The vapor pressure of Substance X is measured at several temperatures:temperature (C) vapor pressure (atm)34 0.23644 0.29254 0.355Use this information to calculate the enthalpy of vaporization of X. Round your answer to 2 significant digits. Be sure your answer contains a correct unit symbol. Clears your work. Undoes your last action. Provides information about entering answers.

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:

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.

The intermolecular forces between a formaldehyde molecule and a hydrogen sulfide molecule are dipole-dipole forces and London dispersion forces. Both forces occur due to the polar nature of these molecules and the temporary creation of dipoles.

The types of intermolecular forces that act between a formaldehyde (H2CO) molecule and a hydrogen sulfide (H2S) molecule are dipole-dipole forces and London dispersion forces. In this case, both molecules are polar, meaning they have uneven charge distribution. Therefore, the positive end of one molecule is attracted to the negative end of the other, resulting in dipole-dipole forces. Both molecules, being non-ideal gases, exhibit London dispersion forces, which are weak, temporary attractions occurring when the electrons in two adjacent atoms occupy positions that make the atoms form temporary dipoles.

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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.

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:

h = 0.346 m

Explanation:

mercury barometer:

∴ Pa = 0.455 atm = 46102.875 Pa = 46102.875 Kg/ms²

∴ ρ = 13600 Kg/m³

∴ g = 9.80 m/s²

⇒ h = (46102.875 Kg/ms²) / (13600 Kg/m³ )(9.80 m/s²)

⇒ h = 0.346 m

The height of the mercury column in a barometer on top of a mountain, where the atmospheric pressure is 0.455 atm, would be around 33 centimeters.

To compute the height of the mercury column, we use the formula h = P/(gρ). Here, h represents height, P is the atmospheric pressure, g is the acceleration due to gravity and ρ is the density of the fluid (in this case, mercury). Given that P = 0.455 atm, ρ = 13.6*10^3 kg/m^3, and the standard value for g is approximated as 9.81 m/s^2, we then convert atmospheres to pascal by multiplying by 101325, since 1 atm = 101325 Pa.

This gives us P = 0.455 atm * 101325 Pa/atm = 46107.875 Pa. Using these values in our formula, we get h = 46107.875 Pa /(9.81 m/s^2 x 13.6*10^3 kg/m^3) which simplifies to approximately 0.33 meters or 33 centimeters. Hence, on top of a high mountain, where the atmospheric pressure is 0.455 atm, the height of the mercury column would be around 33 cm.

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Answer:

1) ΔG°r(298 K) = - 28.619 KJ/mol

2) ΔG°r will decrease with decreasing temperature

Explanation:

1) ΔG°r = ∑νiΔG°f,i

⇒ ΔG°r(298 K) = ΔG°CO2(g) + ΔG°H2(g) - ΔG°H2O(g) - ΔG°CO(g)

from literature, T = 298 K:

∴ ΔG°CO2(g) = - 394.359 KJ/mol

∴ ΔG°CO(g) = - 137.152 KJ/mol

∴ ΔG°H2(g) = 0 KJ/mol........pure substance

∴ ΔG°H2O(g) = - 228.588 KJ/mol

⇒ ΔG°r(298 K) = - 394.359 KJ/mol + 0 KJ/mol - ( - 228.588 KJ/mol ) - ( - 137.152 KJ7mol )

⇒ ΔG°r(298 K) = - 28.619 KJ/mol

2) K = e∧(-ΔG°/RT)

∴ R = 8.314 E-3 KJ/K.mol

∴ T = 298 K

⇒ K = e∧(-28.619/(8.314 E-3)(298) = 9.624 E-6

⇒ ΔG°r = - RTLnK

If T (↓) ⇒ ΔG°r (↓)

assuming T = 200 K

⇒ ΔG°r(200 K) = - (8.314 E-3)(200)Ln(9.624E-3)

⇒ ΔG°r (200K) = - 19.207 KJ/mol < ΔG°r(298 K) = - 28.619 KJ/mol

Final answer:

The ΔG° at 298 K for the given reaction is 130.0 kJ/mol. Lowering the temperature will decrease ΔG° because the reaction is exothermic.

Explanation:

Delta G (ΔG°) at 298 K can be calculated using the equation ΔG° = ΔH° - TΔS°, where ΔH° is the standard enthalpy change, T is the temperature in Kelvin, and ΔS° is the standard entropy change of the reaction. Substitute the given values into the equation to obtain ΔG°. For the given reaction, CO(g) + H2O(g) → H2(g) + CO2(g), ΔG° = 177.8 kJ/mol - (298 K * 0.1605 kJ/K mol) = 130.0 kJ/mol.

