Using the data: C2H4(g), = +51.9 kJ mol-1, S° = 219.8 J mol-1 K-1 CO2(g), = ‑394 kJ mol-1, S° = 213.6 J mol-1 K-1 H2O(l), = ‑286.0 kJ mol-1, S° = 69.96 J mol-1 K-1 O2(g), = 0.00 kJ mol-1, S° = 205 J mol-1 K-1 calculate the maximum amount of work that can be obtained, at 25.0 °C, from the process: C2H4(g) + 3 O2(g) → 2 CO2(g) + 2 H2O(l)

Answer:

a) NH3/NH4Cl (pKa = 9.25)

b) [tex]\frac{[Base]}{[Acid]} =0.708[/tex]

c)

Explanation:

Hello,

a) In this case, for a buffering capacity, if we want to select the best buffer, we should ensure that the buffer's pKa approaches the desired pH, therefore, since the buffer NH3/NH4Cl has a pKa of 9.25 that is very close to the desired pH of 9.10, we can pick it as the best choice.

b) In this case, we use the Henderson-Hasselbach equation in order to compute the molar ratio:

[tex]pH=pKa+log(\frac{[Base]}{[Acid]} )\\\\log(\frac{[Base]}{[Acid]} )=9.10-9.25=-0.15\\\\\frac{[Base]}{[Acid]} =10^{-0.15}\\\\\frac{[Base]}{[Acid]} =0.708[/tex]

c) Finally, for the ratio of masses, we use the molar mass of both ammonia as the base (17 g/mol) and ammonium chloride as the acid (53.45 g/mol) to compute it, assuming 1.00 L as the volume of the solution:

[tex]\frac{m_{Base}}{m_{Acid}} =0.708\frac{molBase}{molAcid}*\frac{17gBase}{1molBase} *\frac{1molAcid}{53.45gAcid}\\ \\\frac{m_{Base}}{m_{Acid}} =0.225[/tex]

Regards.

The best choice to create a buffer with pH 9.10 is NH₃/NH₄Cl (pKa=9.25), ratio of molarities and masses for NH₃/NH₄Cl are 0.708 & 0.225 respectively.

pH of buffer solution will be calculated by using the Henderson Hasselbalch equation as:

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

        log([NH₃]/[NH₄Cl]) = 9.10 - 9.25 =

        log([NH₃]/[NH₄Cl]) = -0.15

       [NH₃]/[NH₄Cl] = [tex]10^{-0.15}[/tex] = 0.708

        M = n/V, where

        M = molarity

        V = volume = 1L

        n = moles = W(mass) / M(molar mass)

        Mass(NH₃)/Mass(NH₄Cl) = 0.708 {(mol of NH₃×17g of NH₃NH₃) /          

                                                              (mol of NH₄Cl×53.45g of NH₄Cl)

       Mass(NH₃)/Mass(NH₄Cl) = 0.225

Hence required values are calculated above.

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Answer:

B, the number of molecules in 1 mole of a substance

Answer:

Bronsted lowry base = Proton acceptor = H2O

Bronsted lowry acid = Proton donor = HF-

Explanation:

The equation is given as;

HF-(aq) + H2O(l) --> F-(aq) + H3O(aq)

A bronsted lowry base is any specie that can accept hydrogen ion (proton) from another molecule.

Basically a bronsted lowry base is  a proton acceptor while a bronsted lowry acid is a proton donor.

In the reaction above, upon comparing both the reactants and products;

Bronsted lowry base = Proton acceptor = H2O

Bronsted lowry acid = Proton donor = HF-

Explanation:

a. Esters

b. Carboxylic acids

c. Esters (ethyl hexanoate smells like pineapple)

d. Carboxylic acids (produces a carboxylic salt)

For the given phrases the following description is better.

a. Esters

b. Carboxylic acids

c. Esters (ethyl hexanoate smells like pineapple)

d. Carboxylic acids (produces a carboxylic salt)

An ester is a synthetic compound got from a corrosive in which somewhere around one - OH hydroxyl bunch is supplanted by an - O-alkyl (alkoxy) bunch, as in the replacement response of a carboxylic acid and a liquor.

