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:

A process that involves rearrangement

Explanation:

A chemical reaction is the process that involves rearrangement of the molecular or ironic structure of a substance, as a distinct from a change in physical form or a nuclear reaction.

Answer:

Explanation:

Chemistry

The chemical reaction H2(g) + ½ O2(g) → H2O(l) describes the formation of water from its elements.

The reaction between iron and sulfur to form iron(II) sulfide is another chemical reaction, represented by the chemical equation:

8 Fe + S8 → 8 FeS

Answer:

See explanation

Explanation:

When we talk about electrophilic substitution, we are talking about a substitution reaction in which the attacking agent is an electrophile. The electrophile attacks an electron rich area of a compound during the reaction.

The five membered furan ring is aromatic just as benzene. This aromatic structure is maintained during electrophilic substitution reaction. The attack of the electrophile generates a resonance stabilized intermediate whose canonical structures have been shown in the image attached.

The words given are not clear, so the clear question is as follows:

Match the words below to the appropriate blanks in the sentences. Make certain each sentence is complete before submitting your answer:

A. β-1,4- and α-1,6-glycosidic

B. α-1,4-glycosidic

C. α-1,4-galactose

D. an unbranched glucose

E. a branched fructose

F. α-1,6-glycosidic

Amylose is ......... polymer of ....... units joined by ........ bonds.

Amylopectin is ....... polymer of .......units joined by ........ bonds.

Answer:

D. an unbranched glucose

C. α-1,4-galactose

B. α-1,4-glycosidic

E. a branched fructose

A. β-1,4- and α-1,6-glycosidic

F. α-1,6-glycosidic

Explanation:

Amylose and amylopectin are two types of polysaccharides that can be found in starch granules.

Amylose is linear or unbranched glucose polymer of α-1,4-galactose units that are joined by α-1,4-glycosidic.

Amylopectin is a branched fructose polymer of β-1,4- and α-1,6-glycosidic units joined by α-1,6-glycosidic bonds.

Hence, the correct answers in the sequential order are:

Amylose:

D. an unbranched glucose

C. α-1,4-galactose

B. α-1,4-glycosidic

Amylopectin:

E. a branched fructose

A. β-1,4- and α-1,6-glycosidic

F. α-1,6-glycosidic

Answer:

A. as an agglomeration metal

B. as a native metal

D. in sulfide ore minerals

Explanation:

Copper is a metal with symbol Cu and atomic number 29. It has a pinkish-orange color and is malleable, ductile and has a high thermal and electrical conductivity. This is why it is often used in electrical appliances.

Copper exists as an agglomeration metal, as a native metal or in sulfide ore minerals such as Cu2S.

The examples of copper deposits are  agglomeration metal, as a native metal or in sulfide ore minerals. Option A, B, and D are correct.

Copper is a metal with high thermal and electrical conductivity. hence, it is often used in electrical appliances.

Copper found as an agglomeration metal, as a native metal or in sulfide ore minerals such as [tex]\bold { Cu_2S.}[/tex]

Therefore, the examples of copper deposits are  agglomeration metal, as a native metal or in sulfide ore minerals. Option A, B, and D are correct.

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Answer:

The oxidation state of silver in [tex]\rm Ag_2O[/tex] is [tex]+1[/tex].

The oxidation state of sulfur in [tex]\rm SO_2[/tex] is [tex]+4[/tex].

Explanation:

The oxidation states of atoms in a compound should add up to zero.

There are two silver [tex]\rm Ag[/tex] atoms and one oxygen [tex]\rm O[/tex] atom in one formula unit of [tex]\rm Ag_2O[/tex]. Therefore:

[tex]\begin{aligned}&\rm 2 \times \text{Oxidation state of $\rm Ag$}+ \rm 1 \times \text{Oxidation state of $\rm O$} = 0\end{aligned}[/tex].

