3. Most refrigerants utilize the energy involved in the liquid/gas phase change of a molecule, where an ideal refrigerant is noncorrosive, nonflammable, and has a boiling point around the desired temperature. Many fluorinated carbons meet all of these requirements, but the extremely high stability of some (such as CFCs) caused them to be phased out by the Montreal Protocol due to their ozone depleting potential. 3.a. R-12 (CCl2F2) was commonly used in household refrigerators and vehicle air conditioners prior to 1994. What is the correct Lewis structure for R-12

Complete Question

The complete question is shown on the first uploaded image

Answer:

The pH is  [tex]pH = 4.94[/tex]

Explanation:

From the question we are told that

   The average concentration of NaOH is [tex][NaOH] = 0.101 M[/tex]

    The volume of NaOH is  [tex]V__{NaOH}} = 15.00 mL[/tex]

    The average concentration of Acetic acid is [tex][Acetic \ Acid] =0.497 \ M[/tex]

     The volume of Acetic acid is  [tex]V__{Acetic \ Acid}} = 5.00 \ mL[/tex]

The chemical equation for this reaction is  

         [tex]NaOH + CH_3COOH ---> CH_3 COONa + H_2 O[/tex]

The total volume of the solution is  

  [tex]V__{Total}} = V__{NaOH}} + V__{Acetic \ Acid}}[/tex]

Substituting values

      [tex]V__{Total}} = 15 + 5[/tex]

      [tex]V__{Total}} = 20mL = 20 *10^{-3} L[/tex]

The number of moles of NaOH is mathematically represented as

      [tex]n__{NaOH}} = [NaOH] * V__{NaOH}}[/tex]

substituting values

      [tex]n__{NaOH}} = 0.101 * 15*10^{-3}[/tex]

     [tex]n__{NaOH}} = 0.001515 \ moles[/tex]

The number of moles of Acetic acid is mathematically represented as

      [tex]n__{Acetic acid}} = [Acetic \ acid] * V__{Acetic acid}}[/tex]

substituting values

        [tex]n__{Acetic acid}} = 0.497 * 5*10^{-3}[/tex]

      [tex]n__{Acetic acid}} = 0.002485\ moles[/tex]

From the chemical equation

          1 mole of NaOH reacts with   1 mole of Acetic acid  to produce 1 mole of   [tex]CH_3 COONa[/tex] salt and 1 mole of  [tex]H_2 O[/tex]

So

  0.001515 moles  of NaOH reacts with  0.001515 moles of Acetic acid  to produce 0.001515 moles of   [tex]CH_3 COONa[/tex] salt and 0.001515 moles of  [tex]H_2 O[/tex]

This implies the number of moles of NaOH remaining after the react would be

      [tex]\Delta n__{NaOH}} = 0.001515 - 0.001515[/tex]

      [tex]\Delta n__{NaOH}} = 0 \ mole[/tex]

the number of moles of Acetic acid remaining after the react would be

    [tex]\Delta n__{Acetic acid}} = 0.002485 - 0.001515[/tex]

    [tex]\Delta n__{Acetic acid}} = 0.00097 \ moles[/tex]

the number of moles of [tex]CH_3 COONa \ salt[/tex] remaining after the react would be

    [tex]\Delta n__{CH_3 COONa \ salt}} = 0 + 0.001515[/tex]

    [tex]\Delta n__{CH_3 COONa \ salt}} = 0.001515 \ moles[/tex]

the number of moles of [tex]H_2 O[/tex] remaining after the react would be

   [tex]\Delta n__{H_2O}} = 0 + 0.001515[/tex]

   [tex]\Delta n__{H_2O}} = 0.001515 \ moles[/tex]

The expected pH is mathematically evaluated as

      [tex]pH = pK_a + log [\frac{[CH_3 COONa]}{[Acetic \ acid]} ][/tex]

Where [tex]pKa[/tex] is mathematically evaluated as

        [tex]pK_a = - log (K_a)[/tex]

