The diffusion flux is defined as: a) Mass, per unit surface area, per unit time b) Surface area, per unit time c) Time, per unit surface area. d) The product of time and surface area, per unit mass

Answer:

11.39

Explanation:

Given that:

[tex]pK_{b}=4.82[/tex]

[tex]K_{b}=10^{-4.82}=1.5136\times 10^{-5}[/tex]

Given that:

Mass = 1.805 g

Molar mass = 82.0343 g/mol

The formula for the calculation of moles is shown below:

[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]

Thus,

[tex]Moles= \frac{1.805\ g}{82.0343\ g/mol}[/tex]

[tex]Moles= 0.022\ moles[/tex]

Given Volume = 55 mL = 0.055 L ( 1 mL = 0.001 L)

[tex]Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}[/tex]

[tex]Molarity=\frac{0.022}{0.055}[/tex]

Concentration = 0.4 M

Consider the ICE take for the dissociation of the base as:

                                  B +   H₂O    ⇄     BH⁺ +        OH⁻

At t=0                        0.4                          -              -

At t =equilibrium     (0.4-x)                        x           x            

The expression for dissociation constant is:

[tex]K_{b}=\frac {\left [ BH^{+} \right ]\left [ {OH}^- \right ]}{[B]}[/tex]

[tex]1.5136\times 10^{-5}=\frac {x^2}{0.4-x}[/tex]

x is very small, so (0.4 - x) ≅ 0.4

Solving for x, we get:

x = 2.4606×10⁻³  M

pOH = -log[OH⁻] = -log(2.4606×10⁻³) = 2.61

pH = 14 - pOH = 14 - 2.61 = 11.39

Answer:

single bond

Explanation:

Single bond is the chemical bond in which 2 valence electrons are involved. These are the sigma bond which are formed formed by the head-on overlapping between the orbitals of the two atoms involving in the bond. The bond lies on the inter nuclear axis and thus , the bond can be rotated without the breaking of bond.

On the other hand, double and triple bonds contains one sigma bond and also bonds which are formed by the sideways overlapping of the orbitals and thus, these do not lie on inter nuclear axis and breaks on rotation.

Answer:

a.

silver iodide BC

manganese(II) hydroxide A

b.

silver iodide Ksp = S²

manganese(II) hydroxide Ksp = 4S³

Explanation:

Ksp (Solubility products) are the equilibrium constants for poorly soluble compounds. As every equilibrium constant, it is formed by the product of the products raised to their stoichiometric coefficients divided by the product of reactants raised to their stoichiometric coefficients. We only include in the constant gases and aqueous species. So, to solve this task, we need to write each reaction and its Ksp.

1. Silver Iodide

AgI(s) ⇄ Ag⁺(aq) + I⁻(aq)

When AgI is put in water, the concentration of Ag⁺ and I⁻ that actually dissolve is known as solubility (S). So the concentration of both Ag⁺ and I⁻ would be S.

AgI(s) ⇄ Ag⁺(aq) + I⁻(aq)

                 S             S

We can replace this in the Ksp expression:

Ksp = [Ag⁺].[I⁻] = S.S = S²

We can follow the same steps to find out the relationship between Ksp and S for each compound.

2. Mn(OH)₂(s) ⇄ Mn²⁺(aq) + 2OH⁻(aq)

                                S             2S

In this case, the concentration of OH⁻ is 2S because 2 moles are produced along with 1 mole of Mn²⁺.

Ksp = [Mn²⁺].[OH⁻]² = S.(2S)² = 4S³

A. Fe(OH)₂(s) ⇄ Fe²⁺(aq) + 2OH⁻(aq)

                              S              2S

Ksp = [Fe²⁺].[OH⁻]² = S.(2S)² = 4S³

B. CaSO₃(s) ⇄ Ca²⁺(aq) + SO₃²⁻(aq)

                             S              S

Ksp = [Ca²⁺].[SO₃²⁻] = S.S = S²

C. NiCO₃(s) ⇄  Ni²⁺(aq) + CO₃²⁻(aq)

                            S              S

Ksp = [Ni²⁺].[CO₃²⁻] = S.S = S²              

D. Ba₃(PO₄)₂(s) ⇄ 3 Ba²⁺(aq) + 2 PO₄³⁻(aq)

                                   3S              2S

Ksp = [Ba²⁺]³.[PO₄³⁻]²= (3S)³.(2S)²= 108S⁵

a. The salts that can be compared using the same Ksp expressions are:

silver iodide BC

manganese(II) hydroxide A

b.

silver iodide Ksp = S²

manganese(II) hydroxide Ksp = 4S³

None of the listed salts can be compared to silver iodide using Ksp values. Manganese(II) hydroxide can be compared to Fe(OH)2. The Ksp expressions in terms of solubility, s, are s^2 for silver iodide and 4s^3 for manganese(II) hydroxide.

