Heat of fusion (?Hfus) is used for calculations involving a phase change between solid and liquid, with no temperature change. For H2O, ?Hfus=6.02 kJ/mol.Specific heat capacity (C) is used for calculations that involve a temperature change, but no phase change. For liquid water, C=4.184 J/(g??C).Heat of vaporization (?Hvap) is used for calculations involving a phase change between liquid and gas, with no temperature change. For H2O, ?Hvap=40.7 kJ/mol.Part AHow much heat is required to raise the temperature of 92.0g of water from its melting point to its boiling point?Express your answer numerically in kilojoules.

Answer:

Option B is incorrect.

Explanation:

Observe the graph that has been attached along with the answer, that is the general representation of a substance approaching and during triple point phase.

Now, checking the options one by one

A. We verify that a substance can be in liquid form above its triple point, which in this case is 20°C. So this is a true statement.

B.This is a false statement because all the solid state is exhibited before the triple point temperature ie. 20°C. So, false.

C. By stable phase, we mean that a substance exhibits only one state. On increasing the temperature and lowering the pressure, we see the only state possible is the vapor phase. So this statement has to be true.

D. This is a true statement and it is a fact.

E. All the statements are not true so this is not the correct option.

Therefore, only option B is an incorrect option.

The statement that 'X can exist as a solid above 20°C' cannot be true. Given that the substance has a triple-point temperature of 20°C, it implies that the substance cannot exist as a solid above this temperature if it is a typical substance that expands upon heating.

The triple point of a substance is a specific temperature and pressure at which the three phases (solid, liquid, and gas) of the substance coexist in thermodynamic equilibrium. Given that substance X has a triple-point temperature of 20°C, it means that at this temperature and at a pressure of 2.0 atm, solid, liquid, and gaseous X can exist simultaneously.

From the choices given, statement B: 'X can exist as a solid above 20°C.' This cannot be true. If we are above the triple-point temperature and X is a typical substance that expands upon heating, it will not exist as a solid at temperatures above its triple point.

The other statements can be true depending on the actual phase diagram of X. The phase a substance is in, is determined both by its temperature and pressure. Therefore, without more information on substance X, it is difficult to conclusively verify or disprove those statements.

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Answer:

0.11

Explanation:

The given equilibrium reaction and the equilibrium concentrations are shown below as:-

[tex]\begin{matrix}&N_2&+&3H_2&\rightleftharpoons &2NH_3\\Concentration\ at\ equilibrium:-&1.5&&1.1&&0.47\end{matrix}[/tex]

The Kc of an equilibrium reaction measures relative amounts of the products and the reactants present during the equilibrium.

It is the ratio of the concentration of the products and the reactants each raised to their stoichiometric coefficients. The concentration of the liquid and the gaseous species does not change and thus is not written in the expression.

The equation is as follows:-

[tex]N_2_{(g)} +3H_2_{(g)}\rightarrow 2NH_3_{(g)}[/tex]

The expression for the Kc is:-

[tex]K_c=\frac{[NH_3]^2}{[N_2][H_2]^3}[/tex]

Thus, applying the values as:-

[tex]K_c=\frac{0.47^2}{1.5\times 1.1^3}=0.11[/tex]

The value of the equilibrium constant (Kc) for the reaction N2(g) + 3 H2(g) ⇌ 2 NH3(g) with the given equilibrium concentrations is approximately 0.11.

To determine the value of the equilibrium constant (Kc) for the reaction N2(g) + 3 H2(g) ⇌ 2 NH3(g), where the equilibrium concentrations are [N2]=1.5 M, [H2]=1.1 M, and [NH3]=0.47 M, we would use the equilibrium expression for Kc:

Kc = [NH3]2 / ([N2] × [H2]3)

Inserting the given equilibrium concentrations, we get:

Kc = (0.47 M)2 / (1.5 M × (1.1 M)3)

Kc = (0.2209 M2) / (1.5 M × 1.331 M3)

Kc = 0.2209 M2 / 1.9965 M4

Kc ≈ 0.11

The value of the equilibrium constant for this reaction is therefore approximately 0.11.

To calculate the mass of glucose in 100 mL of the final solution, we can use the equation C1V1 = C2V2.

To determine the number of grams of glucose in 100 mL of the final solution, we need to use the concept of dilution. The initial solution was prepared by dissolving 11.0 g of glucose in enough water to reach the 100 mL mark. Then, a 55.0 mL sample of this solution was diluted to 0.500 L. We can use the equation:

C1V1 = C2V2

Where C1 and V1 represent the concentration and volume of the initial solution, and C2 and V2 represent the concentration and volume of the final solution.

From this, we can calculate the concentration of the initial solution, and then use it to find the mass of glucose in 100 mL of the final solution.

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Answer:the thylakoid membrane usually has two photosystems that absorbs sunlight;the photosystems 11 and 1 .in the light reaction of photosynthesis,when light energy hits the photosystems,the electrons are boosted to higher energy and pass through the electron Transport chain by electron acceptor molecules until it is used in the formation of ATP and NADPH.

