A reaction occurs between sodium carbonate and hydrochloric acid producing sodium chloride, carbon dioxide, and water. The correct set of coefficients, respectively, for the balanced reaction is:

Answer:

Option A - When |ΔHsolute| > |ΔHhydration|

Explanation:

A solution is defined as a homogeneous mixture of 2 or more substances that can either be in the gas phase, liquid phase, solid phase.

The enthalpy of solution can either be positive (endothermic) or negative (exothermic).

Now, we know that enthalpy is amount of heat released or absorbed during the dissolving process at constant pressure.

Now, the first step in thus process involves breaking up of the solute. This involves breaking up all the intermolecular forces holding the solute together. This means that the solute molecules are separate from each other and the process is always endothermic because it requires energy to break interaction. Thus;

The enthalpy ΔH1 > 0.

Thus, the enthalpy of the solute has to be greater than the enthalpy of hydration.

An endothermic ΔHsolution occurs when |ΔH solute| < |ΔH hydration|.

A substance dissolves in water when the solute - solvent interaction exceeds the solute - solute solute interaction. The energy required to break the bonds between solutes is the ΔHsolute and the energy released when solute - solvent interaction take place is called the ΔHhydration.

We know that when  |ΔH solute| < |ΔH hydration|, energy is required to break up the solute - solute interaction and  ΔHsolution is endothermic.

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Answer:

56.5 m/s

Explanation:

Step 1: Convert the temperature to Kelvin

We will use the following expression.

K = °C + 273.15

K = -15.0°C + 273.15 = 258.2 K

Step 2: Calculate the root-mean-square velocity

The root-mean-square velocity is the square root of the average square velocity. We can calculate it using the following expression.

[tex]v_{rms} = \sqrt{\frac{3RT}{M} }[/tex]

where,

[tex]v_{rms} = \sqrt{\frac{3 \times \frac{8.314J}{mol.K} \times 258.2K }{2.02g/mol} } = 56.5m/s[/tex]

The root-mean-square velocity for a hydrogen molecule at the given temperature is 1,793.74 m/s.

The given parameters;

The root-mean-square velocity for a hydrogen molecule is calculated as follows;

[tex]v_{rms} = \sqrt{\frac{3RT}{m} } \\\\[/tex]

where;

[tex]v_{rms} = \sqrt{\frac{3 \times 8.314 \times (-15+273)}{2\times 10^{-3}} } \\\\v_{rms} = 1,793.74 \ m/s[/tex]

Thus, the root-mean-square velocity for a hydrogen molecule at the given temperature is 1,793.74 m/s.

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Answer:

[H₂A] = 5.0409x10⁻⁷M

[HA⁻] = 0.001951M

[A²⁻] = 0.234

11.29 = pH

Explanation:

When Na₂A = A²⁻is in equilibrium with water, the reactions that occurs are:

A²⁻(aq) + H₂O(l) ⇄ HA⁻(aq) + OH⁻(aq)

By definition of Ka of reaction, you can find Kb (the inverse basic reaction), thus:

Kb1 = KwₓKa2 =

Where Kw is K of equilibrium of water, 1x10⁻¹⁴

1x10⁻¹⁴/ 6.19x10⁻⁹ =

1.6155x10⁻⁶ = [HA⁻] [OH⁻] / [A²⁻]  

And HA⁻ will be in equilibrium as follows:

HA⁻(aq) + H₂O(l) ⇄ H₂A(aq) + OH⁻(aq)

Kb2 = KwₓKa1 = 1x10⁻¹⁴/ 7.68x10⁻⁵ =

1.3021x10⁻¹⁰ = [H₂A] [OH⁻] / [HA⁻]

Thus, the Na₂A has 2 equilibriums, for the first reaction, concentrations wil be:

[HA⁻] = X

[OH⁻] = X

[A²⁻] = 0.236M - X

X as how much will react, Reaction coordinate

Replacing in Kb1:

1.6155x10⁻⁶ = [HA⁻] [OH⁻] / [A²⁻]

1.6155x10⁻⁶ = [X] [X] / [0.236-X]

3.8126x10⁻⁶ - 1.6155x10⁻⁶X = X²

3.8126x10⁻⁶ - 1.6155x10⁻⁶X - X² = 0

Solving for X :

X = -0.00195 → False solution. There is no negative concentrations

X = 0.001952.