We can directly compare salts by considering their solubility product constants, or Ksp values

. 1. Silver iodide: No salts on the right share the same ions with silver iodide, so none can be directly compared.
2. Manganese(II) hydroxide: It contains Mn2+ and OH- ions. Thus, it can be compared to the Fe(OH)2 salt which contains Fe2+ and OH- ions, making the matching answer A.

For the solubility expressions for each salt, when Multiplied we put the number first and then combine all exponents:

For silver iodide (AgI), the equilibrium is AgI ↔ Ag+ + I-, so the Ksp expression is Ksp = s x s = s^2.

For manganese(II) hydroxide (Mn(OH)2), the equilibrium is Mn(OH)2 ↔ Mn2+ + 2OH-, so the Ksp expression is Ksp = 4s^3.

https://brainly.com/question/1419865

#SPJ11

Answer:

Ea = 177x10³ J/mol

ko = [tex]1.52x10^{19}[/tex] J/mol

Explanation:

The specific reaction rate can be calculated by Arrhenius equation:

[tex]k = koxe^{-Ea/RT}[/tex]

Where k0 is a constant, Ea is the activation energy, R is the gas constant, and T the temperature in Kelvin.

k depends on the temperature, so, we can divide the k of two different temperatures:

[tex]\frac{k1}{k2} = \frac{koxe^{-Ea/RT1}}{koxe^{-Ea/RT2}}[/tex]

[tex]\frac{k1}{k2} = e^{-Ea/RT1 + Ea/RT2}[/tex]

Applying natural logathim in both sides of the equations:

ln(k1/k2) = Ea/RT2 - Ea/RT1

ln(k1/k2) = (Ea/R)x(1/T2 - 1/T1)

R = 8.314 J/mol.K

ln(2.46/47.5) = (Ea/8.314)x(1/528 - 1/492)

ln(0.052) = (Ea/8.314)x(-1.38x[tex]10^{-4}[/tex]

-1.67x[tex]10^{-5}[/tex]xEa = -2.95

Ea = 177x10³ J/mol

To find ko, we just need to substitute Ea in one of the specific reaction rate equation:

[tex]k1 = koxe^{-Ea/RT1}[/tex]

[tex]2.46 = koxe^{-177x10^3/8.314x492}[/tex]

[tex]1.61x10^{-19}ko = 2.46[/tex]

ko = [tex]1.52x10^{19}[/tex] J/mol

Answer : The formula of limiting reagent is, [tex]Mg_3N_2[/tex].

The mass of [tex]Mg(OH)_2[/tex] is, 174 grams.

The mass of excess reactant is, 36 grams.

Solution : Given,

Mass of [tex]Mg_3N_2[/tex] = 101 g

Mass of [tex]H_2O[/tex] = 144 g

Molar mass of [tex]Mg_3N_2[/tex] = 101 g/mole

Molar mass of [tex]H_2O[/tex] = 18 g/mole

Molar mass of [tex]Mg(OH)_2[/tex] = 58 g/mole

First we have to calculate the moles of [tex]Mg_3N_2[/tex] and [tex]H_2O[/tex].

[tex]\text{ Moles of }Mg_3N_2=\frac{\text{ Mass of }Mg_3N_2}{\text{ Molar mass of }Mg_3N_2}=\frac{101g}{101g/mole}=1moles[/tex]

[tex]\text{ Moles of }H_2O=\frac{\text{ Mass of }H_2O}{\text{ Molar mass of }H_2O}=\frac{144g}{18g/mole}=8moles[/tex]

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

[tex]Mg_3N_2+6H_2O\rightarrow 3Mg(OH)_2+2NH_3[/tex]

From the balanced reaction we conclude that

As, 1 mole of [tex]Mg_3N_2[/tex] react with 6 mole of [tex]H_2O[/tex]

So, given 1 moles of [tex]Mg_3N_2[/tex] react with 6 moles of [tex]H_2O[/tex]

From this we conclude that, [tex]H_2O[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]Mg_3N_2[/tex] is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of [tex]Mg(OH)_2[/tex]

From the reaction, we conclude that

As, 1 mole of [tex]Mg_3N_2[/tex] react to give 3 mole of [tex]Mg(OH)_2[/tex]

So, given 1 mole of [tex]Mg_3N_2[/tex] react to give 3 moles of [tex]Mg(OH)_2[/tex]

Now we have to calculate the mass of [tex]Mg(OH)_2[/tex]

[tex]\text{ Mass of }Mg(OH)_2=\text{ Moles of }Mg(OH)_2\times \text{ Molar mass of }Mg(OH)_2[/tex]

[tex]\text{ Mass of }Mg(OH)_2=(3moles)\times (58g/mole)=174g[/tex]

The mass of [tex]Mg(OH)_2[/tex] is, 174 grams.

