Calculate the free energy change for the reaction:
2 NO(g) + O2(g) → 2 NO2(g)
(A) - 9.3 kcal
(B) + 24.9 kcal
(C) + 9.3 kcal
(D) - 16.6 kcal
(E) + 16.6 kcal​

Explanation:

methyl orange turns pink in an acidic solution

yellow in basic and orange in neutral

Answer: The substance would be a strong acid.

Explanation:

Methyl orange turns orange when it is added to a medium acid and turns red when it is added to a strong acid. The pH of the acid would be above 4 because it is a strong acid. Methyl orange turns yellow when it is added to a medium base.

Answer:

The equilibrium temperature of water is 25.6 °C

Explanation:

Step 1: Data given

Mass of the sample of brass = 25.0 grams

The specific heat capacity = 0.375 J/g°C

Mass of water = 250.0 grams

Temperature of water = 25.0 °C

The initial temperature of the brass is 96.7°C

Step 2: Calculate the equilibrium temperature

Heat lost = heat gained

Q(sample) = -Q(water)

Q = m*C* ΔT

m(sample)*c(sample)*ΔT(sample) = - m(water)*c(water)*ΔT(water)

⇒m(sample) = the mass of the sample of brass = 25.0 grams

⇒with c(sample) =The specific heat capacity = 0.375 J/g°C  

⇒with ΔT = the change of temperature = T2 - T1 =T2 - 96.7 °C

⇒with m(water) = the mass of the water = 250.0 grams

⇒with c(water) = the specific heat capacity = 4.184 J/g°C

⇒with ΔT(water) = the change of temperature of water = T2 - T1 = T2 - 25.0°C

25.0 * 0.375 * (T2 - 96.7) = - 250.0 * 4.184 J/g°C * (T2 - 25.0°C)

9.375T2 - 906.56 = -1046T2 + 26150

9.375T2 + 1046T2 = 26150 + 906.56

1055.375T2 = 27056.26

T2 = 25.6 °C

The equilibrium temperature of water is 25.6 °C

Answer:

d

Explanation:

it's called the hydrocarbon

Answer:

A hydrocarbon

Explanation:

Answer: EXTINCTION

Explanation:

Answer:

extinction

Explanation:

Answer:

ai) Rate law,  [tex]Rate = k [CH_3 Cl] [Cl_2]^{0.5}[/tex]

aii) Rate constant, k = 1.25

b) Overall order of reaction = 1.5

Explanation:

Equation of Reaction:

[tex]CH_{3} Cl (g) + 3 Cl_2 (g) \rightarrow CCl_4 (g) + 3 HCl (g)[/tex]

If [tex]A + B \rightarrow C + D[/tex], the rate of backward reaction is given by:  

[tex]Rate = k [A]^{a} [B]^{b}\\k = \frac{Rate}{ [A]^{a} [B]^{b}}\\k = \frac{Rate}{ [CH_3 Cl]^{a} [Cl_2]^{b}}[/tex]

k is constant for all the stages

Using the information provided in lines 1 and 2 of the table:

[tex]0.014 / [0.05]^a [0.05]^b = 00.029/ [0.100]^a [0.05]^b\\0.014 / [0.05]^a [0.05]^b = 00.029/ [2*0.05]^a [0.05]^b\\0.014 / = 0.029/ 2^a\\2^a = 2.07\\a = 1[/tex]

Using the information provided in lines 3 and 4 of the table and insering the value of a:

[tex]0.041 / [0.100]^a [0.100]^b = 0.115 / [0.200]^a [0.200]^b\\0.041 / [0.100]^a [0.100]^b = 0.115 / [2 * 0.100]^a [2 * 0.100]^b\\[/tex]

[tex]0.041 = 0.115 / [2 ]^a [2]^b\\ \[[2 ]^a [2]^b = 0.115/0.041\\ \[[2 ]^a [2]^b = 2.80\\\[[2 ]^1 [2]^b = 2.80\\\[[2]^b = 1.40\\b = \frac{ln 1.4}{ln 2} \\b = 0.5[/tex]

