What is the pH of a buffer that consists of 0.254 M CH3CH2COONa and 0.329 M CH3CH2COOH? Ka of propanoic acid, CH3CH2COOH is 1.3 x 10-5 Enter your answer with two decimal places.

Answer: The mass of water produced in the reaction is 97.2 grams

Explanation:

We are given:

Moles of calcium hydroxide = 2.70 moles

The chemical equation for the reaction of calcium hydroxide and HCl follows:

[tex]Ca(OH)_2+HCl\rightarrow CaCl_2+2H_2O[/tex]

By Stoichiometry of the reaction:

1 mole of calcium hydroxide produces 2 moles of water

So, 2.70 moles of calcium hydroxide will produce = [tex]\frac{2}{1}\times 2.70=5.40mol[/tex] of HCl

To calculate mass for given number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]

Molar mass of water = 18 g/mol

Moles of water = 5.40 moles

Putting values in above equation, we get:

[tex]5.40mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(5.40mol\times 18g/mol)=97.2g[/tex]

Hence, the mass of water produced in the reaction is 97.2 grams

The correct answer is 54.0 grams of water will be produced from 2.70 moles of Ca(OH)2 reacting with HCl.

 To solve this problem, we need to consider the stoichiometry of the reaction between calcium hydroxide (Ca(OH)2) and hydrochloric acid (HCl). The balanced chemical equation for this reaction is:

[tex]Ca(OH)_2(s) + 2HCl(aq) \rightarrow CaCl_2(aq) + 2H_2O(l)[/tex]

 From the equation, we can see that 1 mole of Ca(OH)2 produces 2 moles of H2O upon reacting with excess HCl.

 Given that we have 2.70 moles of Ca(OH)2, we can calculate the moles of water produced using the stoichiometric ratio:

[tex]2.70 \ moles \ Ca(OH)_2 \times (2 \ moles \ H_2O / 1 \ mole \ Ca(OH)_2) = 5.40 \ moles \ H_2O[/tex]

 Now, we need to convert the moles of water to grams using the molar mass of water (H2O), which is approximately 18.015 grams per mole:

[tex]5.40 \ moles \ H_2O \times (18.015 \ grams \ H_2O / 1 \ mole \ H_2O) = 97.281 \ grams \ H_2O[/tex]

 However, this calculation is incorrect because it does not take into account the correct stoichiometry of the reaction. The balanced equation shows that 1 mole of Ca(OH)2 produces 2 moles of H2O, not 1 mole. Therefore, the correct calculation should be:

[tex]2.70 \ moles \ Ca(OH)_2 \times (2 \ moles \ H_2O / 1 \ mole \ Ca(OH)_2) \times (18.015 \ grams \ H_2O / 1 \ mole \ H_2O) = 54.0 \ grams \ H_2O[/tex]

 Thus, the mass of water produced from 2.70 moles of Ca(OH)2 reacting with HCl is 54.0 grams.

Answer:

the answer 37

Explanation:

Answer:

See explaination

Explanation:

The invariant mass of an electron is approximately9. 109×10−31 kilograms, or5. 489×10−4 atomic mass units. On the basis of Einstein's principle of mass–energy equivalence, this mass corresponds to a rest energy of 0.511 MeV.

Check attachment for further solution to the exercise.

Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem.

Final answer:

The question involves ranking solutions of weak acids and bases by their pH. After assessing the ionization and hydrolysis of the compounds, the order from lowest to highest pH is 0.1 M NH4Br, 0.1 M NaCN, 0.1 M KF, and 0.1 M KBr.

Explanation:

To rank the solutions in order of increasing pH, we must consider the ionization of each compound in water and the corresponding Ka or Kb values. A lower Ka or a higher Kb value usually results in a higher pH.

0.1 M KF: KF will form from the dissolution of a weak acid (HF) and a strong base. The resulting F- ions will hydrolyze to form OH- ions, creating a basic solution with a pH higher than 7 but not as high as that of a strong base.

Therefore, the order from lowest to highest pH should be b, a, c, d.

This question is about the Macro and Micro views of the PhET simulation for solutions and the concepts related to dissolving salts and sugars in water.

1. Pure water contains only water molecules that interact strongly with each other due to their partial charges, which are graphically depicted as δ+ and δ−.

2. Solutions are formed when a solute like a salt or sugar becomes homogeneously distributed in a solvent like water, and this distribution can be viewed in the Micro view.

3. When salts dissolve, they separate into individual ions that strongly interact with the water molecules.

4. Binary salts are made up of two elements at varying ratios, where one element is a charged cation, and the other is a negatively charged anion.

5. In the Micro view, each shake of the container releases 6 molecules of its respective solute, but 6 molecules of the salt NaCl actually produce more ions in solution than 6 molecules of the salt CaCl2.