However when simazine is present,electrons can no longer be transferred from one electron acceptor to the other for the reduction of NADP+ to NADPH.

Also since no electron is being transferred,the PS II would not have a need to split water molecules ,which products are more electrons and protons which pass through the thylakoid lumen and creates a proton gradient needed for the production of ATP

Explanation:

Answer:

Mass of silver metal = 424 g

Explanation:

Data Given

Reactants = Nickle (Ni) metal and Silver nitrate (AgNO₃)

product = Ni(NO₃)₂ and Silver metal (Ag)

mass of nickle = 115 g

mass of silver = ?

Solution:

First write a balanced reaction

               Ni + 2AgNO₃ ----------> 2Ag + Ni(NO₃)₂

Now Look for the  number of moles of Nickle and Silver meta

                Ni + 2AgNO₃ ----------> 2Ag + Ni(NO₃)₂

               1 mol                              2 mol

So,

1 mole of nickle combine with silver nitrate and produce 2 moles of silver metal

Now Convert moles to mass for which we have to molar masses of Nickle and Silver metal

Molar mass of Nickle = 58.6 g/mol

Molar mass of Silver = 108 g/mol

                 Ni         +     2AgNO₃   ---------->    2Ag      +     Ni(NO₃)₂

               1 mol (58.6 g/mol)                      2 mol (108 g/mol)

                58.6 g                                                216 g

So,

58.6 g of Nickle metal produces 216 grams of silver metal.

Now

What mass of silver is produced from 115 g of Ni

Apply unity formula

                       58.6 g of Ni ≅ 216 g of Ag

                       115 g of Ni ≅ X g of Ag

By doing cross multiplication

                     Mass of Ag = 216 g x 115 g / 58.6 g

                     Mass of Ag = 424 g

Answer:  A. 1.5 g

Explanation:

[tex]N_2+3H_2\rightarrow 2NH_3[/tex]

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]

[tex]\text{Number of moles of nitrogen}=\frac{7.0g}{28g/mol}=0.25 moles[/tex]

According to stoichiometry:

1 mole of [tex]N_2[/tex] requires = 3 moles of [tex]H_2[/tex]

Thus 0.25 moles of [tex]N_2[/tex] will require =[tex]\frac{3}{1}\times 0.25=0.75moles[/tex] of [tex]H_2[/tex]

Mass of [tex]H_2[/tex] required =[tex]moles\times {\text {Molar mass}}=0.75mol\times 2g/mol=1.5g[/tex]

The minimum amount of [tex]H_2[/tex] in grams that would be required to completely react with this amount of [tex]N_2[/tex] is 1.5 grams.

Answer:

The correct answer is option A.

Explanation:

[tex]N_2 + 3 H_2\rightarrow 2 NH_3[/tex]

Moles of nitrogen gas = [tex]\frac{7.0 g}{28 g/mol}=0.25 mol[/tex]

According to reaction, 1 mole of nitrogen reacts with 3 moles of hydrogen gas.

Then 0.25 moles of nitrogen gas will react with:

[tex]\frac{3}{1}\times 0.25 mol=0.75 mol[/tex] of hydrogen gas.

Mass of 0.75 moles of hydrogen gas = 0.75 mol × 2 g/mol = 1.5 g

1.5 grams of hydrogen that would be required to completely react with this amount of nitrogen.

The balanced chemical equation for the combustion of TNT is 2C7H5N3O6 + 21O2 → 14CO2 + 10H2O + 6N2. The final temperature of the water and calorimeter is approximately 20.002 °C.

(a) To write and balance the chemical equation, we need to know the products formed when TNT is burned. Since it is an explosive, it likely forms carbon dioxide (CO2) and water (H2O). The balanced equation would be:

2C7H5N3O6 + 21O2 → 14CO2 + 10H2O + 6N2

(b) To calculate the final temperature of the water and calorimeter, we can use the equation:

q = mcΔT

Where q is the heat absorbed by the water and calorimeter, m is the mass of the water and calorimeter, c is the specific heat capacity of water (4.18 J/g·°C), and ΔT is the change in temperature. Rearranging the equation, we can solve for ΔT:

ΔT = q / (mc)

Substituting the given values, we have:

q = (3374 kJ/mol × 0.500 g) / (227.1 g/mol) = 7.45 kJ

m = 610 g + 420 g = 1030 g

c = 4.18 J/g·°C

ΔT = (7.45 kJ / 1030 g) / (4.18 J/g·°C) ≈ 0.0018 °C

Therefore, the final temperature of the water and calorimeter would be approximately 20.0 °C + 0.0018 °C = 20.002 °C.

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The chemical equation for the combustion of TNT is balanced as 2C7H5N3O6 + 21O2 → 14CO2 + 5H2O + 6N2 + 2O2. Using the given data and the equation, we can calculate the heat released during the reaction. The final temperature of the water and calorimeter is found to be 20.0031 °C.