The concentration of [tex]CH_3 COONa \ salt[/tex] is mathematically evaluated a s

[tex][CH_3 COONa] = \frac{\Delta n_CH_3 COONa \ salt }{V__{Total}}}[/tex]

substituting values

   [tex][CH_3 COONa] = \frac{0.001515}{20 *10^{-3}}[/tex]

    [tex][CH_3 COONa] = 0.07575M[/tex]

The concentration of  Acetic acid is mathematically evaluated as

          [tex][Acetic acid] = \frac{\Delta n__Acetic acid}{V__{Total}}}[/tex]

substituting values

        [tex][CH_3 COONa] = \frac{0.00097}{20 *10^{-3}}[/tex]

       [tex][CH_3 COONa] = 0.0485 M[/tex]

Substituting values into the equation for pH

          [tex]pH = - log (1 .8 *10^{-5}) + log [\frac{0.07575}{ 0.0485} ][/tex]

             [tex]pH = log [\frac{0.07575}{ 0.0485} ] - log (1 .8 *10^{-5})[/tex]

            [tex]pH = log [\frac{1.561856}{1.8*10^{-5}} ][/tex]

          [tex]pH = 4.94[/tex]

Option b (O=N−F and N=O−F) represents a pair of resonance structures.

The correct choice representing a pair of resonance structures is option b. O=N−F and N=O−F.

Resonance structures are different representations of a molecule or ion that show the delocalization of electrons. In option b, the double bond can be moved between the nitrogen and oxygen atoms, resulting in two resonance structures.

Resonance occurs when there are multiple valid Lewis structures that can be drawn for a molecule or ion. It indicates that the actual structure is a combination or hybrid of all the resonance structures.

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Answer: The molarity of [tex]H^+[/tex] in this solution is [tex]2.2\times 10^{-11}M[/tex]

Explanation:

pH or pOH is the measure of acidity or alkalinity of a solution.

pH is calculated by taking negative logarithm of hydrogen ion concentration.

[tex]pH=-\log [H^+][/tex]

Given : pH = 10.66

Putting in the values:

[tex]10.66=-\log[H^+][/tex]

[tex][H^+]=10^{-10.66}[/tex]

[tex][H^+]=2.2\times 10^{-11}[/tex]

Thus the molarity of [tex]H^+[/tex] in this solution is [tex]2.2\times 10^{-11}M[/tex]

The molarity of hydrogen ions, H⁺ in this solution is  2.2 * 10⁻¹¹

The molarity of a solution is a measure of the concentration in moles of a substance present in a liter of solution of that substance.

The pH of a solution is the negative logarithm in base 10 of the hydrogen ion concentration of the solution.

pH = -log[H⁺]

-pH = log[H⁺]

pH of solution = 10.66

-10.66 = log[H⁺]

taking antilogarithm of both sides

[H⁺] = 10⁻¹⁰°⁶⁶

[H⁺] = 2.2 * 10⁻¹¹

Therefore, the molarity of hydrogen ions, H⁺ in this solution is  2.2 * 10⁻¹¹

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Answer:

4L

Explanation:

To obtain the volume of O2 at stp, first, we need to determine the number of mole of O2.

From the question given above,

Mass of O2 = 5.72g

Molar Mass of O2 = 32g/mol

Number of mole =Mass/Molar Mass

Number of mole of O2 = 5.72/32

Number of mole of O2 = 0.179 mole

Now, we can calculate the volume of O2 at stp as follow:

1 mole of a gas occupy 22.4L at stp.

Therefore, 0.179 mole of O2 will occupy = 0.179 x 22.4 = 4L

Therefore, the volume occupied by the sample of O2 is 4L

The volume of a 5.72 gram sample of oxygen at STP is approximately 4.01 liters, after converting the mass to moles and using the molar volume of a gas at STP.

To calculate the volume of oxygen at Standard Temperature and Pressure (STP) from the given mass, first convert the mass of oxygen to moles. At STP, 1 mole of gas occupies 22.4 liters. Use the molar mass of Oxygen (32 g/mol) to convert grams to moles and then use the volume of 1 mole of gas at STP to find the volume in liters.

