A 23.2 g sample of an organic compound containing carbon, hydrogen and oxygen was burned in excess oxygen and yielded 52.8 g of carbon dioxide and 21.6 g of water. Determine the empirical formula of the compound.

Answer:

The answer is 18.12KJ is required to vaporise 48.7 g of dichloromethane at its boiling point

Explanation:

To solve the above question we have the given variable as follows

ΔHvap = heat of vaporisation of dichloromethane per mole = 31.6KJ/mole

However since the heat of vaporisation is the heat to vaporise one mole of dichloromethane, then, for 48.7 grams of dichloromethane, we have.

The number of moles of dichloromethane present = 48.7/84.93 = 0.573 moles

Therefore, the amount of heat required to vaporise 48.7 grams of dichloromethane at its boiling point is 31.6KJ/mole×0.573moles =18.12KJ

Answer: A. Perform a gravity filtration with a stemless funnel.

Explanation: Gravity filtration with a stemless funnel prevent blocking up the funnel. Undissolved sample may come out in the stem when the solution cools down thus will be blocking the funnel. Using stemless funnel prevent this problem.

Answer:

The right answer to this question is no, all of the liquid will not evaporate, there will be 8.61 ×10⁻⁴ moles or 6.22×10⁻² grams left in the container

At equilibrium the pressure in the container will be the vapor pressure of liquid pentane which is = 100. mm Hg

Explanation:

To solve this we list the known values as follows

vapor pressure of liquid pentane = 100 mmHg = 13.33 KPa

Temperature T = 260 K

Volume of container = 350 mL = 0.00035 m³

The number of moles of liquid pentane = n

The universal gas constant = R = 8.314 J/(mol·K)

Thus From the ideal gas equation PV = nRT →

Thus plugging in the values in the above equateion we have

n = [tex]\frac{PV}{RT} = \frac{(13330)(0.00035)}{(8.314)(260)}[/tex] = 2.16×10⁻³ moles

Hence the number of moles in 0.218 g sample of liquid   pentane C₅H₁₂ with molar mass = 72.15 g/mol = 0.218/72.15 = 3.02×10⁻³ moles

Hence the number of moles present in the sample placed in the closed evacuated container = 3.02×10⁻³ moles

However number of moles to completely evaporate at 100 mmHg and 260 K is 2.16×10⁻³ moles hence, 3.02×10⁻³ moles - 2.16×10⁻³ moles,  or 8.61 ×10⁻⁴ moles will be left in the container

converting the value in moles to mass we have number of moles, n = mass/(molar mass)

Therefore the mass = number of moles × molar mass = 8.61 ×10⁻⁴ × 72.15 = 6.22 × 10⁻² grams left in the container

The pressure in the container at equilibrium will be vapor pressure of liquid pentane C₅H₁₂, or 100. mm Hg

In the described closed system, all of the liquid pentane will evaporate due to the vapor pressure. The final pressure when equilibrium is reached will be the vapor pressure, which is 100.0 mm Hg.

The subject of the question is the vapor pressure of pentane, C5H12, which is a concept from Chemistry.

The given vapor pressure is 100.0 mm Hg at 260 K. In this closed system the liquid and vapor will come to equilibrium at the given temperature, at which point the vapor pressure will be equal to the given 100.0 mm Hg.

The sample mass of 0.218 g doesn't exceed the amount needed for evaporation at the equilibrium pressure. Therefore, all of the liquid pentane will evaporate.

The final pressure in the container when equilibrium is reached will be the vapor pressure, which is 100.0 mm Hg.

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Answer: The average potential energy of the PIB is 0 irrespective of the wave function.

Explanation:

⟨H⟩=⟨KE⟩+⟨V⟩

the nn quantum number

⟨KE⟩=(π^2 ℏ^2)/(2mL^2 )

the average kinetic energy of the wavefunction is dependent on

⟨V⟩=∫sin(kx)0sin(kx)dx=0

The average potential energy of the PIB is 0 irrespective of the wave function.