The rate law is: [tex]Rate = k [CH_3 Cl] [Cl_2]^{0.5}[/tex]

The rate constant [tex]k = \frac{Rate}{ [CH_3 Cl]^{a} [Cl_2]^{b}}[/tex] then becomes:

[tex]k = 0.014 / ( [0.050] [0.050]^(0.5) )\\k = 1.25[/tex]

b) Overall order of reaction =  a + b

Overall order of reaction = 1 + 0.5

Overall order of reaction = 1.5

The rate law is for this reaction is    [tex]r = 1.25 M^{-0.5} s^{-1} [CH_3Cl] [Cl_2]^{0.5}[/tex]

where the rate constant is k = [tex]1.25 M^{-0.5} s^{-1}[/tex] and the overall order of the reaction is 1.5.

Let's consider the following reaction.

CH₃Cl(g) + 3 Cl₂(g) → CCl₄(g) + 3 HCl(g)

The rate law for a chemical reaction is an equation that relates the reaction rate with the concentrations of the reactants.

The  rate law for this reaction is:

r = k [CH₃Cl]ᵃ [Cl₂]ᵇ

where,

If we write the ratio r₂/r₁, we get:

r₂/r₁ = k [CH₃Cl]₂ᵃ [Cl₂]₂ᵇ / k [CH₃Cl]₁ᵃ [Cl₂]₁ᵇ

r₂/r₁ = {[CH₃Cl]₂/ [CH₃Cl]₁}ᵃ

(0.029 M/s)/(0.014 M/s) = {0.100M/0.050 M}ᵃ

a ≈ 1

If we write the ratio r₃/r₂, we get:

r₃/r₂ = k [CH₃Cl]₃ᵃ [Cl₂]₃ᵇ / k [CH₃Cl]₂ᵃ [Cl₂]₂ᵇ

r₃/r₂ = {[Cl₂]₃/[Cl₂]₂}ᵇ

(0.041 M/s)/(0.029 M/s) = {0.100 M/0.050 M}ᵇ

b ≈ 0.5

So far, the rate law is:

[tex]r = k [CH_3Cl] [Cl_2]^{0.5}[/tex]

Let's use the values of the first experiment to find the value of the rate constant.

[tex]k = \frac{r}{[CH_3Cl][Cl_2]^{0.5} } = \frac{0.014 M/s}{(0.050 M)(0.050 M)^{0.5} } = 1.25 M^{-0.5} s^{-1}[/tex]

The final rate law is:

[tex]r = 1.25 M^{-0.5} s^{-1} [CH_3Cl] [Cl_2]^{0.5}[/tex]

It is the sum of the individual orders of reaction.

a + b = 1 + 0.5 = 1.5

The rate law is for this reaction is    [tex]r = 1.25 M^{-0.5} s^{-1} [CH_3Cl] [Cl_2]^{0.5}[/tex]

where the rate constant is k = [tex]1.25 M^{-0.5} s^{-1}[/tex] and the overall order of the reaction is 1.5.

Learn more about the rate law here: https://brainly.com/question/14945022

Answer:

1. Increase in the temperature of the water

2. Increasing the surface area of the lithium

Explanation:

1. Increase in the temperature of the water

The activation energy for the lithium water reaction is +161 kJ/mol while the activation energy for the sodium is +109 kJ/mol, hence for increased reaction rate, the water temperature will be raised to enable more lithium atoms enter into reaction with the water molecules as their energy is increased lowering the activation energy required for the reaction.

2. Increasing the surface area of the lithium

As the lithium floats on the water, due to its low temperature and the heat evolved from the reaction of lithium with the cold water is below the melting point of lithium, the reaction rate can be increased by increasing the surface area of lithium sample by grinding so as to increase the number of lithium water reaction sites.

Answer:

The correct answer is 596.5 kJ.