Now we have to calculate the moles of excess reactant [tex](H_2O)[/tex].

Moles of excess reactant = 8 - 6 = 2 moles

Now we have to calculate the mass of excess reactant.

[tex]\text{ Mass of }H_2O=\text{ Moles of }H_2O\times \text{ Molar mass of }H_2O[/tex]

[tex]\text{ Mass of }H_2O=(2moles)\times (18g/mole)=36g[/tex]

The mass of excess reactant is, 36 grams.

Answer:

The weight percent of NaBr is 25,7%

Explanation:

Molality is a way to express the concentration of a chemical in terms of moles of substances per kg of solution. Weight percent is other way to express concentration in terms of mass of substance per mass of solution per 100

Thus, to obtain weight percent you must convert moles of NaBr to grams with molar mass and these grams to kilograms, thus:

2,50 mol / kg solution × (102,9 g / 1mol) × (1 kg / 1000 g) =

[tex]\frac{0,257 kg NaBr}{kg solution }[/tex] × 100 = 25,7 % (w/w)

I hope it helps!

Explanation:

The given data is as follows.

       n = 2 mol,         P = 1 atm,         T = 300 K

        Q = +34166 J,         W= -1216 J (work done against surrounding)

       [tex]C_{v}[/tex] = [tex]\frac{3R}{2}[/tex]

Relation between internal energy, work and heat is as follows.

      Change in internal energy ([tex]\Delta U[/tex]) = Q + W

                                   = [34166 + (-1216)] J

                                   = 32950 J

Also,  [tex]\Delta U = n \times C_{v} \times \Delta T[/tex]

                      = [tex]3R \times (T_{2} - T_{1})[/tex]

                 32950 J = [tex]3 \times 8.314 J/mol K \times (T_{2} - 300 K)[/tex]

                [tex]\frac{32950}{24.942} = T_{2} - 300 K[/tex]

                            1321.06 K + 300 K = [tex]T_{2}[/tex]    

                                       [tex]T_{2}[/tex] = 1621.06 K

Thus, we can conclude that the final temperature of the gas is 1621.06 K.

Answer: The number of moles of nitrogen gas is 0.505 moles.

Explanation:

To calculate the number of moles of nitrogen gas, we use ideal gas equation, which is:

[tex]PV=nRT[/tex]

where,

P = pressure of the gas = 4.27 atm

V = Volume of the gas = 2.96 L

T = Temperature of the gas = [tex]32.0^oC=[32.0+273]K=305K[/tex]

R = Gas constant = [tex]0.0821\text{ L. atm }mol^{-1}K^{-1}[/tex]

n = number of moles of gas = ?

Putting values in above equation, we get:

[tex]4.27atm\times 2.96L=n\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 305K\\n=0.505mol[/tex]

Hence, the number of moles of nitrogen gas is 0.505 moles.

Explanation:

The given data is as follows.

     Volume = 2.96 L,       Temperature = [tex]32.0^{o}C[/tex] = (32 + 273) K = 305 K,

        Pressure = 4.27 atm,       n = ?

And, according to ideal gas equation, PV = nRT

           [tex]4.27atm \times 2.96 L= n \times 0.0821 Latm/mol K \times 305K[/tex]    

                n = [tex]\frac{12.64 atm L}{25.04 Latm/mol}[/tex]

                         = 0.505 mol

Thus, we can conclude that the number of moles of nitrogen gas in the container is 0.505 mol.  

Answer:

salt mass=102.5gCrCl3.6H2O

Explanation:

Hello !

To know how much CrCl3.6H2O we need, we follow the steps below

First we have to know what the mass of the compound is

Cr = 52g / mol

Cl = 35.5g / mol

H = 1g / mol

O = 16g / mol

We calculate the mass

52+ (35.5 * 3) + 6 * (2 * 1 + 16) = 266.5g

We have 20g Cr, so we calculate the amount of salt:

20gCr * (266.5g salt / 52gCr) = 102.5gCrCl3.6H2O

salt mass=102.5gCrCl3.6H2O

Answer:

The correct option is: E. No precipitate will form.

Explanation:

A solubility chart refers to the list of solubility of various ionic compounds. It shows the solubility of the various compounds in water at room temperature and 1 atm pressure.