Replacing, concentrations for the first equilibrium are:

Now, in the second equilibrium:

[HA⁻] = 0.001952 - X

[OH⁻] = X

[H₂A] = X

Replacing in Kb1:

1.3021x10⁻¹⁰ = [H₂A] [OH⁻] / [HA⁻]

1.3021x10⁻¹⁰ = [X] [X] / [0.001952 - X]

2.5417x10⁻¹³ - 1.3021x10⁻¹⁰X = X²

2.5417x10⁻¹³ - 1.3021x10⁻¹⁰X - X² = 0

Solving for X :

X = -5.04x10⁻⁷ → False solution. There is no negative concentrations

X = 5.0409x10⁻⁷

Replacing, concentrations for the second equilibrium are:

Thus, you have concentrations of H2A, HA−, and A2−

Now, for pH, you need [OH⁻] concentration that is:

[OH⁻] = 0.0019525

pOH = -log[OH⁻] = 2.709

As 14 = pH+ pOH

14-2.709=pH

Answer:

1. Hexanol (C6H14O) > Carbon tetrachloride (CCl4) > Dichloromethane (CH2Cl2) > butene (C4H8) >  propane (C3H8)

2. a. b and e

Explanation:

1. Arrangement of given compounds in order from highest to lowest boiling point are as following:

hexanol (C6H14O): 157 degree celcius

carbon tetrachloride (CCl4):  76.72 degree celcius

dichloromethane (CH2Cl2): 40 degree celcius

butene (C4H8): -6.3 degree celcius

propane (C3H8): -42 degree celcius

Hexanol (C6H14O) > Carbon tetrachloride (CCl4) > Dichloromethane (CH2Cl2) > butene (C4H8) >  propane (C3H8).

2. True statements are:

a. Hydrocarbons have only dispersion forces because hydrocarbons are very non-polar and are highly symmetric molecules.

b. Stronger intermolecular forces require higher boiling points because the bond between the molecules is very strong and require high energy and boiling point to break that strong bond.

e. Because the size of molecules increases with increase in number of electrons and dispersion force increases the attraction of the molecules to each other. Hence require high boiling point to separate them from each other.

Answer: 13,68 ML

Explanation: In the problem density and mass are give & you have to figure out the volume of mercury. So use this formula:

Mass = Volume  × Density

Volume = Mass ÷ Density

Volume = 186 g ÷ 3.6 g/ml

             = 13.68 ml

Answer:

Molecular formula for the gas is: C₄H₁₀

Explanation:

Let's propose the Ideal Gases Law to determine the moles of gas, that contains 0.087 g

At STP → 1 atm and 273.15K

1 atm . 0.0336 L = n . 0.082 . 273.15 K

n = (1 atm . 0.0336 L) / (0.082 . 273.15 K)

n = 1.500 × 10⁻³ moles

Molar mass of gas = 0.087 g / 1.500 × 10⁻³ moles = 58 g/m

Now we propose rules of three:

If 0.580 g of gas has ____ 0.480 g of C _____ 0.100 g of C

58 g of gas (1mol) would have:

(58 g . 0.480) / 0.580 = 48 g of C  

(58 g . 0.100) / 0.580 = 10 g of H

 48 g of C / 12 g/mol = 4 mol

 10 g of H / 1g/mol = 10 moles

The molecular formula of the compound is C4H10.

At STP;

P = 1 atm

T = 273 K

V = 33.6 mL or 0.0336 L

R = 0.082 atmLK-1mol-1

n = ?

Hence;

n = PV/RT

n = 1 atm × 0.0336 L/0.082 atmLK-1mol-1 × 273 K

n = 0.0015 moles

Number of moles = mass/molar mass

Molar mass= Mass/Number of moles

Molar mass = 0.087 g/0.0015 moles

Molar mass = 58 g/mol

Mass of carbon = (58 g × 0.480) / 0.580 = 48 g of C  

Mass of hydrogen = (58 g × 0.100) / 0.580 = 10 g of H

Number of moles of carbon = 48 g of C / 12 g/mol = 4 mol

Number of moles of hydrogen = 10 g of H / 1g/mol = 10 moles

Formula of the compound must then be C4H10.

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Answer:

Na + Cl ⇒ NaCl

Explanation:

Skeleton equation: Na + Cl ⇒ NaCl

This is already balanced so that is the answer.