Explanation:

The mass of ethanol or C2H5OH mentioned in the question is 20 gm.  

The molar mass of ethanol is 46 g/mol.  

The moles of the compound can be determined by using the formula,  

n = weight of the compound/molar mass

= 20/46 = 0.435 moles

It is mentioned in the question that standard heat of combustion of ethanol is 1372 kJ/mole, that is, one mole of ethanol is producing 1372 kilojoules of energy at the time of combustion.  

Therefore, the energy liberated by completely burning the 20 grams of ethanol is 0.435*1372 = 596.5 kJ.  

Since the standard heat of combustion of ethanol is -1372 kJ/mol, the heat released by the combustion of a 20.0 g sample is -593 kJ.

Let's consider the thermochemical equation for the combustion of ethanol.

C₂H₅OH(l) + 3 O₂(g) → 2 CO₂(g) + 3 H₂O(l)   ΔH°c = -1372 kJ/mol

We can calculate the heat released by burning 20.0 g of ethanol considering the following relationships.

[tex]20.0 g EtOH \times \frac{1molEtOH}{46.07gEtOH} \times \frac{(-1367kJ)}{1molEtOH} = -593 kJ[/tex]

Since the standard heat of combustion of ethanol is -1372 kJ/mol, the heat released by the combustion of a 20.0 g sample is -593 kJ.

Learn more: https://brainly.com/question/2874342

The standard heat of combustion of ethanol, C₂H₅OH, is -1372 kJ/mol ethanol. How much heat (in kJ) would be liberated by completely burning a 20.0 g sample.

Answer: The molecular formula for the given organic compound X is [tex]C_6H_{8}O_7[/tex]

Explanation:

We are given:

Mass of [tex]CO_2=4.13g[/tex]

Mass of [tex]H_2O=1.13g[/tex]

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 4.13 g of carbon dioxide, =[tex]\frac{12}{44}\times 4.13=1.13g[/tex] of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 1.13 g of water, [tex]\frac{2}{18}\times 1.13=0.125g[/tex] of hydrogen will be contained.

Mass of oxygen in the compound = (3.00) - (1.13+ 0.125) = 1.75 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon =[tex]\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{1.13g}{12g/mole}=0.094moles[/tex]

Moles of Hydrogen =[tex]\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.125g}{1g/mole}=0.125moles[/tex]

Moles of Oxygen =[tex]\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.75g}{16g/mole}=0.109moles[/tex]

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles

For Carbon = [tex]\frac{0.094}{0.094}=1[/tex]

For Hydrogen = [tex]\frac{0.125}{0.094}=1.33[/tex]

For Oxygen = [tex]\frac{0.109}{0.094}=1.16[/tex]

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1: 1.33: 1.16

Converting them into whole number ratios by multiplying by 6:

The ratio of C : H : O = 6: 8: 7

Hence, the empirical formula for the given compound is [tex]C_6H_8O_7[/tex]

Empirical mass = [tex]6\times 12+8\times 1+7\times 16=192g[/tex]

The equation used to calculate the valency is :

[tex]n=\frac{\text{molecular mass}}{\text{empirical mass}}[/tex]

Putting values in above equation, we get:

[tex]n=\frac{192g/mol}{192g/mol}=1[/tex]

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

[tex]C_6H_8O_7\times 1=C_6H_{8}O_7[/tex]

Thus molecular formula for the given organic compound X is [tex]C_6H_{8}O_7[/tex]

Answer:

The interaction between sodium ion and the partial negative charge on the oxygen is stronger.

Explanation:

The predominant interaction that exists between sodium ion and the partial negative charge on the oxygen is ion-dipole interaction.

The predominant interaction that exists between two ethanol molecules is hydrogen bonding interaction.

The order of strength of the intermolecular interactions in decreasing order:

Ionic bond > ion-dipole interaction > hydrogen bonding > dipole-dipole interaction > ion-induced dipole interaction > induced dipole- dipole interaction > london force

So, ion-dipole interaction is stronger than hydrogen bonding.

