The standard free energy change for a reaction can be calculated using the equation ΔG∘′=−nFΔE∘′ ΔG∘′=−nFΔE∘′ where nn is the number of electrons transferred, FF is Faraday's constant, 96.5 kJ·mol−1·V−1, and ΔE∘′ΔE∘′ is the difference in reduction potential. For each of the given reactions, determine the number of electrons transferred (n)(n) and calculate standard free energy (ΔG∘′)(ΔG∘′) . Consider the half-reactions and overall reaction for reaction 1. half-reactions:fumarate 2−+2H+CoQH2↽−−⇀succinate−↽−−⇀CoQ+2H+ half-reactions:fumarate−+2H+↽−−⇀succinate2−CoQH2↽−−⇀CoQ+2H+ overall reaction:fumarate2−+CoQH2↽−−⇀succinate2−+CoQΔE∘′=−0.009 V

Answer: [tex]\Delta S[/tex] = 473.92J/K.mol

Explanation: In physics, Entropy is defined as a degree of disorder in a system. Entropy change is given by the sum of all the products multiplied by their respective coeficients minus the sum of all the reagents multiplied by their respective coeficients:

[tex]\Delta S = m\Sigma product - n\Sigma reagent[/tex]

The balanced reaction:

[tex]H_{2}S_{(g)}+2H_{2}O_{(l)}=>3H_{2}_{(g)}+SO_{2}_{(g)}[/tex]

gives the proportion reagents react to form products, so, if 1.6 moles of [tex]H_{2}S_{(g)}[/tex]:

3.2 moles of water is used;

4.8 moles of hydrogen gas is formed;

1.6 moles of sulfur dioxide is also formed;

Calculating entropy change:

[tex]\Delta S = (4.8*131+1.6*248.8)-(1.6*205.6+3.2*70)[/tex]

[tex]\Delta S=628.8+398.08-328.96-224[/tex]

[tex]\Delta S[/tex] = 473.92J/K.mol

Entropy change for the given chemical reaction is [tex]\Delta S[/tex] = 473.92J/K.mol

Answer:

d. Oxidation and reduction

Explanation:

For this question we have to remember the definition of each type of reaction:

-) Hydrogenation

In this reaction, we have the addition of hydrogen to a molecule. Usually, an alkene or alkyne. In the example, molecular hydrogen is added to a double bond to produce an alkane.

-) Alkylation

In this reaction, we have the addition of a chain of carbon to another molecule. In the example, an ethyl group is added to a benzene ring.

-) Hydrolysis

In this reaction, we have the breaking of a bond by the action of water. In the example, a water molecule can break the C-O bond in the ester molecule.

-) Halogenation

In this reaction, we have the addition of a halogen (atoms on the VIIIA group). In the example, "Cl" is added to the butene.

-) Ammoniation

In this reaction, we have the addition of the ammonium ion ([tex]NH_4^+[/tex]). In the example, the ammonium ion is added to an acid.

-) Oxidation and reduction

In this reaction, we have opposite reactions. The oxidation is the loss of electrons and the reduction is the gain of electrons. For example:

[tex]Ag^+~+~e^-~->~Ag[/tex] Reduction

[tex]Al~->~Al^+^3~+~3e^-[/tex] Oxidation

Answer:

Explanation:

The equation for the decay is given as;

¹⁴₆C  --> X + ⁰₋₁e

For conservation of matter, the mass number and atomic number has to be the same in both the reactant and product side f he equation;

Mass number;

14 = x + 0

x = 14

Atomic Number;

6 = x + (-1)

x = 6 + 1 =7

¹⁴₆C  --> ¹⁴₇N + ⁰₋₁e

The name of the product nuclide with atomic number of 7 is Nitrogen. The symbol is; ¹⁴₇N

Answer:

The correct answer is 1332 KJ.

Explanation:

Based on the given information,  

ΔH°f of C2H4 is 51.9 KJ/mol, ΔH°O2 is 0.0 KJ/mol, ΔH°f of CO2 is -394 KJ/mol, and ΔH°f of H2O is -286 KJ/mol.  