6. Not all soluble molecules are salts, e.g., a covalent species like sugar readily dissolves in water without forming ions.

7. The reason they dissolve is because their partially charged atoms are able to associate with the partially charged atoms of water molecules, and these attractive forces occur as long as they are between atoms with opposite charges.

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This answer explains the properties of solutions, including partial charges and the formation of ions. It also provides examples of different solutes and their behavior in water.

1. Pure water contains only water molecules that interact strongly with each other due to their partial charges, which are graphically depicted as δ+ and δ−.

2. Solutions are formed when a solute like a salt or sugar becomes homogeneously distributed in a solvent like water, and this distribution can be viewed in the Micro view.

3. When salts dissolve, they separate into individual ions that strongly interact with the water molecules.

4. Binary salts are made up of two elements at varying ratios, where one element is a charged cation, and the other is a negatively charged anion.

5. In the Micro view, each shake of the container releases 6 molecules of its respective solute, but 6 molecules of the salt NaCl actually produce more ions in solution than 6 molecules of the salt sugar.

6. Not all soluble molecules are salts, e.g., a covalent species like sugar readily dissolves in water without forming ions.

7. The reason they dissolve is because their partially charged atoms are able to associate with the partially charged atoms of water molecules, and these attractive forces occur as long as they are between atoms with opposite charges.

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Answer:

Entropy change is zero

Explanation:

∆S=S2-S1=0

We're S is entropy

Answer:

49.544 g.

Explanation:

The balanced equation of reaction is given below;

Hg(ClO3)2 (aq) + Na2S (aq) --------> 2 NaClO3 (aq) + HgS (s).

So, the parameters given in the question are; Mass of Hg(ClO3) = 118.08 and the mass of Na2S = 16.642 g.

Therefore, the first thing we are going to be looking at is the reactant which is the limiting reagent from the number of moles

(1). For Hg(ClO3), the molar mass = 367.5 g/mol. Therefore, the number of moles, n = mass/ molar mass.

Number of moles = 118.08/ 367.5.

Number of moles = 0.321 moles.

(2). For Na2S, the molar mass = 78.05 g /mol.

The number of moles = 16.643 / 78.05.

The number of moles= 0.213.

Therefore, the limiting reagent = Na2S.

This means that the excess reagent is Hg(ClO3).

==> In the balanced equation of reaction above, the solid precipitate = HgS.

Hence, the mass of HgS formed = 0.213 × 232.6 g/mol. = 49.544 g.

This problem involves calculating the mass of a precipitate from a doubly replacement reaction. Sodium sulfide is the limiting reactant and by using stoichiometry, the moles of sodium sulfide will amount to the same moles of HgS, the precipitate. Multiplying the moles of HgS by its molar mass gives the mass of the precipitate.

This question requires a two-step process that starts with writing balanced chemical equations for the reaction and then using stoichiometry to calculate the mass of the precipitate. The balanced reaction is 2Na2S + Hg2(ClO3)2 -> 2NaClO3 + Hg2S2. Using stoichiometry and details given in the question, you can calculate the molar mass of the substances and find the limiting reactant.

Mercury(II) chlorate (Hg2(ClO3)2) has a molar mass of about 511.63 g/mol. Sodium sulfide (Na2S) has a molar mass of about 78.04 g/mol. Using stoichiometry, it's clear that sodium sulfide is the limiting reactant.

Using the moles of sodium sulfide calculated, you find that the reaction will produce the same number of moles of HgS as a precipitate. By multiplying this number by the molar mass of HgS (232.66 g/mol), you obtain the mass of the precipitate that will be formed. Deduce the exact values yourself for a better understanding of this chemical reaction involving stoichiometry and solubility product.

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Answer:

1. C9H20(l) + 14O2(g) —> 9CO2(g) + 10H2O(g)

2. 213.70L of CO2

Explanation:

1. The equation for the combustion of liquid nonane into gaseous carbon dioxide and gaseous water is given below:

C9H20(l) + O2(g) —> CO2(g) + H2O(g)

The equation can be balanced as follow:

There are 9 atoms of C on the left side and 1 atom on the right side. It can be balance by putting 9 in front of CO2 as shown below:

C9H20(l) + O2(g) —> 9CO2(g) + H2O(g)

There are 20 atoms of H on the left side and 2 atoms on the right side. It can be balance by putting 10 in front of H2O as shown below:

C9H20(l) + O2(g) —> 9CO2(g) + 10H2O(g)

Now, there are a total of 28 atoms of O on the right side and 2 atoms on the left side. It can be balance by putting 14 in front of O2 as shown below:

C9H20(l) + 14O2(g) —> 9CO2(g) + 10H2O(g)

Now the equation is balanced.

2. Let us convert 0.130 kg of nonane (C9H20) to mole. This is illustrated below:

Molar Mass of C9H20 = (12x9) + (20x1) = 108 + 20 = 128g/mol

Mass of C9H20 from the question =

0.130kg = 0.130 x 1000 = 130g

Mole of C9H20 =?