Hence, the interaction between sodium ion and the partial negative charge on the oxygen is stronger.

Explanation:

Ag+(aq) + 2NH3 (aq) ⇄ Ag(NH3)2 (aq)

When it comes to question of this sort, the LeChatelier principle should come to mind. The LeChatelier principle states that changes in the temperature, pressure, volume, or concentration of a system will result in predictable and opposing changes in the system in order to achieve a new equilibrium state.

increasing the concentration of Ag+

This would lead to an increase in concentration of reactants, the system would annul or oppose this change by moving towrds the right, favouring the forward reaction.

decreasing the concentration of NH3

This would lead to a decrease in concentration of reactants, the system would annul this change by moving to the left, favourin the backward reaction.

increasing the concentration of Ag(NH3)2

This would lead to an increase in concnetration of products, the system would annul or oppose this change by moving towards the left, favouring the backward reaction.

Explanation:

Moles=mass/molar mass

moles × molar mass = mass

0.206 x 119= mass

Mass= 24. 51grams

Answer:

24.51 g to the nearest hundredth

Explanation:

1 mole of SOCl2 =  32 + 16 + 2 *35.5 = 119 g

0.206 mol = 0.206 * 119

=  24.51 g.

Answer:2.2059

Explanation:find their total ram is(14+(1×3))

Find the portion occupied by

H2 which is three

Divide h2 by total ram then multiply by mass

Answer:

Plum pudding model

Explanation:

Thomson’s model of the atom called: PLUM PUDDING MODEL

Answer:

The volume of base required is = 25.66 mL

Explanation:

Volume of titrant required is given by:

Final titre value - Initial titre value

=26.78 - 1.12

= 25.66 mL

Note: repeat and average volumes for accuracy of result.

Answer:

Explanation:

2 HgCl₂ + C₂O₄²⁻   =  2 Cl⁻ + Hg₂Cl₂ + 2CO₂

1 )

Rate of reaction [tex]= k[HgCl_2]^m[C_2O_4^{-2}]^n[/tex]

             [HgCl₂]        [C₂O₄²⁻ ]           Rate  

1 .              .124             .115               1.61 x 10⁻⁵

2 .             .248             .115             3.23 x 10⁻⁵

3 .              .124             .229              6.4 x 10⁻⁵

4 .              .248             .229            1.28 x 10⁻⁴

comparing 1 and 3 , when concentration of HgCl₂ remains constant and concentration of C₂O₄²⁻  becomes twice , rate becomes 4 times so rate is proportional to square of concentration of C₂O₄²⁻  .

Hence n = 2

comparing 1 and 2 , when concentration of HgCl₂ becomes twice  and concentration of C₂O₄²⁻  remains constant  , rate becomes 2 times so rate is proportional to simply  concentration of C₂O₄²⁻  .

Hence m = 1

Putting the data of  1 in the rate equation found

 1.61 x 10⁻⁵ = k x .124 x  .115²

k = 11.3 x 10⁻⁴ M⁻² s⁻¹

Answer:

I believe the answer would be: D

an independent variable that is affected by the other

experimental variable

Correct me if im wrong

Explanation:

Answer:

The  pKa  of  Glu is   [tex]pK_a = 4.25[/tex]

Explanation:

From the question we are told that

  The [tex]K_a[/tex] value of  Glu is [tex]K_a = 5.62*10^{-5}[/tex]

The [tex]pKa[/tex] is mathematically represented as

         [tex]pK_a = -log (K_a )[/tex]

         substituting values

          [tex]pK_a = -log (5.62 *10^{-5} )[/tex]

          [tex]pK_a = 4.25[/tex]

The question is incomplete, the complete question is;

Gaseous methane (CH4) reacts with gaseous oxygen gas (O2) to produce gaseous carbon dioxide (CO2) and gaseous water (H2O). If 28.2 g of carbon dioxide is produced from the reaction of 15.1 g of methane and 81.2 g of oxygen gas, Calculate the percent yield of carbon dioxide. Be sure your answer has the correct number of significant digits in it.

