What is one disadvantage of a unicellular organisms?

Given what we know, we can confirm that since Mr. Summers has to test a hypothesis, his next step should be to design an experiment.

Therefore, we can confirm that Mr. Summers must design an experiment given that this is the best way to gather data in order to be analyzed in the future and draw a valid conclusion.

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Answer:

Chloroform has a boiling point of 61.15 degrees Celsius and a vapor pressure of 9.5 kPa at 20 degrees Celsius. Glycerol, on the other hand, has a boiling point of 290 degrees Celsius and a vapor pressure of 0.0002 kPa at 20 degrees Celsius. Therefore, chloroform has a much lower boiling point and a much higher vapor pressure than glycerol. This means that chloroform is more volatile and evaporates more easily than glycerol.

Explanation:

Chloroform has a boiling point of 61.15 degrees Celsius and a vapor pressure of 9.5 kPa at 20 degrees Celsius. Glycerol, on the other hand, has a boiling point of 290 degrees Celsius and a vapor pressure of 0.0002 kPa at 20 degrees Celsius. Therefore, chloroform has a much lower boiling point and a much higher vapor pressure than glycerol. This means that chloroform is more volatile and evaporates more easily than glycerol.

Answer:

A. Electrons all have approximately the same energy.

C. Electrons move freely among atoms (delocalized).

Next Question:

B. Electrons move among orbitals of different energies.

C. Electrons move freely among atoms (delocalized).

Explanation:

Electron sea model

Metallic Band theory

Conclusively, we can say Electron sea model

Metallic Band theory

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Answer:

B

Explanation:

It adds so there are 4 aluminum on both sides as well as 6 oxygen on both sided.

Answer:

B. would be the correct option.

Explanation:

To balance the equation, both the reactant and product must have the same number so with 4Al+30 with two on the bottom, you would get a total of 4 Al atoms and 6 O atoms. Now, the other side must have 4 Al atoms and 6 O atoms, so multiply them by 2 and that would be your answer.

Answer: n= 1.02 moles

Explanation: relative atomic mass of Fe is 55.85.

Amount of substance n= m/M = 56.7 g /55.85 g/mol

Answer: .5m

Explanation:

Answer:

The answer is E) none of the above.

In all of the given reactions, mass and charge are conserved. The law of conservation of mass states that the mass of reactants must be equal to the mass of the products in a chemical reaction. The law of conservation of charge states that the total charge of the reactants must be equal to the total charge of the products.

In reaction A, the water molecule dissociates into a hydrogen ion (H+) and a hydroxide ion (OH-), but the total mass and charge are still conserved.

In reaction B, the reduction of CIO2 to Cl- is balanced by the oxidation of water to form OH-. The electrons and charge are conserved.

In reaction C, H2SO4 reacts with NaOH to form Na2SO4, H+ and OH-. The mass and charge are conserved.

In reaction D, ZnCl reacts with water to form ZnOH, H+ and Cl-. The mass and charge are also conserved.

The correct order of the increasing polarity of the analyte functional group isEthers < Esters.

The given statement is "Polarities of analyte functional group increase in the order of hydrocarbon ethers < esters."  The order of polarities of functional groups is the order of their increasing polarity (i.e., less polar to more polar) based on their electron-donating or withdrawing ability from the rest of the molecule.Polarity of analyte: The analyte's polarity is directly proportional to the dipole moment of the functional group, which is associated with a difference in electronegativity between the atoms that make up the functional group.The electronegativity of an element is its ability to attract electrons towards itself. The greater the difference in electronegativity between two atoms, the more polar their bond, and hence the greater the polarity of the molecule.

To find the correct order of the increasing polarity of the analyte functional group, let's first compare the two groups: hydrocarbon ethers and esters. Here, esters have a carbonyl group while ethers have an oxygen atom with two alkyl or aryl groups. The carbonyl group has more electronegative oxygen, which pulls electrons away from the carbon atom, resulting in a polar molecule. On the other hand, ethers have a less polar oxygen atom with two alkyl or aryl groups, making them less polar than esters. Therefore, the correct order of the increasing polarity of the analyte functional group isEthers < Esters.