Now the balanced equation is:  

C2H4 (g) + 3O2 (g) ⇔ 2CO2 (g) + 2H2O (l)

ΔH°rxn = 2 × ΔH°f CO2 + 2 × ΔH°fH2O - 1 × ΔH°fC2H4 - 3×ΔH°fO2

ΔH°rxn = 2 (-394) + 2(-286) - 1(51.9) - 3(0)

ΔH°rxn = -1411.9 KJ

Now, the given ΔS°f of C2H4 is 219.8 J/mol.K, ΔS°f of O2 is 205 J/mol.K, ΔS°f of CO2 is 213.6 J/mol.K, and ΔS°f of H2O is 69.96 J/mol.K.  

Now based on the balanced chemical reaction,  

ΔS°rxn = 2 × ΔS°fCO2 + 2 ΔS°fH2O - 1 × ΔS°f C2H4 - 3 ΔS°fO2

ΔS°rxn = 2 (213.6) + 2(69.96) - 1(219.8) -3(205)

ΔS°rxn = -267.68 J/K or -0.26768 KJ/K

T = 25 °C or 298 K

Now putting the values of ΔH, ΔS and T in the equation ΔG = ΔH-TΔS, we get

ΔG = -1411.9 - 298.0 × (-0.2677)

ΔG = -1332 KJ.  

Thus, the maximum work, which can obtained is 1332 kJ.  

Answer:

2.77 mol/kg

Explanation:

Molality is a sort of concentration that indicates the moles of solute in 1kg of solvent. In this case our solvent is water and, if we consider water's density as 1g/mL, we determine that the mass of solvent is 750 g.

We convert the mass to kg → 750 g . 1kg /1000g = 0.750 kg

Our solute is the NaOH → 83 g.

We convert the mass to moles → 83 g . 1mol /40g = 2.075 mol

Molality (mol/kg) = 2.075 mol / 0.75kg = 2.77 m

Answer:

a. Moles in the vessel = 1.85 moles of the gas

b. 1.11x10²⁴ molecules are in the vessel

Explanation:

a.It is possible to determine moles of a gas using the general law of gases:

PV = nRT

Where P is pressure: 5.00atm; V is volume = 9.00L; R is gas constant: 0.082atmL/molK; T is absolute temperature: 273.15K +24.0 = 297.15K

Computing the values:

PV / RT = n

5.00atm* 9.00L / 0.082atmL/molK*297.15K = n

b. With Avogadro's number we can convert moles of any compound to molecules thus:

Avogadro's number = 6.022x10²³ molecules / mole

1.85moles ₓ (6.022x10²³ molecules / mole) =

Explanation:

According to your question,

no. a. ans would be like; chemical formula is defined as the an expression which determines no. and type of molecule of a compound.

b. no. ans; it tells that what type of compound is formed with the type and no. of atoms present in the atom.

c. no ans; the formulation of h2so4 states that it is acid named as hydrochloric acid which is formed by reacting of hydrogen (2 atoms ) ,sulpher (*1atom) and oxygen(4atoms).

Hope it helps...

Answer: Hợp chất CTHH 0 °C 10 °C 20 °C 30 °C 40 °C 50 °C 70 °C

Actini(III) hydroxide Ac(OH)3   0,0022    

Amonia NH3 1176 900 702 565 428 333 188

Amoni azua NH4N3 16  25,3  37,1  

View 42 more rows  

                    hehe

Answer:

Explanation:

ok

The structure of amylase deteriorates due to high temperature of the solution.

This experiment shows that the structure of amylase deteriorates due to high temperature which prevents this amylase from performing its function properly.

At high temperatures the amylase will break starch down very slowly or not at all due to denaturation of the enzyme's active site due to which it can't perform its function properly so we can conclude that high temperature denatures amylase enzyme.

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Nakula's conclusion was- "My results show that boiling destroys amylase"

Amylase is an enzyme that breaks down starch molecules into sugar molecules. He boiled the solution in tube P, and when he checked tube P for sugar, there wasn't any. He didn't boil the solution in tube Q and he found sugar in it.