Number of mole = Mass/Molar Mass

Mole of C9H20 = 130/128 = 1.016 mole

The equation for the reaction is:

C9H20(l) + 14O2(g) —> 9CO2(g) + 10H2O(g)

From the balanced equation above,

1 mole of C9H20 produced 9 moles of CO2.

Therefore, 1.016 mole of C9H20 will produce = 1.016 x 9 = 9.144 moles

Now, let us calculate the volume of CO2 formed. This is illustrated below:

Data obtained from the question include:

P (pressure) = 1atm

T (temperature) = 12°C = 285K

n (number of mole of CO2) = 9.144 moles

R (gas constant) = 0.082atm.L/Kmol

V (volume of CO2) =?

Using the ideal gas equation PV = nRT, the volume of CO2 can be obtained as follow:

PV = nRT

1 x V = 9.144 x 0.082 x 285

V = 213.70L

Therefore, 213.70L of CO2 is produced

The combustion of nonane, C9H20(l) + 14O2(g) → 9CO2(g) + 10H2O(g), produces carbon dioxide and water. Using stoichiometry and the ideal gas law, the volume of carbon dioxide produced from 0.130 kg of nonane is calculated to be approximately 216.5 L.

The first step is to write a balanced chemical equation for the combustion of nonane. The complete combustion of hydrocarbons, like nonane, in the presence of oxygen (O2) yields carbon dioxide (CO2) and water (H2O). The correct balanced equation is:

C9H20(l) + 14O2(g) → 9CO2(g) + 10H2O(g).

To calculate the volume of carbon dioxide produced from the amount of nonane combusted, one must apply the concepts of stoichiometry and gas laws. According to the equation, 9 moles of CO2 is produced from 1 mole of nonane. As the molar mass of nonane is 128.26 g/mol, 0.130 kg or 130 g of nonane is approximately 1.013 moles. Therefore, we will have around 9.117 moles of carbon dioxide produced.

Applying the ideal gas law, PV = nRT, at standard temperature (12°C = 285.15 K) and pressure (1 atm), and R = 0.0821 L.atm / K.mol, we can calculate the volume as V = nRT/P. The volume of CO2 produced would be approximately 216.5 L, with correct significant digits.

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Answer:

They are probably archeabacteria because the colony lives in an extreme environment.

Explanation:

Archeabacteria are singled cell microorganisms that live in extreme environment. They are found in hot springs, salt lakes, oceans, soils and marshlands. They posses different shapes like rods, spheres, spiral and plates. Thermophiles, halophiles, and methanogens are the three types of archeabacteria.

Eubacteria are microorganisms that are found in most of the earth's habitats like soil, water, etc. They are have different shapes like cocci, bacilli, filaments, vibro,etc. They do not live in extreme environment unlike the archeabacteria. This is the major difference between the archeabacteria and eubacteria.

Both archeabacteria and eubacteria

are prokaryotes. Archeabacteria can both be autotrophic or heterotrophic and can live in places without oxygen. Some eubacteria are autotrophs and some are heterotrophs.

Answer:

They are probably archaebacteria because the colony lives in an extreme environment.

Explanation:

Answer:

Cr2O72- + 5Fe2+ +14H+ -> 2Cr3+ + 5Fe3+ +7H2O

Explanation:

Fe2+ is oxidized to Fe3+ and Cr6+ is reduced to Cr3+

Final answer:

Explains the balanced redox reaction between dichromate ion and iron(II) to yield iron(III) and chromium(III) in acidic solution.

Explanation:

Balanced Equation: 7H₂O(l) + 6Fe²+(aq) + Cr₂O72-(aq) → 6Fe³+(aq) + 2Cr³+(aq) + 14H+(aq)

Explanation: The reaction between dichromate ion and iron(II) to yield iron(III) and chromium(III) in acidic solution is represented by this balanced redox equation. By balancing the atoms and charges on both sides, the equation shows the transformation of Fe²+ to Fe³+ and Cr2O7²- to Cr³+.

Example: In a similar process, Cr atoms are oxidized to form Cr³+ while Fe atoms are reduced to Fe³+, showcasing the transfer of electrons in the redox reaction.