The oxidation state of oxygen in most compounds (with the exception of peroxides and fluorides) is [tex]-2[/tex]. Silver oxide [tex]\rm Ag_2O[/tex] isn't an exception. Therefore:

[tex]\begin{aligned}&\rm 2 \times \text{Oxidation state of $\rm Ag$}+ \rm 1 \times \text{Oxidation state of $\rm O$} = 0\\ &\rm 2 \times \text{Oxidation state of $\rm Ag$}+ \rm 1 \times (-2) = 0\end{aligned}[/tex].

Solve this equation for the (average) oxidation state of [tex]\rm Ag[/tex]:

[tex]\text{Oxidation state of $\rm Ag$} = 1[/tex].

Similarly, because there are one sulfur [tex]\rm S[/tex] atom and two oxygen [tex]\rm O[/tex] atoms in each [tex]\rm SO_2[/tex] molecules:

[tex]\begin{aligned}&\rm 1\times \text{Oxidation state of $\rm S$}+ \rm 2 \times \text{Oxidation state of $\rm O$} = 0\end{aligned}[/tex].

The oxidation state of [tex]\rm O[/tex] in [tex]\rm SO_2[/tex] is also [tex]-2[/tex], not an exception, either.

Therefore:

[tex]\begin{aligned}&\rm 1 \times \text{Oxidation state of $\rm S$}+ \rm 2 \times \text{Oxidation state of $\rm O$} = 0\\ &\rm 1 \times \text{Oxidation state of $\rm S$}+ \rm 2 \times (-2) = 0\end{aligned}[/tex].

Solve this equation for the oxidation state of [tex]\rm S[/tex] here:

[tex]\text{Oxidation state of $\rm S$} = 4[/tex].

Answer: A=1, B=3, C=12

Explanation:

For this problem, you will need to know your unit conversions. There are 3 ft in 1 yard. Knowing this, we can find A, B, C.

For A and B, we know that we want to cancel out ft so the answer can be in yards. To do so, we need to put B=3 and A=1.

Now that we know the unit conversion, we can directly solve.

36 ft×(1 yd/3 ft)=12 yd

Our final answer is A=1, B=3, C=12.

Answer:

A=1,000, B=1, C=5,400

Explanation:

the question was 5.4L x AmL / BmL = CmL

Answer:

[tex]m_{AgBr}=1.48gAgBr[/tex]

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

[tex]KBr(aq)+AgNO_3(aq)\rightarrow AgBr(s)+KNO_3(aq)[/tex]

Thus, since the potassium bromide and silver nitrate are in a 1:1 mole ratio, the first step is to identify the limiting reactant, by considering the reacting volumes of reactants in order to compute the available moles of potassium bromide and the moles of potassium bromide consumed by the 15.4 mL of 0.512-M solution of silver nitrate:

[tex]n_{KBr}=0.0599L*0.282\frac{molKBr}{L} =0.0169molKBr\\\\n_{KBr}^{consumed}=0.0154L*0.512\frac{molAgNO_3}{L} *\frac{1molKBr}{1molAgNO_3}=0.00788molKBr[/tex]

In such a way, since less moles are consumed than available, we infer that silver nitrate is the limiting reactant, for which the resulting grams of silver bromide (molar mass 187.8 g/mol) result:

[tex]m_{AgBr}=0.00788molAgNO_3*\frac{1molAgBr}{1molAgNO_3} *\frac{187.8gAgBr}{1molAgBr} \\\\m_{AgBr}=1.48gAgBr[/tex]

Best regards.

Answer:

0.1957 moles of NaF

Explanation:

The Pb²⁺ and F⁻ are in equilibrium with PbF₂ as follows:

PbF₂(s) ⇄ Pb²⁺(aq) + 2F⁻(aq)

Where Ksp expression is:

Ksp = 4.0x10⁻⁸ = [Pb²⁺] [F⁻]²

A saturated solution contains the maximum possible amount of Pb²⁺ and F⁻. That is:

PbF₂(s) ⇄ Pb²⁺(aq) + 2F⁻(aq)

PbF₂(s) ⇄ X + 2X

Where X is amount of ions presents in solution

4.0x10⁻⁸ = [Pb²⁺] [F⁻]²

4.0x10⁻⁸ = [X] [2X]²

4.0x10⁻⁸ = 4X³

4.0x10⁻⁸/4 = X³

1.0x10⁻⁸ = X³

2.15x10⁻³M = X

That means initial concentration of Pb²⁺ is = X = 2.15x10⁻³M and [F⁻] = 2X = 4.30x10⁻³M