⟨H⟩=⟨KE⟩=(π^2 ℏ^2)/(2mL^2 )

Answer:

                      ΔT  =  20.06 °C

Explanation:

The equation used for this problem is as follow,

                                    Q  =  m Cp ΔT   ----- (1)

Where;

           Q  =  Heat  =  1.17 kJ = 1170 J

           m  =  mass  =  24.1 g

           Cp  =  Specific Heat Capacity  =  2.42 J.g⁻¹.°C⁻¹

           ΔT  =  Change in Temperature  =  ??

Solving eq. 1 for ΔT,

                                ΔT  =  Q / m Cp

Putting values,

                                ΔT  =  1170 J / 24.1 g × 2.42 J.g⁻¹.°C⁻¹

                                ΔT  =  20.06 °C

Answer:

The answer to this question is 45.33 moles of aluminum sulfate is produced when 125 moles of aluminum hydroxide and 136 moles of sulfuric acid react

Explanation:

To solve this, we write out the chemical equation of he reaction thus

Al(OH)3(s) + 3 H2SO4(aq) -----> Al2 (SO4)3(aq) + 6 H2O(l)

here it is seen that one moles of aluminum hydroxide reacts with three moles of sulfuric to produce one mole of aluminum sulfate and six moles of water

hence

136 moles of sulfuric acid reacts with 136/3 or 45.33 moles of aluminum hydroxide to produce 136/3 or 45.33 moles of aluminum sulfate and 2× 136 moles of water

Hence the amount in moles of aluminum sulfate produced is 45.33 moles

Final answer:

Using the balanced chemical equation 3 Al(OH)3 + 3 H2SO4 → Al2(SO4)3 + 6 H2O, and knowing that aluminum hydroxide is the limiting reactant, 125 moles of aluminum hydroxide will produce 41.67 moles of aluminum sulfate.

Explanation:

The question asks how many moles of aluminum sulfate will be produced when reacting 125 moles of aluminum hydroxide with 136 moles of sulfuric acid. To answer this, we need the balanced chemical equation:

3 Al(OH)3 + 3 H2SO4 → Al2(SO4)3 + 6 H2O

The stoichiometry of the reaction shows that 3 moles of aluminum hydroxide react with 3 moles of sulfuric acid to produce 1 mole of aluminum sulfate. Since there are more moles of sulfuric acid present, aluminum hydroxide is the limiting reactant. Therefore, we can calculate the moles of aluminum sulfate produced by dividing the moles of aluminum hydroxide by 3, which gives us:
125 moles Al(OH)3 ÷ 3 = 41.67 moles Al2(SO4)3 (rounded to two decimal places as per the significant figures in the provided moles of reactants).

Answer:

65.4 is the mass for 1.9×10²⁴ atomsof Pb

Explanation:

1mol of atoms of Pb has → NA (6.02×10²³ atoms) and weighs → 207.2 g

Therefore 1.9×10²⁴ atomsof Pb may weigh (1.9×10²⁴ . 207.2) / NA = 65.4 g

Answer:

the correct answer is d

Explanation:

The HA bar on the right must be converted completely to H3O+ and A-. Strong acids completely dissociate in solution. Complete dissociation would mean that there is no HA bar left on the right of the arrow.

To represent a strong acid in the graph, the 'HA' bar has to be almost non-existent while the H3O+ and A- bars significantly increase reflecting the fact that strong acids completely disassociate in water. The correct answer is 'The HA bar on the right must be converted completely to H3O+ and A-'.

The correct answer to the given question is option D).

In the context of this question, the graph represents a moderately weak acid and we're asked to determine what changes would occur in the graph for a relatively strong acid.

Here, the 'HA' would represent the weak acid that partially disassociates into H3O+ (hydronium ions) and A- (the conjugate base). One characteristic of a strong acid is that it completely disassociates in water.

Therefore, to represent a strong acid, the 'HA' bar on the right would need to be much lower or even non-existent (to indicate complete disassociation). In turn, the H3O+ and A- bars on the right would need to increase significantly/acquire all the 'HA' bar's original height to represent the products of the strong acid's complete disassociation.

So, the correct answer would be (D) 'The HA bar on the right must be converted completely to H3O+ and A-'.