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Answer:

13.1 L

Explanation:

Using the combined gas law:

P1V1/T1 = P2V2/T2

Where;

P1 = initial pressure (atm)

P2 = final pressure (atm)

V1 = initial volume (L)

V2 = final volume (L)

T1 = initial temperature (K)

T2 = final temperature (K)

Based on the information provided in this question;

P1 = 1.95 atm

P2 = 0.65 atm

V1 = 5.0 L

V2 = ?

T1 = 20 °C = 20 + 273 = 293K

T2 = -15 °C = -15 + 273 = 258K

Using P1V1/T1 = P2V2/T2

1.95 × 5/293 = 0.65 × V2/258

9.75/293 = 0.65V2/258

0.0333 = 0.00252V2

V2 = 0.0333 ÷ 0.00252

V2 = 13.1 L

Answer: Velocity

Explanation:

Answer: B velocity

Explanation: I did the review

B)

A given substance that has a melting point of -7C and a boiling point of 59C exists as a liquid at room temperature conditions.

The states of matter refer to the separation of the molecules of a given material in certain temperature conditions.

The states of the matter are three and include the gaseous state, liquid state and solid-state.

In conclusion, a given substance that has a melting point of -7C and a boiling point of 59C exists as a liquid at room temperature conditions.

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To find the steady-state diffusion current density at x=5μm into the sample, we can use the formula for diffusion current density:

Jn = q * Dn * (δn / Lp)

Where:
Jn is the diffusion current density
q is the charge of an electron (1.6 x 10^-19 C)
Dn is the minority carrier diffusion coefficient
δn is the excess minority carrier concentration
Lp is the minority carrier diffusion length

First, let's calculate the diffusion coefficient using the Einstein relation:

Dn = μn * kb * T

Where:
μn is the minority carrier mobility
kb is Boltzmann's constant (1.38 x 10^-23 J/K)
T is the temperature in Kelvin

We are given:
δn = 3.3 x 10^13 cm^-3 (excess minority carrier concentration)
Lp = 7.5 μm (minority carrier diffusion length)

Substituting the values into the equation, we get:

Jn = (1.6 x 10^-19 C) * (Dn) * (3.3 x 10^13 cm^-3) / (7.5 μm)

Now, let's convert the units:
1 μm = 10^-4 cm
1 A = 10^2 mA

Jn = (1.6 x 10^-19 C) * (Dn) * (3.3 x 10^13 cm^-3) / (7.5 x 10^-4 cm)

Simplifying the equation, we have:

Jn = (1.6 x 10^-19 C) * (Dn) * (3.3 x 10^13 cm^-3) / (7.5 x 10^-4 cm)
  = (1.6 x 10^-19 C) * (Dn) * (3.3 x 10^13 cm^-3) * (1 / 7.5 x 10^-4 cm)
  = (1.6 x 10^-19 C) * (Dn) * (3.3 x 10^13 cm^-3) * (1.33 x 10^3 cm)

Finally, let's calculate the diffusion current density:

Jn = (1.6 x 10^-19 C) * (Dn) * (3.3 x 10^13 cm^-3) * (1.33 x 10^3 cm)
  = (5.28 x 10^-6 C * Dn)
As a result, we cannot determine the correct option from the given choices (a), (b), (c), (d), or (e).

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We find the diffusion current density to be 126 \(\frac{mA}{cm^{2} }\). The correct answer is (b) 126 \(\frac{mA}{cm^{2} }\).

To determine the steady-state diffusion current density at x=5μm into the sample, we can use the equation:

Jn = qDn * (dn/dx)

Where Jn is the diffusion current density, q is the charge of an electron (1.6 × \(10^{-19}\) C), Dn is the diffusion coefficient of the minority carrier, and (dn/dx) is the gradient of the minority carrier concentration.

First, let's calculate the diffusion coefficient using the Einstein relationship:

Dn = k * T * μn

Where k is Boltzmann's constant (1.38 × \(10^{-23}\) J/K), T is the temperature in Kelvin (300 K), and μn is the minority carrier mobility.