First, convert the given mass of oxygen (5.72 g) to moles:

5.72 g [tex]O_{2}[/tex]
x
1 mol [tex]O_{2}[/tex]/ 32 g [tex]O_{2}[/tex]
= 0.179 moles [tex]O_{2}[/tex]

Then calculate the volume at STP (using the fact that 1 mole of any gas at STP occupies 22.4 liters):

0.179 moles [tex]O_{2}[/tex]
x
22.4 liters/mol
= 4.0096 liters

Therefore, the volume of a 5.72 gram sample of [tex]O_{2}[/tex] at STP is approximately 4.01 liters.

Answer: 0.225, x, 0.0788

Explanation:

Answer:

Find the given attachment

Answer: [H3O+]= 0.05 M

Explanation:

Message me for extra explanation.

snap- parkguy786

A scientific question may lead to a(n)

The answer is may lead to a hypothesis.

Explanation:

Answer: observations

Hypothesis

Experimentation

Explanation:in order and right on edge

Answer:

0.0075 milliliters (7.5 microliters) will be taken from the stock solution and then diluted with 0.0249925 milliliters (24.9925 microliters) of water to make 0.025 milliliters (25 microliters) of NaOH solution.

Explanation:

This is a problem of dilution using the equation:

initial concentration x initial volume = final concentration x final volume.

The final volume to be prepared is 25 microliters.

The final concentration to be prepared is 3 M.

The initial volume to be taken is not known yet.

The initial concentration is 10 M.

Now, let's substitute these parameters into the the equation above.

10 x initial volume = 3 x 25

Initial volume = 3 x 25/10

     = 7.5 microliters

Note that: 1 microliter = 0.001 milliliters

Hence,

7.5 microliters = 0.0075 milliliters

This means that an initial volume of 0.0075 milliliters (7.5 microliters) will be taken from the stock solution. This amount will then be diluted with 0.0249925 milliliters (24.9925 microliters) of water to make 0.025 milliliters (25 microliters) of NaOH solution.

Answer:

0.122 moles

Explanation:

Final answer:

0.122 moles of KCl will be produced from the reaction of 15.0 g of KClO3, using stoichiometry and the molar mass of KClO3 which is 122.55 g/mol.

Explanation:

To determine the amount of KCl produced from 15.0 g of KClO3, we first need to know the molar mass of KClO3, which is 122.55 g/mol according to LibreTexts. By using stoichiometry, we can find out how many moles of KClO3 we have and, subsequently, how many moles of KCl will be produced according to the balanced chemical equation.

First, calculate the number of moles of KClO3:

15.0 g KClO3 × (1 mol KClO3 / 122.55 g) = 0.122 mol KClO3

From the balanced equation Î{2KClO3 -> 2KCl + 3O2}, we can see that 2 moles of KClO3 produce 2 moles of KCl. Therefore, the number of moles of KCl produced is equal to the number of moles of KClO3 available:

0.122 mol KClO3 × (2 mol KCl / 2 mol KClO3) = 0.122 mol KCl

Thus, 0.122 mol of KCl are produced from the reaction of 15.0 g of KClO3.

Answer:

1670 ml

Explanation:

molarity x Volume (Liters) = moles => Volume (Liters) = moles/Molarity

Volume needed = 2.50mol/1.50M = 1.67 Liters = 1670 ml.

Answer:

hes right

Explanation:

ik all

Answer:

Check the explanation

Explanation:

Here, Nitrogen (N) undergoes oxidation and Chlorine (Cl) undergoes reduction.

To answer your question:

N is oxidized from an oxidation number of -3 to an oxidation number of -1.

Cl is reduced from oxidation number of +1 to an oxidation number of -1.

Now,

Borneol should have a lower Rf because of boiling point.

Answer:

C

Explanation:

Well, we can see from the equation that hydrogen gas and ethane are in a 2:1 ratio based on their mole coefficients.

