What bonds to cytosine (C) in DNA.

Answer is: The heat is unusable and lost to the surroundings.

Gasoline is a mixture  of many different hydrocarbons: alkanes (paraffins), cycloalkanes  and alkenes (olefins).

Balanced chemical reaction of gasoline combustion:

C₈H₁₈ + 25/2O₂ → 8CO₂ + 9H₂O + energy.

One part of energy produced in combustion of gasoline is used for car ungine (kinetic energy) and other part of that energy (in form of heat) is lost to the surronding and engine does not use it.

The statements that describe phase changes are particles in a liquid need to move more slowly in order to freeze and attractive forces between the particles in a liquid are broken when a liquid boils.

For better understanding let's explain what it means

From the above we can therefore say that the answer The statements that describe phase changes are particles in a liquid need to move more slowly in order to freeze and attractive forces between the particles in a liquid are broken when a liquid boils, is correct

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The correct statements about phase changes are Option 1 & 3 : particles in a liquid need to move more slowly to freeze, and attractive forces between particles in a liquid are broken when the liquid boils.

Phase changes refer to the transformation of a substance from one state of matter to another: solid, liquid, or gas. Let's examine the given statements:

Based on the explanation, the correct statements are 1 and 3.

Answer:c mobile home

Explanation:

A mobile home is preconstructed structure, that can be run and transported to other location according to the choice of living place. This can be prepared from wood or other kind of cheap material which can keep the weight of mobile home low.

Thus mobile home is the lowest maintainace cost home for a single family.

The balanced chemical equation for the reaction between phosphoric acid and cesium hydroxide to produce cesium dihydrogen phosphate and water is 3 CsOH(aq) + H3PO4(aq) → Cs3PO4(aq) + 3 H2O(l).

The balanced chemical equation for the reaction between aqueous solutions of phosphoric acid (H3PO4) and cesium hydroxide (CsOH) to produce aqueous cesium dihydrogen phosphate (CsH2PO4) and liquid water (H2O) is as follows:

3 CsOH(aq) + H3PO4(aq) → Cs3PO4(aq) + 3 H2O(l)

This reaction is an example of an acid-base neutralization, resulting in the formation of a salt (cesium dihydrogen phosphate) and water.

You couldn't substitute 3M H2SO4 for concentrated HNO3 in Part 1 of the experiment due to the significant differences in their reactivity and properties. HNO3 is a powerful oxidizing agent and provides specific reactions, while H2SO4 is a strong acid but not as strong an oxidizing agent.

The reason you couldn't substitute 3M H2SO4 for concentrated HNO3 in Part 1 of the experiment is due to the difference in their reactivity and properties. Nitric Acid (HNO3) is a powerful oxidizing agent. This means it has the ability to oxidize other substances, bringing about specific chemical reactions. On the other hand, Sulfuric Acid (H2SO4), while still a strong acid, is not as strong an oxidizing agent and would not yield the same results if used as a substitute for HNO3 in this experiment.

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The spectator ions for the reaction between lithium sulfide and copper nitrate are [tex]\boxed{{\text{c}}{\text{. L}}{{\text{i}}^ + }{\text{ and NO}}_3^ - }[/tex].

Further Explanation:

Spectator ions:

These are the ions that exist in the same form on both sides of the reaction. These ions do not affect the equilibrium of any reaction.

The three types of equations that are used to represent the chemical reaction are as follows:

1. Molecular equation

2. Total ionic equation

3. Net ionic equation

The reactants and products remain in undissociated form in the molecular equation. In the case of the total ionic equation, all the ions that are dissociated and present in the reaction mixture are represented while in the case of the overall or net ionic equation only the useful ions that participate in the reaction are represented.

The steps to write the net ionic reaction are as follows:

Step 1: Write the molecular equation for the reaction with the phases in the bracket.