Next, let's find the gradient of the minority carrier concentration:

(dn/dx) = (Δn/Δx)

Given that the minority carrier concentration at x=0 is 3.3×\(10^{13}\)  \(cm^{-3}\) and the minority carrier diffusion length is 7.5μm, we can find the concentration gradient:

Δn = 3.3×\(10^{13}\)  \(cm^{-3}\)  - 5×\(10^{15}\)  \(cm^{-3}\) (uniform doping)
Δx = 5μm - 0μm

Now, substitute the values into the equations and calculate the diffusion current density:

Dn = k * T * μn
Δn = 3.3×\(10^{13}\)  \(cm^{-3}\) - 5×\(10^{15}\)  \(cm^{-3}\)
Δx = 5μm - 0μm
Jn = qDn * (dn/dx)

By plugging in the values and solving the equation, we find the diffusion current density to be:

Jn ≈ 126 \(\frac{mA}{cm^{2} }\)

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Answer:

See Below

Explanation:

As this question is about the type of transformer, i am not going to discuss the detail the construction of it, rather the underlying principle. In practices transformer has two component namely primary and secondary. For an ideal transformer energy must conserved.

\(P_p=P_s\)

\(\\\Rightarrow V_{p}I_{p} & = & V_{s}I_{s}\qquad\text{as}\qquad \boxed{P=VI}\\\text{or},\frac{V_{p}}{V_{s}} & = & \frac{I_{s}}{I_{p}}\tag{1}\end{eqnarray}\)(1)

Where, \(V_p\), \(I_p\) are the voltage and current in the primary circuit and \(V_s\), \(I_s\) are that for secondary circuit respectively. If consider the number of turns of the coil in primary (\(N_p\)) and secondary circuit \((N_s)\) , then expression (1) further extends to

\(\frac{V_{p}}{V_{s}} = \frac{I_{s}}{I_{p}}=\frac{N_{p}}{N_{s}}\tag{2}\)       (2)

Equation (2) is the main equation for transformer.

Now consider

\(\frac{V_{p}}{V_{s}} = \frac{N_{p}}{N_{s}}\)

Case-1: If \(N_{p} > N_{s}\) then \(V_{s} < V_{p}\) . This the step-down transformer. Where the number of turns in the primary is greater than that of secondary.

Case-2:If \(N_{s} > N_{p}\)  then \(V_{s} > V_{p}\). This the step-up transformer. Where the number of turns in the primary is less than that of secondary.

A transformer is an electrical device that uses electromagnetic induction to transmit electrical energy between two or more circuits. This induction produces a force across the conductor, which is subsequently subjected to varying magnetic fields. In a power application, transformers typically reduce or enhance alternating current voltages.

This is where a step down transformer comes in, to increase or decrease an alternating current current. The primary voltage is larger than the secondary voltage in this sort of device. In a 220v application, a step down transformer will be required to use a 110v product. These transformers frequently rely on magnetic induction between coils. This is the component that transforms voltage and/or current levels. As a result, you'll have two (or more) insulated wire coils twisted around an iron core. When you add voltage to the primary (one of the coils), the core is magnetized and voltage is induced in the secondary (other coil). The voltage reduction is determined by the ratio of turns in the two sets of windings. As a result, if you have 200 turns on the main and 100 on the secondary, your ratio will be 2:1. The voltage ratio of a single transformer remains constant throughout all usage of that transformer.

To recap, a step down transformer transforms low current, high voltage electricity to high current, low voltage power. It is also possible to use a step down transformer as a reverse connection. To do this, a single phase step down transformer of 1 kva or greater is required.

The primary reason we may want a step down transformer in the first place is to conserve energy. When electrons move down a metal wire, they do not follow a straight, smooth route. They jostle around, wasting energy and heating up the wire. However, greater voltage and lower current consume less energy. This is why power plants deliver extremely high voltages down the line to your home, workplace, and so on.

Another reason for such high voltages is for applications that require them, like as industrial facilities. Their massive, powerful machinery may demand this voltage and do not need a step down transformer. These circumstances may necessitate the use of a step-up transformer, which may be accomplished, as previously stated, by utilizing a step-down in reverse. Because it is not the most efficient method, it is preferable to invest in a genuine step up transformer after researching your particular voltage needs. A step up transformer is one that raises the voltage from its main to secondary power source. In this sort of transformer, the secondary coil has more turns than the main coil, hence the induced secondary coil voltage is greater than the applied voltage on the primary coil.