The effect of lowering the temperature on ΔG° can be determined by understanding how changes in temperature affect the equilibrium constant (K) of the reaction. According to Le Chatelier's principle, if the reaction is exothermic (negative ΔH°), a decrease in temperature will cause the equilibrium to shift towards the products, leading to a decrease in ΔG°. On the other hand, if the reaction is endothermic (positive ΔH°), a decrease in temperature will cause the equilibrium to shift towards the reactants, leading to an increase in ΔG°. In this case, since the reaction is exothermic, ΔG° will decrease with decreasing temperature.

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:

OH⁻ and H₂O

Explanation:

From the equation given in question ,

NH₃ (aq) + H₂O (l) ⇌ NH₄⁺ (aq) + OH⁻ (aq)

Acids are the species which furnish hydrogen ions in the solution or is capable of forming bonds with electron pair species as they are electron deficient species.

When an acid donates a proton, it changes into a base which is known as its conjugate base.

Bases are the species which furnish hydroxide ions in the solution or is capable of forming bonds with electron deficient species as they are electron rich species. When a base accepts a proton, it changes into a acid which is known as its conjugate acid.

The acid and the base which is only differ by absence or presence of the proton are known as acid conjugate base pair.

Hence , from the equation given in the question ,

The conjugate acid - base pair is -

OH⁻ and H₂O

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:

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:

D) Oxygen is oxidized and hydrogen is reduced.

Explanation:

In the electrolysis of water, an electric current passes through an electrolytic solution (e.g. aqueous NaCl), leading to the following redox reaction.

H₂O(l) → H₂(g) + 1/2 O₂(g)

The corresponding half-reactions are:

Reduction:  2 H₂O(l) + 2 e⁻ → H₂(g) + 2 OH⁻

Oxidation:  2 H₂O(l) → O₂(g) + 4 H⁺(aq) + 4 e⁻

As we can see, H in water is reduced (its oxidation number decreases from 1 to 0), while O in water is oxidized (its oxidation number increases from -2 to 0).

The reaction would progress faster and the activation energy would be lowered when the enzyme gets added.

Enzymes are proteins that basically speed up the chemical reaction without being used. Enzymes are usually specific for a particular substrate. The substrate in the reaction bind to the active site of the enzyme which is present on the surface of the enzymes forming the enzyme-substrate complex.

Performing the enzyme-substrate complex the enzyme changes the shape slightly so that the substrate can fit tightly to its active site. Then this enzyme-substrate Complex undergoes a reaction to form a product. Enzymes lower the activation energy of a reaction i.e the required amount of energy needed for a reaction to occur.They do this by binding to a substrate and holding it in a way that allows the reaction to happen more efficiently.

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:

The amount of energy required to transport hydrogen ions from a cell into the stomach is 37.26KJ/mol.

Explanation:

The free change for the process can be written in terms of its equilibrium constant as:

ΔG° = [tex]-RTInK_(eq)[/tex]

where:

R= universal gas constant

T= temperature

[tex]K_eq[/tex]= equilibrum constant for the process

Similarly, free energy change and cell potentia; are related to each other as follows;

ΔG= -nFE°

from above;

F = faraday's constant

n = number of electrons exchanged in the process; and  

E = standard cell potential

∴ The amount of energy required for transport of hydrogen ions from a cell into stomach lumen can be calculated as:

ΔG° = [tex]-RTInK_(eq)[/tex]

where;

[texK_eq[/tex]=[tex]\frac{[H^+]_(cell)}{[H^+(stomach lumen)]}[/tex]

For transport of ions to an internal pH of 7.4, the transport taking place can be given as:

[tex]H^+_{inside}[/tex] ⇒ [tex]H^+_{outside}[/tex]

Equilibrum constant for the transport is given as:

[tex]K_{eq}=\frac{[H^+]_{outside}}{[H^+]_{inside}}[/tex]

[tex]=\frac{[H^+]_{cell}}{[H^+]_{stomach lumen}}[/tex]

[tex][H^+]_{cell}[/tex]= 10⁻⁷⁴

=3.98 * 10⁻⁸M

[tex][H^+]_{stomach lumen}[/tex] = 10⁻²¹

=7.94 * 10⁻³M

Hence;

[tex]K_{eq}=\frac{[H^+]_{cell}}{[H^+]_{stomachlumen}}[/tex]