The missing part of the question is shown in the image attached

Answer:

C10H22(l) + 31/2 O2 (g)-----> 10CO2(g) + 11H2O(l)

V= 70.4L of CO2

Explanation:

Equation of the reaction is:

C10H22(l) + 31/2 O2 (g)-----> 10CO2(g) + 11H2O(l)

Number of moles of decane = mass/ molar mass

Molar mass of decane= 122gmol-1

n= 0.370×10^3g/122gmol-1= 3.0 moles

T= 13°C +273=286K

P= 1atm

R= 0.082 atmLK-1mol-1

From :

PV= nRT

V= nRT/P

V= 3.0×0.082×286/1

V= 70.4L of CO2

The balanced chemical equation for the combustion of liquid decane [tex](C\_{10}H\_{22}\))[/tex] into gaseous carbon dioxide [tex](CO\_{2}\)[/tex]and gaseous water [tex](H\_{2}O\) is: \[ 2C_{10}H_{22(l)} + 31O_{2(g)} \rightarrow 20CO_{2(g)} + 22H_2O_{(g)} \][/tex]

To calculate the volume of carbon dioxide gas produced when 0.500 moles of decane are burned, we use the stoichiometry of the balanced equation and Avogadro's law, which states that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules (or moles). From the balanced equation, 1 mole of decane produces 20 moles of CO\_{2}\. Therefore, 0.500 moles of decane will produce:

[tex]\[ 0.500 \text{ moles } C_{10}H_{22} \times \frac{20 \text{ moles } CO_2}{2 \text{ moles } C_{10}H_{22}} = 5.00 \text{ moles } CO_2 \][/tex]

According to Avogadro's law, at the same temperature and pressure, the volume of a gas is directly proportional to the number of moles of the gas. Since the pressure and temperature are not specified in the question, we cannot calculate the exact volume of CO\_{2}\ produced. However, if we assume standard temperature and pressure (STP) conditions, where 1 mole of an ideal gas occupies 22.414 liters, the volume of CO\_{2}\ produced would be:

[tex]\[ 5.00 \text{ moles } CO_2 \times \frac{22.414 \text{ L}}{1 \text{ mole } CO_2} = 112.07 \text{ L} \][/tex]

Since the question specifies that the pressure is exactly 1 atmosphere but does not specify the temperature, we cannot use the value of 22.414 L/mol for the molar volume of CO\_{2}\. We need the temperature to determine the molar volume of CO\_{2}\ at that specific condition. If the temperature is also at STP (0°C or 273.15 K), then the volume calculated above is correct. If the temperature is different, we would need to use the ideal gas law, PV = nRT, to find the volume V, where P is the pressure, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature in Kelvin.

Since the temperature is not provided, we can only provide the volume of [tex]CO\_{2}\[/tex] at STP, which is 112.07 L, rounded to four significant digits as 112.1 L. If the temperature were provided, we would use the ideal gas law to calculate the volume at that specific temperature and pressure.

Answer:

carbon dioxide

Explanation:

Answer:

False

Explanation:

Metals are made up of atoms bonded to each other by metallic bonds. Such bonds have a sea of delocalized electrons, which are free to flow. Electricity requires an unobstructed flow of electrons in order to be able to be conducted. Because all metals provide this inherently, then all metals can conduct electricity, making the statement false.

Hope this helps!

Answer:

The solution is basic

Explanation:

Low pH means that a solution is acidi while low pOH means that a solution is basic.

Answer:

Check the explanation

Explanation:

Going by the question above for parta, the feed flow rate is stated to be 9072g/hr.& result is offered for part a considering the same. if the unit is kg/hr, then kindly let me know & i shall make slight modification that might be required in solution which in that case, area & steam consumption answer shall differ.

In the b part of the question, reduced flow rate is higher than the initial flow rate therefore same has been considered in gram /hr.

Kindly check the attached images below to see the step by step solution to the above question.

In this exercise we have to use the knowledge of mechanics to calculate the steamused of the solution, in this way we will have to:

a) [tex]M_S=\frac{H_S}{\lambda_S} = 8.43 KG/h[/tex] , [tex]A= 0.0526 m^2[/tex]

b) We can assume that all water shall evaporate and produd shell contain solid only.

So from the data informed in the exercise we have that:

So apply overall mass balance, will be:

[tex]F=L+V\\9072=1814.4-V\\V=7.2576 kg/h[/tex]

Now, for enthalpy balance we shall make assumption that, all liquid phase enthalphy is that heet capacity of water is constant, so the data will be:

So, find the enthalpy of liquid, we have;

[tex]H_L= L*C_P*(T_P-T_F)\\=(1.8144)(4.18)(60.0586-15.6)\\H_v=V*(P(t_P-T_F)+\lambda)\\=(7.2576)(4.18(60.0586-15.6)+2357.5477))H_s=H_L+H_V\\H_S= (1.8144)(4.18)(60.0586-15.6)+(7.2576)(4.18(60.0586-15.6)+2357.5477))\\=18796.051 KJ/h[/tex]

Now, we can find the mass flow rate, that will be;

[tex]M_S=\frac{H_S}{\lambda_S} = 8.43 KG/h[/tex]

Now we can find the area of evaporator, that will be:

[tex]H_S=UA=(T_S-T_P)\\A=0.0526m^2[/tex]

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Answer:

The equilibrium constant for the reversible reaction = 0.0164

Explanation:

At equilibrium the rate of forward reaction is equal to the rate of backwards reaction.