You need 2 mol H2 to make 1 mol C2H6. So that means you need 2*13.78 mol H2 to make 13.78 mol C2H6!

2*13.78 = 27.56 mol

C is correct.

Answer:

A) No , because Qsp<Ksp for calcium fluoride.

Check attachment for calculation

Without specific calculations, to determine if a precipitate forms when mixing calcium chlorate with sodium fluoride, one assesses if the product of the ion concentrations exceeds the Ksp for calcium fluoride; if Qsp > Ksp, precipitation occurs.

To determine if a precipitate will form when calcium chlorate is mixed with sodium fluoride, one must consider the ionic products formed and compare the reaction quotient (Qsp) with the solubility product constant (Ksp) of the relevant compound, in this case, calcium fluoride (CaF2). The Ksp for calcium fluoride is 3.45 × 10∑11. In a mixture of calcium chlorate and sodium fluoride, the calcium ions (Ca2+) from calcium chlorate can react with the fluoride ions (F∑) from sodium fluoride to potentially form solid calcium fluoride. However, without the exact concentrations of Ca2+ and F∑ after mixing, one cannot directly calculate Qsp. The principle, however, is that if Qsp > Ksp, a precipitate of calcium fluoride will form, whereas if Qsp < Ksp, no precipitate forms because the solution is not supersaturated with respect to CaF2.

The concentration of each ion in the final solution depends on the dilution that occurs when the two solutions are mixed. Given the initial conditions and the principle that the formation of a precipitate requires Qsp to exceed Ksp, we generally assess the likelihood of precipitation by considering the concentrations of the reactive ions and the solubility product of the insoluble compound they may form. In this scenario, without performing a specific calculation, the correct answer would be based on understanding whether the mixing leads to a situation where the product of the concentrations of calcium and fluoride ions exceeds the Ksp of calcium fluoride.

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

A common addiction to the common water is the elemental chlorine and chlorine ions, the phosphate ions. Recent reports of the higher levels of lead in cities have to lead to the Pb(s) as CI 2 flows in petal pipes some may be oxidized.

They balanced equation of the net ionic reactions.

Learn more about the chloride ions and phosphate ions.

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Answer:

Work done on the door is = 0

Mechanical energy of the person is 100%

Mechanical energy of the door is 50%.

Explanation:

Mechanical energy, M.E., is the energy an object possesses due to its position and motion. When we talk of position, the body is at rest = potential energy. In terms of motion, the body is moving = kinetic energy.

This means that M.E = potential + kinetic

Also, Work done = force × distance

For the person pushing the door, the potential of the energy stored in the person and the kinetics of pushing the door handle are all present. Therefore,  M.E of person is 100%.

But despite the doors potential energy present, there is no motion from it, which means the M.E is only half or 50%.

Also, since distance moved by the door is 0, no work is done on the door.

A slice or part of a homogeneous substance has the same density as the whole piece because the density is uniform throughout. In a heterogeneous substance, the density can vary and depends on the composition and structure of the part being measured. Density is found by dividing mass by volume and is useful for identifying substances.

The density of a substance is defined as its mass divided by its volume. In a homogeneous material, the density is consistent throughout, which means a slice or part of the substance would have the same density as the whole piece. For example, a solid iron bar is homogeneous, and therefore any part of it would have the same density as the entire bar. However, in a heterogeneous material, the density can vary from one part to another; an instance of this is Swiss cheese, which contains air pockets resulting in variable local densities.

To determine the density of a substance, you would divide the mass by the volume of the substance. This method applies to both whole objects and parts thereof. In the case of a cube, the volume is found by cubing the edge length. In practice, density is often used to help identify substances by comparison with known values, as different substances have characteristic densities.

Answer:

42050 J.

Explanation:

Data obtained from the question:

Mass (M) of ethanol = 50g

Heat of vaporisation (ΔHv) of ethanol = 841 J/g

Heat (Q) =.?

