A researcher places a reactant for decomposition in an expandable reaction chamber and purges the air from the vessel with nitrogen gas. The 500mL reaction vessel is sealed at a pressure of 1.00atm and 390K. If the decomposition reaction was triggered by an electrical shock, producing 3.1g of oxygen gas, what would the volume (L) of the reaction vessel be if the temperature and pressure were kept constant

Answer:

The heat absorbed by the sample of water is 3,294.9 J

Explanation:

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

The sensible heat of a body is the amount of heat received or transferred by a body when it undergoes a temperature variation (Δt) without there being a change of physical state (solid, liquid or gaseous). Its mathematical expression is:

Q = c * m * ΔT

Where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.

In this case:

Replacing:

Q= 4.184 [tex]\frac{J}{g*C}[/tex] * 45 g* 17.5 C

Solving:

Q=3,294.9 J

The heat absorbed by the sample of water is 3,294.9 J

Answer:

KNO₃ - Potassium nitrate

Fe(OH)₃ - Iron(III) hydroxide

H₂SO3 - Sulfurous acid

Al₂(SO₄)₃ - Aluminum sulfate

Hope that helps.

Answer:

OH⁻ is the Bronsted-Lowry base.

Explanation:

A Bronsted-Lowry base is a substance that accepts protons. In the chemical equation, OH⁻ accepts a proton from H₂PO₄⁻ to become H₂O. H₂PO₄⁻ would be a Bronsted-Lowry acid because it donates a proton to OH⁻ and becomes HPO₄²⁻.

Hope that helps.

The representative elements are the elements in the tall columns of the periodic table.

The representative elements that can also be referred to as the main group elements. They can be used to represent the chemistry of the group to which they belong.

Hence, the representative elements are the elements in the tall columns of the periodic table.

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Answer:

The concentration takes 25.360 seconds to decrease to one-ninth of its initial value.

Explanation:

The decomposition of the compound has an exponential behavior and process can be represented by this linear first-order differential equation:

[tex]\frac{dc}{dt} = -\frac{1}{\tau}\cdot c(t)[/tex]

Where:

[tex]\tau[/tex] - Time constant, measured in seconds.

[tex]c(t)[/tex] - Concentration of the compound as a function of time.

The solution of the differential equation is:

[tex]c(t) = c_{o} \cdot e^{-\frac{t}{\tau} }[/tex]

Where [tex]c_{o}[/tex] is the initial concentration of the compound.

The time is now cleared in the result obtained previously:

[tex]\ln \frac{c(t)}{c_{o}} = -\frac{t}{\tau}[/tex]

[tex]t = -\tau \cdot \ln \frac{c(t)}{c_{o}}[/tex]

Time constant as a function of half-life is:

[tex]\tau = \frac{t_{1/2}}{\ln 2}[/tex]

Where [tex]t_{1/2}[/tex] is the half-life of the composite decomposition, measured in seconds.

If [tex]t_{1/2} = 8\,s[/tex], then:

[tex]\tau = \frac{8\,s}{\ln 2}[/tex]

[tex]\tau \approx 11.542\,s[/tex]

And lastly, given that [tex]\frac{c(t)}{c_{o}} = \frac{1}{9}[/tex] and [tex]\tau \approx 11.542\,s[/tex], the time taken for the concentration to decrease to one-ninth of its initial value is:

[tex]t = -(11.542\,s)\cdot \ln\frac{1}{9}[/tex]

[tex]t \approx 25.360\,s[/tex]

The concentration takes 25.360 seconds to decrease to one-ninth of its initial value.

Answer:

1- yes

  HBr--hydrogen bromide

2- no

  BaBr₂----barium bromide  

3- yes

NCl----- nitrogen chlorine  

Hydrogen ,bromine and nitrogen , chlorine are the pair of elements which  will form a molecular compound by covalent bond as they have 1, 7,5, 7 valence electrons respectively.

Covalent bond is defined as a type of bond which is formed by the mutual sharing of electrons to form electron pairs between the two atoms.These electron pairs are called as bonding pairs or shared pair of electrons.

Due to the sharing of valence electrons , the atoms are able to achieve a stable electronic configuration . Covalent bonding involves many types of interactions like σ bonding,π bonding ,metal-to-metal bonding ,etc.

Sigma bonds are the strongest covalent bonds while the pi bonds are weaker covalent bonds .Covalent bonds are affected by electronegativities of the atoms present in the molecules.Compounds having covalent bonds have lower melting points as compared to those with ionic bonds.