Answer:

5.56 × 10⁻⁸

Explanation:

Step 1: Given data

Step 2: Calculate the concentration of H⁺

We will use the following expression.

pH = -log [H⁺]

[H⁺] = antilog -pH = antilog -3.99 = 1.02 × 10⁻⁴ M

Step 3: Calculate the acid dissociation constant (Ka)

We will use the following expression.

[tex]Ka = \frac{[H^{+}]^{2} }{Ca} = \frac{(1.02 \times 10^{-4})^{2} }{0.187} = 5.56 \times 10^{-8}[/tex]

Answer:

The density would be larger than the correct value.

Explanation:

First off, the realtionship between denisty and volume is given in the equation below;

Density = Mass / Volume

From this equation, Density is inversely proportional to volume. This means as the volume increases, the density decreases and as the volume decreases the density increases.

Assuming all thing's being normal;

Mass = 2g

Volume = 10ml

Density = 2 / 10 = 0.2 g/ml

Second case scenario;

'your pipet was incorrectly calibrated so that it transferred less than 10.00 mL"

Lets have a value of 8ml for our volume. Mass remains constant.

Density = 2 / 8 = 0.25 g/ml

The density would be larger than the correct value.

Answer: The density would be larger than the correct value.

First off, the relationship between density and volume is given by:

Density = Mass / Volume

From this equation, Density is inversely proportional to volume. This means as the volume increases, the density decreases and as the volume decreases the density increases.

Assuming all thing's being normal;

Mass = 2g

Volume = 10ml

Density = [tex]\frac{2}{10}=0.2[/tex]   g/ml

Second case scenario;

'your pipet was incorrectly calibrated so that it transferred less than 10.00 mL"

Lets have a value of 8ml for our volume. Mass remains constant.

Density = [tex]\frac{2}{8}= 0.25[/tex]  g/ml

The density would be larger than the correct value.

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Answer:

[tex]H_{comb}=-4406kJ/mol[/tex]

Explanation:

Hello,

In this case, the enthalpy of combustion is understood as the energy released when one mole of fuel, in this case octene, is burned in the presence of oxygen and is computed with the enthalpies of formation of the fuel, carbon dioxide and water as shown below (oxygen is circumvented as it is a pure element):

[tex]H_{comb}=8*\Delta _fH_{CO_2}+5\Delta _fH_{H_2O}-\Delta _fH_{C_8H_{10}}[/tex]

Thus, since we already know the enthalpy of combustion of the fuel, for carbon and water we have -393.5 and -241.8 kJ/mol respectively, thereby, the enthalpy of combustion turns out:

[tex]H_{comb}=8*(-393.5kJ/mol)+5(-241.8kJ/mol)-49.0kJ/mol\\\\H_{comb}=-4406kJ/mol[/tex]

Best regards.

Answer:

The molarity of the sulfuric acid in the solution is 1.77 M.

Explanation:

The molarity of the sulfuric acid in the solution can be found using the following equation:

[tex] C_{i}V_{i} = C_{f}V_{f} \rightarrow C_{f} = \frac{C_{i}V_{i}}{V_{f}} [/tex]    

Where:

[tex]C_{i}[/tex]: is the initial concentration of the acid

[tex]V_{i}[/tex]: is the initial volume of the solution = 22.6 cm³

[tex]V_{f}[/tex]: is the final volume of the solution = 88.5 cm³

The initial concentration of the H₂SO₄ is:

[tex] C_{i} = \frac{n}{V} = \frac{m}{M*V} = \frac{d*\% ^{m}_{m}}{M} [/tex]

Where:

n: is the number of moles

m: is the mass

M: is the molar mass = 98.079 g/mol

d: is the density of the acid = 1.39 g/cm³

%: is the percent by mass = 49.0 %

[tex] C_{i} = \frac{1.39 \frac{g}{cm^{3}}*\frac{1000 cm^{3}}{1 L}*\frac{49 g}{100 g}}{98.079 \frac{g}{mol}} = 6.94 M [/tex]          

Finally, the final concentration of H₂SO₄ after the dilution is:

[tex] C_{f} = \frac{6.94 M*22.6 cm^{3}}{88.5 cm^{3}} = 1.77 M [/tex]

Therefore, the molarity of the sulfuric acid in the solution is 1.77 M.