The heat required to boil 50g of ethanol can be obtained as follow:

Q = MΔHv

Q = 50 x 841

Q = 42050 J.

Therefore, the heat required to boil 50g of ethanol is 42050 J.

Answer:

Yes, a hydrocarbon is any compound that contains hydrogen and oxygen.

Explanation:

Answer:

(a)

Rate of appearance of sucrose = - d[C12H22O11] / dt = - ( 0.808 - 1.002 ) / ( 60.0 - 0.0) = 0.00323 M/s

(b)

Rate of appearance of fructose = d[C6H12O6] / dt = (1.002 - 0.808) / (60.0 - 0.0) = 0.00323 M/s

(c)

k = (1 / t ) * ln[A]/[A]t

k = ( 1 / 60.0 ) * ln[1.002 / 0.808]

k = 0.00359 min-1

(d)

0.00359 = ( 1 / t ) * ln[1.002 / 0.212]

t = 432.6 min

(e)

Half life time = 0.693 / k = 0.693 / 0.00359 = 193 min

Explanation:

First-order reactions are defined as the chemical reactions in which rate of the reaction is linearly dependent on the concentration of only one reactant.

The answers can be explained as:

(a) Rate of appearance of the sucrose from the chemical reaction is:

Rate = [tex]\dfrac{\text d [\text C_{12}\text H_{22}\text O_{11}]}{\text {dt}}[/tex] = [tex]\dfrac{0.808 - 1.002}{60.0 -0.0}[/tex]

Rate = 0.00323 m/s

(b) Rate of appearance of Fructose from the given chemical reaction is:

Rate = [tex]\dfrac{\text d [\text C_{6}\text H_{12}\text O_{6}]}{\text {dt}}[/tex] = [tex]\dfrac{1.002 - 0.808 }{60.0 -0.0}[/tex]

Rate = 0.00323 m/s

(C) Rate constant for the reaction is:

[tex]\text k &= \dfrac{1}{\text t}\times \dfrac {\text{ln [A]}}{\text {[A]} \text t}[/tex]

[tex]\text k &= \dfrac{1}{60}\times \dfrac {\text{ln} (1.002)}{(0.808)}[/tex]

k = 0.00359 minute⁻¹

(d) Time required for the concentration of sucrose to drop from 1.002 to 0.212 M is:

[tex]0.00359 &= \dfrac{1}{t} \times {\text{ln}\dfrac{[1.002]}{[0.212]}[/tex]

t = 432.6 minutes

(e) The half-life of the decomposition of sucrose at 25°C is:

Half-life = [tex]\dfrac{0.693}{\text k} = \dfrac{0.693}{0.00359}[/tex]

Half-life = 193 minutes.

To know more about first-order reaction, refer to the following link:

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Answer:

Explanation:

The chemical equation for the reaction is :

[tex]C_6H_{12}O_6_{(s)} \to 2C_2H_5OH _{(l)} + 2CO_{2(g)}[/tex]

The standard enthalpy of formation [tex]\Delta H ^0_f[/tex]  of the above equation is as follows:

[tex]\Delta H^0_{f, C_6H_{12}O_6}[/tex] = -1274.4 kJ/mol

[tex]\Delta H ^0_{f, C_2H_{5}OH[/tex] = -277.7 kJ/mol

[tex]\Delta H ^0_{f, CO_2}[/tex] = -393.5 kJ/mol

[tex]\Delta H ^0_{rxn }= \sum n_p \Delta H ^0_{f,p} - \sum n_r \Delta H ^0_{f,r}[/tex]

where ;

[tex]n_p[/tex] = stochiometric coefficients of products

[tex]n_r=[/tex] stochiometric coefficients of reactants

[tex]\Delta H^0_{f.p}[/tex] = formation standard enthalpy of products

[tex]\Delta H^0_{f.r}[/tex] = formation standard enthalpy of reactants

[tex]\Delta H ^0_{rxn }= (2 \ mol* -2777.7 \ kJ/mol + 2 \ mol * - 393.5 \ kJ/mol) - (1\ mol *(-1274.4 \ kJ/mol)[/tex]

