Answer:
False
Explanation:
The first reaction is;
NO(g) + 1/2O2(g) ---->NO2(g)
K= [NO2]/[NO] [ O2]^1/2
The second reaction is;
2NO(g) + O2(g) ---->2NO2(g)
K'= [NO2]^2/[NO]^2 [O2]
It now follows that;
K'= K^2
Hence the statement in the question is false
When the equation,
O2 + __C 10H 22 →
CO2 +
H2O is balanced, the coefficient of O2 is?
Please help!
A sample of helium gas at a pressure of 0.778 atm and a temperature of 20.2 °C, occupies a volume of 574 mL. If the gas is allowed to expand at constant temperature until its pressure is 0.373 atm, the volume of the gas sample will be mL.
Answer:
I don't know ❤️
Five kg of carbon dioxide (CO2) gas undergoes a process in a well-insulated piston-cylinder assembly from 2 bar, 280 K to 20 bar, 520 K. If the carbon dioxide behaves as an ideal gas, determine the amount of entropy produced, in kJ/K.
This question is incomplete, the complete question is;
Five kg of carbon dioxide (CO2) gas undergoes a process in a well-insulated piston-cylinder assembly from 2 bar, 280 K to 20 bar, 520 K. If the carbon dioxide behaves as an ideal gas, determine the amount of entropy produced, in kJ/K. Assuming;
a) constant specific heats Cp = 0.939 kJ/Kg K
b) variable specific heats
Answer:
a) the amount of entropy produced is 0.731599 kJ/K
b) the amount of entropy produced is 0.69845 kJ/K
Explanation:
Given the data in the question;
5 kg of carbon dioxide (CO₂) gas undergoes a process in a well-insulated piston-cylinder assembly.
m = 5 kg
Molar mass M = 44.01 g/mol
P₁ = 2 bar, P₂ = 20
T₁ = 280 K, P₂ = 520 K
Since its insulated { q = 0 } ( kinetic and potential energy effects = 0 )
Now,
a) the amount of entropy produced, in kJ/K, Assuming constant specific heats with Cp = 0.939 kJ/Kg K
S[tex]_{Generation[/tex] = m × ((Cp × In( T₂/T₁) - R × In( P₂/p₁ ))
we substitute
S[tex]_{Generation[/tex] = 5 × (( 0.939 × In( 520/280) - 0.1889 × In( 20/2 ))
= 5 × ( 0.5812778 - 0.434958 )
= 5 × 0.1463198
= 0.731599 kJ/K
Therefore, the amount of entropy produced is 0.731599 kJ/K
b) the amount of entropy produced, in kJ/K, Assuming variable specific heats.
Now, from Table A-23: Ideal Gas Properties of Selected Gases;
T₁,T₂ : s₁⁰ = 211.376 kJ/kmol-K, s₂⁰ = 236.575 kJ/kmol-K
now, s₁ = s₁⁰ / M and s₂ = s₂⁰ / M
we substitute
s₁ = s₁⁰ / M = 211.376 / 44.01 = 4.8029 kJ/kg
s₂ = s₂⁰ / M = 236.575 / 44.01 = 5.37548 kJ/kg
S[tex]_{Generation[/tex] = m × (( s₂ - s₁ ) - R × In( p₂ / p₁ ))
we substitute
S[tex]_{Generation[/tex] = 5 × (( 5.37548 - 4.8029 ) - 0.1880 × In( 20 / 2 ))
= 5 × ( 0.57258 - 0.432885997 )
= 5 × 0.13969
= 0.69845 kJ/K
Therefore, the amount of entropy produced is 0.69845 kJ/K
Titanium is a metal often used as an alloying agent to provide materials that are strong, lightweight, and temperature-resistant Which of the following represents the correct ground-state configuration for a neutral atom of titanium?
A) 1s 2s 2p 3s 3p 48°30°
B) 1s 2s 2p 3s 3p 4s3d
C) 15*2s2p 3s 3p 4s
D) 15°2s 2p 3s 3p 3d
Answer:B) 1s 2s 2p 3s 3p 4s 3d
Explanation:
The ground state electron configuration shows how the electrons in the atomic orbitals of an atom are in their lowest , most stable energy arrangements and since Electrons must be filled following the Aufbau's principle(electrons fill lowest energy shells first)
Now, Titanium lies in period IV and group 4 of the periodic table with 22 as its atomic number
Thus, the ground-state electron configuration of a neutral atom of titanium is 1s²2s²2p⁶3s²3p⁶4s²3d².