When the relationship between voltage and turns in each coil is shown, it looks like this:

(Primary coil voltage minus secondary coil voltage) = (Primary coil turns minus secondary coil turns)

___________________

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20.87 mL of the 0.384 M NaOH solution is required to reach the equivalence point.

In the titration of hydrocyanic acid with sodium hydroxide, the balanced chemical equation is:

HCN + NaOH → NaCN + H2O

This is a one-to-one reaction, which means that the number of moles of NaOH needed to neutralize the HCN will be equal to the number of moles of HCN present in the solution.

First, we need to calculate the number of moles of HCN in the solution:

moles of HCN = M × V

moles of HCN = 0.353 mol/L × (22.7/1000) L

moles of HCN = 0.008014 moles

Since the reaction is one-to-one, we know that 0.008014 moles of NaOH will be needed to reach the equivalence point.

Next, we need to calculate the volume of NaOH solution required to reach the equivalence point:

moles of NaOH = M × V

V = moles of NaOH / M

V = 0.008014 mol / 0.384 mol/L

V = 0.02087 L or 20.87 mL

Therefore, 20.87 mL of the 0.384 M NaOH solution is required to reach the equivalence point.

At the equivalence point, the number of moles of NaOH added is equal to the number of moles of HCN originally present in the solution, so the remaining solution will be a salt solution of NaCN and water.

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Answer:Super-giants

Explanation:

Answer:

464.29mL de solvente y 35.71mL de ácido nítrico concentrado deben agregarse.

Explanation:

El ácido nítrico concentrado viene al 70% v/v por temas de estabilidad. El volumen de ácido nítrico que se debe agregar si se quieren hace 500mL al 5% de HNO3 es:

500mL * (5mL / 100mL) = 25mL de ácido nítrico se deben agregar.

Como el ácido nítrico está al 70%:

25mL ácido nítrico * (100mL / 70mL ácido nítrico) = 35.71mL de ácido nítrico concentrado deben agregarse.

Y el volumen de solvente debe ser:

500mL - 35.71mL = 464.29mL

The dry gas would occupy 1.46 L at standard conditions.

When gas is collected over water, the vapor pressure of the water affects the total pressure measured. To account for this, we need to use Dalton's law of partial pressure, which states that the total pressure of a gas mixture is the sum of the partial pressures of each gas component.

First, we need to calculate the partial pressure of the collected gas. We can do this by subtracting the vapor pressure of water at 20 degrees Celsius (17.5 mm Hg) from the total pressure measured:

Partial pressure of gas = total pressure - vapor pressure of water
Partial pressure of gas = 622.0 mm Hg - 17.5 mm Hg
Partial pressure of gas = 604.5 mm Hg

Next, we can use the ideal gas law (PV = nRT) to calculate the volume of the dry gas at standard conditions (0 degrees Celsius and 1 atm):

PV = nRT
V = nRT/P

where P is the partial pressure of the gas (604.5 mm Hg converted to atm), n is the number of moles of gas (which we can calculate using the volume of the collected gas and the known molar volume of a gas at STP), R is the gas constant, and T is the temperature in Kelvin (273 K).

V = (40 L)(0.0821 L·atm/mol·K)(293 K)/(0.793 atm)
V = 1.46 L

Therefore, the dry gas would occupy 1.46 L at standard conditions.

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Answer:

and earthquake epicenters are related to tectonic plate boundaries. causes Earth's plates to move. Most volcanoes and earthquakes are caused by the motion and interaction of Earth's plates. The way Earth's plates interact at boundaries is an important factor in the locations of earthquakes and volcanoes

Explanation:

Answer: 2.5 grams

Explanation:

Expression for rate law for first order kinetics is given by:

\(t=\frac{2.303}{k}\log\frac{a}{a-x}\)

where,

k = rate constant

t = age of sample

a = let initial amount of the reactant

a - x = amount left after decay process  

a) to calculate the rate constant:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

\(t_{\frac{1}{2}}=\frac{0.693}{k}\)

\(k=\frac{0.693}{2years}=0.346years^{-1}\)

b) to find amount left after 4 years

\(4=\frac{2.303}{0.346}\log\frac{10}{a-x}\)

\((a-x)=2.5g\)

Thus 2.5 g of cesium would remain from a 10 g sample after 4 years

Rusting is an electrochemical reaction. Iron rusts faster when alloyed with cobalt than when alloyed with manganese because, in the iron-manganese alloy, manganese is rendered the anode and iron is rendered the cathode

An alloy is a combination of two metals. There are various reasons for producing alloys such as greater tensile strength, corrosion resistance and improved aesthetic appearance.