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The net ionic equation for the reaction of strontium chloride and potassium sulfate, forming strontium sulfate solid, is Sr2+(aq) + SO42-(aq) → SrSO4(s)

When an aqueous solution of strontium chloride is mixed with an aqueous solution of potassium sulfate, strontium sulfate precipitates out and potassium and chloride ions remain in the solution. The balanced net ionic equation for this reaction is as follows:

Sr2+(aq) + SO42-(aq) → SrSO4(s)

This equation represents the change where strontium ions from strontium chloride and sulfate ions from potassium sulfate are combined to form strontium sulfate solid. The charges are balanced with the positive 2 charge of the strontium ion balancing the negative 2 charge of the sulfate ion.

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Answer:

The answer to this is the concentration of glucose (C6H12O6) in mg/dL = 95.48 mg/dL

Explanation:

To solve this we list out the known variables thus

Measured concentration of  glucose  (C6H12O6) = 5.3mM per liter

The molar mass of glucose  = 180.156 g/mol

From the above, it is seen that one mole of glucose contains 180.156 grams of  C6H12O6 therefore 5.3 mM which is 5.3 × 10⁻³ moles contains

5.3 × 10⁻³ moles × 180.156 g/mol = 0.9548 grams of glucose

Also 1 d L = 0.1 L  or 1 L = 10 dL and 1 mg = 1000 g, hence

thus 0.9548 grams per liter is equivalent to 1000/10 × 0.9548 milligrams per dL or 95.48 mg/dL

Answer:

option d

Explanation:

Molecular sizes of gaseous molecules are very less. Volume occupied by the all the molecules of the gases are very less or negligible as compared to the container in which it is kept. Therefore, most of the volume occupied by gaseous molecules are negligible.

Volume occupied by the gaseous molecules are actually the volume of the container and its does not depend upon the amount, molecular mass or dipole moment of the gaseous molecules.

Therefore, the correct option is d ‘Because most of the volume occupied by the substance is empty space.’

Answer:

weaker and longer

Explanation:

Since there are 3 bonds in ethyne in comparision with the 2 bonds of ethyne between carbon atoms, they are attracted more to each other → the bond gets shorter . And since there are one more bond that supports the union → the bond gets stronger

thus the carbon-carbon double bond in ethene is weaker and longer than the carbon-carbon triple bond in ethyne

Phosphoric acid is a triprotic acid that can donate three protons in solution, forming three anions. The pH of a 0.100 M solution is approximately 1.0, assuming it only ionizes once. The concentrations of the formed species are estimated to be highest for H2PO4⁻ and much lower for HPO4²⁻ and PO4³⁻.

Phosphoric acid, H3PO4 is a triprotic acid, meaning it can donate three hydrogen ions in a solution. This results in the formation of three different species:  H2PO4⁻, HPO4²⁻, and PO4³⁻.

The estimated pH of a 0.100 M phosphoric acid solution will depend on the degree of dissociation, but for the first ionization, we can approximate it using the expression pH=-log[H⁺], where [H⁺] is the hydronium ion concentration. Given that a 0.100 M phosphoric acid solution ionizes mostly once, we have [H⁺]≈0.100 M, leading to an estimated pH around 1.0.

Next, the concentrations of the species in equilibrium can be calculated without exact Kb or Ka values as long as we make the approximation that dissociation after the first hydrogen ion is minimal in a dilute solution like 0.100 M. In this case, we will assume [H2PO4⁻]≈0.100 M and [HPO4²⁻] and [PO4³⁻] will be much less than [H2PO4⁻].

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Answer:

Explanation:

If our entire Solar System were the size of a quarter, the planets and the sun is now a tiny speck of dust. The flat disc of the coin can represent the orbits of the planets.

Using this scale, the diameter of our Milky Way galaxy will be about the size of North America.

The nearest star, other than our own Sun, is about four light years away. That means it takes four years for its light to reach us. Since light travels at a speed of 3.0 x 10^8 meters per second, each light year is such a great distance. Proxima Centauri Milky way would be another quarter, two soccer fields away.

A much further star, Deneb is actually 1,800 light years away, the nearest star would be about 24,000 miles away.

Answer:

[tex]\beta=63.85^{\circ}[/tex] from the vertical OR [tex]16.15^{\circ}[/tex] from horizontal.