I hope it helps you!

Answer:

4.76

Explanation:

In this case, we have to start with the buffer system:

[tex]CH_3COOH~->~CH_3COO^-~+~H^+[/tex]

We have an acid ([tex]CH_3COOH[/tex]) and a base ([tex]CH_3COO^-[/tex]). Therefore we can write the henderson-hasselbach reaction:

[tex]pH~=~pKa+Log\frac{[CH_3COO^-]}{[CH_3COOH]}[/tex]

If we want to calculate the pH, we have to calculate the pKa:

[tex]pH=-Log~Ka=4.76[/tex]

According to the problem, we have the same concentration for the acid and the base 0.1M. Therefore:

[tex][CH_3COO^-]=[CH_3COOH][/tex]

If we divide:

[tex]\frac{[CH_3COO^-]}{[CH_3COOH]}~=~1[/tex]

If we do the Log of 1:

[tex]Log~1=~zero[/tex]

So:

[tex]pH~=~pKa[/tex]

With this in mind, the pH is 4.76.

I hope it helps!

Answer:

As K >>> 1, the reaction will shift to the products

Explanation:

To know the direction of any reaction you must calculate the equilibrium constant, K. If K is < 1, the reaction will shift to the reactants and if k > 1 the reaction will shift to the products.

With the reactions:

HSO₄⁻ ⇄ SO₄²⁻ + H⁺ Ka = 1.26x10⁻²

And:

NH₄⁺ ⇄ NH₃⁺ + H⁺ Ka = 5.6x10⁻¹⁰

The inverse reaction:

NH₃⁺ + H⁺ ⇄ NH₄⁺ 1/Ka = 1.8x10⁹

The sum of the reactions:

HSO₄⁻ + NH₃⁺ + H⁺ ⇄ NH₄⁺ + SO₄²⁻ + H⁺ K = 1.26x10⁻² ₓ 1.8x10⁹ = 2.3x10⁷

Answer:

I think it would be the Dead Sea

Explanation:

Because the dead sea is already usually in the warmer temperatures, the boiling point of the water would be lower than the rest.

Answer:

pressure is = 54.9802atm

Explanation:

using ideal gas equation

PV=nRT

Answer: 1) Endothermic

2) Yes, absorbed.

3) 166.86 kJ will be absorbed.

Explanation:

1) To determine if a reaction is endothermic (heat is absorbed by the system) or exothermic (heat is released by the system), first calculate its change in Enthalpy, which is given by:

ΔH = [tex]H_{products} - H_{reagents}[/tex]

For the reaction 2Fe₂O₃(s) ⇒ 4FeO(s) + O₂(g):

Enthalpy of Reagent (Fe₂O₃(s))

Enthalpy of formation for Fe₂O₃(s) is - 822.2 kJ/mol

The reaction needs 2 mols of the molecule, so:

H = 2(-822.2)

H = - 1644.4

Enthalpy of Products (4FeO(s) + O₂(g))

Enthalpy of formation of O₂ is 0, because it is in its standard state.

Enthalpy of formation of FeO is - 272.04 kJ/mol

The reaction produces 4 mols of iron oxide, so:

H = 4(-272.04)

H = -1088.16

Change in Enthalpy:

ΔH = [tex]H_{products} - H_{reagents}[/tex]

ΔH = - 1088.16 - (-1644.4)

ΔH = + 556.2 kJ/mol

The change in enthalpy is positive, which means that the reaction is absorving heat. Then, the chemical reaction is Endothermic.

2) When Fe₂O₃(s) reacts, heat is absorbed because it is an endothermic reaction.

3) Calculate how many mols there is in 94.2 g of Fe₂O₃(s):

n = [tex]\frac{mass}{molar mass}[/tex]

n = [tex]\frac{94.2}{160}[/tex]

n = 0.6 mols

In the reaction, for 2 mols of Fe₂O₃(s), 556.2 kJ are absorbed. Then:

2 mols --------------- 556.2 kJ

0.6 mols ------------- x

x = [tex]\frac{0.6*556.2}{2}[/tex]

x = 167 kJ

It will be absorbed 167 kJ of energy, when 94.2 g of Fe₂O₃(s) reacts.

Answer:

4 and 6 would work for this

Answer:

the nucelolo is not part of the atom

Explanation:

The nucleolus is NOT a part of the atom. The nucleolus has its definition in biology as an element of the cell.