Also, according to the solubility rules, the salts of chlorides, bromides and iodides are generally soluble and mostly all salts of sulfate are soluble.

Since, all the compounds formed in this double replacement reaction are soluble in water. Therefore, no precipitate will be formed.

ZnSO₄ (aq) + MgCl₂ (aq) → ZnCl₂ (aq) + MgSO₄ (aq)    

Explanation:

(a)   According to the mole concept, 1 mole of an atom contains [tex]6.022 \times 10^{23}[/tex] atoms.

Hence, number of atoms present in [tex]4.48 \times 10^{-23}[/tex] g will be as follows.

                 [tex]4.48 \times 10^{-23} \times 6.022 \times 10^{23}[/tex]

                 = 26.97 g

or,              = 27 g

This means that 1 mole of aluminium atoms contain 27 g.

(b)     Mass of air plane is 5000 kg of [tex]5000 \times 1000 g[/tex] (as 1 kg = 1000 g).

As mass of 1 mole aluminium is calculated as 27 g and mass of air plane is given as 5000000 g.

Therefore, calculate the number of moles of aluminium atoms as follws.

               No. of moles of Al atoms = [tex]\frac{\text{mass of small air plane}}{\text{mass of 1 mol Aluminium}}[/tex]

                                             = [tex]\frac{5000000g}{27 g}[/tex]

                                              = 185185.18

So, the answer in three significant digits will be 185000 moles of aluminum atoms have a mass equal to the mass of a small airplane.

The study of chemicals and bonds is called chemistry. There are two types of elements and these are metals and nonmetals.

The correct answer is described below

According to the question, the answer of the first question is:-

The number of atoms will be:-

[tex]4.48*10^{-23}*6.022*10^{23}\\\\=26.97[/tex]

The correct answer is 27

The answer to the second question is as follows:-

The mass of the air is [tex]5000*1000g[/tex]. hence, As the mass of 1-mole aluminum is calculated as 27 g and mass of airplane is given as 5000000 g.

No moles will be

:- [tex]\frac{500000}{27} \\\\=185185.18[/tex]

Hence, the correct answer is mentioned above.

For more information about the moles, refer to the link:-

https://brainly.com/question/16759172

Answer:

8.125 atm

Explanation:

Hello,

Boyle's law states that at constant temperature:

[tex]P_1V_1=P_2V_2[/tex]

In this case:

[tex]P_1=1 atm, V_1=27.3L, V_2=3.36L[/tex]

Solving for [tex]P_2[/tex]:

[tex]P_2=\frac{P_1V_1}{V_2}=\frac{1atm*27.3L}{3.36L}\\P_2=8.125atm[/tex]

Best regards!

Answer:

23.4 mph

Explanation:

The conversion factors are multiplied so that the units cancel out:

(100 meters / 9.58 sec) x (1 mile / 1609 meters) x (60 sec / min) x (60 min /1 hr) = 23.4 mph

Answer:

1000m2 / s2

Explanation:

Hello! In order to verify this, we have to do unit conversion. We also have to know that J (Joule) = kg * m2 / s2

Then we can start with the test.

1kJ / kg * (1000J / 1kJ) = 1000J / kg

1000J / kg = 1000kg * m2 / kg * s2

In this step we can simplify "kg".

So the result is

1000m2 / s2

Final answer:

To demonstrate that 1 kJ/kg equals 1000 m²/s², we recognize that the joule (J) is defined as kg-m²/s², and by converting kJ to J and canceling out the kg units, we affirm the equality.

Explanation:

To show that 1 kJ/kg is equal to 1000 m²/s², we start by recognizing that the unit of energy, the joule (J), is defined as 1 kilogram-meter²/second² (kg-m²/s²). Therefore, when we talk about energy per unit mass, we are effectively dividing energy by mass, leaving us with units of m²/s².

Given that 1 joule is 1 kg·m²/s², 1 kJ is 1000 joules (since the prefix 'kilo' means 1000). So when we have 1 kJ/kg, it's the same as saying 1000 J/kg. When we divide each term (kg·m²/s²) by kg, the kilograms cancel out, leaving us with m²/s². Thus, 1 kJ/kg is indeed equivalent to 1000 m²/s².