Explanation:

Given:

Thrust of launching the rocket, [tex]F=5\times 10^6\ N[/tex]

angle of launch from the horizontal, [tex]\theta=37^{\circ}[/tex]

mass of the rocket, [tex]m=200000\ kg[/tex]

Now the direction of acceleration due to the thrust force is in the direction of force:

[tex]a=\frac{F}{m}[/tex]

[tex]a=\frac{5000000}{200000}[/tex]

[tex]a=25\ m.s^{-2}[/tex]

And the acceleration due to gravity is always directed towards the center of the earth i.e. vertically downwards.

Since acceleration is a vector quantity we, approach accordingly:

[tex]\tan\beta=\frac{a\cos\theta}{g}[/tex]

[tex]\tan\beta=\frac{25\cos37^{\circ}}{9.8}[/tex]

[tex]\beta=63.85^{\circ}[/tex] from the vertical OR [tex]16.15^{\circ}[/tex] from horizontal.

Answer: The correct statement is, It behaves as the electron donor.

Explanation:

According to the Lewis concept:

A Lewis-acid is defined as a substance that accepts electron pairs.

A Lewis-base is defined as a substance which donates electron pairs.

For example : Acid + Base ⇄ Acid-base adduct

[tex]H^++NH_3\rightleftharpoons NH_4^+[/tex]

As per question, the characteristic of a Lewis base is that it behaves as the electron donor.

Hence, the correct statement is, It behaves as the electron donor.

Answer:

A

Explanation:

It behaves as the electron donor

Answer:

The rate at which [tex]P_4[/tex] is being produced is 0.0228 M/s.

The rate at which [tex]PH_3[/tex] is being consumed is 0.0912 M/s.

Explanation:

[tex]4PH_3\rightarrow P_4(g)+6H_2(g)[/tex]

Rate of the reaction : R

[tex]R=\frac{-1}{4}\frac{d[PH_3]}{dt}=\frac{1}{6}\frac{d[H_2]}{dt}=\frac{1}{1}\frac{d[P_4]}{dt}[/tex]

The rate at which hydrogen is being formed = [tex]\frac{d[H_2]}{dt}=0.137 M/s[/tex]

[tex]R=\frac{1}{6}\frac{d[H_2]}{dt}[/tex]

[tex]R=\frac{1}{6}\times 0.137 M/s=0.0228 M/s[/tex]

The rate at which [tex]P_4[/tex] is being produced:

[tex]R=\frac{1}{1}\frac{d[P_4]}{dt}[/tex]

[tex]0.0228 M/s=\frac{1}{1}\frac{d[P_4]}{dt}[/tex]

The rate at which [tex]PH_3[/tex] is being consumed :

[tex]R=\frac{-1}{4}\frac{d[PH_3]}{dt}[/tex]

[tex]0.0228 M/s\times 4=\frac{-1}{1}\frac{d[PH_3]}{dt}[/tex]

[tex]\frac{-1}{1}\frac{d[PH_3]}{dt}=0.912 M/s[/tex]

The rate at which P4 is being produced is 0.034 M/s and the rate at which PH3 is being consumed is 0.2055 M/s.

The given reaction is 4PH3(g) → P4(g) + 6H2(g). We are given the rate at which molecular hydrogen is being formed, which is 0.137 M/s. To find the rate at which P4 is being produced, we need to consider the stoichiometry of the reaction. From the balanced equation, we can see that for every 4 moles of PH3 consumed, 1 mole of P4 is produced. Therefore, the rate at which P4 is being produced is 0.137/4 or 0.034 M/s.

Similarly, to find the rate at which PH3 is being consumed, we can use the stoichiometry of the reaction. From the balanced equation, we can see that for every 4 moles of PH3 consumed, 6 moles of H2 is produced. Therefore, the rate at which PH3 is being consumed is (6/4) * 0.137 or 0.2055 M/s.

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Answer:

Percent composition of the solution is 26 % of sucrose and 74 % of water

Explanation:

Percent composition is the mass of solute, either of solvent in 100 g of solution.