Now, using again Ksp, if you want a [Pb²⁺] = 1.0x10⁻⁶M, the [F⁻] you need is:

4.0x10⁻⁸ = [Pb²⁺] [F⁻]²

4.0x10⁻⁸ = [1.0x10⁻⁶M] [F⁻]²

0.04M = [F⁻]²

0.2M = [F⁻]

You need a final concentration of 0.2M of F⁻. As initial concentration was 4.30x10⁻³M and volume of the buffer is 1.00L, the moles of F⁻ = moles of NaF you must add are:

0.2M - 4.30x10⁻³M =

Answer:

pH = 3.49

Explanation:

We have a buffer system formed by a weak acid (HNO₂) and its conjugate base (NO₂⁻ coming from KNO₂). We can calculate the pH  of a buffer ssytem using the Henderson-Hasselbach equation.

pH = pKa + log [base] / [acid]

pH = -log Ka + log [NO₂⁻] / [HNO₂]

pH = -log 4.50 × 10⁻⁴ + log 0.290 M / 0.210 M

pH = 3.49

The pH of the solution containing 0.210 M nitrous acid (HNO₂) and 0.290 M potassium nitrite (KNO₂) is 3.49

We'll begin by calculating the the pKa of acid. This can be obtained as follow:

Acid dissociation constant (Ka) = 4.50×10¯⁴

pKa = –Log Ka

pKa = –Log 4.50×10¯⁴

pKa = 3.35

Concentration of HNO₂, [HNO₂] = 0.210 M

Concentration of KNO₂, [KNO₂] = 0.290 M

The pH of the solution can obtain by using the Henderson-Hasselbach equation as illustrated below:

pH = pKa+ log [NO₂⁻] / [HNO₂]

pH = 3.35 + log (0.290 / 0.210)

Thus, the pH of the solution is 3.49

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Answer:

211.47 grams

Explanation:

We need to set up a dimensional analysis to solve this problem by converting from moles to grams.

First, find the molar mass of HCl. Since the molar mass of H (hydrogen) is 1.01 g/mol and the molar mass of Cl (chlorine) is 35.45 g/mol, then the molar mass of HCl is:

1.01 + 35.45 = 36.46 g/mol

We have 5.8 moles of HCl, so multiply by its molar mass:

(5.8 mol) * (36.46 g/mol) = 211.468 ≈ 211.47 g

The answer is thus 211.47 grams.

~ an aesthetics over

Answer:

[tex]\large\boxed{211.47}\\[/tex] grams

Explanation:

First, you need to gather the atomic masses of the elements involved in the compound - hydrogen and chlorine. Referencing a modern periodic table will give you this information.

[tex]\large\boxed{211.47}[/tex] is the final answer.

Answer:

0.8261 M.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

HC2H3O2 + NaOH —> NaC2H3O2 + H2O

From the balanced equation above, we obtained the following:

Mole ratio of the acid, HC2H3O2 (nA) = 1

Mole ratio of the base, NaOH (nB) = 1

Data obtained from the question include the following:

Volume of acid, HC2H3O2 (Va) = 10 mL

Molarity of acid, HC2H3O2 (Ma) =..?

Volume of base, NaOH (Vb) = 16.95 mL Molarity of base, NaOH (Mb) = 0.4874 M

Finally, we shall determine the molarity of the acid solution, as follow:

MaVa/MbVb = nA/nB

Ma x 10 / 0.4874 x 16.95 = 1

Cross multiply

Ma x 10 = 0.4874 x 16.95

Divide both side by 10

Ma = (0.4874 x 16.95) /10

Ma = 0.8261 M.

Therefore, the molarity of the vinegar solution (HC2H3O2) is 0.8261 M.

Answer:

7.38 g/cm³ is the density of the metal

Explanation:

In a Face-centered cubic unit cell you have 4 atoms. Also, the edge length is √8×r (r is radius of the atom).