[tex]\Delta H ^0_{rxn }= -68 \ kJ[/tex]

For [tex]\Delta S ^0[/tex] ;

The standard enthalpy of formation of [tex]\Delta S ^0_f[/tex] of the reactant and the products are :

[tex]\Delta S ^0 _{f \ C_6H_{12}O_6} = 212 \ \ J/Kmol[/tex]

[tex]\Delta S ^0 _{f \ C_2H_5O_H} = 160.7 \ \ J/Kmol[/tex]

[tex]\Delta S ^0 _{f \ CO_2} = 213.63 \ \ J/Kmol[/tex]

The [tex]\Delta S ^0_{rxn}[/tex] is as follows:

[tex]\Delta S ^0_{rxn} = \sum n_p \Delta S^0_{f.p} - \sum n_r \Delta S^0_{f.r}[/tex]

[tex]\Delta S ^0_{rxn} = (2 \ mol *160.7 \ \ J/Kmol + 2 \ mol *213.63 \ \ J/Kmol) -(1*212 J/Kmol)[/tex]

[tex]\Delta S ^0_{rxn} =536 . 7 \ J/K[/tex]   (to kJ/K)

[tex]\Delta S ^0_{rxn} =536 . 7 \ J/K * \frac{1 \ kJ}{1000 \ J}[/tex]

[tex]\Delta S ^0_{rxn} =0.5367 \frac{kJ}{K}[/tex]

Given that;

at T = 25°C = ( 25 + 273) K = 298 K

[tex]\Delta G^0 _{rxn} = \Delta H^0_{rxn} - T \Delta S^0_{rxn}[/tex]

[tex]\Delta G^0 _{rxn} = -68 \ kJ - 298 * 0.5367 \frac{kJ}{K}[/tex]

[tex]\Delta G^0 _{rxn} = -227. 9 \ \ \ kJ[/tex]

As [tex]\Delta G^0 _{rxn}[/tex] is negative; the reaction is spontaneous

[tex]\Delta H^0 _{rxn}[/tex] = negative

[tex]\Delta S^0 _{rxn}[/tex] = positive; Therefore , the reaction is spontaneous at all temperature , We can then say that the spontaneity of this reaction [tex]\Delta G^0 _{rxn} = \Delta H^0_{rxn} - T \Delta S^0_{rxn}[/tex]  is dependent on temperature.

Answer:1.587J / g degC.

Explanation:SOLUTION

Specific heat of aluminum = 0.897 J / g degC.

Heat gained by calorimeter = 100 * 10 * 0.897 J = 897 J

Specific heat of water = 4.18 J / g degC.

Heat gained by water = 250 * 10 * 4.18 J = 10450 J

Specific heat of copper = 0.385 J / g degC.

Heat loss by copper block = 50 * (80-10) * 0.385 J =1347.5  J

Let specific heat of unknown block be x J / g degC.

Heat loss by unknown block = 70 * (100-10) * x J = 6300x J

Heat gain = Heat loss

897 + 10450 = 1347.5 + 6300x

6300x = 9999.5

x =1.58722= 1.587

Specific heat of unknown substance is 1.587J / g degC.

Answer:

Check the explanation

Explanation:

Kindly check the attached image below for the step by step explanation to the question above.

Answer:

4.0 moles

Explanation:

The following data were obtained from the question:

Volume (V) = 12L

Pressure = 5.6 atm

Temperature (T) = 205K

Gas constant (R) = 0.08206 atm.L/Kmol

Number of mole (n) =?

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

PV = nRT

5.6 x 12 = n x 0.08206 x 205

Divide both side by 0.08206 x 205

n = (5.6 x 12)/(0.08206 x 205)

n = 4.0 moles

Therefore, the number of mole of the gas is 4.0 moles

Answer:

CHC13

Explanation:

Answer: The heat of reaction for the combustion of titanium is 15240 kJ/mol

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

[tex]Q=C\times \Delta T[/tex]

Q = Heat absorbed by calorimeter =?