Answer: Work done for the process is -390 J

Explanation:

The chemical equation for the reaction of zinc metal with sulfuric acid follows:

[tex]Zn+H_2SO_4\rightarrow ZnSO_4+H_2[/tex]

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 zinc = 10.0 g

Molar mass of zinc = 65.38 g/mol

Putting values in above equation, we get:

[tex]\text{Moles of zinc}=\frac{10.0g}{65.38g/mol}=0.153mol[/tex]

The equation given by ideal gas follows:

[tex]P\Delta V=nRT[/tex]

where, P = pressure of the gas

[tex]\Delta V[/tex] = Change in volume of the gas

T = Temperature of the gas = 298 K

R = Gas constant = 8.314 J/mol.K

n = number of moles of gas = 0.153 mol

Putting values in above equation, we get:

[tex]P\Delta V=0.153mol\times 8.314J/mol.K\times 298K\\\\P\Delta V=397J[/tex]

To calculate the work done, we use the equation:

[tex]\text{Work done}=-P\Delta V\\\\W=-390J[/tex]

Hence, work done for the process is -390 J

Answer:

ΔHf = 3301 kJ/mol

Explanation:

The standard enthalpy of formation of a compound is the change of enthalpy during the formation of 1 mole of the substance from its constituent elements, with all substances in their standard states.

In a reaction ΔHf = ∑ n ΔHproducts - ∑ n ΔHreagents

Where n are moles

For the reaction:

1 C₆H₆ + 7.5 O₂ → 6 CO₂ + 3 H₂O

The ΔHf = -285,8 kJ/mol × 3 - 393,5 kJ/mol×6 - [82,6 kJ/mol×1 -0kJ/mol×7,5]

ΔHf = 3301 kJ/mol

I hope it helps!

Answer:

The concentration of chloride ions in the final solution is 3 M.

Explanation:

The number of moles present in a solution can be calculated as follows:

number of moles = concentration in molarity * volume

In 100 ml of a 2 M KCl solution, there will be (0.1 l * 2mol/l) 0.2 mol Cl⁻

For every mol of CaCl₂, there are 2 moles of Cl⁻, then, the number of moles of Cl⁻ in 50 l of a 1.5 M solution will be:

number of moles of Cl⁻ = 2 * number of moles of  CaCl₂

number of moles of Cl⁻ = 2 ( 50 l * 1.5 mol / l ) = 150 mol Cl⁻

The total number of moles of Cl⁻ present in the solution will be (150 mol + 0.2 mol ) 150.2 mol.

Assuming ideal behavior, the volume of the final solution will be ( 50 l + 0.1 l) 50.1 l. The molar concentration of chloride ions will be:

Concentration = number of moles of Cl⁻ / volume

Concentration = 150.2 mol / 50.1 l = 3.0 M

To find the molar concentration of chloride ions in the solution, calculate the moles of chloride ions in each solution and add them together, then divide by the total volume of the final solution.

To determine the molar concentration of chloride ions in the solution, we need to calculate the number of moles of chloride ions and divide it by the total volume of the solution.

First, calculate the moles of chloride ions in KCl solution:

Moles of chloride ions = molarity of KCl solution * volume of KCl solution

Next, calculate the moles of chloride ions in CaCl2 solution:

Moles of chloride ions = (molarity of CaCl2 solution * volume of CaCl2 solution) * 2 (since CaCl2 has 2 chloride ions per formula unit)

After finding the moles of chloride ions in each solution, add them together and divide by the total volume of the final solution to get the molar concentration of chloride ions.

https://brainly.com/question/31973745

#SPJ11

Final answer:

The importance of an energy balance in a chemical process facility lies in ensuring efficient energy use, maintaining safety, optimizing chemical processes for increased yield, and reducing environmental impact.

Explanation:

An energy balance is crucial in a chemical process facility because it ensures the efficient use of energy, minimizes waste, and helps maintain the safety of the operation. Since energy is neither created nor destroyed (law of conservation of energy), it's important to know where energy is being consumed and produced within chemical processes. This understanding allows engineers to optimize the process, increase the yield of the desired product, and reduce environmental impact by minimizing wasted energy and reducing unwanted by-products.

An energy balance aids in maintaining the operation within the desired range of conditions, preventing unsafe levels of energy which could lead to accidents. Moreover, certain reactions require specific amounts of energy to produce raw materials or to synthesize products, and tracking energy use is essential for economic and environmental reasons.

Answer:

20.7

Explanation:

20.7298014 rounded off to 3 sig fig =20.7

Answer:

the equilibium changes since Q > Kc, by increasing the volume, therefore, the reaction will try to use some of the excess product and favor the reverse reaction to reach equilibrium.