Carboxylic acid is any of a class of natural mixtures in which a carbon (C) particle is clung to an oxygen (O) molecule by a twofold bond and to a hydroxyl bunch (―OH) by a solitary bond.

Find more information about esters here:

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Answer:

Water

Explanation:

When a base react to and hydrogen ion, we can produce water.

According to these equation

H⁺  +  OH⁻   ⇄   H₂O       Kw: 1×10¹⁴

Remember that OH⁻ is determined by a strong base.

This reaction is called neutralization. You can also produce water with a weak base, because OH⁻ are released. For example, let's mention ammonia which is a weak base, it takes protons from water (H⁺)

NH₃  +  H₂O  ⇄  OH⁻  +  NH₄⁺          Kb

When the ammonium ion (acid), reacts to a base, you produce water.

NH₄⁺   +  NaOH →   NH₃ +  H₂O  +  Na⁺

Answer:

310 mL

Explanation:

Step 1: Given data

Step 2: Calculate the initial volume

We have a concentrated HBr solution and we want to prepare a diluted one. We can do so using the dilution rule.

C₁ × V₁ = C₂ × V₂

V₁ = C₂ × V₂ / C₁

V₁ = 0.400 M × 5.50 L / 7.10 M

V₁ = 0.310 L = 310 mL

Answer:

(1) -12 Kcal/mol

Explanation:

Our answer options for this question are:

(1) -12 Kcal/mol

(2) -13 Kcal/mol

(3) -15 Kcal/mol

(4) -16 Kcal/mol

With this in mind, we can start with the chemical reaction (Figure 1). In this reaction, two bonds are broken, a C-H and a Br-Br. Additionally, a C-Br and a H-Br are formed.

If we want to calculate the enthalpy value, we can use the equation:

ΔH=ΔHbonds broken-ΔHbonds formed

If we use the energy values reported, its possible to calculate the energy for each set of bonds:

ΔHbonds broken

C-H = 94.5 Kcal/mol

Br-Br = 51.5 Kcal/mol

Therefore:

105 Kcal/mol + 53.5 Kcal/mol = 146 Kcal/mol

ΔHbonds formed

C-Br = 70.5 Kcal/mol

H-Br = 87.5 Kcal/mol

Therefore:

70.5 Kcal/mol + 87.5 Kcal/mol = 158 Kcal/mol

ΔH of reaction

ΔH=ΔHbonds broken-ΔHbonds formed=(146-158) Kcal/mol = -12 Kcal/mol

I hope it helps!

Answer:

To prepare vegetables for finishing by grilling, sautéing, pan frying, deep frying, or stewing, you should parboil them to cook them to partial doneness.

The correct answer is 3 liters

Explanation:

If a substance has a solubility of [tex]12 \frac{grams}{liter}[/tex], this means in 1 liter, the grams that can be dissolved are 12 grams. Now, considering Justin and Ellie need to dissolve 36 grams to calculate the number of liters just divide the total of grams into 12 as each liter dissolves only 12 grams. The process is shown below:

36 grams (the amount that will be dissolved) ÷ 12 (grames dissolved per liter) = 3 liters (liters to dissolved 36 grams)

Answer:

It would be 3 liters

Explanation:

Answer:

The anawer of this question is 0.024 m/h

Explanation:

Other explanations of the question are additional.

Answer:

12.5g

Explanation:

Half life = 2.4 Minutes.

The half life of a compound is the time it takes to decay to half of it's original concentration or mass.

Time lapsed= 7.2 minutes. This is equivalent to 3 half lives ( 3 * 2.4)

Initial mass = 100g

First half life;

100g --> 50g

Second half life;

50g --> 25g

Third half life;

25g --> 12.5g

The amount of Zn-71 that remains after 7.2 mins has elapsed is 12.5 g

We'll begin by calculating the number of half-lives that has elapsed. This can be obtained as follow:

Half-life (t½) = 2.4 mins

Time (t) = 7.2 mins

[tex]n = \frac{t}{t_{1/2}} \\\\n = \frac{7.2}{2.4} \\\\[/tex]

Thus, 3 half-lives has elapsed.