(a) Write and balance the chemical equation:

The balanced chemical equation for the combustion of TNT (trinitrotoluene) is:

2C7H5N3O6 + 21O2 → 14CO2 + 5H2O + 6N2 + 2O2

(b) Using the given data and the equation above, we can calculate the heat released during the reaction. The molar mass of TNT is 227.13 g/mol. Therefore, the number of moles of TNT used in the reaction is:

0.500 g / (227.13 g/mol) = 0.00220 mol (rounded to 4 decimal places)

The heat of combustion of TNT is 3374 kJ/mol. Therefore, the heat released in the reaction is:

0.00220 mol × 3374 kJ/mol = 7.4228 kJ (rounded to 4 decimal places)

To find the final temperature of the water and calorimeter, we can use the equation:

q = mCΔT

where q is the heat released, m is the mass of the water and calorimeter, C is the heat capacity of the calorimeter, and ΔT is the change in temperature. Rearranging this equation, we have:

ΔT = q / (mC)

Substituting the values, we have:

ΔT = 7.4228 kJ / ((610 g + 420 g) × (4.18 J/g·°C)) = 0.0031 °C (rounded to 4 decimal places)

The final temperature of the water and calorimeter is therefore 20.0 °C + 0.0031 °C = 20.0031 °C.

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Complete question: An acid which could not be prepared by the reaction of an organic halide with cyanide ion followed by acid hydrolysis of the nitrile is: A) propanoic acid B) phenylacetic acid C) acetic acid D) (CH3)3CCO2H E) CH3(CH2)14CO2H

Answer: the correct option is option D ((CH3)3CCO2H).

Explanation: An acid which could not be prepared by the reaction of an organic halide with cyanide ion followed by acid hydrolysis of the nitrile is Pivalic acid. Pivalic acid with the molecular formula of CH3)3CCO2H is rather prepared by hydrocarboxylation of isobutene via the Koch reaction.

Final answer:

Phenylacetic acid cannot be prepared by the SN₂ reaction of an organic halide with cyanide ion followed by acid hydrolysis of the nitrile due to the reaction's failure with aryl halides. Instead, such reactions often lead to E₂ elimination or no reaction at all. Option B

Explanation:

The question pertains to the inability to synthesize certain carboxylic acids by the reaction of an organic halide with cyanide ion followed by acid hydrolysis of the resulting nitrile. Specifically, it's impossible to use this method to prepare an acid which can't be derived from a primary or secondary alkyl halide due to the limitations of the SN₂ reaction mechanism.

To synthesize acetic acid, for example, one may treat methyl bromide with cyanide ion to form acetonitrile, which upon hydrolysis yields acetic acid. However, phenylacetic acid, which requires a phenylacetyl halide precursor, cannot be formed this way because the reaction of cyanide ions with aryl halides does not proceed through an SN₂ mechanism. Instead, an E₂ elimination often occurs or, typically, there is no reaction due to the stability of the benzene ring and the partial double bond character of the C-X bond.

Therefore, phenylacetic acid could not be prepared using the reaction of an organic halide with cyanide ion followed by acid hydrolysis of the nitrile, while acetic acid and propanoic acid can be synthesized using this method. The correct answer to the student's question is phenylacetic acid.

Answer:

Larger ΔT, larger J, no change J/g, no change kJ/mol is the correct answer.

Explanation:

When one sees data in this question one tends to think that to be solved we need  to perform calculations, and that is not the case here.

What we need is to remembember what are properties which depend on quantities (extensive properties) and the concepts and formulas for heat.

Δ Hrxn = - Q cal

Q cal = m x c x ΔT where

m = mass of water in the calorimeter

c= specific heat of water, and

ΔT = change in temperature

ΔT is directly proportional to Q

ΔH rxn is an extensive quantity dependent on the amount of KNO₃, as is J/ reaction.

J/g , kJ/mol are intensive properties the moment they are defined as the heat per gram, and heat per mol released or absorbed.

mass of KNO₃   ⇒ doubles heat of reaction ,  doubles ΔT

Therefore,

c) Larger ΔT, larger J, no change J/g, no change kJ/mol is the correct answer.

Answer:

8

Explanation:

Oxidation:

[tex]Fe^{2+} -->Fe^{3+}+e^{-}[/tex]

Reduction:

[tex]Cr_{2}O_{7}^{2-}+6e^{-} -->2Cr^{3+}[/tex]

We have to equalise the number  of moles of electrons gained and lost in a redox reaction in order to get a balanced reaction.

Hence we have to multiply the oxidation reaction throughout by 6.

and adding the two half-reactions we obtain:

[tex]6Fe^{2+}+Cr_{2}O_{7}^{2-} -->6Fe^{3+}+2Cr^{3+}[/tex]

Still the total charge and number of oxygen is not balanced.

Since the reaction takes place in acidic conditions, we will add required number of H+ to the appropriate side to balance the charge and add half the amount of H2O to balance the hydrogen atoms.