Amylase had broken down starch molecules to sugar molecules in tube Q. Tube P's solution had been boiled, and this showed that when he boiled it, it destroyed the amylase, that is why the starch molecules hadn't been broken down into sugar molecules.

Answer:

mol LiCl = 4.83 m

Explanation:

GIven:

Solution of LiCl in water XLiCl = 0.0800

Mol of water in kg = 55.55 mole

Find:

Molality

Computation:

mole fraction = mol LiCl / (mol water + mol LiCl)

0.0800 = mol LiCl / (55.55 mol + mol LiCl)

0.0800 mol LiCl + 4.444 mol = mol LiCl

mol LiCl - 0.0800 mol LiCl = 4.444 mol

0.92 mol LiCl = 4.444 mol

mol LiCl = 4.83 m

1) Based on the crystal field strength, Cl ligand would give the longest d-d transition when complexed with Ti(III). as this is the weak field igand and would cause minimum splitting of d orbitals.

Answer:

HBr, CH3C(OH)2 and CH3COOH, Br-

Explanation:

The conjugate acid-base pairs acid reacts with base to form a conjugate acid and conjugate base.

Conjugate acid is formed when a bases receives a proton  (H+) and a conjugate base is formed when an acid losses a proton (H+).

From the given equation:

HBr, CH3C(OH)2 and CH3COOH, Br- are conjugate acid-base pair, where HBr is an acid and CH3C(OH)2 is a conjugate acid while CH3COOH and Br- is the conjugate base.

Answer:

5.92 g

Explanation:

Convert milliliters to liters.

217 mL = 0.217 L

Since molarity (M) is moles per liter(mol/L), multiply the molarity by the volume to find out how many moles you will need.

0.217 L × 0.246 M = 0.05338 mol

Now, convert the moles to grams using the molar mass.  The molar mass of calcium chloride is 110.98 g/mol.

0.05338 mol × 110.98 g/mol = 5.924 g ≈ 5.92 g

You will need 5.92 g of calcium chloride.

Answer:

[tex]\large \boxed{1.503 \times 10^{23}\text{ molecules of Cu(OH)}_{2}}$}[/tex]

Explanation:

You must calculate the moles of Cu(OH)₂, then convert to molecules of Cu(OH)₂.

1. Moles of Cu(OH)₂[tex]\text{Moles of Cu(OH)}_{2} = \text{24.35 g Cu(OH)}_{2} \times \dfrac{\text{1 mol Cu(OH)}_{2}}{\text{97.562 g Cu(OH)}_{2}} = \text{0.2496 mol Cu(OH)}_{2}[/tex]

2. Molecules of Cu(OH)₂[tex]\text{No. of molecules} = \text{0.2496 mol Cu(OH)}_{2} \times \dfrac{6.022 \times 10^{23}\text{ molecules Cu(OH)}_{2}}{\text{1 mol Cu(OH)}_{2}}\\\\= \mathbf{1.503 \times 10^{23}}\textbf{ molecules Cu(OH)}_{2}\\\text{There are $\large \boxed{\mathbf{1.503 \times 10^{23}}\textbf{ molecules of Cu(OH)}_{2}}$}[/tex]

The number of molecules of copper (II) hydroxide in 23.45 g sample has been [tex]\rm \bold {1.45\;\times\;10^2^3}[/tex].

According to the Avogadro number, the number of molecules in a mole of atom has been equivalent to the Avogadro constant. The value of Avogadro constant has been [tex]\rm 6.023\;\times\;10^2^3[/tex].

The moles of a compound has been given as:

[tex]\rm Moles=\dfrac{Mass}{Molar\;mass}[/tex]

The moles in 23.45 g copper (II) hydroxide has been:

[tex]\rm Moles=\dfrac{23.45}{97.562} \\Moles=0.24\;mol[/tex]

The moles of copper (II) hydroxide has been 0.24 mol.