Original amount (N₀) = 100 g

Number of half-lives (n) = 3

[tex]N = \frac{N_{0}}{2^{n}} \\\\N = \frac{100}{2^{3 }}\\\\N = \frac{100}{8}\\\\[/tex]

Thus, the amount of Zn-71 that remains after 7.2 mins is 12.5 g

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Answer:

[Ag⁺] = 5.0x10⁻¹⁴M

Explanation:

The product solubility constant, Ksp, of the insoluble salts PbI₂ and AgI is defined as follows:

Ksp(PbI₂) = [Pb²⁺] [I⁻]² = 1.4x10⁻⁸

Ksp(AgI) = [Ag⁺] [I⁻] = 8.3x10⁻¹⁷

The PbI₂ just begin to precipitate when the product  [Pb²⁺] [I⁻]² = 1.4x10⁻⁸

As the initial [Pb²⁺] = 0.0050M:

[Pb²⁺] [I⁻]² = 1.4x10⁻⁸

[0.0050] [I⁻]² = 1.4x10⁻⁸

[I⁻]² = 1.4x10⁻⁸ / 0.0050

[I⁻]² = 2.8x10⁻⁶

So, as the [I⁻] concentration is also in the expression of Ksp of AgI and you know concentration in solution of I⁻ = 1.67x10⁻³M:

[Ag⁺] [I⁻] = 8.3x10⁻¹⁷

[Ag⁺] [1.67x10⁻³] = 8.3x10⁻¹⁷

Answer:

[tex]M=0.684M[/tex]

Explanation:

Hello,

In this case, considering that the solution is formed by NaCl as the solute and water as the solvent, we can compute the molarity as shown below:

[tex]M=\frac{mol_{solute}}{V_{solution}}[/tex]

Whereas the volume of the solution must be in liters. In such a way, since the addition of sodium chloride does not significantly changes the volume of the solution we can say it remains in 100 mL (0.100 L) and the moles of sodium chloride are computed by using its molar mass (58.45 g/mol):

[tex]mol_{solute}=4g*\frac{1mol}{58.45g} =0.0684mol[/tex]

Therefore, the molarity is:

[tex]M=\frac{0.0684mol}{0.100L} \\\\M=0.684\frac{mol}{L}=0.684M[/tex]

Regards.

Answer:

When equation for neutralization of HBr by Ca(OH)₂ is correctly balanced, 1.2046*10²⁴ molecules of water will be formed

Explanation:

A neutralization reaction is one in which an acid (or acidic oxide) reacts with a base (or basic oxide). In the reaction a salt is formed and in most cases water is formed. A Salt is an ionic compound formed by the union of ions and cations through ionic bonds.

In the reactions of a strong acid (those substances that completely dissociate) with a strong base (they dissociate completely, giving up all their OH-), the complete neutralization of the species is carried out:

2 HBr (aq) + Ca(OH)₂ (s) → CaBr₂ (aq) + 2 H₂O (l)

The reaction is already balanced, complying with the law of conservation of matter. This law states that since no atom can be created or destroyed in a chemical reaction, the number of atoms that are present in the reactants must be equal to the number of atoms present in the products.

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), 2 moles of water H₂O are formed.

On the other hand, Avogadro's Number or Avogadro's Constant is called the number of particles that make up a substance (usually atoms or molecules) and that can be found in the amount of one mole of said substance. Its value is 6.023 * 10²³ particles per mole. Avogadro's number applies to any substance.

Then you can apply the following rule of three: if 1 mole of H₂O contains 6.023*10²³ molecules, 2 moles of H₂O, how many molecules does it contain?