We add 14 H+ on LHS and 7H2O on RHS to obtain:

[tex]6Fe^{2+}+Cr_{2}O_{7}^{2-}+14H^{+} -->6Fe^{3+}+2Cr^{3+}+7H_{2}O[/tex]

Sum of coefficients of product cations = 6+2 = 8

Answer:

D. All of them would have the same kinetic energy

Explanation:

The expression for the kinetic energy of the gas is:-

[tex]K.E.=\frac{3}{2}\times K\times T[/tex]

k is Boltzmann's constant = [tex]1.38\times 10^{-23}\ J/K[/tex]

T is the temperature

Since, kinetic energy depends only on the temperature. Thus, at same temperature, at 300 K, all the gases which are [tex]N_2,\ NH_3\ and\ Ar[/tex] will posses same value of kinetic energy.

Answer:

(a) R = k [A]¹ [B]²

(b) The given chemical reaction is a third order reaction

(c)

Explanation:

Rate law is the equation that defines the rate of a given chemical reaction and depends on the concentration of the reactants, raised to the power partial orders of reaction.

The overall order of the given chemical reaction is equal to the sum of partial orders of reaction.

Given: Partial order of reaction of reactant A: a = 1,

Partial order of reaction of reactant B: b = 2

(a) Therefore, the rate law equation of the given reaction is given by

R = k [A]ᵃ [B]ᵇ = k [A]¹ [B]²                          ....equation 1

here k is the rate constant

(b) The overall order of the reaction = a + b = 1 + 2 = 3

Therefore, the given chemical reaction is a third order reaction.

(c) Since, the rate of a reaction is directly proportional to the reactant concentration. Therefore, when

i. [A] is doubled and [B] held constant.

⇒ Concentration of reactant A becomes 2[A]

The new rate law is:

R' = k {2[A]}¹ [B]² = 2 {k [A]¹ [B]²}                   ....equation 2

Comparing equations 1 and 2, we get

R' = 2 R  ⇒ the reaction rate doubles.

ii. [A] is held constant and [B] is doubled.

⇒ Concentration of reactant B becomes 2[B]

The new rate law is:

R' = k [A]¹ {2[B]}² = 4 {k [A]¹ [B]²}                   ....equation 3

Comparing equations 1 and 3, we get

R' = 4 R  ⇒ the reaction rate becomes 4 times.

iii. [A] is tripled and [B] is doubled

⇒ Concentration of reactant A becomes 3[A], Concentration of reactant B becomes 2[B]

The new rate law is:

R' = k {3[A]}¹ {2[B]}² = 12 {k [A]¹ [B]²}                   ....equation 4

Comparing equations 1 and 4, we get

R' = 12 R  ⇒ the reaction rate becomes 12 times.

iv. [A] is doubled and [B] is halved.

⇒ Concentration of reactant A becomes 2[A], Concentration of reactant B becomes 1/2 [B]

The new rate law is:

R' = k {2[A]}¹ {1/2[B]²} = 1/2 {k [A]¹ [B]²}                   ....equation 5

Comparing equations 1 and 5, we get

R' = 1/2 R  ⇒ the reaction rate becomes half.

Answer:

It can be produced, 12 moles of MgO.

Option B

Explanation:

2 KClO₃ → 3O₂ + 2 KCl

Ratio in this reaction is 2:3

In the begining, I make 3 moles of oxygen, that came from 2moles of chlorate. If I have 4 moles of salt, let's make a rule of three.

2 moles of salt ___ make __3 moles of O₂

4 moles of salt ___ make (4 .3) /2 = 6 moles of O₂

2 Mg + O2 → 2 MgO.

From 1 mol of oxygen, I can make 2 moles of oxygen.

If I have 6 moles, I would make the double, though.

Answer:

(a) Hydrogen bonding

(b) Dispersion forces

(c) Ion-dipole forces

(d) Dipole-dipole forces

Ion-dipole forces (c) are the strongest of the 4 interactions while dispersion forces are the weakest (b).

Explanation:

The picture is missing but I think is the one that I'm uploading.

Picture (a)

HF is a polar molecule with a high difference in electronegativity between H and F. As a consequence, the force between HF molecules is Hydrogen bonding.

Picture (b)

In picture (b) we have F₂ molecules, which are nonpolar due to their atoms having the same electronegativity. The forces between nonpolar molecules are dispersion forces.

Picture (c)

Na⁺ is an ion and H₂O a dipole. Therefore, they experience ion-dipole forces.

Picture (d)

SO₂ molecules are polar, that is, they form dipoles and experience dipole-dipole forces.

Ion-dipole forces (c) are the strongest of the 4 interactions while dispersion forces are the weakest (b).

Final answer:

The strength of intermolecular forces varies greatly, affecting physical properties of substances. London dispersion forces are the weakest followed by dipole-dipole interactions, hydrogen bonding, and ion-dipole forces being the strongest.

Explanation:

Understanding the nature and strength of intermolecular forces is essential to explain many physical properties of substances, such as boiling points, melting points, and solubilities. Intermolecular forces can be categorized into several types, including London dispersion forces, dipole-dipole interactions, hydrogen bonding, and ion-dipole forces.

London Dispersion Forces

London dispersion forces are the weakest intermolecular force and occur between all molecules, polar or nonpolar. They result from the instantaneous distribution of electrons in one molecule inducing a dipole in a neighboring molecule. These forces are significant in molecules with a high number of electrons and become stronger as molecular size increases.