Let's remember that the atom is made up of three fundamental elements that are: electrons, protons and neutrons. Protons and neutrons are found in the nucleus while electrons are orbiting around the nucleus.

Answer:

core

Explanation:

there is no such thing as a core in an atom.

The middle is known as the nucleus

Answer:

D2 has 1 neutron

and 1 proton

Explanation:

Before proceeding, let's understand what thermosphere and mesosphere are;

The mesosphere is the third layer of the atmosphere, directly above the stratosphere and directly below the thermosphere.

The thermosphere is the layer in the Earth's atmosphere directly above the mesosphere and below the exosphere.

The meteors make it through the exosphere and thermosphere without much trouble because those layers don't have much air. But when they hit the mesosphere, there are enough gases to cause friction and create heat.

Temperatures in the mesosphere decrease with altitude. Because there are few gas molecules in the mesosphere to absorb the Sun's radiation, the

heat source is the stratosphere below. The mesosphere is extremely cold.

Satellites and the International Space Station orbit the Earth within the thermosphere.

Mesosphere;

- is directly above the stratosphere

- acts as a barrier to meteoroids

- is very cold

Thermosphere;

- reaches out into space

- is where satellites travel

Answer:mesosphere: directly above stratosphere, is very cold, acts as a barrier to meteoroids

Explanation:

Answer:

5727 years or 5730 (rounded to match 3 sig figs) whichever one your teacher prefers

Explanation:

First Order decay has a half life formula of Half Life = Ln (2) / k = 0.693/K

Half-life = 0.693/k = 0.693/1.21 x10-4 =  5727 years or 5730 (rounded to match 3 sig figs)

This should be correct because if you google the half-life of 14 C it is ~ 5700 years

Answer:

53.9g of KC₆H₅CO₂ is the mass the student should dissolve.

Explanation:

Full question is:

A chemistry graduate student is given 125.mL of a 1.00M benzoic acid HC6H5CO2 solution. Benzoic acid is a weak acid with =Ka×6.3x10−5. What mass of KC6H5CO2 should the student dissolve in the HC6H5CO2 solution to turn it into a buffer with pH =4.63?

It is possible to find pH of a buffer by using H-H equation:

pH = pka + log₁₀ [A⁻] / [HA]

Where pKa is -log Ka (4.2), [A⁻] is concentration of conjugate base, KC₆H₅CO₂ and [HA] concentration of the weak acid, HC₆H₅CO₂

-This concentration could be replaced per moles of each compound-

Moles of 125mL = 0.125L of a 1.00M of benzoic acid are:

0.125L × (1.00mol / L) = 0.125 moles benzoic acid.

Replacing in H-H equation, for a pH of 4.63:

4.63 = 4.2 + log₁₀ [A⁻] / [0.125mol]

0.43 = log₁₀ [A⁻] / [0.125mol]

2.6915 = [A⁻] / [0.125mol]

0.336 moles = [A⁻]

That means the student needs 0.336 moles of KC₆H₅CO₂ to obtain the buffer with pH = 4.63.

As molar mass of KC₆H₅CO₂ is 160.21g/mol, mass of 0.336 moles are:

0.336 moles × (160.21g / mol) =

The given question is incomplete as it lacks the appropriate options, however, the options related to this question is as follows:

a)Test a composite material in the lab to see how it can be improved.

b)Look into trends of various diseases and their effects on society.

c)Research available drugs for treating illnesses such as cancer and heart disease, as well as the side effects of these drugs.

d)Make a plan for how to conquer a disease with drugs.

Answer:

The correct answer is - b)Look into trends of various diseases and their effects on society.

Explanation:

The young scientist who wants to focus on creating drugs to treat disease must think and look for the diseases that are needed to be treated at the moment or affecting the population most and how it affects the society and then plan how to do it.

So, the scientist needs to check the trends of various disease and their effect on society for developing understanding which disease needs to be cure first and how it affects.

Thus, the most appropriate answer is - option B.

Answer:

b)Look into trends of various diseases and their effects on society.