[tex]amount of molecules=\frac{2moles*6.023*10^{23}molecules }{1 mole}[/tex]

amount of molecules= 1.2046*10²⁴ molecules

When equation for neutralization of HBr by Ca(OH)₂ is correctly balanced, 1.2046*10²⁴ molecules of water will be formed

Answer:

B. 58°C

Explanation:

Hello,

In this case, the relationship among heat, mass, specific heat and temperature for water is mathematically by:

[tex]Q=mCp\Delta T=mCp(T_2-T_1)[/tex]

In such a way, solving for the final temperature [tex]T_2[/tex] we obtain:

[tex]T_2=T_1+\frac{Q}{mCp}[/tex]

Therefore, we final temperature is computed as follows, considering that the involved heat is negative as it is lost for water:

[tex]T_2=85\°C+\frac{-15kJ*\frac{1000J}{1kJ} }{135g*4.184\frac{J}{g\°C} }\\\\T_2=58\°C[/tex]

Thereby, answer is B. 58°C .

Regards.

Answer:

a. [tex]3CH_4+2H_2O+CO_2\rightarrow 4CH_3OH[/tex]

b. [tex]2N_2O_5\rightarrow 2N_2 + 5O_2[/tex]

c. [tex]2H_2+2CO_2+O_2\rightarrow 2H_2CO_3[/tex]

Explanation:

Hello,

In this case, since those rate relationships have the stoichiometric coefficient at the denominators of the fractions preceding each ratio and the negative terms account for reactants and positive for products, we have:

a. [tex]3CH_4+2H_2O+CO_2\rightarrow 4CH_3OH[/tex]

b. [tex]2N_2O_5\rightarrow 2N_2 + 5O_2[/tex]

c. [tex]2H_2+2CO_2+O_2\rightarrow 2H_2CO_3[/tex]

Best regards.

Answer:

27°C or 300K

Explanation

We were told that the pressureof the system decreased by 10 times implies that P2= P1/10

Where P2=final pressure

P1= initial pressure

Wew were also told that the volume of the system increased by 5 times this implies that V2= 5×V1

Where T2= final temperature =-123C= 273+(-123C)=150K

T1= initial temperature

But from gas law

PV=nRT

As n and R are constant

P1V1/T1 = P2V2/T2

T1= P1V1T2/P2V2

T1=2×T2

T1=2×150

T1=300K

=300-273

=27°C

the initial temperature (°C) of a system is 27°C

Answer:

131836.43 pounds

Explanation:

one kilogram is 2.20462 pounds. multiply 2.20462 by 59800

Answer: 131836.43

Formula: Multiply the mass value by 2.205

59800×2.205=131836.43

Answer:

[tex][H^+]=0.000123M[/tex]

[tex]pH=3.91[/tex]

Explanation:

Hello,

In this case, dissociation reaction for acetic acid is:

[tex]CH_3COOH\rightleftharpoons CH_3COO^-+H^+[/tex]

For which the equilibrium expression is:

[tex]Ka=\frac{[CH_3COO^-][H^+]}{[CH_3COOH]}[/tex]

Which in terms of the reaction extent [tex]x[/tex] could be written as:

[tex]1.74x10^{-5}=\frac{x*x}{[CH_3COOH]_0-x}=\frac{x*x}{0.0010M-x}[/tex]

Thus, solving by using a solver or quadratic equation we obtain:

[tex]x_1=0.000123M\\\\x_2=-0.000141M[/tex]

And clearly the result is 0.000123M, which also equals the concentration of hydronium ion in solution:

[tex][H^+]=0.000123M[/tex]

Now, the pH is computed as follows:

[tex]pH=-log([H^+])=-log(0.000123)\\\\pH=3.91[/tex]

Best regards.

Answer:

the answer is 2,000 nickels.

Explanation:

we multiplied 100 by 100, because there are 100 cents in a dollar, and we divided 10,000 by 5, because there are 5 cents in a nickel.

Answer:

We can have: Calcium, strontium, or barium

Explanation:

In this case, we have to remember the solubility rules for sulfate [tex]SO_4~^-^2[/tex] and the chloride [tex]Cl^-[/tex]:

Sulfate

All sulfate salts are SOLUBLE-EXCEPT those also containing: Calcium, silver, mercury (I), strontium, barium or lead.([tex]Ca^+^2~,Ag^+~,Hg_2^+^2~,Sr^+2~,Ba^+^2~,Pb^+^2[/tex]), which are NOT soluble.

