What would be the volume of a balloon containing 64g of oxygen gas at STP (standard temperature and pressure)? You should be able to obtain this answer by calculation or by using logical reasoning.

Answer:

C = 4.60x10⁻⁵ M

Explanation:

The concentration of the compound can be calculated using Beer-Lambert Law:

[tex]A = \epsilon*C*l[/tex]

Where:

A: is the absorbance of the ethanol = 0.58  

ε: is the molar absorptivity of the ethanol = 12600 M⁻¹cm⁻¹

C: is the concentration of the compound =?

l: is the optical path length = 1 cm

Hence, the concentration of the compound is:

[tex]C = \frac{A}{\epsilon*l} = \frac{0.58}{12600 M^{-1}cm^{-1}*1 cm} = 4.60 \cdot 10^{-5} M[/tex]

Therefore, the concentration of the compound is 4.60x10⁻⁵ M.

I hope it helps you!

Answer:

60.288 grams of O₂ gas are contained in 890.0 mL at 21.0 degrees Celsius and 750.0 psi.

Explanation:

Ideal gases are a simplification of real gases that is done to study them more easily. It is considered to be formed by point particles, do not interact with each other and move randomly. It is also considered that the molecules of an ideal gas, in themselves, do not occupy any volume.

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:  

P*V = n*R*T

where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas.

In this case:

Replacing:

51.0345 atm*0.890 L= n*0.082 [tex]\frac{atm*L}{mol*K}[/tex] *294 °K

Solving:

[tex]n=\frac{51.0345 atm*0.890 L}{0.082\frac{atm*L}{mol*K} *294 K}[/tex]

n=1.884 moles

Being 32 g / mole the molar mass of oxygen O₂, the following rule of three applies: if 1 mole contains 32 grams, 1,884 moles, how much mass does it have?

[tex]mass=\frac{1.884 moles*32 grams}{1mole}[/tex]

mass= 60.288 grams

60.288 grams of O₂ gas are contained in 890.0 mL at 21.0 degrees Celsius and 750.0 psi.

Answer:

The mass of the O2 gas is 60.16 grams

Explanation:

Step 1: Data given

Volume of O2 gas = 890.0 mL = 0.890 L

Temperature of the O2 gas = 21.0 °C = 294 K

Pressure = 750.0 psi = 51.03 atm

Step 2: Calculate moles of O2 gas

p*V = n*R*T

⇒with p = the pressure of the O2 gas = 51.03 atm

⇒with V = the volume = 0.890 L

⇒with R = the gas constant == 0.08206 L*atm/mol*K

⇒with T = the temperature = 294 K

⇒with n = the number of moles of O2 gas

n = (p*V) / (R*T)

n = (51.03 atm * 0.890 L) / (0.08206L*atm/mol*K * 294K)

n = 1.88 moles of O2

Step 3: Calculate the mass of O2 gas

Mass = moles * molar mass O2

Mass O2 gas = 1.88 moles * 32.0 g/mol

Mass O2 gas = 60.16 grams

The mass of the O2 gas is 60.16 grams

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:

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".

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:

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.

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:

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:

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.

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 right conjugate of [tex]OH^-[/tex] is [tex]H_2O[/tex]

Explanation:

According to the Bronsted-Lowry conjugate acid-base theory, an acid is defined as a substance which looses donates protons and thus forming conjugate base and a base is defined as a substance which accepts protons and thus forming conjugate acid.

For the given chemical equation:

[tex]OH^-(aq)+HCO_3^-(aq)\rightleftharpoons H_2O(l)+CO_3^{2-}(aq)[/tex]

Here, [tex]OH^-[/tex] is gaining a proton, thus it is considered as a brønsted-lowry base and after gaining a proton, it forms [tex]H_2O[/tex] which is a conjugate acid.

Thus the right conjugate of [tex]OH^-[/tex] is [tex]H_2O[/tex]

The question is incomplete; the complete question is;

Which of the following statements is/are true for a 0.10 M solution of a weak acid HA?

a. [ H+] >> [ A−]

b. [ H+] = [ A−]

c. The pH is 1.00.

d. The pH is less than 1.00.

Answer:

b. [ H+] = [ A−]

Explanation:

Given the acid as HA, we know that being a weak acid, its dissociation in water can never be 100%. If it were a strong acid, then it could have undergone a 100% dissociation in solution. The conjugate base of a weak acid is a always a weak base hence A^- is expected to act as a weak base. At the same concentration, weak acids have a higher pH value than strong acids. Hence if the pH of a strong acid HA is 1, then the pH of a weak acid HA must be greater than 1.

But, we look at the equation for the dissociation of the weak acid HA

HA(aq)⇄H^-(aq) + A^-(aq). This implies that the HA dissociates in a 1:1 ratio therefore; [H+] = [ A−], hence the answer given above.

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:

Scientific notation is a system in which quantities are too big or too tiny to compose in decimal form.

Key words:

1: Scientific

2: Quantities

3: Decimal

Please mark brainliest

Hope this helps.

Answer:

Plum pudding model

Explanation:

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

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:

vacuum

Explanation:

vacuum is the form by which matter cant exist

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: 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:

ΔH°c = -2219.9 kJ

Explanation:

Let's consider the combustion of propane.

C₃H₈(g) + 5 O₂(g) ⟶ 3 CO₂(g) + 4 H₂O(l)

We can find the standard enthalpy of the combustion (ΔH°c) using the following expression.

ΔH°c = [3 mol × ΔH°f(CO₂(g)) + 4 mol × ΔH°f(H₂O(l))] - [1 mol × ΔH°f(C₃H₈(g)) + 5 mol × ΔH°f(O₂(g))]

ΔH°c = [3 mol × (-393.5 kJ/mol) + 4 mol × (-285.8 kJ/mol)] - [1 mol × (-103.8 kJ/mol) + 5 mol × (0 kJ/mol)]

ΔH°c = -2219.9 kJ

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:

.76

Explanation:

Answer:

there are 5 ways this could happen

Autosomal dominant inheritance:  a child recieves a normal gene from one parent and a defective gene from the other parent.

can occur on any of the 22 non-sex chromosomes and have a 50% inheritence rate, gender is not a factor, and disorder differs with inheritance.

examples:  Huntington's disease, neurofibromatosis, achondroplasia, familial hypercholesterolemia

Autosomal recessive inheritance:  both parents carry the defective gene but they are not affected by the disorder.

there is a 25% chance of defective gene from both parents, a 50% chance of inheriting one gene to become a carrier, gender is not a factor in the pattern of the defective gene.

examples:  Tay-Sachs disease, sickle cell anemia, cystic fibrosis, phenylketonuria (PKU)

X-linked (sex-linked) recessive inheritance:  mother carries the affective gene on one of the two X chromosomes.

males inherite X chromosomes from their mothers and Y from their father; which gives the son a 50% chance of inheriting the disorder.

daughters have a 50% chance, but they are not affected by the disorder.

examples:  Hemophilia A, Duchenne muscular dystrophy

X-linked Dominant:  females are affected more so than males; more common for males if they are in the same generation if the mom is affected (because females have two X-chromosomes)

example:  Hypophatemic rickets (Vitiamin Dresistant rickets, ornithine transcarbamylase deficiency.

Mitochondrial:  can affect both males and femlaes, can only be passed by females due to all mitochondria of all children is from the mother, and can appear in every generation.

examples:  Lebrer's hereditary optic neuropathy and Kearns-Sayre syndrome

Explanation:

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 formula is as follows

2Cu + O2 ---> 2CuO

When two copper atoms react with a diatonic oxygen, they form copper oxide (rust)

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.

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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