Difficulty(Level:1)
Which option below defines the law of conservation of energy
-Energy is created but some is always lost when doing so
-Energy or Chakra must be conserved to accomplish difficult ninjitsus
-Energy constantly moves toward a form of disorder
-Conserve Energy or the world will end
-Energy is never created or destroyed, it just changes forms.
Answer:
I say the last option
Explanation:
I hope this help let me know if you have any questions
The first step of electrophilic aromatic substitution involves the formation of the arenium ion intermediate.
a. True
b. Fasle
Answer:
True
Explanation:
Aromatic compounds undergo substitution rather than addition reactions because the aromatic structure is maintained.
Electrophilic aromatic substitution begins with attack of the electrophile on the aromatic ring to yield a delocalized intermediate called the arenium intermediate. Loss of hydrogen from this intermediate yields the final product.
PLEASE HELP ASAP
Use the equation below to answer the following questions.
2Al(s) + 3Cu(NO3)2(aq) 3Cu(s) + 2Al(NO3)3(aq)
Determine the oxidation state of the atoms in the equation's reactants and products: (6 points)
Oxidation state of Al in reactant:
in product:
Oxidation state of Cu in reactant:
in product:
Oxidation state of N in reactant:
in product:
Oxidation state of O in reactant:
in product:
Explain why this is a redox reaction.
Thank you!
Answer:
See explanation
Explanation:
A redox reaction equation shows a gain/loss of electrons from left to right in a reaction. In a redox reaction, a specie looses electrons while another specie gains electrons.
Considering the equation; 2Al(s) + 3Cu(NO3)2(aq) -----> 3Cu(s) + 2Al(NO3)3(aq)
The oxidation state of Al reactant is zero since the oxidation state of all uncombined elements is zero. The oxidation state of All in the product is +3
The oxidation state of the copper in the reactant is +2. The oxidation state of copper in the product is zero.
The oxidation state of N in the reactant and product is +5.
The oxidation state of oxygen in the reactant and product is (-2).
This is a redox reaction because from left to right, Al was oxidized (oxidation number increased from zero to +3) while Cu was reduced (oxidation number decreased from +2 to zero).
Which is a statement of cell theory? All cells are made up of living molecules. All plants are made of cells. All animals are made of cells. All cells are produced from other cells.
Answer:
all cells are produced from other preexisting cells through cell division
Is a 4p S 4s transition allowed in sodium? If so, what is its wavelength? If not, why not? b. Is a 3d S 4s transition allowed in sodium? If so, what is its wavelength? If not, why not? g
Answer:
a) 4p ⇒ 4s transition is Allowed
b) 3d ⇒ 4s transition not allowed
Explanation:
a) 4p ⇒ 4s transition
This transition is allowed because for a 4p state; l = 1 and for a 4s state I = 0
hence Δl = 1 - 0 = 1
Energy of 4p ( Ei ) = 3.75eV
Energy of 4s ( E2 ) = 3.19 eV
where : λ = 1240 eV nm / ( E₂ - E₁ )
= 2214 nm ≈ 2.214 μm
b) 3d ⇒ 4s transition
This transition is not allowed
a 3d state , l = 2 while for 4s state l = 0
hence Δl = 2 - 0 = 2
therefore the transition is not allowed
Predict the reactants of this chemical reaction. That is, fill in the left side of the chemical equation. Be sure the equation you submit is balanced. (You can edit both sides of the equation to balance it, if you need to.)
Note: you are writing the molecular, and not the net ionic equation.
NaClO3(aq) + H2O(l)
Answer:
HClO₃(aq) + NaOH(aq) ⇒ NaClO₃(aq) + H₂O(l)
Explanation:
We have the products of a reaction and we have to predict the reactants. Since the products are salt and water, this must be a neutralization reaction. In a neutralization reaction, an acid reacts with a base. To form NaClO₃, the acid must be HClO₃(aq) and the base NaOH(aq). The balanced chemical equation is:
HClO₃(aq) + NaOH(aq) ⇒ NaClO₃(aq) + H₂O(l)
How many grams of CO2 are formed if 44.7 g C5H12 is mixed with 108 g O2?