To solve this problem, we need first to calculate the volume of the unit cell and then, with molar mass calculate the mass of 4 atoms. As density is the ratio between mass and volume we can obtain this value.

Volume = a³

a = √8×r

(r = 198x10⁻¹²m)

a = 5.6x10⁻¹⁰ m

Volume = 1.756x10⁻²⁸ m³

1m = 100cm → 1m³ = (100cm)³:

1.756x10⁻²⁸ m³× ((100cm)³ / 1m³) =

There are 4 atoms of gold:

4 atoms × (1mol / 6.022x10²³ atoms) = 6.64x10⁻²⁴ moles of gold

As 1 mole weighs 195.08g:

6.64x10⁻²⁴ moles of gold × (195.08g / mol) =

1.296x10⁻²¹g / 1.756x10⁻²² cm³ =

The density of the metal is 7.40 g/cm³

In cubic crystal system, face-centered cubic FFC is the name given to sort of atom arrangement observed in which structure is made up of atoms organized in a cube with a portion of an atom in each corner and six extra atoms in the center of each cube face.

It is expressed by using the formula:

[tex]\mathbf{\rho = \dfrac{Z \times M}{N_A\times a^}}[/tex]

where;

For face-centered cubic FFC;

The edge length  [tex]\mathbf{a =2 \sqrt{2}\times r }[/tex]

[tex]\mathbf{a =2 \sqrt{2}\times 198 \ pm }[/tex]

[tex]\mathbf{a =560.0285 \ pm }[/tex]

a = 5.60 × 10⁻⁸ cm

Replacing it into the previous equation, we have:

[tex]\mathbf{\rho = \dfrac{4 \times 195.8}{6.022 \times 10^{23} \times( 5.60 \times 10^{-8} )^3}}[/tex]

[tex]\mathbf{\rho = 7.40\ g/cm^3 }[/tex]

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Answer:

50.

Explanation:

We can write Graham's Law of Diffusion as:

(Rate 1)^2 = Molecular Mass 2

--------------    -------------------------

(Rate 2)^2    Molecular Mass 1

So using the Given Information:

1^2  / (1/5)^2 = Molecular Mass of unknown gas / 2, so:

25 = M/2

M = 50.

Answer:

- Aluminium sulfate Al2(SO4)3 dissociates in two aluminium ions and three sulfate ions, therefore, van't Hoff factor is 5 (incorrect).

Explanation:

Hello,

In this case, since the van't Hoff factor is related with the species that result from the ionization of a chemical compound, we can see that that

- Aluminium sulfate Al2(SO4)3 dissociates in two aluminium ions and three sulfate ions, therefore, van't Hoff factor is 5 (incorrect).

- Ammonium nitrate NH4NO3 dissociates in one ammonium ions and one nitrate ion, therefore, van't Hoff factor is 2 (correct).

- Sodium sulfate Na2SO4 dissociates in two sodium ions and one sulfate, therefore, van't Hoff factor is 3 (correct).

- Sucrose is not ionized, therefore, van't Hoff factor is 1 (correct).

Best regards.

Answer:

a. spontaneous

Explanation:

Hello,

In this case, since the Gibbs free energy is a metric that allows us to know whether a chemical reaction is spontaneous (Gibbs free energy less than 0) or nonspontaneous (Gibbs free energy greater than 0) we can mathematically define it as:

[tex]\Delta G=\Delta H-T\Delta S[/tex]

Thus, if the enthalpy is negative and the entropy is negative, the subtraction become always negative, for which the Gibbs free energy is negative as well, therefore, based on the aforementioned, any process with a negative change in enthalpy and a positive change in entropy will be: a. spontaneous.

Best regards.

Answer:

A. Disturbing the equilibrium by adding NO2Cl decreases Qc to a value less than Kc.

B. To reach a new state of equilibrium, Qc therefore increases which means that the denominator of the expression for Qc decreases.