Explanation:

CO(g) + 3H2(g) ↔ CH4(g) + H2O(g)

∴ Kc = [ H2O ] * [ CH4 ] / [ H2 ]³ * [ CO ] = 3.93.....equilibrium, V = 2.00L

PV = nRT; assuming P,T a standart conditions ( 1 atm, 298 K )

⇒ n / V = P / RT

∴ V = 8.00L

∴ R = 0.082 atm.L/K.mol

⇒ mol CO(g) = 0.327 mol = mol H2O = mol CH4

⇒ mol H2(g) = 0.327 mol CO * ( 3mol H2 / mol CO) = 0.981 mol

⇒ [ CO ] = 0.041 = [ CH4 ] = [ H2O]

⇒ [ H2 ] = 0.123 M

∴ Q =  [ H2O ] * [ CH4 ] / [ H2 ]³ * [ CO ]

⇒ Q = ( 0.041² ) / (( 0.123³ ) * ( 0.041 ))

⇒ Q = 22.03

⇒ we have more product present than we would have in the equilibrium.

Answer:

Q = 62.9

Q > Kc, so the reaction will proceed to the left so that Q takes the value of Kc.

Explanation:

Let's consider the following reaction.

CO(g) + 3 H₂(g) ⇄ CH₄(g) + H₂O(g)

The equilibrium constant (Kc) is:

[tex]Kc= 3.93 =\frac{[CH_{4}].[H_{2}O]}{[CO].[H_{2}]^{3} }[/tex]

If the volume is multiplied by 4 (2.00 L → 8.00 L), the concentrations will be divided by 4. The reaction quotient (Q) is:

[tex]Q=\frac{(0.25[CH_{4}]).(0.25[H_{2}O])}{(0.25[CO]).(0.25[H_{2}])^{3} }=16\frac{[CH_{4}].[H_{2}O]}{[CO].[H_{2}]^{3} }=16Kc=16 \times 3.93 = 62.9[/tex]

Q > Kc, so the reaction will proceed to the left so that Q takes the value of Kc.

Answer:

The ratio [G1P]/[G6P] = 5.7 . 10⁻².

Explanation:

Let us consider the reaction G1P ⇄ G6P, with ΔG° = -7.1 kJ/mol. According to Hess's Law, we can write the inverse reaction, and Gibbs free energy would have an opposite sign.

G6P ⇄ G1P        ΔG° = 7.1 kJ/mol

This is the reaction for which we want to find the equilibrium constant (the equilibrium ratio of [G1P] to [G6P]):

[tex]Kc=\frac{[G1P]}{[G6P]}[/tex]

The equilibrium constant and Gibbs free energy are related by the following expression:

[tex]Kc=e^{-\Delta G\si{\textdegree}/R.T } } =e^{-7.1kJ/mol/8.314.10^{-3}kJ/mol.K.298K} } }=5.7.10^{-2}[/tex]

where,

R is the ideal gas constant (8.314 . 10⁻3 kJ/mol.K)

T is the absolute temperature (in kelvins)

Final answer:

To calculate the equilibrium ratio of [G1P] to [G6P], the Gibbs free energy equation is used with ΔG', universal gas constant R, and the temperature T substituted. The result is that the concentration of G6P is approximately 1.331 times that of G1P at equilibrium and at 25°C.

Explanation:

The student is asking about the equilibrium ratio of concentrations of glucose-1-phosphate ([G1P]) to glucose-6-phosphate ([G6P]) at 25°C when the standard free energy change (ΔG') for the isomerization reaction is given as -7.1 kJ/mol. To calculate this ratio, we can use the Gibbs free energy equation for the equilibrium constant (Keq):

ΔG' = -RT ln(Keq)

Where ΔG' is the standard free energy change, R is the universal gas constant (8.314 J/mol K), T is the temperature in Kelvin (25°C + 273.15 = 298.15 K), and Keq is the equilibrium constant which for this reaction is [G6P]/[G1P].

Substituting the values into the equation we get:

-7100 J/mol = -(8.314 J/mol K)(298.15 K) ln([G6P]/[G1P])

Now, we solve for ln([G6P]/[G1P]):

ln([G6P]/[G1P]) = ΔG' / (-R * T)

ln([G6P]/[G1P]) = -7100 J/mol / (-(8.314 J/mol K)(298.15 K))

ln([G6P]/[G1P]) = 0.286

Exponentiating both sides to remove the natural logarithm, we get:

[G6P]/[G1P] = e0.286 = 1.331

Therefore, at equilibrium and at 25°C, the concentration of G6P is approximately 1.331 times that of G1P.

