SO2 we need 4.00 / 2 = 2.00 moles O2

Step 4: Calculate volume of O2

For 1 mol we have a volume of 22.4 L

V = (n*R*T)/ p

V = (2.00 * 0.08206 * 273)/p

V = 44.8 L

For 2.00 moles we have a volume of 2*22.4 = 44.8 L

44.8 L of O2 will react (option D)

8 0
4 years ago
In k4[fe(cn)6], how many 3d electrons does the iron atom have?
san4es73 [151]
First let's find out the oxidation number of Fe in K₄[Fe(CN)₆] compound.

The oxidation number of cation, K is +1. Hence, the total charge of the anion, [Fe(CN)₆] is -4. CN has charge has -1. There are 6 CN in anion. Let's assume the oxidation number of Fe is 'a'.

Sum of the oxidation numbers of each element  = Charge of the compound
                                                          a + 6 x (-1) = -4
                                                                     a -6 = -4
                                                                         a = +2

Hence, oxidation number of Fe in [Fe(CN)₆]⁴⁻ is +2. 

Now Fe has the atomic number as 26. Hence, number of electrons in Fe at ground state is 26. 
Electron configuration = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s² = [Ar] 3d⁶ 4s²

When making Fe²⁺, Fe releases 2 electrons. Hence, the number of electrons in Fe²⁺ is 26 - 2 = 24.
Hence, the electron configuration of Fe²⁺ = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶
                                                                      = [Ar] 3d⁶

Hence, the number of 3d electrons of Fe in K₄[Fe(CN)₆] compound is 6.
3 0
3 years ago
A closed vessel system of volume 2.5 L contains a mixture of neon and fluorine. The total pressure is 3.32 atm at 0.0°C. When th
gayaneshka [121]

Answer:

moles Ne = 0.154 mol

moles F₂ = 0.217 mol

Explanation:

Step 1: Data given

Volume of the vessel system = 2.5 L

Total pressure = 3.32 atm at 0.0 °C

The mixture is heated to 15.0 °C

The entropy of the mixture increases by 0.345 J/K

The heat capacity of monoatomic gas = 3/2R and that for a diatomic gas = 5/2R

Step 2: Define the gas

Neon is a monoatomic gas, composed of Ne atoms

 ⇒ Cv(Ne) ≅ (3/2)R

Fluorine is a diatomic gas, composed of F₂ molecules.  

⇒ Cv(F₂) ≅ (5/2)R

Step 3: Calculate moles of gas

p*V = n*R*T

⇒ with p = the total pressure = 3.32 atm

⇒ with V = the total volume = 2.5 L

⇒ with n = the number of moles of gas

⇒ with R = the gas constant = 0.08206 L*atm/K*mol

⇒ with T = the temperature = 273.15 Kelvin

n(total) = p*V/RT = (3.32 atm*2.5 L)/(0.08206 L*atm/mol•K*273.15) = 0.3703 mol

Step 4: Calculate moles of Ne and F2

For one mole heated at constant volume,  

∆S = Cv*ln(288.15/273.15) = 0.05346*Cv

⇒ ∆S for 0.3703 mol,  

∆S = (0.3703 mol)(0.05346)Cv = 0.345 J/K

 ⇒ Cv = 17.43 J/mol*K for the Ne/F₂ mixture.

For pure Ne, Cv = (3/2)R = 1.5*8.314 J/mol*K = 12.471 J/mol*K

For pure F₂, Cv = (5/2)R = 2.5 * 8.314 J/mol*K = 20.785 J/mol*K

if X is the mole fraction of Ne, we can find X by:

17.43 J/mol*K = X* 12.471 J/mol*K + (1 – X) * 20.785 J/mol*K

 ⇒ 20.875 – 8.314 * X = 17.43

X = 0.415 , 1 – X = 0.585

moles Ne = (0.415)(0.3703 mol) = 0.154 mol

moles F₂ = (0.585)(0.3703 mol) = 0.217 mol

4 0
4 years ago
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