Explanation:

ap3x

Answer:

[tex]1) NH_{4}IO_{3}\\2) Pb(IO_{3})_{4} \\3) NH_{4}(C_{2}H_{3}O_{2})\\4) Pb(C_{2}H_{3}O_{2})_{4}[/tex]

Explanation:

[tex]1) NH_{4}^{+}IO_{3}^{-} ---> NH_{4}IO_{3}\\2) Pb^{4+}(IO_{3}^{-})_{4} --->Pb(IO_{3})_{4} \\3) NH_{4}^{+}(C_{2}H_{3}O_{2})^{-} ---> NH_{4}(C_{2}H_{3}O_{2})\\4) Pb^{4+}(C_{2}H_{3}O_{2})^{-} _{4} --->Pb(C_{2}H_{3}O_{2})_{4}[/tex]

Answer:is increase and more collisions :)

Explanation:

Answer:

[tex]\mathbf{H_2PO^{-}_4}[/tex]  and [tex]\mathbf{HPO^{2-}_4}[/tex] are the amphoteric species.

Explanation:

Phosphorus is a non-metallic element belonging to Group 5 of the Periodic table. As a triprotic acid (i.e an acid that has three dissociable protons that undergo stepwise ionization) . Phosphorus reacts vigorously with hot water to form tetraoxosulphate(V) acid.

i.e

[tex]P_4O_{10(s)} + 6 H_2O_{l} \to4H_3PO_{4(aq)}[/tex]

SO;

As a triprotic acid;

The ionization equations are as follows:

NOTE: Instead of the amphoteric species  to be circle as instructed , due to the fact that we are using the LaTex Editor ; we will just need to let them be in bold format which will differentiate them from the rest.

[tex]H_3PO_{4} \leftrightarrow H^+ + H_2PO^{-}_4[/tex]

[tex]\mathbf{H_2PO^{-}_4} \leftrightarrow H^+ + HPO^{2-}_4[/tex]

[tex]\mathbf{HPO^{2-}_4} \leftrightarrow H^+ + PO^{3-}_4[/tex]

Thus; [tex]\mathbf{H_2PO^{-}_4}[/tex]  and [tex]\mathbf{HPO^{2-}_4}[/tex] are the amphoteric species.

Answer:

A

Explanation:

The products are the stuff on the right side of the arrow (yield sign). In this case that would be C₆H₁₂O₂ + 6O₂.

Answer:

dimethoxy(phenyl)methanol

Explanation:

For this question, we have to remember the mechanism of the Grignard reaction. In this case, phenylmagnesium bromide is our nucleophile, a carbo-anion is produced (step 1). Then this carbo-anion can attack the carbonyl group in the dimethyl carbonate, the double bond is delocalized into the oxygen producing a negative charge (step 2). Finally, with the addition of the hydronium ion ([tex]H_3O^+[/tex]), the anion can be protonated to produce the alcohol (dimethoxy(phenyl)methanol) (step 3).

See figure 1

I hope it helps!

Answer:

[tex]0.48~\frac{J}{g~^{\circ}C}[/tex]

Explanation:

In this question, we have to remember the relationship between Q (heat) and the specific heat (Cp) the change in temperature (ΔT), and the mass (m).

[tex]Q=m*Cp*ΔT[/tex]

The next step is to identify what values we have:

[tex]Q~=~1870~J[/tex]

[tex]m~=~85.01~g[/tex]

[tex]ΔT~=~45.2~^{\circ}C[/tex]

[tex]Cp~=~X[/tex]

Now, we can plug the values and solve for "Cp":

[tex]1870~J=~85.01~g~*Cp*45.2~^{\circ}C[/tex]

[tex]Cp=\frac{1870~J}{85.01~g~*45.2~^{\circ}C}[/tex]

[tex]Cp=0.48~\frac{J}{g~^{\circ}C}[/tex]

The unknow metal it has a specific value of [tex]0.48~\frac{J}{g~^{\circ}C}[/tex]

I hope it helps!

Answer:

[tex]\Delta S=6045.8\frac{J}{K}[/tex]

Explanation:

Hello,

In this case, we can compute the change in the entropy for vaporization processes in term of the enthalpy of vaporization as shown below:

[tex]\Delta S=\frac{m*\Delta H}{T}[/tex]

Whereas the temperature is in Kelvins. In such a way, the entropy results:

[tex]\Delta S=\frac{1.00kg*2256x10^3\frac{J}{kg} }{(100+273.15)K}\\\\\Delta S=6045.8\frac{J}{K}[/tex]

Best regards.