Answer:

[tex]t=5.84\times 10^{-5}s[/tex]

Explanation:

[tex]ln(\frac{A}{B})=-kt[/tex]

or, [tex]t=\frac{1}{-k}\times ln(\frac{A}{B})[/tex]

Now plug-in all given values:

[tex]t=\frac{1}{-4.80\times 10^{4}s^{-1}}\times ln(\frac{1.00\times 10M}{1.65\times 10^{2}M})[/tex]

So, [tex]t=5.84\times 10^{-5}s[/tex]

Answer:

a) pH = 2,23

b) pH = 3,26

c) pH = 3,74

d) pH = 7,98. Here we have the equivalence point of the titration

Explanation:

In a titration of a strong base (NaOH) with a weak acid (HCOOH) the reaction is:

HCOOH + NaOH → HCOONa + H₂O

a) Here you have just HCOOH, thus:

HCOOH ⇄ HCOO⁻ + H⁺ where ka =1,8x10⁻⁴ and pka = 3,74

When this reaction is in equilibrium:

[HCOOH] = 0,200 -x

[HCOO⁻] = x

[H⁺] = x

Thus, equilibrium equation is:

1,8x10⁻⁴ = [tex]\frac{[x][x] }{[0,200-x]}[/tex]

The equation you will obtain is:

x² + 1,8x10⁻⁴x - 3,6x10⁻⁵ = 0

Solving:

x = -0,006090675 ⇒ No physical sense. There are not negative concentrations

x = 0,005910674

As x = [H⁺] and pH = - log [H⁺]

pH = 2,23

b) Here, it is possible to use:

HCOOH + NaOH → HCOONa + H₂O

With adition of 5,00 mL of 1,00M NaOH solution the initial moles are:

HCOOH = [tex]0,100 L.\frac{0,200 mol}{L} =[/tex] = 2,0x10⁻² mol

NaOH = [tex]0,005 L.\frac{1,00 mol}{L} =[/tex] = 5,0x10⁻³ mol

HCOO⁻ = 0.

In equilibrium:

HCOOH = 2,0x10⁻² mol - 5,0x10⁻³ mol = 1,5x10⁻² mol

NaOH = 0 mol

HCOO⁻ = 5,0x10⁻³ mol

Now, you can use Henderson–Hasselbalch equation:

pH = 3,74 + log [tex]\frac{5,0x10^{-3} }{1,5x10^{-2} }[/tex]

pH = 3,26

c) With adition of 10 mL of 1,00M NaOH solution the initial moles are:

HCOOH = [tex]0,100 L.\frac{0,200 mol}{L} =[/tex] = 2,0x10⁻² mol

NaOH = [tex]0,010 L.\frac{1,00 mol}{L} =[/tex] = 1,0x10⁻² mol

HCOO⁻ = 0.

In equilibrium:

HCOOH = 2,0x10⁻² mol - 1,0x10⁻² mol = 1,0x10⁻² mol

NaOH = 0 mol

HCOO⁻ = 1,0x10⁻² mol

Now, you can use Henderson–Hasselbalch equation:

pH = 3,74 + log [tex]\frac{1,0x10^{-2} }{1,0x10^{-2} }[/tex]

pH = 3,74

d) With adition of 20 mL of 1,00M NaOH solution the initial moles are:

HCOOH = [tex]0,100 L.\frac{0,200 mol}{L} =[/tex] = 2,0x10⁻² mol

NaOH = [tex]0,020 L.\frac{1,00 mol}{L} =[/tex] = 2,0x10⁻² mol

HCOO⁻ = 0.

Here we have the equivalence point of the titration, thus, the equilibrium is:

HCOO⁻ + H₂O ⇄ HCOOH + OH⁻ kb = kw/ka where kw is equilibrium constant of water = 1,0x10⁻¹⁴; kb = 5,56x10⁻¹¹

Concentrations is equilibrium are:

[HCOOH] = x

[HCOO⁻] = 0,1667-x

[OH⁻] = x

Thus, equilibrium equation is:

5,56x10⁻¹¹ = [tex]\frac{[x][x] }{[0,01667-x]}[/tex]

The equation you will obtain is:

x² + 5,56x10⁻¹¹x - 9,27x10⁻¹³ = 0

Solving:

x = -9,628361x10⁻⁷⇒ No physical sense. There are not negative concentrations

x = 9,627806x10⁻⁷

As x = [OH⁻] and pOH = - log [OH⁻]; pH = 14 - pOH

pOH = 6,02

pH = 7,98

I hope it helps!