=[tex]\frac{3.98*10^{-8}}{7.94*10{-3}}[/tex]

= 5.012 × 10⁻⁶

Furthermore, free energy change for this reaction is related to the equilibrium concentration given as:

ΔG° = [tex]-RTInK_(eq)[/tex]

If temperature T= 37° C ; in kelvin

=37° C + 273.15K

=310.15K; and

R-= 8.314 j/mol/k

substituting the values into the equation we have;

ΔG₁ = [tex]-(8.314J/mol/K)(310.15)TIn(5.0126*10^{-6})[/tex]

= 31467.93Jmol⁻¹

≅ 31.47KJmol⁻¹

If the potential difference across the cell membrane= 60.0mV.

Energy required to cross the cell membrane will be:

ΔG₂ = [tex]-nFE°_{membrane}[/tex]

ΔG₂ = [tex]-(1 mol)(96.5KJ/mol/V)(60*10^{-3})[/tex]

= 5.79KJ

Therefore, for one mole of electron transfer across the membrane; the energy required is 5.79KJmol⁻¹

Now, we  can calculate the total amount of energyy required to transport H⁺ ions across the membrane:

Δ [tex]G_{total} = G_{1}+G_{2}[/tex]

= (31.47+5.79) KJmol⁻¹

= 37.26KJmol⁻¹

We can therefore conclude that;

   The amount of energy required to transport ions from cell to stomach lumen is 37.26KJmol⁻¹

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:

6.00986

Explanation:

Answer:

a) H2O2 is the limiting reactant

b) There will remain 0.450 moles of N2H4

c) There will be produced 0.250 moles of N2 and 1 mol of H2O

Explanation:

Step 1: Data given

Number of moles of N2H4 = 0.7 moles

Number of moles H2O2 = 0.500 moles

Molar mass of N2H4 = 32.05 g/mol

Molar mass of H2O2 = 34.01 g/mol

Step 2: The balanced equation

N2H4 + 2H2O2 → N2 + 4H2O

Step 3: Calculate the limiting reactant

For 1 mol of N2H4 we need 2 moles of H2O2 to produce 1 mol of N2 and 4 moles of H2O

H2O2 is the limiting reactant. It will completely be consumed. (0.500 moles).

N2H4 is in excess. There will react 0.500/2 = 0.250 moles of N2H4

There will remain 0.700 - 0.250 moles = 0.450 moles of N2H4

Step 4: Calculate moles of products

For 1 mol of N2H4 we need 2 moles of H2O2 to produce 1 mol of N2 and 4 moles of H2O

For 0.500 moles of H2O2. we'll have 0.250 moles of N2 and 1 mol of H2O

Answer:

Explanation:

(a). Hydrogen peroxide is the limiting reactant, because from the rxn the mole ratio is 1:2, therefore, 0.7mol of N2H4 is supposed to react with 1.4mol of H2O2.

(b). From the mole ratio, 0.5mol of H2O2 will react with 0.25mol of N2H4. Therefore, the unused mol of N2H4 will be (0.7-0.25)mol

=0.45mol

(c). 0.25mol of N2 and 1.0mol of H20 are the products formed based on the mole ratio from the rxn.

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:

B

Explanation:

That electron is found in the 3d orbital.

ml=-2,-1,0,1,2

n=3

l=2

ml=0

ms=-1/2

Since it must be of opposite spin according to Pauli exclusion principle.

Answer : The value of [tex]\Delta G_{rxn}[/tex] is -24.9 kJ/mol

Explanation :

First we have to calculate the value of 'Q'.

The given balanced chemical reaction is,

[tex]Fe_2O_3(s)+3CO(g)\rightarrow 2Fe(s)+3CO_2(g)[/tex]

The expression for reaction quotient will be :

[tex]Q=\frac{(p_{CO_2})^3}{(p_{CO})^3}[/tex]

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Now put all the given values in this expression, we get

[tex]Q=\frac{(2.1)^3}{(1.4)^2}[/tex]

[tex]Q=3.375[/tex]

Now we have to calculate the value of [tex]\Delta G_{rxn}[/tex].

The formula used for [tex]\Delta G_{rxn}[/tex] is:

[tex]\Delta G_{rxn}=\Delta G^o+RT\ln Q[/tex]    ............(1)

where,

[tex]\Delta G_{rxn}[/tex] = Gibbs free energy for the reaction  = ?