The number of molecules in 0.24 mol sample has been driven by:

[tex]\rm 1\;mol=6.023\;\times\;10^2^3\;molecules\\0.24\;mol=0.24\;\times\;10^2^3\;molecules\\0.24\;mol=1.45\;\times\;10^2^3\;molecules[/tex]

The number of molecules of copper (II) hydroxide in 23.45 g sample has been [tex]\rm \bold {1.45\;\times\;10^2^3}[/tex].

For more information about molecules in a mole of sample, refer to the link:

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Answer:

2.8x10⁻¹⁵ M.

Explanation:

Hello,

In this case, the dissociation reaction for silver phosphate is:

[tex]Ag_3PO_4(s)\rightleftharpoons 3Ag^+(aq)+PO_4(aq)[/tex]

Therefore, the equilibrium expression is:

[tex]Ksp=[Ag^+]^3[PO_4^-][/tex]

In such a way, since the initial solution contains an initial concentration of silver ions (from silver nitrate) of 0.10M, we can write the equilibrium expression in terms of the reaction extent [tex]x[/tex]:

[tex]2.8x10^{-18}=(0.10+3x)^3*(x)[/tex]

Thus, solving for [tex]x[/tex] we have:

[tex]x=2.8x10^{-15}M[/tex]

Thus, the molar solubility of silver phosphate is 2.8x10⁻¹⁵ M.

Regards.

Answer:

0.0061650760770388 mole

Answer:

1.12 × 10⁻⁴ M

Explanation:

Step 1: Write the reaction for the solution of Mg(OH)₂

Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2 OH⁻(aq)

Step 2: Make an ICE chart

We can relate the solubility product constant (Ksp) with the solubility (S) through an ICE chart.

       Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2 OH⁻(aq)

I                                0                    0

C                              +S                +2S

E                                S                  2S

The solubility product constant is:

Ksp = 5.61 × 10⁻¹² = [Mg²⁺] × [OH⁻]² = S × (2S)² = 4S³

S = 1.12 × 10⁻⁴ M

Answer:

There are required 70.1 years for the activity of a sample of cesium-137 to fall to 20.0 percent of its original value

Explanation:

The radioactive decay follows always first-order kinetics where its general law is:

Ln[A] = -Kt + ln[A]₀

Where [A] is actual concentration of the atom, k is rate constant, t is time and [A]₀ is initial concentration.

We can find rate constant from half-life as follows:

Rate constant:

t(1/2) = ln 2 / K

As half-life of Cesium-137 is 30.2 years:

30.2 years = ln 2 / K

K = 0.02295 years⁻¹

Replacing this result and with the given data of the problem:

Ln[A] = -Kt + ln[A]₀

Ln[A] = -0.02295 years⁻¹* t + ln[A]₀

Ln ([A] / [A₀]) = -0.02295 years⁻¹* t

As you want time when [A] is 20% of [A]₀, [A] / [A]₀ = 0.2:

Ln (0.2) = -0.02295 years⁻¹* t

70.1 years = t

Answer:

1 L

Explanation:

ppm means parts per million. Generally the relationship between mass and litre is given as;

1 ppm = 1 mg/L

This means that 1 ppm is equivalent to 1 mg of a substance dissolved in 1 L of water.

In those portions, the heat supplied to the substance by the heater does not lead to a temperature rise as intermolecular forces are broken.

When a substance is heated, we normally expect that its temperature will rise as a consequence.

However, heat may be supplied to a substance but its temperature does not rise owing to the fact that the heat energy supplied is used to break intermolecular bonds.

This occurs during fusion and boiling. The heat supplied at these point does not result in temperature rise since it is used to break intermolecular bonds. The temperature remains steady during these processes as shown by the two flat portions on the graph in the image attached to the question. This heat supplied is known as the latent heat.