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Answer:

-471 Kj/mole acrylic acid

Explanation:

THIS IS THE COMPLETE QUESTION BELOW;

There are two steps in the usual industrial preparation of acrylic acid, the immediate precursor of several useful plastics. In the first step, calcium carbide and water react to form acetylene and calcium hydroxide: CaC (s) + 2 H2O(g) - CH (9) + Ca(OH),(s) AH -414. kJ In the second step, acetylene, carbon dioxide and water react to form acrylic acid: 6 C H (9) + 3 CO2(9) + 4H2O(g) - SCH,CHCO,H) AH-132. kJ Calculate the net change in enthalpy for the formation of one mole of acrylic acid from calcium carbide, water and carbon dioxide from these reactions. Round your answer to the nearest kJ. x 5 ?

The two equations from the reaction can be written as;

a)CaC₂(s) + 2H₂O(l) ------->C₂H₂(g) + CaOH₂(s)

Δ H= -414Kj ........................ equation (a)

b)6C₂H₂(g) +3CO₂(g)+4H₂O(g) -------> 5CH₂CHCO₂H(g) Δ H= 132Kj ...................... equation (b)

In equation (b)acrylic acid was produced by the reaction between Acetylene carbon dioxide and water

Then we can multiply equation(a) by factor of 6 and the ΔH Then we have (6× -414Kj)= ΔH= -2484Kj.

6CaC₂(s) + 12H₂O(l) ------->6C₂H₂(g) + 6CaOH₂(s)

Δ H= -2484Kj.................. equation (c)

6C₂H₂(g) +3CO₂(g)+4H₂O(g) -------> 5CH₂CHCO₂H(g) Δ H= 132Kj

Then add equation (c) and equation(b) then we have

6CaC₂(s) + 16H₂O(l)+3CO₂(g)------> 5CH₂CHCO₂H(g) + 6CaOH₂(s) ΔH= -2352Kj

ΔH(net)= -2352Kj/5moles

=-471Kj/mole

therefore, net change in enthalpy for the formation of one mole of acrylic acid from calcium carbide, water and carbon dioxide from these reactions. Round your answer to the nearest kJ. x 5 ? is -471Kj/mole acrylic acid

Answer:

ml= 0

Explanation:

The magnetic quantum number usually gives information regarding the directionality of an atomic orbital. The values of the magnetic quantum number ranges between integer values that lie between +l to -l.

However, the n=1 level contains only the 1s orbital which is spherically symmetrical. The s orbital having l=0 also has magnetic quantum number (ml) =0, hence the answer.

Explanation:

This reaction is in equilibrium and would hence obey lechatelier's principle. This principle states that whenever a system at equilibrium undergoes a change, it would react in way so as to annul that change.

Since it is an endothermic reaction, increasing the temperature would cause the reaction to shift towards the right.

This means that it favours product formation and more of the product would be formed.

Answer:

Explanation:

We shall find out volume of air at NTP or at 273 K and 10⁵ Pa ( 1 atm )

Let it be V₂

[tex]\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}[/tex]

[tex]\frac{2\times 10^5\times 5.68}{394} =\frac{10^5\times V_2}{273}[/tex]

V₂ = 7.87 litres

22.4 litres of any gas is equivalent to 1 mole

7.87 litres of air will be equivalent to

7.87 / 22.4 moles

= .35 moles .

Answer:

The two-step mechanism is a slow mechanism and a fast mechanism. When we combine them, the result is

2H₂O₂ (aq) ⇒2H₂O (l) + O₂ (g)

Explanation:

We know that the decomposition of hydrogen peroxide is catalyzed by iodide ion, which means that the iodide ion will react with the hydrogen peroxide. There is a slow mechanism and a fast one:

H₂O₂(aq) + I₋(aq) ⇒ H₂O(l) + IO₋(aq) this is the slow reaction

IO₋(aq) + H₂O₂(aq)⇒ H₂O(l) + O₂(g) + I₋ (aq) this is the fast reaction

If we cancel the same type of molecules and ions, the final result is:

2H₂O₂ (aq) ⇒2H₂O (l) + O₂ (g)

The two-step mechanism represents the slow mechanism and a fast mechanism. At the time of combining them, the result is 2H₂O₂ (aq) ⇒2H₂O (l) + O₂ (g)

The decomposition of hydrogen peroxide should be catalyzed by iodide ion, which represents the iodide ion will react with the hydrogen peroxide. There is a slow mechanism and a fast one