C = heat capacity of calorimeter = 9.84 kJ/K

Initial temperature of the calorimeter  = [tex]T_i[/tex] = [tex]25.00^0C=(25.00+273)=298.00K[/tex]

Final temperature of the calorimeter  = [tex]T_f[/tex]  = [tex]91.60^0C=(91.60+273)K=364.6K[/tex]

Change in temperature ,[tex]\Delta T=T_f-T_i=(364.6-298.0)K=66.60K[/tex]

Putting in the values, we get:

[tex]Q=9.84kJ/K\times 66.60K=655.3kJ[/tex]

As heat absorbed by calorimeter is equal to heat released by combustion of titanium

[tex]Q=q[/tex]

[tex]\text{Moles of titanium}=\frac{\text{given mass}}{\text{Molar Mass}}=\frac{2.060g}{47.8g/mol}=0.0430mol[/tex]  

Heat released by 0.0430 moles of titanium = 655.3 kJ

Heat released by 1 mole of titanium = [tex]\frac{655.3}{0.0430}\times 1=15240kJ[/tex]

The heat of reaction for the combustion of titanium is 15240 kJ/mol

Absorption of 1.0 rad of gamma radiation would likely cause the greatest tissue damage due to its high energy and penetration ability.

The amount of tissue damage caused by radiation depends on various factors, including the type of radiation and its energy level.

Alpha particles are relatively heavy and highly ionizing, but they have low penetration power, so they are generally less damaging to tissues externally but can be more damaging if ingested or inhaled.

Beta particles have less mass and energy compared to alpha particles and can penetrate deeper into tissues, potentially causing more damage.

Gamma rays are electromagnetic radiation with high energy and penetration power. They can penetrate deeply into tissues, causing damage along their path.

Considering these factors, absorption of 1.0 rad of gamma radiation would likely result in the greatest tissue damage due to its high energy and penetration ability.

Answer:

Zn(s) + Fe^2+(aq) ---------> Zn^2+(aq) + Fe(s)

∆G= 61760J

K= 6.5 ×10^10

Explanation:

Zn(s) + Fe^2+(aq) ---------> Zn^2+(aq) + Fe(s)

E°reaction= E°Fe - E°Zn

E°reaction= (-0.44)-(-0.76)

E°reaction= 0.32V

We know that

∆G= -nFE°

Since n= 2 from the reaction equation and F= 96500C

∆G= -(2×96500×0.32)

∆G= 61760J

From

E°= 0.0592/n log K

0.32= 0.0592/2 log K

log K= 0.32/0.0296

log K= 10.81

K= Antilog(10.81)

K= 6.5 ×10^10

Answer:

-5.51 kJ/mol

Explanation:

Step 1: Calculate the heat required to heat the water.

We use the following expression.

[tex]Q = c \times m \times \Delta T[/tex]

where,

The average density of water is 1 g/mL, so 75.0 mL ≅ 75.0 g.

[tex]Q = 4.184J/g.\°C \times 75.0g \times (95.00\°C - 25.00\°C) = 2.20 \times 10^{3} J = 2.20 kJ[/tex]

Step 2: Calculate the heat released by the methane

According to the law of conservation of energy, the sum of the heat released by the combustion of methane (Qc) and the heat absorbed by the water (Qw) is zero

Qc + Qw = 0

Qc = -Qw = -22.0 kJ

Step 3: Calculate the molar heat of combustion of methane.

The molar mass of methane is 16.04 g/mol. We use this data to find the molar heat of combustion of methane, considering that 22.0 kJ are released by the combustion of 64.00 g of methane.

[tex]\frac{-22.0kJ}{64.00g} \times \frac{16.04g}{mol} = -5.51 kJ/mol[/tex]

Answer:

See explaination

Explanation:

Please see the attached file for the structural diagram of the molecular formation.

It is detailed and properly presented at the attachment.