In the reaction, [tex]{\text{L}}{{\text{i}}_2}{\text{S}}[/tex] reacts with [tex]{\text{Cu}}{\left( {{\text{N}}{{\text{O}}_3}} \right)_2}[/tex] to form CuS and [tex]{\text{LiN}}{{\text{O}}_3}[/tex]. The balanced molecular equation of the reaction is as follows:

[tex]{\text{L}}{{\text{i}}_{\text{2}}}{\text{S}}\left( {aq} \right) + {\text{Cu}}{\left( {{\text{N}}{{\text{O}}_3}} \right)_2}\left( {aq} \right) \to {\text{CuS}}\left( s \right) + 2{\text{LiN}}{{\text{O}}_3}\left( {aq} \right)[/tex]

Step 2: Dissociate all the compounds with the aqueous phase to write the total ionic equation. The compounds with solid and liquid phases remain the same. The total ionic equation is as follows:

[tex]2{\text{L}}{{\text{i}}^ + }\left( {aq} \right) + {{\text{S}}^{2-}}\left({aq}\right)+{\text{C}}{{\text{u}}^{2+}}\left( {aq} \right) + 2{\text{NO}}_3^-\left({aq}\right)\to {\text{CuS}}\left(s\right)+2{\text{L}}{{\text{i}}^+}\left({aq}\right)+2{\text{NO}}_3^-\left( {aq}\right)[/tex]

Step 3: The common or spectator ions on both sides of the reaction get cancelled out to get the net ionic equation.

[tex]\boxed{2{\text{L}}{{\text{i}}^ + }\left( {aq} \right)} + {{\text{S}}^{2 - }}\left( {aq} \right) + {\text{C}}{{\text{u}}^{2 + }}\left( {aq} \right) + \boxed{2{\text{NO}}_3^ - \left( {aq} \right)} \to {\text{CuS}}\left( s \right) + \boxed{2{\text{L}}{{\text{i}}^ + }\left( {aq} \right)} + \boxed{2{\text{NO}}_3^ - \left( {aq} \right)}[/tex]

[tex]{\text{L}}{{\text{i}}^ + }[/tex] and [tex]{\text{NO}}_3^ -[/tex] ions are present in the same form on both the reactant and the product side and therefore are known as spectator ions.

Therefore, the net ionic equation is as follows:

[tex]{{\text{S}}^{2 - }}\left( {aq} \right) + {\text{C}}{{\text{u}}^{2 + }}\left( {aq} \right) \to {\text{CuS}}\left( s \right)[/tex]

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Answer details:

Grade: High School

Subject: Chemistry

Chapter: Chemical reaction and equation

Keywords: net ionic equation, spectator ions, CuS, Cu2+, Li+, NO3-, S2-, 2 Li+, 2 NO3-.

[tex]{{\mathbf{K}}^{\mathbf{+}}}[/tex] will be dragged towards the side of [tex]{\mathbf{O}}{{\mathbf{H}}^{\mathbf{-}}}[/tex] while [tex]{\mathbf{B}}{{\mathbf{r}}^{\mathbf{-}}}[/tex] will be dragged towards the side of [tex]{{\mathbf{H}}^{\mathbf{+}}}[/tex]. (Refer to the attached image)

Further explanation:

Electronegativity:

It is defined as the tendency of an atom to attract the shared electrons in the bond towards itself is known as electronegativity. The more electronegative atom will more attract the bonding electrons towards itself than the less electronegative one. Therefore the electrons will spend more time with the more electronegative atom than an electropositive atom. The electronegative atom will acquire the partial negative charge, and the electropositive atom will acquire a partial positive charge.

The polarity of the bond can be estimated by the electronegativity difference [tex]\left({\Delta {\text{EN}}}\right)[/tex]. [tex]\Delta{\text{EN}}[/tex]is the electronegativity difference between the two atoms that are bonded to each other. The formula to calculate [tex]\Delta{\text{EN}}[/tex]in XY bond is as follows:

[tex]{\mathbf{\Delta EN=}}\left({{\mathbf{electronegativity of Y}}} \right){\mathbf{-}}\left({{\mathbf{electronegativity of X}}}\right)[/tex]

Here, X is the electropositive atom and Y is the electronegative atom.