Mass of solution = Mass of solvent + Mass of solute

Mass of solute = 35 g

Mass of solvent = 100 g

As we know, water density = 1g/mL

So 1g/mL . 100 mL = 100 g

35 g + 100 g = 135 g → Mass of solution

(Mass of solute / Mass of solution) . 100 =

(35 g / 135 g) . 100 = 26 %

(Mass of solvent / Mass of solution) . 100 =

(100 g / 135 g) . 100 = 74 %

To calculate the percent composition of sucrose in the solution, divide the mass of sucrose (35 grams) by the total mass of the solution (sucrose plus water, which is 135 grams) and multiply by 100%. The solution has a percent composition of approximately 25.93% sucrose.

The question involves calculating the percent composition of a solution by mass. If a solution contains 35 grams of sucrose (C12H22O11) in 100 mL of water (noting that the density of water is roughly 1 g/mL, so we have 100 grams of water), the total mass of the solution is the sum of the mass of the solute (sucrose) and the solvent (water), which is 35 g + 100 g = 135 g. To find the percent by mass of sucrose in the solution, we use the formula:

Percent by mass of sucrose = (Mass of sucrose / Total mass of solution) × 100%

Inserting the values we have:

Percent by mass of sucrose = (35 g / 135 g) × 100% ≈ 25.93%

Therefore, the percent composition of sucrose in the solution is approximately 25.93%.

Answer:

we only see parts of the lit side as the moon goes around the earth

Explanation:

Unlike the sun, the moon orbits the Earth. This is the reason why we see the different phases of the moon. The reflection of the moon is being illuminated back to us with the help of the sun. So, as the moon circles the Earth, we only see parts of the lit side. Such changes helps us see the moon in different phases such as the Third Quarter, Crescent, New Moon, Full Moon, etc.

For example, during "Full Moon," the moon's entire face is lit up by the sun. Thus, we see the entire moon's lit portion.

Thus, this explains the answer.

Answer:

The complete question is:

Question: What disorder is indicated by these findings? A client comes to the emergency department with status asthmaticus. His respiratory rate is 48 breaths/minute, and he is wheezing. An arterial blood gas analysis reveals a pH of 7.52, a partial pressure of arterial carbon dioxide (PaCO2) of 30 mm Hg, PaO2 of 70 mm Hg, and bicarbonate (HCO3−) of 26 mEq/L.

A. Metabolic acidosis

B. Respiratory acidosis

C. Metabolic alkalosis

D. Respiratory alkalosis

Answer: The correct answer is:

D. Respiratory alkalosis

Explanation:

In Respiratory alkalosis the Partial Pressure of Arterial Carbondioxide (PaCO2) become decreased (i.e. less than 35 mm Hg) and the pH of blood become increased (i.e. more than 7.45). Alveolar hyperventilation causes respiratory alkalosis.

Alveolar hyperventilation occurs when alveolar ventilation is increased than the arterial carbondioxide tension and carbondioxide production.

Alveolar ventilation is the gaseous exchange between alveoli and the external environment.

Whereas, in metabolic acidosis, bicarbonate (HCO3) become decreased (i.e. less than 22 mEq/l and the pH of blood become decreased (i.e. less than 7.35); in respiratory acidosis,  the pH of blood also become decreased (i.e. less than 7.35) and the PaCO2 become increased (i.e. more than 45 mm Hg); and in metabolic alkalosis,  the bicarbonate (HCO3) become increased (i.e. more than 26 mEq/l and the pH become increased (i.e. more than 7.45).

Answer:

D (or E If properly listed to include the active site option)

Explanation:

A. Is true

Enzymes are organically biochemical catalyst and thus they can speed up the rate of chemical reaction in the body

B is true

They are catalysts as said earlier

C is true

They have active sites. An enzyme does not act on all substrates. They have particular group on which they can act. For example, we have carbohydrates enzymes that act on carbohydrates substrate only. This enzymes have no business acting on a protein substrate.

D. Enzymes are proteins

One of the important characteristics of enzymes is that they are protenious in nature

E. This is wrong. Enzymes like any over catalyst are not consumed in the course of the biochemical reaction

Enzymes are biological catalysts that speed up chemical reactions. These are proteins and have an 'active site' where reactions occur. However, they are not consumed during the reaction.