For more about latent heat, see:

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Answer:

The answer for me was "These show where changes of state are occuring"

Explanation:

I got it right

Answer:

hello the molecules are missing from your question below are the Generic molecules : [tex]ABE_{3}[/tex] and [tex]AB_{3} E[/tex]

answer : It can be determined  that both generic molecules are polar

It can be determined that both generic molecules have similar molecular shape

They have different Geometry

They differ in bond angles as well

Explanation:

The two generic molecules : [tex]ABE_{3}[/tex] and [tex]AB_{3} E[/tex]

comparing(similarities) these two generic molecules

It can be determined  that both generic molecules are polar

It can be determined that both generic molecules have similar molecular shape

differences between the generic molecules

They have different Geometry

They differ in bond angles as well

Answer:

1.answer = pushing force

= pulling force

=gravity force

Answer:

(R)-but-3-en-2-ylbenzene

Explanation:

In this reaction, we have a very strong base (sodium ethoxide). This base, will remove a hydrogen producing a double bond. We know that the reaction occurs through an E2 mechanism, therefore, the hydrogen that is removed must have an angle of 180º with respect to the leaving group (the "OH"). This is known as the anti-periplanar configuration.

The hydrogen that has this configuration is the one that placed with the dashed bond (red hydrogen). In such a way, that the base will remove this hydrogen, the "OH" will leave the molecule and a double bond will be formed between the methyl and the carbon that was previously attached to the "OH", producing the molecule (R) -but-3- en-2-ylbenzene.

See figure 1

I hope it helps!

Answer:

2 Na(s) + C(s) + 3/2 O₂(g) → Na₂CO₃(s)

Explanation:

The molar enthalpy of formation of a chemical is defined as the change in enthalpy during the formation of 1 mole of the substance from its constituent elements (Constituent elements are pure elements you have in the periodic table)

For Na₂CO₃ constituent elements are Na(s), C(s) and O₂(g) and the chemical equation that represents the molar enthalpy is:

Answer:

ΔH∘rxn of the reaction is -521.6kJ

Explanation:

Complete question: "Calculate the value of ΔH°rxn for 2F2(g)+2H2O(l)⟶4HF(g)+O2(g)"

You can find the ΔH of a reaction by the algebraic sum of similar reactions (Hess's law) as follows:

(1) H₂(g) + F₂(g) ⟶ 2HF(g) ΔH∘rxn=−546.6 kJ

(2) 2H₂(g)+O₂(g)⟶2H₂O(l) ΔH∘rxn=−571.6 kJ

Subtracting 2ₓ(1) - (2)

2ₓ(1) 2H₂(g) + 2F₂(g) ⟶ 4HF(g) ΔH∘rxn=2ₓ−546.6 kJ = -1093.2kJ

-(2) 2H₂O(l) ⟶ 2H₂(g)+O₂(g) ΔH∘rxn=- (-571.6 kJ) = 571.6kJ

2ₓ(1) - (2) 2F₂(g)+ 2H₂O(l) ⟶ 4HF(g) + O₂(g)

H₂(g) are cancelled because are the same in products and reactants

ΔH∘rxn = -1093.2kJ + 571.6kJ

ΔH∘rxn = -521.6

The two layers are part of the thermosphere are  exosphere and ionosphere.

The Exosphere is the topmost layer of the Earth's atmosphere.

and its gradually disappear into the vacuum of space.

It consist two parts that are:

exosphere and ionosphere.

Thus, option "A" is correct, the rest of the option is not a part of thermosphere.

To learn more about  atmospheric layers click here:

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#SPJ2

Answer:

hi hope your doing great the answer is A

Explanation:

its on Edge 2020

hope i helped :)

Answer:

B) CsCl

Explanation:

The ionic character is formed between two kinds of atoms having a large electronegativity differences e.g metals (like those in groups IA and IIA) and nonmetals (like those in groups VIA and VIIA). The formation of an ionic character involves a transfer of electrons from the less electronegative atom(metal) to the more electronegative atom (non-metal) such that the two kinds of atoms now have completely filled outer shell like the noble gases.

In CsCl, electrons are being transferred from Cs⁺ to Cl⁻ . As a result of this transfer , the atom of the metal becomes positively charged (cation) while that of the non-metal becomes negatively charged (anion).

The highest percentage of ionic character will occur as a result of smaller negatively charged (anion) and larger positively charged (cation). From the options given, CsCl have the highest percentage of ionic character.

Answer:

B

Explanation:

In order to create spontaneity, an endothermic process has to occur along with positive entropy and high temperature