Explanation:
here's the answer to your question
Gu
Magnesium metal is reacted with hydrochloric acid to produce
hydrogen gas. A sample of hydrogen gas is collected over water
in a eudiometer at 28.0°C. The atmospheric pressure is 636
mmHg. Determine the pressure (in atm) of the hydrogen gas
produced
Pressure =
atm
The pressure of hydrogen gas is 607.7 mmHg
According of Dalton's law of Partial pressure, the total pressure of a mixture of gases is the sum of the partial pressures of the individual gases in the mixture.
We can now write;
The for hydrogen collected over water, we have a mixture of hydrogen gas and water vapour.
Total pressure = pressure of hydrogen gas + vapour pressure of water
Pressure of hydrogen gas = Total pressure - vapour pressure of water
Pressure of hydrogen gas = 636 mmHg - 28.3 mmHg
Pressure of hydrogen gas = 607.7 mmHg
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the nutrition label on rice lists the amounts of protein, carbohydrates and fats in one serving. these substances are important for human nutrition
Answer:
Carbohydrates, proteins, and fats are biological macromolecules that are made up of chemical elements which are inherent to chemistry.
Chemistry explain how these macromolecules are bonded together at the molecular level and give an explanation for their behavior.
Explanation:
who much the velocity of a body when it travels 600m in 5 min
Answer:
2 m/s
Explanation:
Applying the formulae of velocity,
V = d/t............. Equation 1
Where V = Velocity of the body, d = distance, t = time
From the question,
Given: d = 600 m, t = 5 minutes = (5×60) = 300 seconds.
Substitute these values into equation 1
V = 600/300
V = 2 m/s.
Hence the velocity of the body when it travels is 2 m/s
g 32.53 g of a solid is heated to 100.oC and added to 50.0 g of water in a coffee cup calorimeter and the contents are allowed to sit until they finally have the same temperature. The water temperature changes from 25.36 oZ to 34.4 oC. What is the specific heat capacity (in J/goC) of the solid
Answer:
0.886 J/g.°C
Explanation:
Step 1: Calculate the heat absorbed by the water
We will use the following expression
Q = c × m × ΔT
where,
Q: heatc: specific heat capacitym: massΔT: change in the temperatureQ(water) = c(water) × m(water) × ΔT(water)
Q(water) = 4.184 J/g.°C × 50.0 g × (34.4 °C - 25.36 °C) = 1.89 × 10³ J
According to the law of conservation of energy, the sum of the energy lost by the solid and the energy absorbed by the water is zero.
Q(water) + Q(solid) = 0
Q(solid) = -Q(water) = -1.89 × 10³ J
Step 2: Calculate the specific heat capacity of the solid
We will use the following expression.
Q(solid) = c(solid) × m(solid) × ΔT(solid)
c(solid) = Q(solid) / m(solid) × ΔT(solid)
c(solid) = (-1.89 × 10³ J) / 32.53 g × (34.4 °C - 100. °C) = 0.886 J/g.°C
consider the following thermochemical reaction for kerosene
2C12H26+37O2=24CO2+15026kj.
a. when 21.3g of CO2 are made, how much heat is released?
b. if 500.00kj of heat are released by thye reaction, how many grams of C12H26 have been consumed.?
c. if this reactionwere being used to generate heat, how many grams of C12H26 would have to be reacted to generate enough heat to raise the temperature of 750g of liquid water from 10 degrees celcius to 90 degrees celcius
Thermochemistry has to do with heat evolved or absorbed in a chemical reactions. Thermochemical equations are equations in which the heat of reaction is included in the reaction equation. The reaction of moles and heat of reaction is important here.
This question has to do with thermochemistry and thermochemical equations.
The answers to each of the questions are shown below;
a) 300.52 KJ
b) 11.39 g
c) 5.78 g
The equation of the thermochemical reaction is;
2C12H26 + 37O2-------> 24CO2 + 15026KJ
Number of moles of CO2 released = 21.3g/44g/mol = 0.48 moles
From the reaction equation;
15026KJ is released when 24 moles of CO2 is released
x KJ is released when 0.48 moles of CO2 is released
x = 15026KJ * 0.48 moles/24 moles
x = 300.52 KJ
b) If 2 moles of C12H26 released 15026KJ of heat
x moles of C12H26 released 500.00KJ
x = 2 * 500.00KJ/15026KJ
x = 0.067 moles
Mass of C12H26 consumed = 0.067 moles * 170 g/mol = 11.39 g
c) Heat gained by water = heat released by combustion of kerosene
Heat gained by water = 0.75 Kg * 4200 * (90 -10)
Heat gained by water = 252 KJ
If 2 moles of C12H26 produced 15026KJ
x moles of C12H26 produces 252 KJ
x = 2 * 252/15026
x = 0.034 moles
Mass of C12H26 = 0.034 moles * 170 g/mol = 5.78 g
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What is the correct ratio of carbon to hydrogen to oxygen in glucose (C6H12O6)?