Now

H₂O₂(aq) + I₋(aq) ⇒ H₂O(l) + IO₋(aq) this is the slow reaction

IO₋(aq) + H₂O₂(aq)⇒ H₂O(l) + O₂(g) + I₋ (aq) this is the fast reaction

Now in case of cancelling, the same type of molecules and ions, the final result is 2H₂O₂ (aq) ⇒2H₂O (l) + O₂ (g)

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Answer:

Complex I:  (1) NADH-ubiquinone(NADH-coenzyme Q oxidoreductase), (8) Electron transfer from NADH to ubiquinone (coenzyme Q)

Complex II:  (3) Electron transfer from succinate to ubiquinone (coenzyme Q) (5) Succinate-coenzyme Q Oxidoreductase (succinate dehydrogenase)

Complex III:  (2) Coenzyme Q-cytochrome c oxidoreductase, (7) Electron transfer from ubiquinol (QH2) to cytochrome c

Complex IV: (4) Electron transfer from cytochrome c to O2, (6) Cytochrome c oxidase

Explanation:

The electron transport chain (ETC) in the mitochondria provides a pathway by which electrons are transferred from NADH and FADH₂ through a series of membrane-bound carriers to  molecular oxygen reducing it to water.

The electron transport chain electron carriers are organized into four complexes, Complexes I - IV.

Complex I : It is also called NADH:ubiquinone reductase. It transfers electrons from NADH to ubiquinone (also known as coenzyme Q)

Complex II : It is also called succinate dehydrogenase. It functions to tranfer electrons from succinate to FAD and then to ubiquinone.

Complex III : It is also called ubiquinone:cytochrome c oxidoreductase. It functions to transfer electrons from ubiquinol (reduced ubiquinone) to cytochrome c.

Complex IV : It is also called cytochrome oxidase. It functions to transfer electrons from cytochrome c to molecular oxygen reducing it to water.

The electron transporter chain is a series of enzymatic reactions to produce and store energy for the organism’s correct functioning. Complex I: 1 and 8. Complex II: 3 and 5. Complex III: 2 and 7. Complex IV: 4 and 6.

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

Electron transporter chain

The electron transporter chain is located in the internal mitochondrial membrane. It constitutes a series of enzymatic reactions to release and save energy for the organism’s correct functioning.  

Along the chain, there are four proteinic complexes in the membrane, I, II, III, and IV, that contain the electrons transporters and the enzymes necessary to catalyze the electrons' transference from one complex to the other.  

Different redox reactions occur to pass electrons along the chain.  

Released energy creates a proton concentration gradient used to synthesize ATP.  

1)  NADH provides electrons to the first complex, Complex I (NADH-  

   ubiquinone or NADH-coenzyme Q oxidoreductase).

From there, electrons go to the coenzyme Q (Ubiquinone) that carries them to complex II and III. Meanwhile, complex I pomp four protons to the intermembrane space.  

2) Complex II (succinate-dehydrogenase) receives electrons from CoQ and also receives electrons from FADH2. Electrons are sent from complex II to ubiquinone Q that carries these electrons to complex III.

3) Complex III (Cytochrome C-reductase) receives electrons from ubiquinone Q and pomps protons to the intermembrane space.

Electrons are transferred to Cytochrome c.  

Electrons travel from cytochrome c to complex IV.

4) Complex IV (Cytochrome C-oxidase)  is the last complex that pomps protons to the intermembrane space. It takes electrons from cytochrome C and sends them to oxygen.

5) Electrons are sent to O₂ molecules, which also receive protons in the matrix to create water molecules.

Four electrons are needed to produce two water molecules from one O₂ molecule.  

The proton gradient is used to produce ATP molecules.

Now, we can join the complexes with the phrases.

Complex I:

1) NADH-ubiquinone (NADH-coenzyme Q oxidoreductase)

8) Electron transfer from NADH to ubiquinone (coenzyme Q)

Complex II:

3) Electron transfer from succinate to ubiquinone (coenzyme Q)

5) Succinate-coenzyme Q Oxidoreductase (succinate dehydrogenase)

Complex III:

2) Coenzyme Q - cytochrome c oxidoreductase

7) Electron transfer from ubiquinol (QH₂) to cytochrom c

Complex IV:

6) Cytochrome C oxidase

4) Electron transfer from cytochrome c to O₂

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

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Answer:

0.1110 mol

Explanation:

Mass = 2g

Molar mass = 18.016 g/mol

moles = ?