Answer:

71.1%

Explanation:

The balanced reaction equation must first be written;

CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (g)

Let us obtain the number of moles of carbon dioxide corresponding to 28.2 g

Number of moles = mass/ molar mass= 28.2/44.01 g/mol = 0.64 moles of CO2

Next, we obtain the limiting reactant. This is the reactant that yields the least amount of product.

For methane;

Number of moles in 15.1 g= mass/molar mass= 15.1/16gmol-1 = 0.9 moles

From the reaction equation;

1 mole of methane yields 1 mole of carbon dioxide

Hence 0.9 moles of methane yields 0.9 moles of carbon dioxide.

For oxygen

Number of moles of oxygen corresponding to 81.2 g of oxygen= mass/ molar mass= 81.2g/32gmol-1 = 2.5 moles of oxygen

From the reaction equation;

2 moles of oxygen gas yields 1 mole of carbon dioxide

2.5 moles of oxygen gas yields 2.5 × 1 /2 = 1.25 moles of carbon dioxide.

Methane is the limiting reactant.

Theoretical yield of carbon dioxide= 0.9 moles

Actual yield of carbon dioxide= 0.64 moles

% yield = actual yield/ theoretical yield × 100

% yield= 0.64/0.9 ×100

% yield = 71.1%

Answer:

Explanation:

Precision generally refers to the repeatability of measurements. That is, the closeness of repeated meaurements in values. Accuracy, on the other hand, refers to the closeness of a measured value to the true value.

For apprentice X, the three measurements taken (26.5, 26.6, 26.4) were close in values with an average of 26.5 and a mean deviation of 0.07. Hence, the apprentice is said to be precise in measurement. However, the average of the measurement is quite far from the true length of meaurement. Hence, the apprentice is inaccurate.

Apprentice Y had a mean measurement value of 27.6 and a mean deviation of 0.2. The measurements are far from each other and the apprentice is therefore said to be far from being precise. The mean measurement value is also far from true value. Hence the apprentice is also inaccurate in his/her measurement.

Apprentice Z has a mean measurement of 27.07 and a mean deviation of 0.11.  The repeated measurements were close enough to be precise while the average is also close to the true length, and hence, the measurement is said to be accurate.

Answer:

vacuum

Explanation:

vacuum is the form by which matter cant exist

Answer:

.76

Explanation:

Answer:

B. The principal quantum number (n) is 6, the angular momentum quantum number (ell) is 2, and there are 5 electrons in the subshell.

Explanation:

Given;

electronic configuration of 6d⁵

where;

6 in the configuration shows that, the principal quantum number (n) is 6

d is the subshell

d orbital has angular momentum quantum number (l) of 2

            s orbital, l = 0

            p orbital, l = 1

            d orbital, l = 2

            f orbital,  l = 3

5 in the configuration means that there are 5 electrons in the subshell

Therefore, 6d⁵ symbol means that "The principal quantum number (n) is 6, the angular momentum quantum number (ell) is 2, and there are 5 electrons in the subshell".

Answer:

Yes it is true

Explanation:

This is because the Clausius theorem states that The cyclic integral always has two defined results. The results include it being less than or equal to zero under certain conditions.

When the system consists of only reversible processes, the cyclic integral is equal to zero. If it consists of and irreversible processes, the integral is usually less than zero.

Answer:

Explanation:

Oxidation In chemistry is the gain in oxygen, loss of hydrogen and loss of electrons. Oxidation occur when an element with gain oxygen, loss hydrogen or loss electrons, which means the element is oxidized. Compared to reduction which is loss ofoxygen, gain of hydrogen and gain of electrons.

From the the question , Ag( silver) is undergoing oxidation.