[tex]\Delta G_^o[/tex] =  standard Gibbs free energy  = -28.0 kJ/mol

R = gas constant = [tex]8.314\times 10^{-3}kJ/mole.K[/tex]

T = temperature = 298 K

Q = reaction quotient = 3.375

Now put all the given values in the above formula 1, we get:

[tex]\Delta G_{rxn}=(-28.0kJ/mol)+[(8.314\times 10^{-3}kJ/mole.K)\times (298K)\times \ln (3.375)[/tex]

[tex]\Delta G_{rxn}=-24.9kJ/mol[/tex]

Therefore, the value of [tex]\Delta G_{rxn}[/tex] is -24.9 kJ/mol

The standard free energy change of the reaction is -25kJ/mol.

The perform the task we must first calculate Kp from the data provided as follows;

P(CO) = 1.4 atm

P(CO2) = 2.1 atm

Kp = (p.CO2)^3/(p.CO)^3

Kp = (2.1)^3/(1.4)^3

Kp = 9.3/2.7

Kp = 3.4

ΔG°rxn =ΔG° + RTlnKp

Where;

R = 8.314 J/Kmol

T =  298 K

ΔG°rxn = -28.0 kJ + (8.314 * 298 * ln 3.4) * 10^-3

ΔG°rxn = -25kJ/mol

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The increase in entropy when the partition is removed is calculated using the entropy change formula ΔS=nRln(Vf/Vi). The total entropy change is the sum of the entropy changes for each gas. For the first case, the total entropy increase is 11.52 J/K; for the 2 moles of gas A case, the increase is 17.29 J/K; and for the case where both compartments contain gas A, the increase is also 11.52 J/K.

For this, we will use the formula for the entropy change, ΔS, which is given by ΔS=nRln(Vf/Vi), where n is the number of moles, R is the gas constant, and Vf and Vi are the final and initial volumes respectively. In the first case, the total entropy change when the partition is removed is the sum of the entropy changes for both gas A and gas B, since both volumes double, so ΔS=(1mol)(8.31 J/mol K)ln(2)+(1mol)(8.31 J/mol K)ln(2)=11.52 J/K.

If there are 2 moles of gas A in the first compartment, the entropy change is ΔS=(2mol)(8.31 J/mol K)ln(2)+(1mol)(8.31 J/mol K)ln(2)=17.29 J/K. For the final case where both compartments initially contain gas A, when the partition is removed, the total volume available to the gas doubles, so each gas experiences an entropy change of ΔS=(1mol)(8.31 J/mol K)ln(2), summed over both gases, gives us again 11.52 J/K.

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Case 1 - For Ideal Gas A and B the value of ΔStotal is 11.52 J/K.

Case 2 - For 2 Moles of Ideal Gas A the value of ΔS is 11.52 J/K.

Case 3 - Both Compartments with Ideal Gas A has the value of ΔS as 11.52 J/K.

Case 1: Ideal Gas A and B

For the first case, where there is 1 mole of ideal gas A in one compartment and 1 mole of ideal gas B in the other, both at 1 atm:

Case 2: 2 Moles of Ideal Gas A

For the second case, with 2 moles of ideal gas A in one compartment:

Case 3: Both Compartments with Ideal Gas A

Now, let’s consider the scenarios in which both compartments contain the same type of gas A:

If one compartment had 2 moles of gas A and the other also had 2 moles, the calculation is the same, but for each set of 2 moles:
ΔS = 2 * 8.314 * ln(2) ≈ 11.52 J/K

The key idea is that the type of molecules and their identical properties simplify calculations using the entropy formula for ideal gases.

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: 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: Option (4) is the correct answer.

Explanation:

Evaporation is defined as a process in which molecules of a substance absorb energy and hence, they gain kinetic energy. As a result, state of the substance changes from liquid phase to vapor phase.

For example, when we boil water then heat is absorbed by molecules of the liquid and hence, they change into vapor state.

Endothermic processes are defined as the process in which heat energy is absorbed by the reactant molecules.

The weak intermolecular forces which can arise either between nucleus and electrons or between electron-electron are known as dispersion forces. These forces are also known as London dispersion forces and these are temporary in nature.

Dispersion forces are present in between both polar and non-polar molecules.

Therefore, we can conclude that evaporation is an endothermic process is the correct statement.

Of the presented statements, the correct one is that evaporation is an endothermic process, as it involves the absorption of heat to overcome intermolecular forces.