The reaction is given as

A ⇌ B

Rate of forward reaction is first order in [A] and the rate of backward reaction is also first order in [B]

The rate of forward reaction = |r₁| = k₁ [A]

The rate of backward reaction = |r₂| = k₂ [B]

(Taking only the magnitudes)

where k₁ and k₂ are the forward and backward rate constants respectively.

k₁ = 0.010 s⁻¹

k₂ = 0.0610 s⁻¹

|r₁| = 0.010 [A]

|r₂| = 0.016 [B]

At equilibrium, the rate of forward and backward reactions are equal

|r₁| = |r₂|

k₁ [A] = k₂ [B] (eqn 1)

Note that equilibrium constant, K, is given as

K = [B]/[A]

So, from eqn 1

k₁ [A] = k₂ [B]

[B]/[A] = (k₁/k₂) = (0.01/0.0610) = 0.0163934426 = 0.0164

K = [B]/[A] = (k₁/k₂) = 0.0164

Hope this Helps!!!

Answer:

_5_ AsO2−(aq) + 3 Mn2+(aq) + _2_ H2O(l) → _5_ As(s) + _3_ MnO4−(aq) + _4_ H+(aq)

Explanation:

Step 1:

The unbalanced equation:

AsO2−(aq) + 3 Mn2+(aq) + H2O(l) → As(s) + MnO4−(aq) + H+(aq)

Step 2:

Balancing the equation.

AsO2−(aq) + 3Mn2+(aq) + H2O(l) → As(s) + MnO4−(aq) + H+(aq)

The above equation can be balanced as follow:

There are 3 atoms of Mn on the left side of the equation and 1 atom on the right side. It can be balance by putting 3 in front of MnO4− as shown below:

AsO2−(aq) + 3Mn2+(aq) + H2O(l) → As(s) + 3MnO4−(aq) + H+(aq)

There are 12 atoms of O on the right side and a total of 3 atoms on the left side. It can be balance by putting 5 in front of AsO2− and 2 in front of H2O as shown below:

5AsO2−(aq) + 3Mn2+(aq) + 2H2O(l) → As(s) + 3MnO4−(aq) + H+(aq)

There are 4 atoms of H on the left side and 1 atom on the right side. It can be balance by putting 4 in front of H+ as shown below:

5AsO2−(aq) + 3Mn2+(aq) + 2H2O(l) → As(s) + 3MnO4−(aq) + 4H+(aq)

There are 5 atoms of As on the left side and 1 atom on the right side. It can be balance by putting 5 in front of As as shown below:

5AsO2−(aq) + 3Mn2+(aq) + 2H2O(l) → 5As(s) + 3MnO4−(aq) + 4H+(aq)

Now the equation is balanced

The balanced redox reaction is

5AsO⁻₂(aq) + 3Mn²⁺(aq) + 2H₂O(l) → 5As(s) + 3MnO₄−(aq) + 4H+(aq)

The unbalanced redox reaction is :

AsO₂−(aq) + 3 Mn²⁺(aq) + H₂O(l) → As(s) + MnO₄−(aq) + H⁺(aq)

The above equation can be balanced  by ensuring the atom of the elements

on the left hand side is equal to those on the right hand side.

We have 3 atoms of Mn on the left side of the equation and 1 atom on the

right hand side.This will be balanced by putting 3 in front of MnO₄− as shown

below:

AsO₂−(aq) + 3 Mn²⁺(aq) + H₂O(l) → As(s) + MnO₄−(aq) + H⁺(aq)

We have 3 atoms of O on the left hand side and 12 atoms of O on the right

hand side. This is  balanced by putting 5 in front of AsO₂− and 2 in front of

H₂O as shown below:

5AsO₂−(aq) + 3 Mn²⁺(aq) + 2H₂O(l) → As(s) + MnO₄−(aq) + H⁺(aq)

We have 4 atoms of H on the left hand side and 1 atom of H on the right

hand side.We can balance by putting 4 in front of H⁺ as shown below:

5AsO₂−(aq) + 3 Mn²⁺(aq) + 2H₂O(l) → As(s) + MnO₄−(aq) + 4H⁺(aq)

We have 5 atoms of As on the left hand side and 1 atom of As on the right

hand side. We can balance by putting 5 in front of As as shown below:

5AsO₂−(aq) + 3 Mn²⁺(aq) + 2H₂O(l) → 5As(s) + MnO₄−(aq) + 4H⁺(aq)

The equation is therefore now balanced as the number of atoms of the

element on the left hand side are equal with those on the right hand side.

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Answer:A and B

Explanation:a stats final stages of life depend on its mass and stars die when they run out of fuel

The intrinsic chemical properties of a reactant can affect the rate of reaction, as seen in the example of food spoiling more quickly on a kitchen counter than in a refrigerator due to temperature differences.