12:12:6
2:1:1
1:2:1
6:6:12
Answer:
Correct ratio of carbon to hydrogen is 2:1:1
Answer:
Its actually 1:2:1
Explanation:
The molecular formula is C6H12O6 because one molecule actually contains 6 C, 12 H, and 6 O atoms. The simplest whole-number ratio of C to H to O atoms in glucose is 1:2:1, so the empirical formula is CH2O.
A solution of acetic acid and water contains 205.0 g L-1 of acetic acid and 820.0 g L-1 of water. Compute the density of the solution ( report your answer in g per mL)
Answer:
[tex]\rho_t=1025000 gmL^{-1}[/tex]
Explanation:
From the question we are told that:
Density of acetic acid [tex]\rho_a=205 gL^{-1}[/tex]
Density of Water [tex]\rho_w=820 gL^{-1}[/tex]
Generally the equation for Solution Density is mathematically given by
[tex]\rho_t= \rho_w+\rho_a[/tex]
[tex]\rho_t=205+820[/tex]
[tex]\rho_t=1025 gL^{-1}[/tex]
[tex]\rho_t=1025000 gmL^{-1}[/tex]
cesium-131 has a half life of 9.7 days. what percent of a cesium-131 sample remains after 60 days?
Explanation:
From the question given above, the following data were obtained:
Half-life (t½) = 9.7 days
Time (t) = 60 days
Percentage remaining after 60 days =?Next, we shall determine the number of half-lives that has elapsed. This can be obtained as follow:
Half-life (t½) = 9.7 days
Time (t) = 60 days
Number of half-lives (n) =?
n = t / t½
n = 60 / 9.7Finally, we shall determine the percentage remaining. This can be obtained as follow:
Let the original amount be N₀
Let the amount remaining be N
Number of half-lives (n) = 60 / 9.7
N = N₀ / 2ⁿ
Divide both side by N₀
N/N₀ = 1/2ⁿ
N/N₀ = 1 / 2⁽⁶⁰÷⁹•⁷⁾
N/N₀ = 0.0137
Multiply by 100 to express in percentage
N/N₀ = 0.0137 × 100
N/N₀ = 1.37%Therefore, the percentage remaining after 60 days is 1.37%
NOTE; N/N₀ is the fraction remaining.Learn more: https://brainly.com/question/14883322
What is the difference between elimination and substitution reaction
Identify the key factors that will determine if a reaction undergoes elimination or substitution mechanism.
Use the following reagents to determine the type of reaction pathway expected and determine the products in each reaction.
a. Tert BuO- in tertbutanol and chlorobutane
b. KOH in water and bromobutane
c. NaI in acetone and bromobutane
Write a conclusion of no more than two paragraphs to summarize your results
Answer:
a) E2
b) SN2
c) SN2
Explanation:
A substitution reaction involves replacement of an atom or group in a molecule by another atom or group. An elimination reaction is the loss of two atoms from the same molecule leading to the formation of a multiple bond in the molecule.
We must note that primary alkyl halides never undergo SN1/E1 reactions. However, the presence of a strong bulky base such as tert BuO- , E2 reactions predominate. In the presence of strong bases such as OH^- and good nucleophiles such as I^-, SN2 mechanism predominates.
If the concentration of products is increased the equilibrium is shifted from * left to right/ to the left/ right to left /down left
Answer:
to the left
Explanation:
If the concentration of products is increased for a reaction that is in equilibrium, the equilibrium would shift to the left side of the reaction (the reactant's side).
For a reaction that is in equilibrium, the reaction is balanced between the reactants and the products. According to Le Cha telier's principle, if one of the constraints capable of influencing the rate of reactions is applied to such a reaction that is in equilibrium, the equilibrium would shift so as to neutralize the effects created by the constraint.