The correct statement among the options provided is that evaporation is an endothermic process. This means that when a liquid turns to a gas, it absorbs heat from the surroundings, which is necessary to overcome the intermolecular forces that hold the molecules together in the liquid phase. Dispersion forces, also known as London dispersion forces, are a type of intermolecular force that exist between all atoms and molecules, whether they are polar or nonpolar. These forces arise from temporary dipoles caused by the fluctuating positions of electrons in atoms or molecules, so it is incorrect to say that only nonpolar, only molecules that can hydrogen bond, or only polar molecules have dispersion forces. Therefore, statements 1, 2, 3, and 5 are incorrect. Statement 6 is not applicable as statement 4 is correct.

Answer:

[tex]T_{C}[/tex] = -4.2°C

[tex]T_{H}[/tex] = 49.4°C

Explanation:

A Carnot cycle is known as an ideal cycle in thermodynamic. Therefore, in theory, we have:

|[tex]\frac{Q_{C} }{Q_{H} }[/tex]| = [tex]\frac{T_{C} }{T_{H} }[/tex]

Similarly,

|[tex]Q_{H}[/tex]| = |[tex]Q_{C}[/tex]| + |[tex]W_{S}[/tex]|

During winter, the value of |[tex]T_{H}[/tex]| = 20°C = 273.15 + 20 = 293.15 K and |[tex]W_{S}[/tex]| = 1.5 kW. Therefore,

|[tex]Q_{H}[/tex]| = 0.75([tex]T_{H}[/tex] -  [tex]T_{C}[/tex])

Similarly,

|[tex]\frac{W_{S} }{Q_{H} }[/tex]| = 1 - [tex]\frac{T_{C} }{T_{H} }[/tex]

1.5/0.75*(293.15-[tex]T_{C}[/tex]) = 1 - ([tex]T_{C}[/tex]/293.15

Further simplification,

[tex]T_{C}[/tex] = -4.2°C

During summer, [tex]T_{C}[/tex] = 25°C = 273.15+25 = 298.15 K, and |[tex]W_{S}[/tex]| = 1.5 kW. Therefore,

|[tex]Q_{C}[/tex]| = 0.75([tex]T_{H}[/tex] -  [tex]T_{C}[/tex])

Similarly,

|[tex]\frac{W_{S} }{Q_{C} }[/tex]| = [tex]\frac{T_{H} }{T_{C} }[/tex] - 1

1.5/0.75*([tex]T_{H}[/tex] - 298.15) = ([tex]T_{H}[/tex]/298.15

Further simplification,

[tex]T_{H}[/tex] = 49.4°C

A) The minimum outside temperature for which the house can be maintained at 20°C is; T_L = -4.21 °C

B) The maximum outside temperature for which the house can be maintained at 20°C is; T_H = 37.21 °C

A) Heat transfer rate; Q'_H = 0.75 kJ/s/°C

Rating of heat pump motor; W'_in = 1.5 kW = 1.5 kJ/s

Temperature inside house; T_H = 20 °C = 293.15 K

Formula to find the minimum outside temperature is;

[tex]\frac{Q_{H}}{W'_{in}} = \frac{T_{H}}{T_{H} - T_{L}}[/tex]

Where [tex]T_{L}[/tex] is the minimum outside temperature.

[tex]{Q_{H}} = {Q'_{H}}({T_{H} - T_{L}})[/tex]

Thus, plugging in the relevant values gives;

(0.75/1.5)(293.15 - [tex]T_{L}[/tex]) = 293.15/(293.15 - [tex]T_{L}[/tex])

0.5(293.15 - [tex]T_{L}[/tex]) = 293.15/(293.15 - [tex]T_{L}[/tex])

(293.15 - [tex]T_{L}[/tex])² = 293.15/0.5

(293.15 - [tex]T_{L}[/tex]) = √586.3

[tex]T_{L}[/tex] = 293.15 - 24.21

[tex]T_{L}[/tex] = 268.94 K

Converting to °C gives

[tex]T_{L}[/tex] = -4.21°C

B) Now, we want to find the maximum outside temperature when T_L = 25°C = 298.15 K

Thus, we will use the formula;

[tex]\frac{Q_{H}}{W'_{in}} = \frac{T_{L}}{T_{H} - T_{L}}[/tex]

(0.75/1.5) * ([tex]T_{H} - T_{L}[/tex]) = [tex]\frac{T_{L}}{T_{H} - T_{L}}[/tex]

0.5 * ([tex]T_{H} - 298.15[/tex]) = 298.15/([tex]T_{H} - 298.15[/tex])

149.075 = ([tex]T_{H} - 298.15[/tex])²

√149.075 = ([tex]T_{H} - 298.15[/tex])

12.21 + 298.15 = [tex]T_{H}[/tex]

T_H = 310.36 K

Converting to °C gives;

T_H = 37.21 °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: check explanation

Explanation:

The balanced equation of reaction is given below;

HCl + NH3 -----------------------> NH4Cl.