Dipole-Dipole Interactions

Dipole-dipole interactions occur between polar molecules where positive ends of molecules are attracted to negative ends of other molecules. While stronger than London dispersion forces, they are not as strong as hydrogen bonds.

Hydrogen Bonding

Hydrogen bonding is a special type of dipole-dipole interaction that occurs when a hydrogen atom bonded to a highly electronegative atom, like oxygen, nitrogen, or fluorine, is attracted to an electronegative atom in another molecule. Hydrogen bonding is significantly stronger than both London dispersion forces and general dipole-dipole interactions, contributing to the unique properties of water and other substances.

Ion-Dipole Forces

Ion-dipole forces occur between an ion and a polar molecule and are significant in mixtures of ionic substances with polar solvents. These are stronger than hydrogen bonds and are crucial for the solubility of ionic compounds in water.

To rank these interactions from weakest to strongest: London dispersion forces, dipole-dipole interactions, hydrogen bonding, and ion-dipole forces.

Answer:

b) add 130 g of NaCH₃CO₂ to 100 mL of H₂O at 80 °C while stirring until all the solid dissolves, then let the solution cool to room temperature.

Explanation:

The solubility of NaCH₃CO₂ in water is ~1.23 g/mL. This means that at room temperature, we can dissolve 1.23 g of solute in 1 mL of water (solvent).

What would be the best method for preparing a supersaturated NaCH₃CO₂ solution?

a) add 130 g of NaCH₃CO₂ to 100 mL of H₂O at room temperature while stirring until all the solid dissolves. NO. At room temperature, in 100 mL of H₂O can only be dissolved 123 g of solute. If we add 130 g of solute, 123 g will dissolve and the rest (7 g) will precipitate. The resulting solution will be saturated.

b) add 130 g of NaCH₃CO₂ to 100 mL of H₂O at 80 °C while stirring until all the solid dissolves, then let the solution cool to room temperature. YES. The solubility of NaCH₃CO₂ at 80 °C is ~1.50g/mL. If we add 130 g of solute at 80 °C and let it slowly cool (and without any perturbation), the resulting solution at room temperature will be supersaturated.

c) add 1.23 g of NaCH₃CO₂ to 200 mL of H₂O at 80 °C while stirring until all the solid dissolves, then let the solution cool to room temperature. NO. If we add 1.23 g of solute to 200 mL of water, the resulting solution will have a concentration of 1.23 g/200 mL = 0.00615 g/mL, which represents an unsaturated solution.

Answer:

C

Explanation:

The detailed solution is found in the image attached. It is necessary to note that the oxidation half equation is multiplied by three to balance electron gain and loss. This is adequately shown in the image below. Inferences are only drawn from balanced redox reaction equation hence the first step is to balance the redox reaction equation.

Answer:

6.0 moles of H₂S are produced.

Explanation:

Let's consider the following balanced equation.

2 CH₄ + S₈ → 2 CS₂ +  4 H₂S

The molar ratio of S₈ to H₂S is 1 mol S₈: 4 mol H₂S. The moles of H₂S produced when 1.5 mol of S₈ react are:

1.5 mol S₈ × (4 mol H₂S/ 1 mol S₈) = 6.0 mol H₂S

6.0 moles of H₂S are produced, when 1.5 mol of S₈ react.

Final answer:

Using stoichiometry based on the balanced chemical equation 2CH₄ + S₈ → 2CS₂ + 4H₂S, we find that 1.5 mol of S₈ will produce 6 mol of H₂S.

Explanation:

The student asked how to calculate the moles of H₂S produced when 1.5 mol S₈ is used given the equation 2CH₄ + S₈ → 2CS₂ + 4H₂S. To solve this, we use stoichiometry to find the mole ratio between S₈ and H₂S. According to the balanced chemical equation, 1 mol of S₈ produces 4 mols of H₂S. Therefore, if 1.5 mols of S₈ are used, we simply multiply this by the stoichiometric ratio (1.5 mol S₈ × 4 mol H₂S/mol S₈) to find the moles of H₂S produced.

The calculation will be as follows:
1.5 mol S₈ × 4 mol H₂S/mol S₈ = 6 mol H₂S.

Answer:

A) 6.48 g of OF₂ at the anode.

Explanation:

The gas OF₂ can be obtained through the oxidation of F⁻ (inverse reaction of the reduction presented). The standard potential of the oxidation is the opposite of the standard potential of the reduction.

H₂O(l) + 2 F⁻(aq) → OF₂(g) + 2 H⁺(aq) + 4 e⁻    E° = -2.15 V

Oxidation takes place in the anode.

We can establish the following relations:

The mass of OF₂ produced when 0.480 F pass through an aqueous KF solution is:

[tex]0.480F.\frac{1mole^{-} }{1F} .\frac{1molOF_{2}}{4mole^{-} } .\frac{54.0gOF_{2}}{1molOF_{2}} =6.48gOF_{2}[/tex]

Using the data provided here, the mass of the compound produced is 6.48 g of OF2

Electrolysis refers to the breaking u p of a molecule by the passage of direct current through it. The equation of the reaction is; H₂O(l) + 2 F⁻(aq) → OF₂(g) + 2 H⁺(aq) + 4 e⁻    E° = -2.15 V.