The option that exemplifies how the intrinsic chemical properties of a reactant affect the rate of reaction is Food spoils more quickly on a kitchen counter than in a refrigerator. This is because the temperature difference between the kitchen counter and the refrigerator affects the rate of chemical reactions. Chemical reactions typically occur faster at higher temperatures, so food spoils more quickly when exposed to the higher temperature of the kitchen counter compared to the lower temperature inside the refrigerator.

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Answer:

The ligand concentration C is 0.274 nM

Explanation:

According to the image, the bulk concentration is inversely proportional to the radius, thus:

[tex]B=\frac{B_{\alpha }r^{2} }{\alpha }[/tex]

Where

Bα = 6.6 nM

r = 14.7 um = 1.47x10⁻⁵m

[tex]B=\frac{6.6*(1.47x10^{-5})^{2} }{5.2x10^{-9} } =0.274nM[/tex]

Answer:

C

Explanation: you have to observe to have an educated guess about something you test it

Answer: The atomic number and mass number of the new element formed is 89 and 228 respectively.

Explanation:

Beta decay is defined as the process in which beta particle is emitted. In this process, a neutron gets converted to a proton and an electron.  The released beta particle is also known as electron.

In this process, the atomic number of the daughter nuclei gets increased by a factor of 1 but the mass number remains the same.

[tex]_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0\beta[/tex]

The chemical equation for the beta decay of Ra-228 follows:

[tex]_{88}^{228}\tyextrm{Ra}\rightarrow _{89}^{228}\textrm{Ac}+_{-1}^0\beta[/tex]

Hence, the atomic number and mass number of the new element formed is 89 and 228 respectively.

Answer:

d)Cells 1 and 2

Explanation:

In a voltaic cell, oxidation occurs at the anode and reduction occurs at the cathode. The half cell that function as anode or cathode in a voltaic cell depends strictly on the reduction potential of the metal ion/metal system in that half cell.

Examining the reduction potentials of the various metal ion/metal systems in the three half cells;

Cu= +0.34 V

Ni= -0.25 V

Zn= -0.76 V

Fe(Fe2+)= -0.44 V

Hence only Zn2+ has a more negative reduction potential than Fe2+. The more negative the reduction potential, the greater the tendency of the system to function as the anode. Thus iron half cell will function as anode in cells 1&2 as explained in the argument above.

The correct answer is e) All three cells.

To determine in which voltaic cell(s) iron acts as the anode, we need to compare the standard reduction potentials (E°) of the half-reactions involved in each cell. The half-reaction for iron in all three cells is:

[tex]\[ \text{Fe}^{2+}(aq) + 2e^- \rightarrow \text{Fe}(s) \quad \text{with} \quad E_{\text{Fe}^{2+}/\text{Fe}}^\circ = -0.44 \, \text{V} \][/tex]

Now, let's consider the half-reactions for the other metals in each cell:

Cell 1:

[tex]\[ \text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s) \quad \text{with} \quad E_{\text{Cu}^{2+}/\text{Cu}}^\circ = +0.34 \, \text{V} \][/tex]

Cell 2:

[tex]\[ \text{Ni}^{2+}(aq) + 2e^- \rightarrow \text{Ni}(s) \quad \text{with} \quad E_{\text{Ni}^{2+}/\text{Ni}}^\circ = -0.25 \, \text{V} \][/tex]

Cell 3:

[tex]\[ \text{Zn}^{2+}(aq) + 2e^- \rightarrow \text{Zn}(s) \quad \text{with} \quad E_{\text{Zn}^{2+}/\text{Zn}}^\circ = -0.76 \, \text{V} \][/tex]

In a voltaic cell, the metal with the more negative standard reduction potential (or less positive) will act as the anode, and the metal with the more positive standard reduction potential will act as the cathode.

Comparing the standard reduction potentials:

- For Cell 1, [tex]\( E_{\text{Cu}^{2+}/\text{Cu}}^\circ > E_{\text{Fe}^{2+}/\text{Fe}}^\circ \)[/tex], so iron has a more negative potential than copper, and iron will act as the anode.

- For Cell 2, \[tex]( E_{\text{Ni}^{2+}/\text{Ni}}^\circ > E_{\text{Fe}^{2+}/\text{Fe}}^\circ \)[/tex], so iron has a more negative potential than nickel, and iron will act as the anode.

- For Cell 3, [tex]\( E_{\text{Zn}^{2+}/\text{Zn}}^\circ < E_{\text{Fe}^{2+}/\text{Fe}}^\circ \)[/tex], so zinc has a more negative potential than iron, and iron will act as the cathode.

However, this last comparison is incorrect because we must consider the sign of the standard reduction potentials. Since both zinc and iron have negative standard reduction potentials, the metal with the less negative value (in this case, iron) will act as the anode, and the metal with the more negative value (zinc) will act as the cathode.