These quantities are realted by the following equation;

Moles = Mass / Molar mass

Substituting the values of the quantities and solving for moles, we have;

Moles = 2 / 18.016 = 0.1110 mol

Answer:

Molarity Cu²⁺ = 0.423M Cu²⁺

Explanation:

40.8g of copper (II) acetate into 200mL of a 0.700M sodium chromate

The reaction of copper acetate with sodium chromate occurs as follows:

Cu(CH₃COO)₂(aq) + Na₂CrO₄(aq) → CuCrO₄(s) + 2CH₃COONa

In water, the Copper(II) acetate dissociates in Cu²⁺ cation.

To know final molarity of Cu²⁺ we need to calculate the moles of Cu²⁺ that don't react with chromate ion, thus:

Moles of 40.8g of copper(II) acetate (Molar mass: 181.63g/mol) are:

40.8g × (1mol / 181.63g) = 0.2246 moles of Copper(II) acetate

Moles of sodium chromate are:

0.200L ₓ (0.700mol / L) = 0.140 moles of sodium chromate.

As 1 mole of Copper(II) acetate reacts per mole of sodium chromate, moles of Copper(II) acetate = Moles of Cu²⁺ that remains after the reaction are:

0.2246mol - 0.140moles = 0.0846 moles of Cu²⁺

Molarity is ratio between moles of solute (Moles Cu²⁺) and volume in liters of solution (200mL = 0.200L):

Molarity Cu²⁺ = 0.0846 moles / 0.200L

Answer:

A. releases a large amount of heat

Explanation:

A reaction is said to be spontaneous if it can proceed on its own without the addition of external energy. A spontaneous reaction is not determined by the length of time, because some spontaneous reactions are completed after a long period of time. They are exothermic in nature. An example is the conversion of graphite to carbon which takes a long period of time to complete. Spontaneous reactions are known to increase entropy in a system. Entropy is the rate of disorder in a system.

In the combustion of fire, energy is released to the surroundings as there is a decrease in energy. This is an example of a spontaneous reaction because it is an exothermic reaction, which causes an increase in entropy and a decrease in energy.

Answer:

The change in the boiling point would be 2.1°C.

Explanation:

1 ) Let us first determine the molarity of this solution :

M = mol / kg,

M = 0.402 mol / 0.200 kg = 2.01 M NaCl

2 ) ΔT = i [tex]*[/tex] K [tex]*[/tex] m

ΔT = 2 [tex]*[/tex] 0.512C/m [tex]*[/tex] 2.01m

ΔT = 2.06C

As you can see, this is none of the answer choices. However the van't Hoff factor i in this case was taken to be 2, but this value is actually less than the predicted ideal solution. This is due to the ion pairing, causing i to be around 1.7 to 1.8. Therefore our solution is option b, 2.1°C.

Answer:

C. They react strongly with water to produce an alkaline solution and hydrogen

Explanation:

All alkali metals react vigorously with cold water. In the reaction, hydrogen gas is given off and the metal hydroxide is produced.

Hope that helps.

Answer:

The hydrogen can be gotten from the added Acid or water during "workup".

Explanation:

Basically we can say that the reaction describe in this question is a Reduction reaction because of the chemical compound called sodium borohydride. In the reaction described above we can see that there is a Reduction of ketone to alcohol by the compound; sodium borohydride.

For the reduction Reaction to occur the C-O bond must break so as to enable the formation of O-H bond and C-H bond.

So, "the reaction mixture is treated with water and H2SO4 to produce the desired alcohol", thus, the oxygen will definitely pick up the hydrogen from H2SO4 or H2O.

Answer:

(a) appended underneath is the inorganic ion shaped in the reaction and the mechanism of its formation  

(b) 2-butyne framed as a minor product is appeared in the connection. It is shaped when the monosodium subordinate of dimethylsulphoxide gets a hydrogen from the propyne and reacts again with monosodium methylsulphoxide.  

(c) The major product framed when diethylsulphoxide is utilized, would be butyne and minor product would be 3-hexyne.

Explanation:

attached below is diagram

Answer:

A) [tex]N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)[/tex].

B) [tex]Kc=0.0933[/tex].

C) 0.9 mol.

D) Increasing both temperature and pressure.