Hence, in this case, if the concentration of the products of a reaction in equilibrium is increased, the equilibrium would shift in such a way that more reactants are formed so as to annul the effects created by the increase in the concentration of the products. Since reactants are always on the left side of chemical equations, it thus means that the equilibrium would shift to the left.
At 445oC, Kc for the following reaction is 0.020. 2 HI(g) <--> H2 (g) + I2 (g) A mixture of H2, I2, and HI in a vessel at 445oC has the following concentrations: [HI] = 1.5 M, [H2] = 2.50 M and [I2] = 0.05 M. Which one of the following statements concerning the reaction quotient, Qc, is TRUE for the above system?
a. Qc = Kc; the system is at equilibrium.
b. Qc is less than Kc; more H2 and I2 will be produced.
c. Qc is less than Kc; more HI will be produced.
d. Qc is greater than Kc; more HI will be produced.
Explanation:
The given balanced chemical equation is:
[tex]2 HI(g) <--> H_2 (g) + I_2 (g)[/tex]
The value of Kc at 445oC is 0.020.
[HI]=1.5M
[H2]=2.50M
[I2]=0.05M
The value of Qc(reaction quotient ) is calculated as shown below:
Qc has the same expression as the equilibrium constant.
[tex]Qc=\frac{[H_2][I_2]}{[HI]^2} \\Qc=(2.50Mx0.05M)/(1.5M)^2\\Qc=0.055[/tex]
Qc>Kc,
Hence, the backward reaction is favored and the formation of Hi is favored.
Among the given options, the correct answer is option d. Qc is greater than Kc; more HI will be produced.
A 13.4 mL sample of CO2 gas was collected in an experiment.
What is this volume in liters (L)? Use significant figures, do NOT include the units.
Explanation:
here's the answer to your question
A runner completes the 300-meter dash in 38 seconds. What is the speed of the runner? Round your answer to the nearest tenth.
Answer:
x= 300m , 38s
answer is 7.9
Answer:
speed=distance/time
Explanation:
speed=300/38
=7.8947368421
to the nearest tenth=7.9
Can steel turn into gas? ( A random question that came to mind. I have no idea if it is or isn't possible.)
Answer:
YES STEEL CAN TURN INTO GAS .
Answer: yes it can!
Explanation:
Nitric acid can be formed in two steps from the atmospheric gases nitrogen and oxygen, plus hydrogen prepared by reforming natural gas. In the first step, nitrogen and hydrogen react to form ammonia: N2 (g) + 3H2 (g) â 2NH3 (g) =ÎHâ92.kJ In the second step, ammonia and oxygen react to form nitric acid and water:
NH3 (g) + 2O2 (g) â HNO3 (g) + H2O (g) =ÎHâ330.kJ
Required:
Calculate the net change in enthalpy for the formation of one mole of nitric acid from nitrogen, hydrogen and oxygen from these reactions.
Answer:
-376 kJ
Explanation:
The first step equation:
[tex]\mathsf{N_{2(g)} + 3H_2{(g)} \to 2NH_3{(g)} \ \ \ \Delta H = -92\ kJ}[/tex] ---- (1)
The second step equation:
[tex]\mathsf{NH_{3(g)} + 2O_2{(g)} \to HNO_3{(g)} +H_2O_{(g)} \ \ \ \Delta H = -330\ kJ}[/tex] ---- (2)
To determine the enthalpy of formation for 1 mole of HNO₃ (nitric acid), we have the following.
From the above equations; let multiply equation (1) by 1 and equation (2) by 2.
[tex]\mathsf{N_{2(g)} + 3H_2{(g)} \to 2NH_3{(g)} \ \ \ \Delta H = -92\ kJ}[/tex] ---- (3)
[tex]\mathsf{2NH_{3(g)} + 4O_2{(g)} \to 2HNO_3{(g)} +2H_2O_{(g)} \ \ \ \Delta H = 2(-330)\ kJ}[/tex] ----- (4)
adding the above two equations, we have:
[tex]\mathsf{N_{2(g)} + 3H_2{(g)}+ 2NH_{3(g)} + 4O_{2(g)} \to 2HNO_{3(g)} + 2NH_3{(g)} +2H_2O_{(g)} \ \ \ \Delta H = (-660 \ kJ -92\ kJ)}[/tex][tex]\mathsf{N_{2(g)} + 3H_2{(g)} + 4O_{2(g)} \to 2HNO_{3(g)} +2H_2O_{(g)} \ \ \ \Delta H = (-752 \ kJ)}[/tex]
Now, from the recent equation, we have:
2 moles of nitric acid = -752 kJ
∴
1 mole of nitric acid will be: = (1 mole × (-752 kJ)) ÷ 2 moles
1 mole of nitric acid will be: = -376 kJ
What is the concentration of HI in the final solution when 65 mL of a 3.0 M HI solution is diluted with pure water to a total volume of 0.15 L g?