Higher the electronegativity difference between the two atoms, more will be the polarity of the bond and vice-versa.

Water is a polar molecule. Hydrogen atom acquires partial positive charge while oxygen atom is partially negatively charged. So the end of the water molecule with hydrogen is positively charged, and that with oxygen is negatively charged.

In the case of a potassium-bromine bond, bromine is more electronegative than potassium. So the electrons will be more attracted towards bromine due to which it develops a partial negative charge. Potassium, being less electronegative than bromine, in turn, acquires a partial positive charge. This separation of charge results in the formation of a polar bond between potassium and bromine.

So [tex]{{\text{K}}^+}[/tex] will be dragged towards the side of [tex]{\text{O}}{{\text{H}}^-}[/tex] while [tex]{\text{B}}{{\text{r}}^-}[/tex] will be dragged towards the side of [tex]{{\text{H}}^+}[/tex].

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Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Covalent bonding and molecular structure

Keywords: electronegativity difference, electropositive, electronegative, electrons, polar, KBr, K, Br, K+, Br-, H+, OH-, potassium, bromine, hydrogen, oxygen.

Answer: 2.4 grams

Explanation:

To calculate the moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]  

For [tex]O_2[/tex]

Given mass = 19 g

Molar mass of  [tex]O_2[/tex] = 32 g/mol

Putting values in above equation, we get:

[tex]\text{Moles of}O_2=\frac{19}{32}=0.60moles[/tex]

[tex]2H_2+O_2\rightarrow 2H_2O[/tex]

1 mole of [tex]O_2[/tex] reacts with 2 moles of [tex]H_2[/tex]

0.60 moles of [tex]O_2[/tex] wil react with =[tex]\frac{2}{1}\times 0.60=1.20[/tex] moles of [tex]H_2[/tex]

mass of [tex]H_2=moles\times {\text {molar mass}}=1.20moles\times 2g/mol=2.4g[/tex]

Thus 2.4 grams of hydrogen will react completely with 19 g of oxygen.

The concentration of drug after 4.5 hoursis [tex]\boxed{{\text{0}}{\text{.25 mg/100 mL}}}[/tex].

Further Explanation:

Radioactive decayis responsible to stabilizeunstable atomic nucleusaccompanied by the release of energy.

Half-life is the duration after which half of the original sample has been decayed and half is left behind. It is represented by [tex]{t_{{\text{1/2}}}}[/tex].

The relation between rate constant and half-life period for first-order reaction is as follows:

[tex]k = \dfrac{{0.693}}{{{t_{{\text{1/2}}}}}}[/tex]                                                                                                    …… (1)

Here,

[tex]{t_{{\text{1/2}}}}[/tex] is half-life period.

k is rate constant.

Substitute 90 min for [tex]{t_{{\text{1/2}}}}[/tex] in equation (1).

[tex]\begin{aligned}k &= \frac{{0.693}}{{90{\text{ min}}}} \\&= 7.7 \times {10^{ - 3}}{\text{ mi}}{{\text{n}}^{ - 1}} \\\end{aligned}[/tex]  

Radioactive decay formula is as follows:

[tex]{\text{A}} = {{\text{A}}_0}{e^{ - kt}}[/tex]                                        …… (2)

Here

A is the concentration of sample after time t.

t is the time taken.

[tex]{{\text{A}}_0}[/tex] is the initial concentration of sample.

k is the rate constant.