Explanation:

Hello,

In this case, given the information, we proceed as follows:

A)

[tex]N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)[/tex]

B) For the calculation of Kc, we rate the equilibrium expression:

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

Next, since at equilibrium the concentration of ammonia is 0.6 M (0.9 mol in 1.5 dm³ or L), in terms of the reaction extent [tex]x[/tex], we have:

[tex][NH_3]=0.6M=2*x[/tex]

[tex]x=\frac{0.6M}{2}=0.3M[/tex]

Next, the concentrations of nitrogen and hydrogen at equilibrium are:

[tex][N_2]=\frac{1.5mol}{1.5L}-x=1M-0.3M=0.7M[/tex]

[tex][H_2]=\frac{4mol}{1.5L}-3*x=2.67M-0.9M=1.77M[/tex]

Therefore, the equilibrium constant is:

[tex]Kc=\frac{(0.6M)^2}{(0.7M)*(1.77M)^3}\\ \\Kc=0.0933[/tex]

C) In this case, the equilibrium yield of ammonia is clearly 0.9 mol since is the yielded amount once equilibrium is established.

D) Here, since the reaction is endothermic (positive enthalpy change), one way to increase the yield of ammonia is increasing the temperature since heat is reactant for endothermic reactions. Moreover, since this reaction has less moles at the products, another way to increase the yield is increasing the pressure since when pressure is increased the side with fewer moles is favored.

Best regards.

Answer:

No, because sulfur does not typically form negative ions or oxidation states less than 2−. The binary compound formed by sulfur and sodium is Na2S

Explanation:

Sulphur is a member of group 16. The oxidation states expected for sulphur in group 16 are -1, -2, +1, +2,+3,+4,+5 or +6. The elements of group 16 usually form negative ions with oxidation number of -2. They do not typically form negative ions with oxidation state less than -2.

The implication of this is that we actually do not expect the existence of a compound in which sulphur forms an S^4- anion. In reality, such an anion does not exist. Rather a binary compound of sulphur and sodium will have the formula Na2S because it contains the S^2- anion.

Answer:

empirical formula = C3H7

molecular formula = C6H14

Answer:

Keq = 0.217

Explanation:

Let's determine the equilibrium reaction.

In gaseous state, water vapor can be decomposed to hydrogen and oxygen and this is a reversible reaction.

2H₂(g)  + O₂(g)  ⇄  2H₂O (g)         Keq

Let's make the expression for the equilibrium constant

Products / Reactants

We elevate the concentrations, to the stoichiometry coefficients.

Keq = [H₂O]² / [O₂] . [H₂]²

Keq = 0.250² / 0.8 . 0.6² =  0.217

Answer:

There are compounds that contain phosphorus-phosphorus bonds with uncommon oxidation states.

Explanation:

Phosphorus is a member of group 15 in the periodic table. Its common oxidation States are -3 and +5. Phosphorus is believed to firm some of its compounds by the participation of hybridized d-orbitals in bonding although this is also disputed by some scientists owing to the high energy of d - orbitals.

Phosphorus form compounds having phosphorus-phosphorus bonds in unusual oxidation states such as diphosphorus tetrahydride, H2P-PH2, and tetraphosphorus trisulfide, P4S hence the answer.

Answer:

a. ΔG : It seems that the proper definition is not given, anyway, it is the free energy change.

b. ΔS : 3. Change in entropy.

c. Keq : 6. Equilibrium constant.

d. ΔH : 7. Change in enthalpy.

e. ΔG° : 9. Standard free energy change.

f. J : 8. Joule.

Explanation:

Hello,

In this case, we can match the symbol with the proper definition as shown below:

a. ΔG : It seems that the proper definition is not given, anyway, it is the free energy change and it uses G since it is better referred to the Gibbs free energy.

b. ΔS : 3. Change in entropy.

c. Keq : 6. Equilibrium constant.

d. ΔH : 7. Change in enthalpy.

e. ΔG° : 9. Standard free energy change.

f. J : 8. Joule.

Best regards.

Answer:

wavelength, λ =  486.6 nm

frequency, f = 6.16 * 10¹⁴ Hz

Explanation:

a. Wavelength

Using the wavelength equation; 1/λ = (1/hc) * 2.18 * 10⁻¹⁸ J * (1/nf² - 1/ni²)

Where nf is the final energy level; ni is the initial energy level; h is Planck's constant = 6.63 * 10⁻³⁴ J.s; c is velocity of light = 3 * 10⁸ m/s

1/λ = 1/(6.63 * 10⁻³⁴ J.s * 3 * 10⁸ m/s) * 2.18 * 10⁻¹⁸ J * (1/2² - 1/4²)

1/λ = 2.055 * 10⁶ m

λ = 4.866 * 10⁻⁷ m

wavelength, λ =  486.6 nm

b.  Frequency

Using f = c/λ

f = (3 * 10⁸ m/s) / 4.866 * 10⁻⁷ m

frequency, f = 6.16 * 10¹⁴ Hz

Answer:

Kc = [H₂CO₃] / [CO₂]

Explanation:

Equilibrium constant expression (Kc) of any reaction is defined as the ratio between molar concentrations in equilibrium of products over reactants.