When iron is alloyed with cobalt, the iron rusts faster than pure iron because iron is rendered the anode and cobalt is rendered the cathode. When the iron is alloyed with manganese, it rusts more slowly than pure iron because in the iron-manganese alloy, manganese is rendered the anode and iron is rendered the cathode.

Missing parts;

An iron object alloyed with cobalt rusts more quickly than a pure iron object. However, an iron object alloyed with manganese rusts less quickly than a pure iron object under the same conditions. This is true because

(1) cobalt is a stronger reducing agent than iron

(2) iron is a stronger reducing agent than manganese

(3) cobalt exhibits more metallic character than either iron or manganese

(4) in the iron-manganese alloy, manganese is rendered the anode and iron is rendered the cathode

(5) in the iron-cobalt alloy, cobalt is rendered the anode and iron is rendered the cathode

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molar solubility of PbBr2 in 0. 500 M KBr solution is 4S3 = 6.60 x 10–6 . PbBr2 ionizes as Pb2+ + 2Br-, molar solubility of PbBr2 in a 0. 500 M Pb(NO3)2 solution is 0.0181 moles/lit..

If molar solubility of PbBr2 is “S”, then solubility of Pb2+ is also “S” but that of Br- would be “2S”. Ksp = [Pb2+] [Br-]2 = (S) (2S)2 = 4S3 = 6.60 x 10–6 4S3 = 6.6 x 10–6,this gives Solubility S = 1.181 x 10–2 = 0.0181 moles/lit. solid's solubility (usually referred to as its molar solubility) is expressed as the concentration of the "dissolved solid" in a saturated solution. This would simply be the concentration of Ag+ or Cl- in the saturated solution for a simple 1:1 solid like AgCl. The other way to express solubility is through molar solubility, which is defined below. It is the number of moles of solute in one litre of saturated solution and is abbreviated with a lower case's'. It is expressed in moles per litre, also known as molarity.

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The pH of the resulting buffer is 4.446.

To find the pH of the resulting buffer, we need to first determine the moles of weak acid and moles of NaOH that have been added to the solution.

Moles of weak acid = (0.200 mol/L) x (0.525 L) = 0.105 mol

Moles of NaOH = (0.150 mol/L) x (0.500 L) = 0.075 mol

Next, we need to determine the moles of weak acid and moles of conjugate base (A-) in the buffer solution.

Since the weak acid is in excess, all of the NaOH will be neutralized by the weak acid to form its conjugate base.

Moles of A- = 0.075 mol

Moles of HA remaining = 0.105 mol - 0.075 mol = 0.03 mol

The initial concentration of weak acid was 0.200 M, so its initial moles were:

Initial moles of HA = (0.200 mol/L) x (0.525 L) = 0.105 mol

The final volume of the solution after mixing the two solutions together is:

Final volume = 0.500 L + 0.525 L = 1.025 L

Using the Henderson-Hasselbalch equation, we can find the pH of the resulting buffer:

pH = pKa + log([A-]/[HA])

pKa = -log(Ka) = -log(8.59×10−5) = 4.066

[A-]/[HA] = 0.075 mol/0.03 mol = 2.5

pH = 4.066 + log(2.5) = 4.446

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The answer is -Energy released after alpha decay is 503.23 GJ.

Energy during a alpha decay is calculated using the formula -\(E =\)Δ\(mC^2\)

Where Δm is the mass defect

C is the speed of light

Calculate the mass defect when  Pu-239 undergoes alpha decay.