The time for the process is to be converted into min. The conversion factor for this is,

[tex]1{\text{ hr}} = 60\min[/tex]  

Therefore time taken can be calculated as follows:

[tex]\begin{aligned}t&= \left( {4.5{\text{ hr}}} \right)\left( {\frac{{60{\text{ min}}}}{{1{\text{ hr}}}}} \right) \\&= 270{\text{ min}} \\\end{aligned}[/tex]  

Substitute 2 mg/100 mL for [tex]{{\text{A}}_0}[/tex] , [tex]7.7 \times {10^{ - 3}}{\text{ mi}}{{\text{n}}^{ - 1}}[/tex] for k and 270 min for t in equation (2).

[tex]\begin{aligned}{\text{A}} &= \left( {\frac{{2{\text{ mg}}}}{{100{\text{ mL}}}}} \right){e^{ - \left( {7.7 \times {{10}^{ - 3}}{\text{ mi}}{{\text{n}}^{ - 1}}} \right)\left( {270{\text{ min}}} \right)}} \\&= \frac{{0.25{\text{ mg}}}}{{100{\text{ mL}}}} \\\end{aligned}[/tex]  

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Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Radioactivity

Keywords: half-life, 0.25 mg/100 mL, 2 mg/100 mL, A, k, t, rate constant, 4.5 h, 270 min, radioactive decay formula.

The answer would be false because the true answer would be astronomy

Answer:

The rate will be doubled by a Factor of 18

Explanation:

Good luck to everyone doing this!! you got this

The rate of the reaction will increase by a factor of 18 if the concentration of C2O42- is tripled and the concentration of HgCl2 is doubled.

The question asks how the rate of the reaction 2 HgCl2(aq) + C2O42-(aq) → 2 Cl-(aq) + 2 CO2 (g) + Hg2Cl2 (s) will change if the concentration of C2O42- is tripled and the concentration of HgCl2 is doubled. Given the rate law rate = k[HgCl2][C2O42-]2, we can predict the effect on the rate. If the rate law is followed, tripling [C2O42-] will increase the rate by a factor of 32 or 9 because the concentration of C2O42- is squared in the rate law. Doubling [HgCl2] will double the rate of reaction. Therefore, tripling [C2O42-] and doubling [HgCl2] together will increase the reaction rate by a factor of 2 * 9 = 18.

In the Ideal Gas law, the letter R represents the universal gas constant. Its value depends on the units being used for pressure, volume, moles, and temperature. The ideal gas law equation is PV=nRT.

In the Ideal Gas law, the letter R stands for the universal gas constant. This constant is derived from the Ideal Gas Equation: R = 0.08206 L atm mol-¹ K-¹ or 8.314 L kPa mol-¹ K-¹. The ideal gas law itself is represented as PV = nRT, where P represents pressure, V is volume, n is the number of moles of the gas, T is temperature in Kelvin, and R is the universal gas constant.

The value of R, 0.08206 L atm mol-¹ K-¹ or 8.314 J/mol · K, depends on the units being used. For instance, when using SI units, R would be 8.31 J/mol · K.

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The benzhydrol radical is more stable than the hydroxyl isopropyl radical because it can delocalize its unpaired electron through resonance across the aromatic ring, while the hydroxyl isopropyl radical lacks such extensive resonance stabilization.

When considering the stability of hydroxyl isopropyl and benzhydrol radicals, it's important to look at resonance stabilization. Radicals that can delocalize their unpaired electron through resonance are generally more stable. In the case of a benzhydrol radical, also known as a benzylic radical, there is significant stabilization due to the possibility of multiple resonance structures. This is because the unpaired electron in the benzhydrol radical can be delocalized through the aromatic ring's system of conjugated pi bonds.

On the other hand, the hydroxyl isopropyl radical, which is an alkyl radical, lacks the extended conjugated system that allows for such delocalization. While it may have some hyperconjugative stabilization, it does not benefit from the extensive resonance stabilization that the benzhydrol radical enjoys. As such, the benzhydrol radical is more stable than the hydroxyl isopropyl radical due to the resonance effects and the subsequent delocalization of the unpaired electron across the aromatic ring.