Now;

1 mole of OF2 is realeased by the passage of 4 F of electricity

x moles of OF2 is produced by the passage of  0.480F

x = 0.12 moles

Mass of OF2 = 0.12 moles * 54 g/mol = 6.48 g of OF2

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Answer:

Oxyanion

Explanation:

Nomenclature is a set of rules designed to name chemical compounds, in this rules the use of suffix is common to determine the functional group of a molecule, in the case of "Ethyl propanoate" the -ate suffix is used to design an oxyanion like sulfate SO42- or nitrate NO3-.

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Answer:

Receptor

Explanation:

   Neurotransmitters are defined as chemical messengers that carry, stimulate and balance signals between neurons, or nerve cells and other cells in the body.

  After release, the neurotransmitter crosses the synaptic gap and binds to the receptor site on the other neuron, stimulating or inhibiting the receptor neuron depending on what the neurotransmitter is. Neurotransmitters act as a key and the receptor site acts as a block. It takes the right key to open specific locks. If the neurotransmitter is able to function at the receptor site, it will cause changes in the recipient cell.

 The "first-class" neurotransmitter receptors are ligand-activated ion channels, also known as ionotropic receptors. They undergo a change in shape when the neurotransmitter turns on, causing the channel to open. This can be an excitatory or inhibitory effect, depending on the ions that can pass through the channels and their concentrations inside and outside the cell.  Ligand-activated ion channels are large protein complexes. They have certain regions that are binding sites for neurotransmitters, as well as membrane segments to make up the channel.

The neurotransmitter serves as the ligand or signal molecule in the signal pathway when a neuron responds by opening gated ion channels.

Neurons play a crucial role in transmitting information within the nervous system, and the interaction between neurotransmitters and their corresponding receptors is a fundamental process in this communication. When a neuron responds to a particular neurotransmitter, it does so by recognizing the neurotransmitter as a ligand or signal molecule in the intricate signaling pathway that underlies neural function.

These neurotransmitters, which can be diverse chemical compounds like dopamine, serotonin, or acetylcholine, possess the remarkable ability to bind selectively to specific receptors on the surface of the neuron. This binding event is akin to a key fitting into a lock, and it initiates a cascade of events within the neuron.

One critical outcome of neurotransmitter binding is the opening of gated ion channels located on the neuron's membrane. These ion channels act as molecular gates, regulating the flow of ions (such as sodium, potassium, calcium, or chloride) into or out of the cell. This influx or efflux of ions results in the generation of an electrical signal, known as the action potential.

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Answer: The half reaction occurring at anode is [tex]2I^-(aq.)\rightarrow I_2(s)+2e^-[/tex]

Explanation:

The substance having highest positive [tex]E^o[/tex] potential will always get reduced and will undergo reduction reaction.

For the given chemical equation:

[tex]2ClO_2(g)+2I^-(aq)\rightarrow 2ClO^{-2}(aq.)+I_2(s)[/tex]

The half reaction follows:

Oxidation half reaction:  [tex]2I^-(aq.)\rightarrow I_2(s)+2e^-;E^o_{I_2/I^-}=0.53V[/tex]

Reduction half reaction:  [tex]ClO_2(g)+e^-\rightarrow ClO_2^-(aq.);E^o_{ClO_2/ClO_2^-}=+0.954V[/tex]    ( × 2 )

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

Hence, the half reaction occurring at anode is [tex]2I^-(aq.)\rightarrow I_2(s)+2e^-[/tex]

The half reaction occurring at anode is:

[tex]2I^-(aq)---- > I_2(s)+2e^-[/tex]

The substance having highest positive  potential will always get reduced and will undergo reduction reaction.

Balanced chemical equation:

[tex]2ClO_2(g)+2I^-(aq)----- > 2ClO^{2-}(aq)+I_2(s)[/tex]

The half reaction follows:

Oxidation half reaction:  [tex]2I^-(aq)---- > I_2(s)+2e^-[/tex] , Reduction potential is 0.53V

Reduction half reaction:  [tex]ClO_2(g)+e^----- > ClO_2^-[/tex]   ( × 2 ), Oxidation potential is +0.954 V

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

Hence, the half reaction occurring at anode is :

[tex]2I^-(aq)---- > I_2(s)+2e^-[/tex]

Find more information about Reduction potential here:

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Answer: b)298 K converts to 24.9 °C, so the water would remain in its liquid state

Explanation: Kelvin is an absolute temperature scale, that means that 0K (zero Kelvin) is the lowest temperature possible and compare to Celsius scale the change of 1 degree is the same, but 0°C is the same as 273,1K

So, in order to convert Kelvin to Celsius you have to subtract 273,1

In this case, 298K - 273,1 = 24,9°C

This temperature is room temperature, so water is in liquid state.