Chloride

All chloride salts as SOLUBLE-EXCEPT those also containing: lead, silver, or mercury (I). ([tex]Pb^+^2~,Ag^+~,Hg_2~^+^2[/tex]), which are NOT soluble.

If we the salt formed a precipitated with the sulfate anion, we will have as possibilities "Calcium, silver, mercury (I), strontium, barium or lead". If We dont have any precipitated with the Chloride anion we can discard "Silver, mercury (I),  lead" and our possibilities are:

"Calcium, strontium, or barium".

I hope it helps!

Answer:

H₂CO₃   H₂O  ⇄  HCO₃⁻  +  H₃O⁺          Ka1

HCO₃⁻  +  H₂O  ⇄  CO₃⁻²  +  H₃O⁺        Ka2

CO₃⁻²  +  H₂O  ⇄  HCO₃⁻   +  OH⁻       Kb1

HCO₃⁻  +  H₂O  ⇄   H₂CO₃  +  OH⁻     Kb2

Explanation:

Formula for carbonic acid is: H₂CO₃

It is a dyprotic acid, because it can release two protons. We can also mention that is a weak one. The equilibrums are:

H₂CO₃   H₂O  ⇄  HCO₃⁻  +  H₃O⁺          Ka1

HCO₃⁻  +  H₂O  ⇄  CO₃⁻²  +  H₃O⁺        Ka2

When the conjugate strong bases, carbonate and bicarbonate take a proton from water, the reactions are:

CO₃⁻²  +  H₂O  ⇄  HCO₃⁻   +  OH⁻       Kb1

HCO₃⁻  +  H₂O  ⇄   H₂CO₃  +  OH⁻     Kb2

Notice, that bicarbonate anion can release or take a proton to/from water. This is called amphoteric,

Answer:

0.0583g

Explanation:

The equation of the reaction is;

2HNO3(aq) + Mg(OH)2(aq) -------> Mg(NO3)2(aq) + 2H2O(l)

From the question, number of moles of HNO3 reacted= concentration × volume

Concentration of HNO3= 0.100 M

Volume of HNO3 = 20.00mL

Number of moles of HNO3= 0.100 × 20/1000

Number of moles of HNO3 = 2×10^-3 moles

From the reaction equation;

2 moles of HNO3 reacts with 1 mole of Mg(OH)2

2×10^-3 moles reacts with 2×10^-3 moles ×1/2 = 1 ×10^-3 moles of Mg(OH)2

But

n= m/M

Where;

n= number of moles of Mg(OH)2

m= mass of Mg(OH)2

M= molar mass of Mg(OH)2

m= n×M

m= 1×10^-3 moles × 58.3 gmol-1

m = 0.0583g

Answer:

245.66g of NH₄Cl is the mass we need to add to obtain the desire pH

Explanation:

The mixture of NH3/NH4Cl produce a buffer. We can find the pH of a buffer using H-H equation:

pH = pKa + log [A⁻] / [HA]

Where [A⁻] is the molar concentration of the base, NH₃, and [HA] molar concentration of the acid, NH₄⁺. This molar concentration can be taken as the moles of each chemical

First, we need to find pKa of NH₃ using Kb. Then, the moles of NH₃ and finally replace these values in H-H equation to solve moles of NH₄Cl we need to obtain the desire pH.

pKb = - log Kb

pKb = -log 1.8x10⁻⁵ = 4.74

pKa = 14 - pKb

pKa = 14 - 4.74

pKa = 9.26

2.00L ₓ (0.200mol NH₃ / L) = 0.400 moles NH₃

pH = pKa + log [NH₃] / [NH₄Cl]

8.20 = 9.26 + log [0.400 moles] / [NH₄Cl]

-1.06 =  log [0.400 moles] / [NH₄Cl]