Answer:

              Volume =  100 cm³

Solution:

             Density measures and compares the amount of mass contained by a substance to its volume. It is expressed as,

                                           Density  =  Mass / Volume

Data Given;

                   Density  =  1.0 g/cm³

                   Mass  =  100 g  

                   Volume  =  ?

Formula Used;

                   Density  =  Mass ÷ Volume

Solving for Volume,

                   Volume  =  Mass ÷ Density  

Putting values,

                   Volume  =  100 g ÷ 1.0 g.cm⁻³

                   Volume =  100 cm³

The Heck reaction involves bond formation between an alkene and an aryl halide. With bromobenzene and propene, two isomers can be obtained, 1-propylbenzene and 2-propylbenzene. These differ in the attachment point of the propyl group to the benzene ring.

The Heck reaction involves a palladium-catalyzed carbon-carbon bond formation between an alkene and an aryl halide. When bromobenzene reacts with propene in a Heck reaction with pd(php3)4 and et3n, two constitutional isomers can be obtained. The difference between the two isomers would lie in the point of attachment of the propyl group to the benzene ring - it could either be at the end (1-propylbenzene) or at the middle (2-propylbenzene).

For 1-propylbenzene, the propyl group is attached to the benzene ring at one end, resulting in a straight chain configuration.

However, for 2-propylbenzene, the propyl group is attached to the benzene ring in the middle, resulting in a branched chain configuration.

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Seafloor spreading provides evidence for D. formation of sedimentary rocks.

In conclusion, when seafloor spreading occurs, there are sediments that overtime accumulates and forms sedimentary rock.

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Answer: Value of A is 2 and value of B is 1.

Explanation:

Lithium is the third element of the periodic table. It belongs to the Period 2 and Group 1 and is considered as an alkali metals.

To write the electronic configuration, we need to find the number of electrons this element contain, which is equal to the atomic number.

Number of electrons = Atomic number = 3

So, the electronic configuration will be: [tex]1s^22s^1[/tex]

The s-orbital in total can contain only 2 electrons.

Hence, the value of A is 2 and value of B is 1.

The periodic table showcases the table of the chemical elements listed in order of atomic number, usually in rows(period) from the left to the right, in such a way that elements with similar atomic structures as well as similar chemical properties can be located in vertical columns (groups).

Lithium is the third chemical element on the periodic table. It can be located on group 1 element also known as the alkali metals. It is soft in nature and has a whitish color.

The electronic configuration explains how electrons are distributed in their atomic orbitals.

The electronic configuration for Lithium can be expressed as:

Li: 1s² 2s¹

Therefore, from the given information, we can conclude that the value for the electronic configuration of A and B  in the Lithium sample given is 2 and 1 respectively

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Examples of matter include;

Keywords: Matter, states of matter, particles, atoms, molecules, elements, compounds  

Level: High school  

Subject: Chemistry  

Topic: Matter and particles of matter  

Sub-topic: Classification of matter

Final answer:

To produce a single molecule of water (H₂O), exactly two hydrogen atoms are required. This 2:1 ratio of hydrogen to oxygen atoms is consistent no matter how many molecules of water are being made.

Explanation:

To produce a single molecule of water, which has the chemical formula H₂O, you need two hydrogen atoms. The subscript '2' in H₂ indicates that there are two hydrogen atoms for every oxygen atom in a water molecule. Therefore, for each molecule of water you create, you must have this 2:1 ratio of hydrogen to oxygen atoms.

For example, if you want to create two water molecules, you would need a total of 4 hydrogen atoms (2 molecules × 2 atoms/molecule). Similarly, to produce five water molecules, you would require 10 hydrogen atoms in all (5 molecules × 2 atoms/molecule). No matter the number of water molecules you are looking to produce, you will always need twice as many hydrogen atoms as you have water molecules because of this consistent ratio.