In context to your question about enzymes, they serve specific roles as biological accelerators or catalysts, significantly boosting the rate of a chemical reaction. This property is showcased under option A. They are indeed classified as proteins (option C), and they do have parts called active sites, where the substrate (the molecule upon which the enzyme acts) binds (option B).

However, option D posits that enzymes are consumed during the reaction, which is incorrect. Unlike many catalysts in non-biological reactions that get consumed during the reaction, enzymes remain unaffected by the reaction. They don't exhaust or alter in a reaction and are available to facilitate other reactions once the process is finished.

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Answer:

True

Explanation:

This are acts or actions that concur with situational expectations.

Answer:

The amount of sucrose that must be added is 1.66 moles

Explanation:

Colligative property of lowering vapor pressure has this formula:

Vapor pressure of pure solvent (P°) - Vapor pressure of solution = P° . Xm

We have both vapor pressure (pure solvent and solution9, so let's determine the ΔP

ΔP = 92.6 Torr - 72 Torr = 20.6 Torr

Let's add the data in the formula

20.6 Torr = 92.6 Torr . Xm

Xm = Mole fraction of solute → (mol of solute/ mol of solute + mol of solvent)

Mol of solvent = 5.83 mol (data from the problem)

Therefore Xm = 20.6 Torr / 92.6 Torr → 0.222

Let's find out the moles of solute (our unknown value)

0.22 = moles of solute / moles of solute + 5.83 moles of solvent

0.222 (moles of solute + 5.83 moles of solvent) = moles of solute

0.222 moles of solute + 1.29 moles of solvent = moles of solute

1.29 moles of solvent = moles of solute - 0.222 moles of solute

1.29 moles = 0.778 moles of solute

1.29 / 0.778 = moles of solute → 1.66 moles

Answer : The pressure of the helium gas inside the container would be, 32.1 atm

Explanation :

To calculate the pressure of the gas we are using ideal gas equation as:

[tex]PV=nRT[/tex]

where,

P = Pressure of [tex]He[/tex] gas = ?

V = Volume of [tex]He[/tex] gas = 7.5 L

n = number of moles [tex]He[/tex] = 10 mole

R = Gas constant = [tex]0.0821L.atm/mol.K[/tex]

T = Temperature of [tex]He[/tex] gas = [tex]20^oC=273+20=293K[/tex]

Putting values in above equation, we get:

[tex]P\times 7.5L=10mole\times (0.0821L.atm/mol.K)\times 293K[/tex]

[tex]P=32.1atm[/tex]

Thus, the pressure of the helium gas inside the container would be, 32.1 atm

Answer:

[tex]-5.226\times 10^{-7} C[/tex] is the charge on each sphere.

Explanation:

Coulomb's law is given as ;

[tex]F=K\times \frac{q_1\times q_2}{r^2}[/tex]

[tex]q_1,q_2[/tex] = Charges on both charges

r = distance between the charges

K = Coulomb constant =[tex]9\times 10^{9} N m^2/C^2[/tex]

We have ;

Charge of ion =[tex]q_1=-q[/tex]

Charge of electron =[tex]q_2=-q[/tex]

[tex]r=26.55 cm =0.2655 m[/tex]

Force between the charges  at r distance will be :  F

F = 0.03500 N

[tex]0.03500 N=9\times 10^{9} N m^2/C^2\times \frac{(-q)\times (-q)}{(0.2655 m)^2}[/tex]

[tex]q=5.226\times 10^{-7} C[/tex]

[tex]-5.226\times 10^{-7} C[/tex] is the charge on each sphere.

Answer:

1. 8.7moles of H2

2. 2.25moles of O2

Explanation:

1. 2NH3 —> N2 + 3H2

From the equation,

2moles of NH3 produce 3 moles of H2.

Therefore, 5.8moles of NH3 will produce Xmol of H2 i.e

Xmol of H2 = (5.8x3)/2 = 8.7moles

2. C3H8 + 5O2 —> 3CO2 + 4H2O

From the equation,

5moles of O2 produced 4moles of H2O.