C. To accomplish this, the concentration of reagents decreases, and the concentration of products increases.

Explanation:

Hello,

In this case, for the equilibrium reaction:

[tex]NO_2Cl(g)+NO(g)\rightleftharpoons NOCl(g)+NO_2(g)[/tex]

Whose equilibrium expression is:

[tex]Kc=\frac{[NO_2][NOCl]}{[NO_2Cl][NO]}[/tex]

The proper matching is:

A. Disturbing the equilibrium by adding NO2Cl decreases Qc to a value less than Kc, since the denominator becomes greater, therefore, Qc decreases.

B. To reach a new state of equilibrium, Qc therefore increases which means that the denominator of the expression for Qc decreases, since the lower the denominator, the higher Qc as it has the concentration of reactants.

C. To accomplish this, the concentration of reagents decreases, and the concentration of products increases, since the reactants must be consumed in order to reestablish equilibrium by shifting the reaction towards the products.

Best regards.

Answer:

1. This reaction is: B. Endothermic.

2. When the temperature is increased the equilibrium constant, K: A. Increases.

3. When the temperature is increased the equilibrium concentration of NO2: A. Increases.

Explanation:

Hello,

In this case, considering the images, we can state that the red color at high temperature is due to the presence of nitrogen dioxide (product) and the lower coloring is due to the presence of dinitrogen tetroxide (reactant) at low temperature.

With the aforementioned, we can conclude that the chemical reaction:

[tex]N_2O_4(g) \rightleftharpoons 2 NO_2(g)[/tex]

Is endothermic since high temperatures favor the formation of the product and the low temperatures favor the consumption of the the reactant. thereby:

1. This reaction is: B. Endothermic.

2. When the temperature is increased the equilibrium constant, K: A. Increases. In this particular case, since the dinitrogen tetroxide has 1 molecule and nitrogen dioxide two molecules in the chemical reaction, the entropy change should be positive, therefore, by increasing the T, the Gibbs free energy of reaction becomes more negative:

[tex]G=H-TS[/tex]

As Gibbs free energy becomes more negative, the equilibrium constant becomes bigger given their relationship:

[tex]K=exp(-\frac{\Delta G}{RT} )[/tex]

3. When the temperature is increased the equilibrium concentration of NO2: A. Increases.

Regards.

Answer:

Organic chemist? I do not know.

Explanation:

Thanks you.

The type of chemist exclusively studies most carbon compounds are organic chemist. Therefore, option C is correct.

The structure, characteristics, and reactivity of compounds containing carbon are studied by organic chemists. Additionally, they create novel organic materials with distinct features and uses.

Analytical capabilities, communication skills, and numeracy skills are three of the most important soft skills for an organic chemist.

Organic chemists often work in research and development in labs at universities, pharmaceutical, industrial, and biotechnology businesses, as well as government agencies, according to the American Chemical Society.

According to one assessment, organic chemistry is the hardest college course. According to certain statistics, almost one out of every two students in organic chemistry leave the course. The hopes of a medical career come tumbling down for those who fit this description. Organic chemistry is undoubtedly challenging.

Thus, option C is correct.

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#SPJ5

Answer: C. the same amount of energy as

Explanation:

A reversible reaction is a chemical reaction where the reactants form products that, in turn, react together to give the reactants back.

Reversible reactions will reach an equilibrium point where the concentrations of the reactants and products will no longer change.

[tex]A+B\rightleftharpoons C+D[/tex]

Thus if forward reaction is exothermic i.e. the heat is released , the backward reaction will be endothermic i.e. the heat is absorbed and in same amount.

The amount of energy released will be equal and opposite in sign to the energy absorbed in that reaction.

Answer:

C.) the same amount of energy as

Explanation:

I got it correct on founders edtell

Answer:

The interpretation of the particular subject is covered in the subsection below in detail.

Explanation:

Answer:

The amount of heat that is absorbed when 3.11 g of water boils at atmospheric pressure is 7.026 kJ.

Explanation:

A molar heat of vaporization of 40.66 kJ / mol means that 40.66 kJ of heat needs to be supplied to boil 1 mol of water at its normal boiling point.

To know the amount of heat that is absorbed when 3.11 g of water boils at atmospheric pressure, the number of moles represented by 3.11 g of water is necessary. Being:

the molar mass of water is:

H₂O= 2* 1 g/mole + 16 g/mole= 18 g/mole

So: if 18 grams of water are contained in 1 mole, 3.11 grams of water in how many moles are present?