Answer:
The concentration of HI in the final solution is 1.3 M.
Explanation:
Dilution is the reduction in concentration of a chemical in a solution. It is achieved by adding more solvent to the same amount of solute.
In other words, in a dilution, the amount of solute does not change, but the volume of the solvent does: as more solvent is added, the concentration of the solute decreases, since the volume (and weight) of the solution increases.
When dealing with dilution you will use the following equation:
C1*V1= C2*V2
C1 = initial concentration V1 = initial volume C2 = final concentration V2 = final volumeIn this case:
C1 = 3 M V1 = 65 mL= 0.065 L (being 1000 mL= 1 L) C2 = ? V2 = 0.15 LReplacing:
3 M* 0.065 L= C2*0.15 L
Solving:
[tex]C2=\frac{3 M*0.065 L}{0.15 L}[/tex]
C2= 1.3 M
The concentration of HI in the final solution is 1.3 M.
Sally has constructed a concentration cell to measure Ksp for MCln. She constructs the cell by adding 2 mL of 0.05 M M(NO3)n to one compartment of the microwell plate. She then makes a solution of MCln by adding KCl to M(NO3)n. She adds 7.903 mL of the resulting mixture to a second compartment of the microwell plate. Sally knows n = +2. She has already calculated [Mn+] in the prepared MCln solution using the Nernst equation. [Mn+] = 8.279 M
Required:
How many moles of [Cl-] must be dissolved in that compartment?
Answer:
0.1309 mol
Explanation:
From the given information:
The metal ion, two ions of [tex]M^{+}[/tex] reacted with Cl⁻ to form [tex]MCl_n[/tex] i.e. the compound formed is [tex]MCl_2[/tex].
The concentration of the metal ion formed [tex][M^+][/tex] = 8.279 M
The concentration of the chlorine ion formed [tex][Cl^-][/tex] = 2 × 8.279 M
= 16.558 M
∴
We know that:
[tex]\mathsf{Molarity = \dfrac{no \ of \ moles }{volume (mL)}}[/tex]
The number of moles of [tex][Cl^-][/tex] = [tex]16.558 \ mol.L^{-1} \times 7.903 \ mL \times \dfrac{1 \ L}{1000 \ mL}[/tex]
= 0.1309 mol
there is 3.5 g of fat in a granola bar. You determine the fat content to be 4.0 G in the lab. What is the percent error
Answer:
[tex]error = 4.0 - 3.5 = 0.5 \\ \\ percent \: error = \frac{0.5}{3.5} \times 100 \\ \\ = 14.29\% [/tex]
Write the complete ground-state electron configuration of I⁻.
Explanation:
here's the answer to your question
Iodine's neutral atom has the following electronic configuration: 1 s² 2 s² 2 p⁶ 3 s² 3 p⁶ 4 s² 3 d¹⁰ 4 p⁶ 5 s² 4 d¹⁰ 5 p⁵. The complete ground-state electron configuration of I⁻ is [Kr]5s² 4d¹⁰ 5p⁶.
What is electronic configuration?Electronic configuration is defined as each electron moves individually within an orbital while being surrounded by an average field produced by all other orbitals. The electron configuration is used to describe an atom's ground state orbitals, but it may also be used to depict an atom that has ionized into a cation or anion by making up for any lost or gained electrons in the orbitals after it.
The ground state electronic configuration is defined as the configuration of lower energy electrons surrounding an atom's nucleus. Ground state is defined as the least energetic state feasible for a physical system. It is important because permits us to determine the location of the excited electrons' departure and return when they release a photon.
Thus, iodine's neutral atom has the following electronic configuration: 1 s² 2 s² 2 p⁶ 3 s² 3 p⁶ 4 s² 3 d¹⁰ 4 p⁶ 5 s² 4 d¹⁰ 5 p⁵. The complete ground-state electron configuration of I⁻ is [Kr]5s² 4d¹⁰ 5p⁶.
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