Answer:

Explanation:

H3PO4(aq) + 3NaOH(aq) → Na3PO4(aq) + 3H2O(l)

mole of NaOH = 23.6 * 10 ⁻³L * 0.2M

= 0.00472mole

let x be the no of mole of H3PO4 required of  0.00472mole of NaOH

3 mole of NaOH required ------- 1 mole of H3PO4

0.00472mole of NaOH ----------x

cross multiply

3x = 0.0472

x = 0.00157mole

[H3PO4] = mole of H3PO4 / Vol. of H3PO4

= 0.00157mole / (10*10⁻³l)

= 0.157M

Answer:

[tex]M_{H_3PO_4}=0.157M[/tex]

Explanation:

Hello,

In this case, the balanced chemical reaction is:

[tex]H_3PO_4(aq) + 3NaOH(aq) \rightarrow Na_3PO_4(aq) + 3H_2O(l)[/tex]

Therefore, we compute the moles of used NaOH by the 0.2000-M solution:

[tex]n_{NaOH}=0.200\frac{mol}{L}*0.0236L=4.72x10^{-3}molNaOH[/tex]

Then, we compute the moles of neutralized phosphoric acid by their 1:3 molar ratio:

[tex]n_{H_3PO_4}=4.72x10^{-3}molNaOH*\frac{1molH_3PO_4}{3molNaOH}=1.57x10^{-3}molH_3PO_4[/tex]

Finally, the concentration:

[tex]M_{H_3PO_4}=\frac{1.57x10^{-3}molH_3PO_4}{0.010L}\\ \\M_{H_3PO_4}=0.157M[/tex]

Best regards.

Answer:

This is true because oxygen belongs to the non-metal part of the periodic table.

Answer:hello

Oxygen is a non-metal it's exactly true.

Explanation:

There's a good rule that if the number of electrons in the last layer is 1,2,3,and sometimes 4 it's a metal but if this number increase it's a non-metal like Oxygen or Nitrogen.

Good luck.

Answer:

  2.3

Explanation:

Your calculator can evaluate this expression. It tells you the pH of lemon juice is 2.3.

Answer:

a) Based on the given information, the volume of the solvent given is 999 ml and the density of water given is 0.9982 gram per ml.  

The mass of solvent can be calculated by the formula mass = density * volume

mass = 0.9982 * 999 = 0.997 Kg

The molarity of the solution given is 2.300 * 10^-2 M

Molality of the glycerol solution can be calculated by using the formula,  

Molality = molarity/solvent (kg) = 2.300 * 10^-2 / 0.997 = 0.023 m

b) Molarity or the moles of the solute given is 2.300 * 10^-2 moles

The moles of solvent can be determined by using the formula, n = mass of solvent/mol.wt = 997/18 (mol.wt of solvent is 18 g/mol), now putting the values we get,  

n = 997/18 = 55.4

The mole fraction of the glycerol will be = 2.300 * 10^-2 M/(2.300 * 10^-2)+55.40

= 4.15 * 10^-4

c) The mass percent of glycerol can be determined by using the formula,  

mass of solute/mass of solution * 100% ----- (i)

The mass of solute can be determined by using the formula,

n = mass of solute/mol.wt of solute

The n or the no. of moles is 2.300 * 10^-2 and the molecular weight of glycerol is 92.09 g/mol. Now putting the values we get,  

mass = 2.300 * 10^-2 * 92.09 = 2.12 grams

Now putting the values in equation (i) we get,  

mass percent = 2.12 / 997 * 100% = 0.21%

d) Based on the above calculation, the mass of solute (glycerol) is 2.12 g or 2.12 * 1000 mg

The volume of water is 999 ml or 999 * 10^-3 L

The concentration of the glycerol solution will be,  

Concentration = 2.12 * 10^3 mg/999 * 10^-3 L  

= 2.12/999 * 10^6 mg/L

= 2122.1 ppm

Considering the solution of molality, mole fraction, mass percentage and ppm, you obtain that:

a) The molality of the glycerol solution is 0.02306 molal.

b) The mole fraction of glycerol in the solution is 4.15×10⁻⁴.

c) The percent by mass is 0.212%.

d) The concentration of the glycerol solution is 2118.07 ppm.