[tex]moles of water=\frac{3.11 grams*1 mole}{18 gramos}[/tex]

moles of water= 0.1728

Finally, the following rule of three can be applied: if to boil 1 mole of water at its boiling point it is necessary to supply 40.66 kJ of heat, to boil 0.1728 moles of water, how much heat is necessary to supply?

[tex]heat=\frac{0.1728 moles*40.66 kJ}{1 mole}[/tex]

heat= 7.026 kJ

The amount of heat that is absorbed when 3.11 g of water boils at atmospheric pressure is 7.026 kJ.

Answer:

see the photo

Explanation:

it was the answer

Answer:its saturated   or  unsaturaded

Explanation:

Answer: The sulphur hexafluoride will have a chemical formula of [tex]SF_6[/tex]

Explanation:

A covalent compound is a compound where the sharing of electrons takes place between two elements which are non-metals.

The naming of covalent compound is given by:

1. The less electronegative element is written first.

2. The more electronegative element is written second. Then a suffix is added with it. The suffix added is '-ide'.

3. If atoms of an element is greater than 1, then prefixes are added which are 'mono' for 1 atom, 'di' for 2 atoms, 'tri' for 3 atoms and so on.

Thus sulphur hexafluoride will have a chemical formula of [tex]SF_6[/tex]

Answer:

D. sf6

Explanation:

Answer:

Volume percent is defined as: v/v % = [(volume of solute)/(volume of solution)] x 100%

Explanation:

Answer:

They get more energy, so they vibrate!

Explanation:

Answer:12.9e-12g or in short 12.9pg

Explanation:as p=1e-12

Answer:

d. HF(g)

Explanation:

Hello,

In this case, the standard entropy S° could be predicted by looking at the amount of bonds the compound has, thus, the fewer the number bonds, the lower the standard entropy, it means that d. HF(g) has lowest value as it has one bond only whereas methane has four bonds, water two bonds and ammonia three bonds.

Best regards.

Answer:

5000g

Explanation:

mass= density × volume

Since the unit of density here is g/mL, we need to convert the volume to mL.

1L= 1000mL

4L= 4 ×1000 = 4000 mL

Mass of object

= 1.25 ×4000

= 5000g

Answer:

5,000 grams

Explanation:

The mass of an object can be found by multiplying the volume by the density.

mass= volume * density

The density is 1.25 g/mL and the volume is 4 L.

First, we must convert the volume to mL. The density is given in grams per milliliter, but the volume is given in liters.

There are 1,000 mL per L. The volume is 4 L. Therefore, we can multiply 4 and 1,000.

4 * 1,000 = 4,000

The volume is 4,000 mL.

Now, find the mass of the object.

mass= volume * density

volume = 4,000

density= 1.25

mass= 4,000 * 1.25 = 5,000

Add the appropriate units for mass, in this case, grams, or g.

mass= 5,000 g

The mass of the object is 5,000 grams.

Answer:

0.0630

Explanation:

The molar mass of urea = 60 g/mol

we all know that:

[tex]\mathtt{number \ of \ moles = \dfrac{mass }{molar \ mass}}[/tex]

Then; the number of moles of urea

= [tex]\mathtt{\dfrac{4.0 \ g}{60 \ g/mol}}[/tex]

= 0.0667 mol

Similarly; the number of moles of methanol

= [tex]\mathtt{\dfrac{32 \ g}{32.04 \ g/mol}}[/tex]

= 0.9988 mol

The total number of moles = (0.0667 + 0.9988) mol

= 1.0655 mol

Finally,the mole fraction of urea  [tex]\mathtt{(X_{urea})}[/tex] = [tex]\mathtt{\dfrac{ n_{urea}}{(n_{urea}+n_{methanol})}}[/tex]

[tex]\mathtt{(X_{urea})}[/tex] = [tex]\mathtt{\dfrac{0.0667 \ mole}{1.0655 \ mole}}[/tex]

= 0.0630