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What Is the noble gas configuration for strontium (atomic number = 38)? Given: He (atomic number = 2) Ne (atomic number = 10) Ar (atomic number = 18) Kr (atomic number = 36) B В A [Kr 45Kr 5​

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.

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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Answer:

See detailed explanation.

Explanation:

Hello there!

In this case, for the given scenario, we will proceed as follows:

1. Here, we infer that the products are iron (III) hydroxide (precipitate) and calcium chloride:

[tex]3Ca(OH)_2+2FeCl_3\rightarrow 3CaCl_2+2Fe(OH)_3[/tex]

2. In this step we firstly calculate the moles of both reactants, by using their molar masses 74.093 and 162.2 g/mol respectively:

[tex]14.0gCa(OH)_2*\frac{1molCa(OH)_2}{74.093gCa(OH)_2}=0.189molCa(OH)_2 \\\\17.0gFeCl_3*\frac{1molFeCl_3}{162.2gFeCl_3}=0.105molFeCl_3[/tex]

Now, we calculate the moles of calcium hydroxide consumed by 0.105 moles of iron (III) chloride by using the 3:2 mole ratio between them:

[tex]0.105molFeCl_3*\frac{3molCa(OH)_2}{2molFeCl_3} =0.157molCa(OH)_2[/tex]

Thus, we infer that calcium hydroxide is in excess as 0.189 moles are available for it but just 0.157 moles react and therefore, iron (III) chloride is the limiting reactant.

3. Here, we use the moles of iron (III) chloride we've just computed, the 2:2 mole ratio with iron (III) hydroxide and its molar mass (106.867 g/mol) as shown below:

[tex]0.105molFeCl_3*\frac{2molFe(OH)_3}{2molFeCl_3} *\frac{106.867gFe(OH)_3}{1molFe(OH)_3} \\\\=11.2gFe(OH)_3[/tex]

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Answer:

See explanation

Explanation:

We must first write the equation of the reaction as follows;

C3H8 + 5O2 ----> 3CO2 + 4H2O

Now;

We obtain the number of moles of C3H8 = 132.33g/44g/mol = 3 moles

So;

1 mole of C3H8 yields 3 moles of CO2

3 moles of C3H8 yields 3 × 3/1 = 9 moles of CO2

We obtain the number of moles of oxygen = 384.00 g/32 g/mol = 12 moles

So;

5 moles of oxygen yields 3 moles of CO2

12 moles of oxygen yields 12 × 3/5 = 7.2 moles of CO2

We can now decide on the limiting reactant to be C3H8

Therefore;

Mass of CO2 produced = 9 moles of CO2 × 44 g/mol = 396 g of CO2

Again;

1 moles of C3H8 yields 4 moles of water

3 moles of C3H8 yields 3 × 4/1 = 12 moles of water

Hence;

Mass of water = 12 moles of water × 18 g/mol = 216 g of water

In order to obtain the percentage yield from the reaction, we have;

b) Actual yield = 269.34 g

Theoretical yield = 396 g

Therefore;

% yield = actual yield/theoretical yield × 100/1

Substituting values

% yield = 269.34 g /396 g × 100

% yield = 68%

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).

Answer:

I don't know ❤️

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

Answer:

The new temperature of the gas is 746 K.

Explanation:

Given that,

The volume of the gas, V₁ = 117 mL

Temperature, T₁ = 100°C = 373

Final volume of the gas, V₂ = 234 mL

We need to find the final temperature. The relation between temperature and volume is given by :

[tex]\dfrac{V_1}{V_2}=\dfrac{T_1}{T_2}\\\\T_2=\dfrac{T_1V_2}{V_1}\\\\T_2=\dfrac{373\times 234}{117}\\\\T_2=746\ K[/tex]

So, the new temperature of the gas is 746 K.

Answer:

True

Explanation:

The reaction between Diisopropyl ether and concentrated aqueous HI forms two initial organic products as shown in the image attached.

The hydrogen of the HI becomes attached to the oxygen in the ether leading to a cleavage of the C-O bond to yield the first compound. The I^- become attached to the other moiety in the original molecule to yield the second compound as shown in the image attached.

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.

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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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:

Answer:

NaCl, Na⁺,Cl⁻.

MgCl₂, Mg²⁺, Cl⁻.

CaO, Ca²⁺, O²⁻.

Li₃P, Li⁺, P³⁻.

Al₂S₃, Al³⁺, S²⁻.

Ca₃N₂, Ca²⁺, N³⁻.

FeCl₃, Fe³⁺, Cl⁻.

FeO, Fe²⁺, O²⁻.

Cu₂S, Cu⁺, S²⁻.

Cu₃N₂, Cu²⁺, N³⁻.

ZnO, Zn²⁺, O²⁻.

Ag₂S, Ag⁺, S²⁻.

K₂CO₃, K⁺, CO₃²⁻.

NaNO₃, Na⁺, NO₃⁻.

Ca(HCO₃)₂, Ca²⁺, HCO₃⁻.

Al(OH)₃, Al³⁺,OH⁻.