We are given the volume in milliliters, let us convert them into Litres;

= 38.30 × 10^-3 Litres.

Here, we have an incomplete question (one parameter is missing- the volume of ammonia,NH3). Therefore, we assume that the volume of Ammonia, NH3 is 10mL(10× 10^-3 Litres).

Step one: we need to calculate the number of moles of HCl.

Number of moles of HCl= molarity × volume.

Number of moles of HCl= 0.507 M × 38.30× 10^-3 L.

Number of moles of HCl= 0.0194181 moles.

From the equation of reaction above, we have that one mole of ammonia is reacting with one mole of Hydrogen Chloride, HCl. Hence, the number of moles of ammonia is equal to the number of moles of Hydrogen Chloride,HCl.

Step two: calculate the molarity of Ammonia, NH3.

The molarity of ammonia= number of moles of ammonia/ volume of Ammonia,NH3.

Molarity of Ammonia= 0.0194181/10× 10^-3 moles NH3.

Molarity of Ammonia= 0.00000194181.

Molarity of Ammonia = 1.94181× 10^-6 M.

The molarity of ammonia after the 2nd addition can be calculated by equating the moles of HCl to the moles of ammonia, as the molar ratio in the reaction equation is 1:1. The equilibrium molarity of ammonia is the moles of ammonia divided by the total volume of the solution (which isn't stated in the question).

In this titration analysis, you've used 0.507 M HCl and 38.30 mL (or 0.03830 L) to neutralize the ammonia. To calculate the molarity of the ammonia, you will be using the reaction equation NH3 + HCl -> NH4Cl. The molar ratio is 1:1, meaning 1 mole of HCl reacts with 1 mole of NH3.

First, we need to find the moles of HCl, which can be obtained by multiplying the HCl molarity by the volume (in liters). So the moles of HCl = 0.507 M * 0.0383 L = 0.01942 moles. This is also the moles of ammonia at equilibrium assuming all of it was neutralized by the HCl.

The equilibrium molarity of ammonia, which is the concentration of ammonia after the 2nd addition, can be calculated by dividing the moles of ammonia by the total volume of the solution. However, the total volume of the solution isn't provided in the question. If it had been stated, you could simply plug in that value (in liters) to get the molarity of ammonia.

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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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Answer: The molarity of NaOH solution is 0.1046 M.

Explanation:

To calculate the number of moles, we use the equation:

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

Given mass of KHP = 0.5516 g

Molar mass of KHP = 204.22 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of KHP}=\frac{0.5516g}{204.22g/mol}=0.0027mol[/tex]

The chemical reaction for the reaction of KHP and NaOH follows

[tex]KHC_8H_4O_4(aq.)+NaOH\rightarrow KNaC_8H_4O_4(aq.)+H_2O(l)[/tex]

By Stoichiometry of the reaction:

1 mole of KHP reacts with 1 mole of NaOH.

So, 0.0027 moles of KHP will react with = [tex]\frac{1}{1}\times 0.0027=0.0027mol[/tex] of NaOH.

To calculate the molarity of NaOH, we use the equation:

[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]

We are given:

Moles of NaOH = 0.0027 moles

Volume of solution = 25.82 mL  = 0.02582L      (Conversion factor:  1L = 1000 mL)

Putting values in above equation, we get:

[tex]\text{Molarity of NaOH }=\frac{0.0027mol}{0.02582L}=0.1046M[/tex]

Hence, the molarity of NaOH solution is 0.1046 M.

Based on the data provided, the balanced equation of the reaction is:

The equation of the reaction between KHP and NaOH is given below as:

From the equation of reaction, mole ratio of KHP and NaOH is 1 : 1

Moles of KHP = mass/molar mass

molar mass of KHP = 204 g/mol

Moles of KHP = 0.5516/204

Moles of KHP = 0.00271 moles

Moles of NaOH = molarity × volume

Volume of NaOH = 25.82 mL = 0.02582 L

0.00271 = molarity × 0.02582

Molarity = 0.00271/0.02582

Molarity of KOH = 0.105 M

Hence, the molarity of NaOH solution is 0.105 M.

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