Therefore, iron acts as the anode in all three cells, which corresponds to option e) All three cells.

Given question is incomplete. The complete question is as follows.

[tex]AB_{2}[/tex] is a molecule that reacts readily with water. Calculate the bond energy of the A–B bond using the standard enthalpy of reaction and the bond energy data provided. Enter a number in kJ to 0 decimal places.

  [tex]2AB_{2}(g) + 2H_{2}O(g) \rightarrow O=O(g) + 4HB(g) + A_{2}[/tex] [tex]\Delta H^{o}[/tex] = –142 kJ

Bond:                             O–H          O=O           H–B    [tex]A \rightarrow A^{+}[/tex]

Bond energy (kJ/mol):   467         498           450         321      

Explanation:

The given reaction is as follows.

         [tex]2AB_{2} + 2H_{2}O \rightarrow O_{2} + 4HB + A_{2}[/tex]

Now, we will calculate the enthalpy of reaction as follows.

   [tex]\Delta H^{o}_{R}[/tex] = -142 kJ

Also, we know that

   [tex]\Delta H^{o}_{R} = \Delta H_{reactants} - \Delta H_{products}[/tex]

                  = [tex][(2 \times 2 (A-B) + 2 \times 2 (O-H)] - [(O=O) + 4(H-B) + (A-A)][/tex]

         -142 = [tex]4(A-B) + 4 \times 467 - 498 - 4(450) - 321[/tex]

    [tex]4(A-B)[/tex] = -142 - 1868 + 498 + 1800 + 321

                    = 609

          (A-B) = 152.25 kJ/mol

Thus, we can conclude that the bond energy of the A–B bond is 152.25 kJ/mol.

Answer:

0.2 mol/kJ

Explanation:

Methane is stored at a temperature of 111.4K, [tex]T_c[/tex]. The heat source to vapourization of methane is ambient air which is at 300 K. [tex]T_H[/tex]

Estimate the vaporization rate at the efficiency of heat engine 60% of its carnot value.

Calculate the vaporization rate from the given data by relation shown below:

[tex]Vaporization rate = \frac{Q_c}{[\frac{\delta H_n^{lv}}{W}]} ..........(1)[/tex]

here,

[tex]Q_c[/tex] is the heat at temperature [tex]T_c, \delta H_n^{lv}[/tex] is the phase transition enthalpy of methane and W is the work

Calculate [tex]Q_c[/tex] from the equation shown below:

[tex]Q_c = Q_n (1 - \eta_{HE}) .............(2)[/tex]

where Q_n is the  heat at temperature of [tex]T_n[/tex] and [tex]\eta_{HE}[/tex] is the efficiency of heat engine

calculate [tex]Q_H[/tex] from the relation shown below:

[tex]Q_H = \frac{W}{\eta_{HE}} ..........(3)[/tex]

calculate the heat engine efficiency from the given carnot engine efficiency as shown below

[tex]\eta_{HE} = 0.6 \times \eta_{carnot} ............(4)[/tex]

here, [tex]\eta_{carnot}[/tex] is the carnot engine efficiency

[tex]\eta{carnot} = 1 - \frac{T_c}{T_H}[/tex]

substituting the values of temperature, we have

[tex]= 1 - \frac{111.4K}{300K}\\= 0.629\\[/tex]

substitute values of [tex]\eta_{carnot}[/tex] in equation 4, we get

[tex]\eta_{HE} =0.6 \times \eta_{carnot}\\ = 0.6 \times 0.629\\ = 0.3774\\\\[/tex]

check the attached file for additional solution

Answer:

85.0 kJ

Explanation:

Calculate the amount of heat needed to melt 160. g of solid octane (C8H18 ) and bring it to a temperature of 99.2 degrees c. Round your answer to 3 significant digits. Also, be sure your answer contains a unit symbol.

To melt 160 g of octane and bring it's temperature to 99.2°C

(from literature)

Heat of fusion of Octane = 20.740 kJ/mol

Melting point of octane = -57°C

Boiling point of Octane = 125.6 °C

Molar mass of octane = 114.23 g/mol

Heat capacity of octane = 255.68 J/K.mol

So, it is evident that Octane is still a liquid at 99.2°C.

So, the required heat is the heat required to melt octane and raise the temperature of Octane liquid to 99.2°C

First, we convert the mass of octane given to number of moles as the heat parameters provided by literature are given in molar units.