Li₃PO₄, Li⁺, PO₄³⁻.

K₂SO₄, K⁺, SO₄²⁻.

Explanation:

Sodium chloride. NaCl, formed by the cation Na⁺ and the anion Cl⁻.

Magnesium chloride. MgCl₂, formed by the cation Mg²⁺ and the anion Cl⁻.

Calcium oxide. CaO, formed by the cation Ca²⁺ and the anion O²⁻.

Lithium phosphide. Li₃P, formed by the cation Li⁺ and the anion P³⁻.

Aluminum sulfide. Al₂S₃, formed by the cation Al³⁺ and the anion S²⁻.

Calcium nitride. Ca₃N₂, formed by the cation Ca²⁺ and the anion N³⁻.

Iron(III)chloride. FeCl₃, formed by the cation Fe³⁺ and the anion Cl⁻.

Iron(II)oxide. FeO, formed by the cation Fe²⁺ and the anion O²⁻.

Copper(I)sulfide. Cu₂S, formed by the cation Cu⁺ and the anion S²⁻.

Copper(II)nitride. Cu₃N₂, formed by the cation Cu²⁺ and the anion N³⁻.

Zinc oxide. ZnO, formed by the cation Zn²⁺ and the anion O²⁻.

Silver sulfide. Ag₂S, formed by the cation Ag⁺ and the anion S²⁻.

Potassium carbonate. K₂CO₃, formed by the cation K⁺ and the anion CO₃²⁻.

Sodium nitrate. NaNO₃, formed by the cation Na⁺ and the anion NO₃⁻.

Calcium bicarbonate. Ca(HCO₃)₂, formed by the cation Ca²⁺ and the anion HCO₃⁻.

Aluminum hydroxide. Al(OH)₃, formed by the cation Al³⁺ and the anion OH⁻.

Lithium phosphate. Li₃PO₄, formed by the cation Li⁺ and the anion PO₄³⁻.

Potassium sulfate. K₂SO₄, formed by the cation K⁺ and the anion SO₄²⁻.

Explanation:

From the question given above, the following data were obtained:

Molarity of stock solution (M₁) = 1.2 M

Molarity of diluted solution (M₂) = 0.5 M

Volume of diluted solution (V₂) = 100 mL

The volume of stock solution needed can be obtained by using the dilution formula as illustrated below:

1.2 × V₁ = 0.5 × 100

1.2 × V₁ = 50

Divide both side by 1.2

V₁ = 50 / 1.2

Thus, 42 mL of the stock solution is needed.

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Answer:

They need 41.7 mL of the original stock solution.

Explanation:

We can use the following equation for dilutions:

Cc x Vc = Cd x Vd

Where Cc and Vc are the concentration and volume values in the concentrated condition, whereas Cd and Vd are the concentration and volume values in the diluted condition.

The concentrated solution is the original stock solution, and it has:

Cc = 1.2 M

The diluted solution must be:

Cd = 0.500 M

Vd = 100 mL

So, we have to calculate Vc. For this, we replace the data in the equation:

[tex]V_{c} = \frac{C_{d} V_{d} }{C_{c} } = \frac{(0.500 M)(100 mL)}{1.2 M} = 41.7 mL[/tex]

Therefore, 41.7 mL of 1.2 M original stock solution are required to make 100  mL of a diluted solution with a concentration of 0.500 M.

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]

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.

Answer:

39.29 mg

Explanation:

Step 1: Calculate the partial pressure of hydrogen

The pressure of the atmosphere is equal to the sum of the partial pressures of the water and the hydrogen. (1 Torr = 1 mmHg)

P = pH₂O + pH₂

pH₂ = P - pH₂O = 758.3 mmHg - 16.5 mmHg = 741.8 mmHg

We will convert it using the conversion factor 1 atm = 760.0 mmHg.

741.8 mmHg × 1 atm/760.0 mmHg = 0.9761 atm

Step 2: Convert 19.0 °C to Kelvin

We will use the following expression.

K = °C + 273.15 = 19.0 + 273.15 = 292.2 K

Step 3: Calculate the mass of hydrogen

First, we will calculate the moles of hydrogen using the ideal gas equation.

P × V = n × R × T

n = P × V / R × T

n = 0.9761 atm × 0.47910 L / (0.08206 atm.L/mol.K) × 292.2 K = 0.01950 mol

The molar mass of hydrogen is 2.015 g/mol. The mass of hydrogen is:

0.01950 mol × 2.015 g/mol = 0.03929 g = 39.29 mg

Answer:

cis

Explanation:

Cis isomers are formed when the substituents on the carbons of the double bond are on the same side of the double bond, forming a U. Trans isomers have substituents on opposite sides of the double bond, forming a sideways Z.

Answer:

a) K2[Ni(CN)4]

b) Na3[Ru(NH3)2(CO3)2]

c) Pt(NH3)2Cl2

Explanation:

Coordination compounds are named in accordance with IUPAC nomenclature.

According to this nomenclature, negative ligands end with the suffix ''ato'' while neutral ligands have no special ending.