0.0087 =  [0.400 moles] / [NH₄Cl]

[NH₄Cl] = 0.400 moles / 0.0087

[NH₄Cl] = 4.59 moles of NH₄Cl we need to add to original solution to obtain a pH of 8.20. In grams (Using molar mass NH₄Cl=53.491g/mol):

4.59 moles NH₄Cl ₓ (53.491g / mol) =

Answer:

Due to the resonance structures

Explanation:

In the question:

"Explain why, when the guanidino group of arginine is protonated, the double-bonded nitrogen is the nitrogen that accepts the proton. There is a scheme of a reversible reaction, where one equivalent of the reactant reacts with two equivalents of H plus"

We have to take into account the structure of the amino acid arginine. In which, we have the amino and the carboxylic groups in the right and the guanidine group in the left.

In this group, we have a central carbon with three nitrogen atoms around and a double bond with the nitrogen on the top. This nitrogen on the top will accept the proton because the structure produced will have a positive charge on this nitrogen. Then, the double bond with the carbon can be delocalized into the nitrogen producing a positive charge in the carbon.

In this structure (the carbocation), we can have several resonance structures. In the blue option, we can produce a double bond with the nitrogen on the right. In the purple option, we can produce a double bond with the nitrogen on the left.

In conclusion, if the nitrogen in the top on the guanidine group accepts an hydrogen atom and we will have several resonance structures that can stabilize the molecule. Due to this, the nitrogen in the top its the best option to accept hydrogens.

See figure 1

I hope it helps!

Answer:

The empirical formulae for the compound is CH3.

Answer:

[tex]H_2SO_4(aq)\rightarrow H^+(aq)+HSO_4^-\\\\HSO_4^-(aq)\rightarrow H^+(aq)+SO_4^{2-}rightarrow[/tex]

Explanation:

Hello,

In this case, given that the sulfuric acid is a diprotic acid (two hydrogen ions) we can identify two ionization reactions, the first one, showing up the dissociation of the first hydrogen to yield hydrogen sulfate ions and the second one, showing up the dissociation of the hydrogen sulfate ions to hydrogen ions and sulfate ions by separated as shown below:

[tex]H_2SO_4(aq)\rightarrow H^+(aq)+HSO_4^-\\\\HSO_4^-(aq)\rightarrow H^+(aq)+SO_4^{2-}[/tex]

They are have one-sensed arrow, since sulfuric acid is a strong acid.

Regards.

The equations that represent the first and second ionization steps for sulfuric acid in water are H₂SO₄→HSO₄+H⁺ & HSO₄⁻→SO₄⁻+H⁺ respectively.

Ionization reactions are those reactions in which atom or molecule will convert into ion by bearing a positive or negative charge on itself.

In water in the following way ionization of sulphuric acid takes place:

        H₂SO₄ → HSO₄⁻ + H⁺

        HSO₄⁻ → SO₄⁻ + H⁺

Hence, two steps are shown above.

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Answer:

Empirical formula is: C₂H₅

Explanation:

The chemical equation of burning of a compound that conatins only Carbon and Hydrogen is:

CₓHₙ + O₂ → XCO₂ + n/2H₂O

That means the moles of CO₂ produced are the moles of Carbon in the compound and moles of hydrogen are twice moles of water. Empirical formula is the simplest ratio between moles of each element in the compound. Thus, finding molse of C and moles of H we can find empirical formula:

Moles C and H:

Moles C = Moles CO₂:

7.851g CO₂ ₓ (1mol / 44g) = 0.1784 moles CO₂ = Moles C

Moles H = 2 Moles H₂O

4.018g H₂O ₓ (1mol / 18.01g) = 0.2231 * 2 = 0.4417 moles H

Ratio C:H

The ratio between moles of hydrogen and moles of Carbon are:

0.4417 moles H / 0.1784 moles C = 2.5

That means there are 2.5 moles of H per mole of Carbon. As empirical formula must be given only in whole numbers,

Answer:

The rms speed of its molecules becomes. (T) has become four times. Therefore, v_(rms) will become two times,...