Therefore, Xmol of O2 will produce 1.8mol of H2O i.e

Xmol of O2 = (5x1.8)/4 = 2.25moles

Answer: carbonyl group C=O

Explanation:

Formaldehyde is an organic compound, it is the simplest form of Aldehydes. It formula is CH2O and has a carbonyl functional group, C=O. The general formula for adehydes is R-COH. The carbon atom is bonded to oxygen with a double bond and one of the two remaining bonds is occupied by hydrogen, and the other by an alkyl group.

One of the properties of adehydes is their solubility in water. The lower members (up to 4 carbons) of aldehydes are soluble in water due to H-bonding. Ofcourse the the higher members are not soluble in water because their hydrophobic long chains.

Aldehydes contain carbonyl group, therefore they undergo reactions like nucleophilic addition reactions, oxidation, reduction, halogenation.

Answer:

molecules

Explanation:

Answer:

Synergism

Explanation:

This is an example of Synergism. Synergism is nothing but working out of two medicines together.

Examples of medical synergies are when doctors treat microbial heart infections with ampicillin and Gentamicin and when people with cancer undergo radiation and chemotherapy or more than one chemotherapy drug at a time.

To calculate the vapor pressure of the solution, we need to consider Raoult's law. According to Raoult's law, the vapor pressure of a solution is equal to the mole fraction of the solvent multiplied by the vapor pressure of the pure solvent. In this case, the solvent is water and the solute is K3PO4. To find the mole fraction of water, we divide the moles of water by the total moles of solute and solvent. Using the given values, the vapor pressure of the solution at 55 °C is 101.0 torr.

To calculate the vapor pressure of the solution, we need to consider Raoult's law. According to Raoult's law, the vapor pressure of a solution is equal to the mole fraction of the solvent multiplied by the vapor pressure of the pure solvent. In this case, the solvent is water and the solute is K3PO4. To find the mole fraction of water, we divide the moles of water by the total moles of solute and solvent. Using the given values, we have:

Moles of water = 0.941 mol

Total moles of solute and solvent = 0.159 mol K3PO4 + 0.941 mol H2O = 1.1 mol

Mole fraction of water = (0.941 mol) / (1.1 mol) = 0.855

Now, we can use the mole fraction of water and the vapor pressure of pure water to calculate the vapor pressure of the solution:

Vapor pressure of the solution = mole fraction of water * vapor pressure of pure water

= 0.855 * 118.1 torr

= 101.0 torr

Therefore, the vapor pressure of the solution at 55 °C is 101.0 torr.

Using Raoult's Law, the vapor pressure of the solution at 55°C is approximately 100.93 torr.

To calculate the vapor pressure of the solution, we will use Raoult's Law, which states:

[tex]\[ P_{\text{solution}} = X_{\text{solvent}} \cdot P^{\star}_{\text{solvent}} \][/tex]

where:

- [tex]\( P_{\text{solution}} \)[/tex] is the vapor pressure of the solution,

- [tex]\( X_{\text{solvent}} \)[/tex] is the mole fraction of the solvent,

- [tex]\( P^{\star}_{\text{solvent}} \)[/tex] is the vapor pressure of the pure solvent.

Given data:

- Mole fraction of water [tex](\( X_{\text{H2O}} \))[/tex] in the solution:

 [tex]\[ X_{\text{H2O}} = \frac{n_{\text{H2O}}}{n_{\text{H2O}} + n_{\text{K3PO4}}}[/tex]

[tex]= \frac{0.941}{0.159 + 0.941} \\\\= \frac{0.941}{1.1} \\ \\ \approx 0.855[/tex]

- Vapor pressure of pure water at 55 °C:

[tex]\[ P^{\star}_{\text{H2O}} = 118.1 \text{ torr} \][/tex]

Now, calculate the vapor pressure of the solution:

[tex]\[ P_{\text{solution}} = X_{\text{H2O}} \cdot P^{\star}_{\text{H2O}} \][/tex]

[tex]\[ P_{\text{solution}} = 0.855 \times 118.1 \text{ torr} \][/tex]

[tex]\[ P_{\text{solution}} \approx 100.93 \text{ torr} \][/tex]

Therefore, the vapor pressure of the solution at 55 °C is approximately 100.93 torr.