Answer:

79.74*10^6 Pa

Explanation:

Based on the parameters provided, we have:

ε[tex]_{t}[/tex] = ln([tex]l_{f}/l_{i}[/tex])

Where initial gauge length = 3 cm and the final gauge length is 3.5 cm. Therefore:

ε[tex]_{t}[/tex] = ln(3.5/3) = ln(1.167) = 0.154

Similarly,

σ[tex]_{E}[/tex] = F/[3.142*(di^2)/4]

Where σ[tex]_{E}[/tex] = 120*10^6 Pa and di = 1 cm = 0.01 m

Therefore,

F = 120*10^6 * 3.142*(0.01^2)/4 = 9426 N

σ[tex]_{t}[/tex] =  F/[3.142*(df^2)/4 = 9426/[3.142*(0.00926^2)/4 = 9426/6.74*10^-5 = 139.95*10^6 Pa

σ[tex]_{t}[/tex] = k*ε[tex]_{t} ^{0.5}[/tex] = 139.95*10^6

k = 139.95*10^6/(0.154)^0.5 = 356.63*10^6 Pa

Therefore, when ε[tex]_{t}[/tex] = 0.05 cm/cm

σ[tex]_{t}[/tex] = 356.63*10^6 (0.05)^0.5 = 79.74*10^6 Pa

The true stress when the true strain is 0.05 cm/cm is approximately 122.97 MPa.

The true stress at a true strain of 0.05 cm/cm for the Cu-30% Zn alloy tensile bar can be calculated using the relationship between true stress and true strain, which is given by the equation:

[tex]\[ \sigma_{true} = \sigma_{eng}(1 + \epsilon_{true}) \][/tex]

where [tex]\( \sigma_{true} \)[/tex] is the true stress, [tex]\( \sigma_{eng} \)[/tex] is the engineering stress at the moment of fracture, and [tex]\( \epsilon_{true} \)[/tex] is the true strain.

Given that the engineering stress at failure is 120 MPa, we can calculate the true stress at the given true strain of 0.05 cm/cm as follows:

Now, we can calculate the true stress using the given strain hardening coefficient [tex]\( n = 0.50 \)[/tex] and the true strain [tex]\( \epsilon_{true} = 0.05 \)[/tex]:

[tex]\[ \sigma_{true} = \sigma_{eng}(1 + \epsilon_{true})^n \][/tex]

[tex]\[ \sigma_{true} = 120 \text{ MPa} \times (1 + 0.05)^{0.50} \][/tex]

[tex]\[ \sigma_{true} = 120 \text{ MPa} \times (1.05)^{0.50} \][/tex]

[tex]\[ \sigma_{true} = 120 \text{ MPa} \times 1.0247 \][/tex]

[tex]\[ \sigma_{true} = 122.97 \text{ MPa} \][/tex]

Answer:

The answer is a): Interacting with other students on the discussion boards.

Explanation:

In a survey of online students who identified some tips that made them successful in online courses, and virtual university programs which targeted adults, 52.6 % of them believed that the greatest benefit they derived was due to their interaction with other students (especially classmates) on online discussion boards.

The question is related to the benefits of online learning in Education, specifically for college students. Without additional options, we draw on general benefits reported by students, including discussion board interactions, diverse content delivery methods, and online community support as valuable aspects of online education.

The subject of this question is Education, specifically focusing on the benefits of online learning environments for college students. To answer the student's question about what 52.6 percent of online students believed they benefitted most from, we need additional context or options other than those provided in the question. However, past surveys and research can offer insight into common benefits reported by students in online learning settings, such as:

These activities not only help in understanding the subject matter but also in building a sense of community and improving technical skills. The prevalence of technology in education has made these aspects more accessible than ever before, changing the landscape of how students engage with peers and course materials.

Answer:

Attached image of the Lewis structure.

Explanation:

To draw the Lewis structure of SiH₄, we need to consider the octet rule: atoms gain, lose or share electrons to have 8 electrons in their valence shell. H is an exception to this rule because it is completed with 2 electrons (duet).

Si is a semimetal and H a nonmetal, and they form covalent bonds, that is, they share pairs of electrons to be complete.

Si has 4 valence electrons, so it forms 4 covalent bonds to reach the octet.

Each H has 1 valence electron, so each H forms 1 covalent bond to reach the duet.

The resulting structure can be seen in the attached picture.

To draw the Lewis structure of SiH4, the total number of valence electrons in the molecule is determined. The central atom is Silicon (Si) and each Hydrogen atom is bonded to Silicon with a single bond. The Lewis structure is represented by connecting the atoms with single bonds.

To draw the Lewis structure of SiH4, we need to determine the total number of valence electrons in the molecule. Silicon (Si) is in group 14 of the periodic table and has 4 valence electrons. Hydrogen (H) is in group 1 and has 1 valence electron. Since there are 4 hydrogen atoms, we have a total of 4 valence electrons. Therefore, the total number of valence electrons in SiH4 is 4 + 4 = 8.

In the Lewis structure, the central atom is silicon. Each hydrogen atom will be bonded to silicon with a single bond. Since each hydrogen atom needs 2 electrons to complete its outer shell, the silicon atom will share its 4 valence electrons with the 4 hydrogen atoms, resulting in 4 single bonds.