Answer:

Plausible structure has been given below

Explanation:

4.04 g of [tex]CO_{2}[/tex] = [tex]\frac{4.04}{44}moles[/tex] [tex]CO_{2}[/tex] = 0.0918 moles of [tex]CO_{2}[/tex]

1 mol of [tex]CO_{2}[/tex] contains 1 mol of C atom

So, 0.0918 moles of [tex]CO_{2}[/tex] contains 0.0918 moles of C atom

1.24 g of [tex]H_{2}O[/tex] = [tex]\frac{1.24}{18}moles[/tex] [tex]H_{2}O[/tex] = 0.0689 moles of [tex]H_{2}O[/tex]

1 mol of [tex]H_{2}O[/tex]  contain 2 moles of H atom

So, 0.0689 moles of [tex]H_{2}O[/tex] contain [tex](2\times 0.0689)moles[/tex] of [tex]H_{2}O[/tex] or 0.138 moles of [tex]H_{2}O[/tex]

Moles of C : moles of H = 0.0918 : 0.138 = 2 : 3

Empirical formula of hydrocarbon is [tex]C_{2}H_{3}[/tex]

So, molecular formula of one of it's analog is [tex]C_{4}H_{6}[/tex]

Plausible structure of [tex]C_{4}H_{6}[/tex] has been given below.

Answer: The molecular formula of glucose is [tex]C_6H_{12}O_6[/tex]

Explanation:

We are given:

Empirical formula of the compound = [tex]CH_2O[/tex]

Empirical mass of the compound = [tex][(1\times 12)+(2\times 1)+(1\times 16)]=30g/mol[/tex]

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

[tex]n=\frac{\text{Molecular mass}}{\text{Empirical mass}}[/tex]

We are given:

Mass of molecular formula = 180.12 g/mol

Mass of empirical formula = 30 g/mol

Putting values in above equation, we get:

[tex]n=\frac{180.12g/mol}{30g/mol}=6[/tex]

Multiplying this valency by the subscript of every element of empirical formula, we get:

[tex]C_{(1\times 6)}H_{(2\times 6)}O_{(1\times 6)}=C_6H_{12}O_6[/tex]

Hence, the molecular formula of glucose is [tex]C_6H_{12}O_6[/tex]

Answer:

At the equivalence point of the titration of a monoprotic weak acid with a strong base: B. the moles of strong base added must equal the moles of weak acid.

Explanation:

In every titration, the equivalence point is defined as the point where the moles of the titrant and analyte are equal. For every acid-base titration, the equivalence point is defined as the point where the moles of the base is equal to the moles of the acid.

If the solutions of the acid and base are at a different concentration the volume added from the buret will not be the same as the volume of the analyte.

Answer:

Monosaccharides are the simplest form of carbohydrates that cannot be hydrolyzed to smaller compounds. Monosaccharides are the basic units of carbohydrates and are also known as simple sugars.  

The monosaccharides are classified on the basis of number of carbon atoms present.

Triose is a type of monosaccharide molecule, which is composed of 3 carbon atoms.

Tetrose is a type of monosaccharide molecule, which is composed of 4 carbon atoms.

Pentose is a type of monosaccharide molecule, which is composed of 5 carbon atoms.

Hexose is a type of monosaccharide molecule, which is composed of 6 carbon atoms.

D-glucose is a hexose sugar and it is the most abundant monosaccharide in the nature.

Answer:

The dissociation equations for NaBr gives Na+ and Br-

The dissociation equations for ZnCl2 gives Zn2+ and 2 Cl-  

Explanation:

The following pictures shows that the dissociation of one particle of NaBr produces one particle of Na+ (sodium cation) and one particle of Br- (bromine anion).

The dissociation of one particle of ZnCl2 produces one particle of Zn+2 (Zinc cation) and two particles of Cl- (chlorine anion).

Answer:

6.113 atm

Explanation:

Data provided in the question:

Molarity of the solution = 0.25 M

Temperature, T = 25° C = 25 + 273 = 298 K

Now,

Osmotic pressure (π) is given as:

π = MRT

where,

M is the molarity of the solution

R is the ideal gas constant = 0.082057 L atm mol⁻¹K⁻¹

on substituting the respective values, we get

π = 0.25 × 0.082057 × 298

or

π = 6.113 atm

Answer:

397 g

Explanation:

From the information given in the question ,

the atomic mass of X = 12 ,

and , the atomic mass of Y = 1 .

For the molecular formula , X₇Y₁₁ , the molecular mass is given as -

X₇Y₁₁ = 7 * atomic mass of X  +  11 *  atomic mass of Y

         = 7 * 12 + 11 * 1

         = 84 + 11  = 95 g / mol

Hence , molecular mass of X₇Y₁₁ = 95 g / mol .

Moles is denoted by given mass divided by the molecular mass ,

Hence ,

n = w / m

n = moles ,

w = given mass ,

m = molecular mass .

From question ,

moles n = 4.18 mol

and ,

m = molecular mass of X₇Y₁₁ = 95 g / mol .

To find the mass of the compound X₇Y₁₁   , the above formula can be used , and putting the respective values ,

n = w / m  

4.18 = w / 95

w = 4.18 * 95 = 397 g