The dissolved particles in a solution containing a molecular solute are the molecules of the solute themselves. When a molecular solute dissolves, its molecules distribute uniformly within the solvent, forming a homogeneous mixture at the molecular level. However, ionic solutes dissociate into individual cations and anions.

The dissolved particles in a solution containing a molecular solute like sugar (C12H22O11) would be molecules. In chemistry, a solution can be thought of as a mixture that is homogeneous at the molecular level. In these solutions, one component (the solvent) is usually present in a significantly larger concentration. Other components (solutes), are present in relatively lesser amounts. When a covalent solid like sugar dissolves in water, its molecules become uniformly distributed among the molecules of water creating an aquaeous solution.

However, not all solutions result in dissolved molecules. Solutions that include ionic compounds like sodium chloride (NaCl) will separate into individual cations and anions when dissolved in water. In contrast to molecular substances, ionic type compounds will break apart into separate ions. Hence, it's crucial to understand the type of solute in your solution.

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Answer:

207.2 g

Explanation:

The mass of 1 mole of lead is given by its molar mass, which  is 207.2 g/mol. This is an average molar mass, corresponding to the weighted average of the 4 lead isotopes, ²⁰⁴Pb, ²⁰⁶Pb, ²⁰⁷Pb, ²⁰⁸Pb, considering their abundance in nature. The mass of 1 mole of Pb is:

1 mol × (207.2 g/mol) = 207.2 g

The alkoxide ion (RO⁻) and the alkyl(aryl) bromide can be used in a Williamson ether synthesis to create ethers. The structural formulas of these compounds depend on the specific alkyl or aryl group involved in the synthesis.

In a Williamson ether synthesis, an alkoxide ion reacts with an alkyl (or aryl) halide to form an ether. To perform this synthesis, let's consider the given ether and determine the structural formulas for the alkoxide ion and the alkyl(aryl)bromide.

Given Ether:

The given ether is not explicitly mentioned in your question. To proceed, let's assume the ether you intend to synthesize has the following structure:

R-O-R

where "R" represents an alkyl or aryl group. This is a general representation for ethers.

Structural Formulas:

1. Alkoxide Ion (RO⁻):

  - The alkoxide ion (RO⁻) is formed by removing a hydrogen atom from an alcohol (R-OH). It has a lone pair of electrons on the oxygen atom. The structural formula is as follows:

    RO⁻

    |

    R

2. Alkyl (Aryl) Bromide:

  - The alkyl bromide or aryl bromide can be represented as "R-Br" or "Ar-Br," where "R" represents an alkyl group, and "Ar" represents an aryl group. The structural formula depends on the specific alkyl or aryl group you are using.

  Example 1 (Alkyl Bromide):

  - For an alkyl bromide, if we use "R" as a general alkyl group, the structural formula is:

    R-Br

  Example 2 (Aryl Bromide):

  - For an aryl bromide, if we use "Ar" as a general aryl group, the structural formula is:

    Ar-Br

In a Williamson ether synthesis, the alkoxide ion (RO⁻) is a strong nucleophile that attacks the alkyl (aryl) halide, leading to the formation of the desired ether by replacing the halogen atom. The specific alkyl or aryl group used will depend on the reactants and the ether you intend to synthesize.

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Norfenefrine (C₈H₁₁NO₂).

We will solve a case related to one of the colligative properties, namely freezing point depression.

The freezing point of the solution is the temperature at which the solution begins to freeze. The difference between the freezing point of the solvent and the freezing point of the solution is called freezing point depression.

[tex]\boxed{ \ \Delta T_f = T_f(solvent) - T_f(solution) \ } \rightarrow \boxed{ \ \Delta T_f = K_f \times molality \ }[/tex]

Given:

A mysterious white powder could be,

When 82 mg of the powder is dissolved in 1.50 mL of ethanol (density = 0.789 g/cm³, normal freezing point −114.6°C, Kf = 1.99°C/m), the freezing point is lowered to −115.5°C.