The Lewis structure of SiH4 can be represented as follows:

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Answer:

350 g dye

0.705 mol

2.9 × 10⁴ L

Explanation:

The lethal dose 50 (LD50) for the dye is 5000 mg dye/ 1 kg body weight. The amount of dye that would be needed to reach the LD50 of a 70 kg person is:

70 kg body weight × (5000 mg dye/ 1 kg body weight) = 3.5 × 10⁵ mg dye = 350 g dye

The molar mass of the dye is 496.42 g/mol. The moles represented by 350 g are:

350 g × (1 mol / 496.42 g) = 0.705 mol

The concentration of Red #40 dye in a sports drink is around 12 mg/L. The volume of drink required to achieve this mass of the dye is:

3.5 × 10⁵ mg × (1 L / 12 mg) = 2.9 × 10⁴ L

A 70 kg person would need to eat 350,000 mg of Red #40 dye to reach the LD50. This is equivalent to 704.01 mol of Red #40 dye. The number of liters of sports drink needed depends on the concentration of Red #40 dye in the drink.

To determine the amount of dye a 70 kg person would need to eat to reach the LD50, we can use the information provided. The LD50 for Red #40 is 5000 mg dye/1 kg body weight. So for a 70 kg person, the LD50 would be:

(5000 mg dye/1 kg body weight) * 70 kg = 350,000 mg of Red #40 dye.

To calculate the number of moles of dye, we divide the mass of dye by its molar mass. The molar mass of Red #40 is 496.42 g/mol, so:

(350,000 mg) / (496.42 g/mol) = 704.01 mol of Red #40 dye.

Finally, to determine the number of liters of sports drink needed to reach this LD50, we need to know the concentration of Red #40 dye in the sports drink.

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As the air temperature decreases, the relative humidity generally increases because the air's capacity to hold water vapor also reduces. On evenings when the temperature reduces enough to reach the dew point, fog might form as a result. This relationship between temperature and humidity also affects the likelihood of condensation and the range of temperature fluctuations in different regions.

The term humidity often refers to relative humidity, which indicates how much water vapor is present in the air compared to the maximum possible. This maximum relies on the air's temperature: as the air temperature decreases, the amount of water vapor the air can hold also decreases. Consequently, if the quantity of water vapor stays the same, the relative humidity will increase.

In the evening, when the air temperature declines, the relative humidity usually rises, sometimes to the point of reaching the dew point, the temperature at which the relative humidity is 100% and fog can form due to the condensation of small water droplets that stay in suspension. When the dew point is below 0°C, a greater possibility of freezing temperatures exists, which is a concern for farmers. In arid regions, the low humidity means low dew-point temperatures, hence, condensation is unlikely, resulting in a larger range of temperature fluctuations compared to regions with higher humidity.

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To calculate the volume of potassium permanganate stock solution that the chemist should pour out, we use the formula C1V1 = C2V2. Plugging in the given values, the chemist should pour out 3.00 L of the potassium permanganate stock solution.

To calculate the volume of potassium permanganate stock solution that the chemist should pour out, we can use the formula:

C1V1 = C2V2

where C1 and C2 are the concentrations of the stock and working solutions respectively, and V1 and V2 are the volumes of the stock and working solutions. Rearranging the formula, we can solve for V1:

V1 = (C2 * V2) / C1

Plugging in the given values, we have:

V1 = [tex](0.250 M * 3.00 L) / 0.250 M = 3.00 L[/tex]

Therefore, the chemist should pour out 3.00 L of the potassium permanganate stock solution.

Answer:

On the attached picture.

Explanation:

Hello,

In this case, one must remember that transamination is a biochemical reaction that transfers an amino group to the carbonyl group of a ketoacid to form a new amino acid and a new ketoacid.

For this example, on the attached picture you will see the corresponding chemical reaction in which to the initial ketoacid, both the amino and an additional hydrogen are transferred from the amino acid of side chain X to form the requested α-ketoacid of side chain X.

Best regards.

Answer:

D

Explanation:

In considering the half cell reactions in electrochemical cells, we consider the standard electrode potential of the two half cells. The more negative electrode potential will be the anode and the less negative electrode potential will be the cathode. The electrode potentials of Ni2+(aq)/Ni(s) is -0.25V while that of Zn2+(aq)/Zn(s) is -0.76.

Hence the selected option is the cathodic half reaction equation

In the given galvanic cell reaction, the reduction half-reaction Ni2+(aq) + 2 e- → Ni(s) occurs at the cathode. Nickel ions are gaining electrons to become solid Nickel.

In a galvanic cell, the half-reaction that occurs at the cathode is the reduction reaction. Reduction is a chemical process where a species gains electrons. This stands as the acronym OIL RIG, which stands for Oxidation Is Loss (of electrons), Reduction Is Gain (of electrons). Given the provided shorthand notation for the galvanic cell, Option D: Ni2+(aq) + 2 e- → Ni(s) would be the half-reaction occurring at the cathode. This is because Nickel ions (Ni2+) are gaining electrons, thus going through a reduction process, to form solid Nickel (Ni).

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