Δm = \(239.052163 -235.0439299 - 4.0026 = 0.0056331\ amu\)

\(E = 0.0056331*931\ MeV = 5.24442\ MeV\)

\(E = 5.24442\ Mev * (6.022 * 10^2^3) = 31.582 *10^2^3\ MeV\)

\(1 MeV = 1.6 *10^-^1^9\ J\)

This, energy released is-

\(E = 31.582 *10^2^3\ MeV * (1.6 *10^-^1^9) = 503.23 * 10^9 J\\E = 503.23\ GJ\)

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Aluminum (Al) is the limiting reagent with 0.601 moles.

To determine the limiting reagent, we need to compare the amount of each reactant with the amount required to completely react with the other reactant.

The balanced chemical equation for the reaction between Fe₂O₃ and Al is:

Fe₂O₃ + 2Al → 2Fe + Al₂O₃

From the equation, we can see that 1 mole of Fe₂O₃ reacts with 2 moles of Al.

To calculate the number of moles of each reactant, we need to divide the mass of each by its molar mass. The molar mass of Fe₂O₃ is 159.69 g/mol and the molar mass of Al is 26.98 g/mol.

The number of moles of Fe₂O₃:

32.0 g / 159.69 g/mol = 0.200 moles

The number of moles of Al:

16.2 g / 26.98 g/mol = 0.601 moles

Since 1 mole of Fe₂O₃ reacts with 2 moles of Aluminum, the limiting reagent is the one with the smallest number of moles. In this case, Al is the limiting reagent with 0.601 moles, while Fe₂O₃ has 0.200 moles. This means that Al will run out first and determine the amount of Fe and Al₂O₃ that can be produced.

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Answer: The equilibrium will shift in the direction of \(NH_3\)

Explanation:

The chemical equation for the aqueous solution of ammonia follows:

\(NH_3+H_2O\rightleftharpoons NH_4^++OH^-\)

According to Le-chtelier's principle:

If there is any change in the variables of the reaction, then the equilibrium will shift in that direction of equilibrium to minimize the effect.

If we add more amount of \(NH_4^+\) to the solution, more of the products will be present. But according to Le-chtelier's principle, to minimize this effect, the equilibrium will shift in the backward direction that in the direction of \(NH_3\)

Hence, the equilibrium will shift in the direction of \(NH_3\)

fluorine, chlorine, bromine, and iodine,

An amide that has a molecular ion with an m/z value of 129.

The molecular formula is C₉H₁₃NO.

To determine the molecular formula of the amide with an m/z value of 129, we need to consider the possible combinations of carbon (C), hydrogen (H), nitrogen (N), and oxygen (O) that would yield that molecular mass.

The m/z value of 129 indicates the mass-to-charge ratio of the molecular ion. Since we're dealing with a neutral molecule, we can assume a charge of +1 for the molecular ion. Therefore, the molecular mass would be equal to 129.

To find the molecular formula, we can consider different combinations of elements that sum up to a molecular mass of 129. Here are a few possibilities:

1. C₈H₁₁NO: In this case, the sum of the atomic masses is (8 × 12.01) + (11 × 1.01) + 14.01 + 16.00 = 128.09, which is close to the desired molecular mass but not exactly 129.

2. C₈H₁₀N₂O: In this case, the sum of the atomic masses is (8 × 12.01) + (10 × 1.01) + (2 × 14.01) + 16.00 = 128.14, which is also close to 129 but not exact.

3. C₉H₁₃NO: In this case, the sum of the atomic masses is (9 × 12.01) + (13 × 1.01) + 14.01 + 16.00 = 129.12, which is very close to 129.

Therefore, the molecular formula that best fits the given m/z value of 129 is C₉H₁₃NO.

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The liquid shown in 'b' need more thermal energy to change to a gas.

Thermal energy is defined as internal energy which is present in a system in state of the thermodynamic equilibrium depending on its temperature. Thermal energy can not be converted into useful work as much easily as the system energy which are not in states of the thermodynamic equilibrium.

As we can see that at same temperature in image 'a' the particle have high kinetic energy than the molecules of image 'b'. This can be shown that the boiling point of image 'a' is lower than the image 'b'. Therefore, at lower temperature it boils faster than the liquid 'b'. Thus we can say that liquid 'a' has less thermal energy than 'b'.

Thus, we concluded that the liquid shown in 'b' need more thermal energy to change to a gas.

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