Question: What is the identity of the white powder?

The Process:

Let us identify the solute, the solvent, initial, and final temperatures.

Prepare masses of solutes and solvents.

We must prepare the solvent mass unit in kg because the unit of molality is the mole of the solute divided by the mass of the solvent in kg.

The molality formula is as follows:

[tex]\boxed{ \ m = \frac{moles \ of \ solute}{kg \ of \ solvent} \ } \rightarrow \boxed{ \ m = \frac{mass \ of \ solute \ (g)}{molar \ mass \ of \ solute \times kg \ of \ solvent} \ }[/tex]

Now we combine it with the formula of freezing point depression.

[tex]\boxed{ \ \Delta T_f =  K_f \times \frac{mass \ of \ solute \ (g)}{molar \ mass \ of \ solute \times kg \ of \ solvent} \ }[/tex]

It is clear that we will determine the molar mass of the solute (denoted by Mr).

We enter all data into the formula.

[tex]\boxed{ \ -114.6^0C - (-115.5^0C) = 1.99 \frac{^0C}{m} \times \frac{0.082 \ g}{Mr \times 0.00118 \ kg} \ }[/tex]

[tex]\boxed{ \ 0.9 = \frac{1.99 \times 0.082}{Mr \times 0.00118} \ }[/tex]

[tex]\boxed{ \ Mr = \frac{0.16318}{0.9 \times 0.00118} \ }[/tex]

We get [tex]\boxed{ \ Mr = 153.65 \ }[/tex]

These results are very close to the molar mass of norfenefrine which is 153.18 g/mol. Thus the white powder is norfenefrine.

Keywords: a mysterious white powder, sugar, cocaine, codeine, norfenefrine, fructose, the solute, the solvent, dissolved, ethanol, normal freezing point, the freezing point depression, the identity

The identity of the white powder is [tex]\boxed{\text{norfenefrine}(\text{C}_{8}\text{H}_{11}\text{NO}_{2})}[/tex] .

Further Explanation:

Colligative properties

These are the properties that depend on the number of solute particles and not on their mass or identities. Following are the four colligative properties:

1. Relative lowering of vapor pressure

2. Elevation in boiling point

3. Depression in freezing point

4. Osmotic pressure

The temperature where a substance in its liquid form is converted into the solid state is known as freezing point. Freezing point depression is a colligative property because it depends on the number of moles of solute particles.

The expression for the freezing point depression is as follows:

[tex]\Delta\text{T}_\text{f}=\text{K}_\text{f}\,\text{m}[/tex]       …… (1)

Here,

[tex]\Delta\text{T}_\text{f}[/tex] is the depression in freezing point.

[tex]\text{K}_\text{f}[/tex] is the cryoscopic constant.

m is the molality of the solution.

The formula to calculate the density of substance is as follows:

[tex]\text{Density of substance}=\dfrac{\text{Mass of substance}}{\text{Volume of substance}}[/tex]                                                      …… (2)

Rearrange equation (2) for the mass of substance.

[tex]\text{Mass of substance}=(\text{Density of substance })(\text{Volume of substance})[/tex]                                 …… (3)

The volume of the substance is to be converted into  . The conversion factor for this is,

[tex]1\,\text{ml}=1\,\text{cm}^3[/tex]  

Therefore the volume of the substance can be calculated as follows:

[tex]\begin{aligned}\text{Volume}&=(1.50\,\text{mL})\left(\frac{1\,\text{cm}^3}{1\,\text{mL}}\right)\\&=1.50\,\text{cm}^3\end{aligned}[/tex]  

Substitute [tex]1.50\,\text{cm}^3[/tex] for the volume of substance and [tex]0.789\,\text{g/gm}^3[/tex] for the density of the substance in equation (3) to calculate the mass of solvent.