Molality is the ratio of the number of moles of any dissolved solute to kilograms of solvent.

The Molality of a solution is determined by the expression:

[tex]molality=\frac{number of moles of solute}{kilogramsof solvent}[/tex]

In this case, you have a 2.300×10⁻² M solution of glycerol (C₃H₈O₃) in water. The sample was created by dissolving a sample of C₃H₈O₃ in water and then bringing the volume up to 1.000 L.

So, being the molarity the number of moles of solute that are dissolved in a certain volume, the number of moles of glycerol can be calculated as:

number of moles of glycerol= 2.300×10⁻² M× 1 L

number of moles of glycerol= 2.300×10⁻² moles

On the other side, the volume of water needed was 999 mL and the density of water at 20.0∘C is 0.9982 [tex]\frac{g}{mL}[/tex]. So, the mass of water needed can be calculated as:

999 mL×0.9982[tex]\frac{g}{mole}[/tex] = 997.2 grams of water= 0.9972 kg of water

Then, the molality of the solution is:

[tex]molality=\frac{2.300x10^{-2} moles}{0.9972 kg}[/tex]

molality= 0.02306 molal

Finally, the molality of the glycerol solution is 0.02306 molal.

The molar fraction is a way of measuring the concentration that expresses the proportion in which a sustance is found with respect to the total moles of the solution.

In this case, the number of moles of glycerol= 2.300×10⁻² moles

Having 997.2 grams of water, the moles of solvent can be determined knowing that the molar mass of water is 18[tex]\frac{g}{mole}[/tex]:

[tex]number of moles of water=997.2 gramsx\frac{1 mole}{18 grams}[/tex]

number of moles of water= 55.4 moles of water

Being the number of total moles the sum of the moles of grycerol and the numbers of moles of water, the mole fraction can be calculated as:

[tex]mole fraction=\frac{2.300x10^{-2} moles}{2.300x10^{-2}moles+55.4 moles}[/tex]

mole fraction= 4.15×10⁻⁴

In summary, the mole fraction of glycerol in the solution is 4.15×10⁻⁴.

The Percentage Composition is a measure of the amount of mass that an element occupies in a compound and indicates the percentage by mass of each element that is part of a compound.

The mass percentage of a component of the solution is defined as the ratio of the mass of the solute to the mass of the solution, expressed as a percentage.

The mass percentage is calculated as the mass of the solute divided by the mass of the solution, the result of which is multiplied by 100 to give a percentage. This is:

[tex]percent by mass=\frac{mass of solute}{mass of solution}x100[/tex]

In this case, remmeber that you have a number of moles of glycerol of 2.300×10⁻² moles .

Being the molar mass glycerol 92.09 [tex]\frac{g}{mole}[/tex], the mass of glycerol can be calculated as:

2.300×10⁻² moles×92.09 [tex]\frac{g}{mole}[/tex]= 2.11807 grams of glycerol

Remembering that you have 997.2 grams of water, the percent by mass is calculated as:

[tex]percent by mass=\frac{mass of glycerol}{mass of glycerol + mass of water}x100[/tex]

Solving:

[tex]percent by mass=\frac{2.11807 grams}{2.11807 grams + 997.2 grams}x100[/tex]

[tex]percent by mass=\frac{2.11807 grams}{999.31807 grams}x100[/tex]

percent by mass= 0.212%

Finally, the percent by mass is 0.212%.

Parts per million (ppm) is a unit of measurement of concentration that measures the number of units of substance in each million units of the whole. In this case, the concentration measurement refers to mg of glycerol per L of solution.

Being the mass of glycerol 2.11807 grams equal to 2118.07 mg (1 g=1000mg), the concentration is:

[tex]concentration=\frac{2118.07 mg}{1 L solution}[/tex]

Solving:

concentration= 2118.07 ppm

In summary, the concentration of the glycerol solution is 2118.07 ppm.

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