The ions written outside the coordination sphere are counter ions. Given the names of the coordination compounds as written in the question, their formulas are provided above.

Answer:

Explanation:

We are to match each land resource to what they are being used for.

Clay             →→→  used to make pottery and tiles

iron ore       →→→  used to make steel

Salt              →→→   used as a flavoring in food

aggregate   →→→   used in construction

graphite       →→→ used to make batteries

Clay is a kind of soil particle that forms as a result of weathering processes. Examples include; pottery clays, glacial clays, and deep-sea clays e.t.c. The presence of one or more clay minerals, as well as variable quantities of organic and detrital components, characterizes all of them. Clay is usually sticky and moist when wet, but hard when dry. They are used in the making of tiles and potteries.

Iron ore:  The iron ore deposits are found in the Earth's crust's sedimentary rocks. They're made up of iron and oxygen that mix during the chemical process in marine and freshwater. iron ores are used to produce almost every iron and steel product that we use today.

Aggregate: are utilized in construction activities. It is a material used to mix cement, gypsum, bitumen, or lime to produce concrete in the construction industry.

Graphite: Graphite is a mineral that occurs in both igneous and metamorphic rocks. It is generally generated on the earth's surface when carbon is exposed to high temperatures and pressures. It is mainly used in the production of batteries and electrodes,

Explanation:

here's the answer to your question

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.

Answer:

Specific gravity of the saturated solution is 2

Explanation:

The specific gravity is defined as the ratio between density of a solution (In this case, saturated solution of potassium iodide, KI) and the density of water. Assuming density of water is 1:

Specific gravity  = Density

The density is the ratio between the mass of the solution and its volume.

In 100mL of water, the mass of KI that can be dissolved is:

100mL * (1g KI / 0.7mL) = 143g of KI

That means all the 100g of KI are dissolved (Mass solute)

As the volume of water is 100mL, the mass is 100g (Mass solvent)

The mass of the solution is 100g + 100g = 200g

In a volume of 100mL, the density of the solution is:

200g / 100mL = 2g/mL.

The specific gravity has no units, that means specific gravity of the saturated solution is 2

This question is incomplete, the complete question is;

Dimethyl ether, a useful organic solvent, is prepared in two steps.

In the first step, carbon dioxide and hydrogen react to form methanol and water:  

CO₂(g) + 3H₂(g)  →  CH₃OH(l) + H₂O(l)                ΔH₁ = -131.kJ

In the second step, methanol reacts to form dimethyl ether and water:

2CH₃OH(l)   →   CH₃OCH₃(g) + H₂O(l)                ΔH₂ = 8.kJ

Calculate the net change in enthalpy for the formation of one mole of dimethyl ether from carbon dioxide and hydrogen from these reactions. Round your answer to the nearest kJ.

Answer:

the net change in enthalpy for the formation of one mole of dimethyl ether is -254 kJ

Explanation:

Given the data in the question;

For the First Step;

CO₂(g) + 3H₂(g)  →  CH₃OH(l) + H₂O(l)                ΔH₁ = -131.kJ

For the First Step;

2CH₃OH(l)   →   CH₃OCH₃(g) + H₂O(l)                ΔH₂ = 8.kJ

Using Hess's Law of Constant Heat Summation;

" regardless of the multiple stages or steps of a reaction, the total enthalpy change for the reaction is the sum of all changes "

we multiply step 1 reaction by the coefficient of 2

2CO₂(g) + 2×3H₂(g)  →  2CH₃OH(l) + 2H₂O(l)                ΔH₁ = 2 × -131.kJ

we have

2CO₂(g) + 6H₂(g)  →  2CH₃OH(l) + 2H₂O(l)                ΔH₁ = -262 kJ

2CH₃OH(l)   →   CH₃OCH₃(g) + H₂O(l)                       ΔH₂ = 8 kJ

{ 2CH₃OH cancels 2CH₃OH }

Hence, we have;

2CO₂ + 6H₂  →  CH₃OCH₃(g) + 3H₂O(l)

So According to Hess's Law;

ΔH[tex]_{sum[/tex] = ΔH₁ + ΔH₂

we substitute

ΔH[tex]_{sum[/tex] = -262 kJ + 8 kJ

ΔH[tex]_{sum[/tex] = -254 kJ

Therefore, the net change in enthalpy for the formation of one mole of dimethyl ether is -254 kJ

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(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

Answer:

Alpha and beta particles originate from the nucleus, protection from radiation is important because if the radiation passes through the body it can damage cells. Exposure to radiation is often limited by increasing the distance from the radioactive source.

Explanation:  

Answer:

A. Reactants change into products

Dynamic equilibrium is showed at the point at which solid liquid and gas intersect.

At the point at which solid liquid and gas intersect represents a system that shows dynamic equilibrium. There is equal amount of reactants and products at the point of dynamic equilibrium because the transition of substances occur between the reactants and products at equal rates, means that there is no net change. Reactants and products are formed at the rate that no change occur in their concentration.

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