[tex]\begin{aligned}\text{Mass of solvent}&=\left(\dfrac{0.789\,\text{g}}{1\,\text{cm}^3}\right)(1.50\,\text{cm}^3)\\&=1.1835\,\text{g}\end{aligned}[/tex]  

This mass is to be converted into kg. The conversion factor for this is,

[tex]1\,\text{g}=10^{-3}\,\text{kg}[/tex]  

Therefore the mass of solvent can be calculated as follows:

[tex]\begin{aligned}\text{Mass of solvent}&=(1.1835\,\text{g})\left(\dfrac{10^{-3}\,\text{kg}}{1\,\text{g}}\right)\\&=0.0011835\,\text{kg}\end{aligned}[/tex]  

The mass of solute is to be converted into g. The conversion factor for this is,

 [tex]1\,\text{mg}=10^{-3}\,\text{g}[/tex]

Therefore the mass of solute can be calculated as follows:

[tex]\begin{aligned}\text{Mass of solute}&=(82\,\text{mg})\left(\dfrac{10^{-3}\,\text{g}}{1\,\text{g}}\right)\\&=0.082\,\text{g}\end{aligned}[/tex]  

The freezing point depression can be calculated as follows:

[tex]\begin{aligned}\Delta\text{T}_\text{f}&=-114.6\,^\circ\text{C}-(-115.5\,^\circ\text{C})\\&=0.9\,\circ\text{C}\end{aligned}[/tex]

The formula to calculate the molality of solution is as follows:

[tex]\text{Molality of solution}=\dfrac{\text{Amount (mol) of solute}}{\text{Mass (kg) of solvent}}[/tex]                                                        …… (4)

The formula to calculate the amount of solute is as follows:

[tex]\text{Amount of solute}=\dfrac{\text{Mass of solute}}{\text{Molar mass of solute}}[/tex]                                                              …… (5)

Incorporating equation (5) into equation (4),

[tex]\text{Molality of solution}=\dfrac{\text{Mass of solute}}{(\text{Mass of solvent})(\text{Molar mass of solute})}[/tex]                               …… (6)

Incorporating equation (6) into equation (1),

[tex]\Delta\text{T}_\text{f}=\text{k}_\text{f}\left(\dfrac{\text{Mass of solute}}{(\text{Mass of solvent})(\text{Molar mass of solute})}\right)[/tex]                                           …… (7)

Rearrange equation (7) to calculate the molar mass of solute.

[tex]\text{Molar mass of solute}=\dfrac{\text{k}_\text{f}}{\Delta\text{T}_\text{f}}\left(\dfrac{\text{Mass of solute}}{\text{Mass of solvent}}\right)[/tex]                                                        …… (8)

Substitute [tex]1.99\,^\circ\text{C/m}[/tex] for [tex]\text{k}_\text{f}[/tex], [tex]0.9\,^\circ\text{C}[/tex] for [tex]\Delta\text{T}_\text{f}[/tex], 0.082 g for the mass of solute and 0.0011835 kg for the mass of solvent in equation (8).

[tex]\begin{aligned}\text{Molar mass of solute}&=\left(\dfrac{1.99}{0.9}\right)\left(\dfrac{0.082}{0.0011835}\right)\\&=153.199\,\text{g/mol}\\&=\approx153.2\,\text{g/mol}\end{aligned}[/tex]  

The molar mass of powdered sugar is 342.3 g/mol.

The molar mass of cocaine is 303.4 g/mol.

The molar mass of codeine is 299.4 g/mol.

The molar mass of norfenefrine is 153.2 g/mol.

The molar mass of fructose is 180.2 g/mol.

The calculated molar mass of solute is similar to that of norfenefrine. So the identity of the white powder is norfenefrine.

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Answer details:

Grade: Senior School

Chapter: Solutions

Subject: Chemistry

Keywords: colligative properties, depression in freezing point, cryoscopic constant, freezing point, 153.199 g/mol, norfenefrine.