Pure solids and liquids don't affect the equilibrium and you don't have to take its concentrations in the equilibrium.

Also, each specie must be powered to its reactant coefficient.

For example, for the reaction:

aA(s) + bB(aq) ⇄ cC(l) + nD(g) + xE(aq)

The equilibrium constant, kc is:

Kc = [D]ⁿ / [B]ᵇ[E]ˣ

You don't take A nor C species because are pure solids and liquids. b, n and x are the reactant coefficients of each substance. Ratio of products over reactants

Thus, for the reaction:

CO₂(aq) + H₂O(l) ⇄ H₂CO₃(aq)

The Kc is:

Answer:

The correct answer is 8.10

Explanation:

Given:

A(g) + 2B(g) ↔ AB₂(g)   Kc = 59 ---- Eq. 1

A(g) + 3B(g) ↔ AB₃(g)   Kc = 478 ----- Eq. 2

We have to rearrange the chemical equations in order to obtain:

AB₂(g) + B(g) ↔ AB₃(g) Kc = ?

AB₂(g) is a reactant, so we have to use the reverse reaction of Eq. 1, in this case Kc= 1/59. Since AB₃(g) is a product, we use the forward reaction of Eq.2, and the constant is the same: Kc= 478.  The following is the sum of rearranged chemical equations, and the compounds in bold and italic are canceled:

 AB₂(g)       ↔   A(g) + 2B(g)          Kc₁= 1/59

A(g) + 3B(g) ↔   AB₃(g)                  Kc₂= 478

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

AB₂(g) + B(g) ↔ AB₃(g)

If we add reactions at equilibrium, the equilibrium constants Kc are mutiplied as follows:

Kc = Kc₁ x Kc₂ = 1/59 x 478 = 478/59 = 8.10

The value of the missing equilibrium constant is 8.10.

The value of the missing equilibrium constant is 8.10

Since

A(g) + 2B(g) ↔ AB₂(g)   Kc = 59 ---- Eq. 1

A(g) + 3B(g) ↔ AB₃(g)   Kc = 478 ----- Eq. 2

Now we have to rearrange the chemical equations in order to obtain:

AB₂(g) + B(g) ↔ AB₃(g) Kc = ?

Here AB₂(g) represents a reactant, so we have to applied the reverse reaction of Eq. 1, in this case Kc= 1/59. Since AB₃(g) is a product, we use the forward reaction of Eq.2, and the constant should be the same: Kc= 478.  

The following is the sum of rearranged chemical equations, and the compounds in bold and italic should be canceled:

AB₂(g)       ↔   A(g) + 2B(g)          Kc₁= 1/59

A(g) + 3B(g) ↔   AB₃(g)                  Kc₂= 478

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

AB₂(g) + B(g) ↔ AB₃(g)

In the case when we add reactions at equilibrium, the equilibrium constants Kc are multiplied as follows:

Kc = Kc₁ x Kc₂ = 1/59 x 478 = 478/59 = 8.10

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Answer:

0.127 M.

Explanation:

The balanced equation for the reaction is given below:

H3PO4 + 3KOH —› K3PO4 + 3H2O

From the balanced equation above, we obtained the following data:

Mole ratio of acid, H3PO4 (nA) = 1

Mole ratio of base, KOH (nB) = 3

Data obtained from the question include:

Volume of acid, H3PO4 (Va) = 78.5 mL

Molarity of acid, H3PO4 (Ma) =...?

Volume of base, KOH (Vb) = 134 mL

Molarity of base, KOH (Mb) = 0.224 M

The concentration of the acid, H3PO4 can be obtained as follow:

MaVa / MbVb = nA/nB

Ma x 78.5 / 0.224 x 134 = 1/3

Cross multiply

Ma x 78.5 x 3 = 0.224 x 134 x 1

Divide both side by 78.5 x 3

Ma = (0.224 x 134 x 1) /(78.5 x 3)

Ma = 0.127 M

Therefore, the concentration of the acid, H3PO4 is 0.127 M.