Number of moles = (mass)/(Molar mass)

Number of moles of octane = (160/114.23) = 1.401 moles

Heat required to melt the octane = nL = (1.401×20.740) = 29.05674 kJ

Heat required to raise the temperature of already melted octane from its melting temperature of -57°C to 99.2°C

= nCΔT

n = 1.401 moles

C = 255.68 J/K.mol

ΔT = (99.2 - (-57)) = 156.2°C (same as a temperature difference of 156.2 K)

Heat required to raise the temperature of already melted octane from -57°C to 99.2°C

= (1.401×255.68×156.2)

= 55,952.04 J = 55.952 kJ

Total heat required to melt the 160 g of Octane and raise its temperature to 99.2°C

= 29.05674 + 55.952 = 85.01 kJ = 85.0 kJ

Hope this Helps!!!

The heat needed to melt the solid octane and bring it to the desired temperature involves a two-step process - melting and then heating. You use the latent heat of fusion and specific heat capacity of octane to calculate the total heat required.

The amount of heat needed to melt the solid octane and bring it to the desired temperature involves a two-step process – melting, and heating. The process should be looked at in separate steps, similar to how the enthalpy of combustion is used to calculate the heat produced in the combustion of 1.00 L of isooctane in the reference provided. Note that different substances have different heat capacities and heat of fusion, so the associated values (likely given in the problem or textbook) for octane specifically would need to be used to get the final answer.

First, you would need to know the latent heat of fusion, which is the amount of energy required to change one kilogram of the substance from solid to liquid without changing its temperature. Then, you would multiply the mass of the octane by the latent heat of fusion to find out how much energy is required for it to go from solid to liquid.

Once the octane is in liquid form, we then need to heat it to the desired temperature. To do this, you use the specific heat capacity formula – the amount of energy required to raise one kilogram of the substance by one degree Celsius. Multiply the mass of the material by the specific heat capacity of the octane and the change in temperature (final temperature – initial temperature).

The added heat required for both of these processes would give you your final answer.

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Answer:

pH = 11.7

Explanation:

pOH= -log [OH]=-log[0.05]

=2.3

pOH+pH= 14

pH= 14-2.3= 11.7

The Ksp of Mn(OH)2 at 25°C is found to be 1.07 x 10^-11 M by analyzing the acids and bases reaction via titration using 0.002M HCl solution.

The subject question deals with an acid-base titration procedure to determine the solubility product constant, Ksp, of manganese(II) hydroxide at 25°C. The concentration of HCl used for titration is 0.002M and it completely reacts with manganese(II) hydroxide in 70.00mL of solution. The molarity of Mn(OH)2 in the saturated solution can be calculated as (0.002M * 4.86mL) / 70.00mL = 1.388 x 10^-4 M.

Since each Mn(OH)2 gives one Mn²⁺ and two OHˉ ions when dissolved, the molarity of OH- ions will be twice that of Mn(OH)2. Therefore, the OH- ion molarity is 2 * 1.388 x 10^-4 M = 2.776 x 10^-4 M.

The Ksp of Mn(OH)2 can be calculated as [Mn²⁺][OHˉ]² = (1.388 x 10^-4 M) * (2.776 x 10^-4 M)² = 1.07 x 10^-11 M. Thus, we can conclude that the Ksp of manganese(II) hydroxide at 25°C is approximately 1.07 x 10^-11 M.

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Answer: The monomer of nylon-6 is caprolactam.

Explanation:

From the question, we are asked to draw the cyclic monomer from the structure of Nylon-6.

Two repeating monomer units can be seen in the polymeric structure given in the problem.

From this we can see and conclude that The monomer of nylon-6 is caprolactam.

attached is a diagrammatic representation of the Structure (Nylon-6 monomer).

cheers i hope this helps!!!!

Final answer:

Nylon-6 is synthesized from the cyclic monomer epsilon-caprolactam through a process called ring-opening polymerization, unlike other nylons that are produced by condensation polymerization. Epsilon-caprolactam features a 6-carbon ring with an amide linkage, making it unique.

Explanation:

Nylon-6 is a unique type of polymer that is derived from a cyclic monomer known as epsilon-caprolactam. Unlike the Nylon 6,6 polymer, which is made from condensation polymerization of hexanedioic acid and 1,6-diaminohexane, Nylon-6 utilizes a process called ring-opening polymerization. This process involves breaking open the ring structure of the epsilon-caprolactam to form linear chains, which then link together head-to-tail to form the polymer. This is a key difference as it does not release a small molecule (like water or hydrogen chloride) during the polymerization process, making it a growth polymerization.

Epsilon-caprolactam, the monomer for Nylon-6, is a 6-carbon cyclic amide. To deduce and draw its cyclic monomer, one should recognize that the repeating unit in Nylon-6 consists of 6 carbons in a chain with an amide linkage connecting the ends. Thus, depicting epsilon-caprolactam involves drawing a ring structure containing 5 methylene (-CH2-) groups and one carbonyl group (=O) adjacent to an NH group, forming a lactam.

Answer:

[OH-] = 10^-4 M

Explanation:

pOH= 14- pH= 14-10=4

[OH-]= antilog (-4)= 10^-4M