Is caffeine optically active

Answer:

D. the splitting of the atomic nucleus into parts

D.The electric force and nuclear forces are in opposition to each other.

Explanation:

Answer:

The correct answer is: D. copper

Explanation:

Copper is a chemical element that belongs to the group 11 and period 4 of the periodic table. It is a d-block element that has atomic number 29. Copper is a pinkish-orange colored metal that has high conductivity.

It readily oxidizes when exposed to atmospheric oxygen to form a greenish-blue layer of copper carbonate. This layer is known as verdigris or patina.

Final answer:

The symbol for aluminum foil is Al. It is a silvery-white, soft, nonmagnetic, and ductile metal.

Explanation:

The symbol for aluminum foil is Al. Aluminum is a chemical element with the symbol Al and atomic number 13. It is a silvery-white, soft, nonmagnetic, and ductile metal in the boron group. Aluminum foil is commonly used in packaging, cooking, and insulation.

A. The formula for Keq is:

Keq = concentration of product [B] / concentration of reactant [A]

where

A = 3 B

Therefore Keq is:

Keq = [B] / [3 B]

Keq = 0.33

B. The formula for ΔG° is:

ΔG° = - R T ln(Keq)

Since Keq is 0.33 and ln Keq = - 1.1, therefore:

ΔG° = - R T (- 1.1)

ΔG° = +

Therefore ΔG° is positive

The value of Keq for the unimolecular reaction with three times as much starting material as product at equilibrium is 1.73. The value of ∆G° is positive.

In a unimolecular reaction with three times as much starting material as product at equilibrium, the value of Keq can be calculated by taking the square root of the concentration ratio of product to reactant. Since there is three times as much starting material as product, the concentration ratio is 3:1. Taking the square root of this ratio gives us a Keq value of 1.73 (rounded to two decimal places).

The value of ΔG° for a reaction can determine whether it is spontaneous or non-spontaneous. If ΔG° is negative, the reaction is spontaneous and can proceed without any outside intervention. If ΔG° is positive, the reaction is non-spontaneous and requires an input of energy to proceed. In this case, since the reaction is at equilibrium and there is three times as much reactant, the value of ΔG° would be positive.

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Final answer:

To prepare a 1.00 M solution from concentrated 98.0 wt% H2SO4, mix 130.4 mL of the acid with water up to 1.000 L. The density of this concentrated sulfuric acid is approximately 1.801 g/mL.

Explanation:

The question involves two distinct parts: (a) calculating the amount of a 98.0 wt% H2SO4 solution required to dilute it to a 1.00 M solution, and (b) determining the density of the concentrated sulfuric acid solution. When diluting a concentrated solution to a specified molarity, To dilute 98.0 wt% H2SO4 to 1.00 M, mix 130.4 mL of the concentrated acid with water. This process involves using the molarity formula and considering the initial concentration of the solution. For calculating the density, understanding that the 98.0 wt% H2SO4 solution means 98 g of H2SO4 is in 100 g of the solution. Given this, with additional calculations and known properties of the solution, the density of 98.0 wt% H2SO4 is around 1.801 g/mL.

(a) Volume of concentrated H₂SO₄ required is 55.6 mL.

(b) Density of 98.0 wt% H₂SO₄ is 1,803.70 g/L

To solve the problems regarding sulfuric acid (H₂SO₄), follow these steps:

(a) Volume of Reagent Required

Calculate the molar mass of H₂SO₄:

[tex]\[ \text{Molar mass of H}_2\text{SO}_4 = 2 \times 1.008 + 32.07 + 4 \times 16.00 = 98.09 \, \text{g/mol} \][/tex]

Determine the amount of H₂SO₄ in 1 liter of 1.00 M solution:

[tex]\[ \text{Moles of H}_2\text{SO}_4 = 1.00 \, \text{M} \times 1.000 \, \text{L} = 1.00 \, \text{mol} \][/tex]

[tex]\[ \text{Mass of H}_2\text{SO}_4 = 1.00 \, \text{mol} \times 98.09 \, \text{g/mol} = 98.09 \, \text{g} \][/tex]

Calculate the volume of 18.0 M H₂SO₄ required to obtain 98.09 grams of H₂SO₄:

[tex]\[ \text{Moles of H}_2\text{SO}_4 \text{ in the concentrated solution} = 18.0 \, \text{M} \][/tex]

[tex]\[ \text{Volume of concentrated solution} = \frac{\text{Moles of H}_2\text{SO}_4}{\text{Concentration}} = \frac{98.09 \, \text{g}}{98.09 \, \text{g/mol}} = 1.00 \, \text{mol} \][/tex]

[tex]\[ \text{Volume} = \frac{1.00 \, \text{mol}}{18.0 \, \text{M}} = 0.0556 \, \text{L} = 55.6 \, \text{mL} \][/tex]

(b) Density of 98.0 wt% H₂SO₄

Determine the mass of H₂SO₄ in 1 liter of solution:

From the molarity and density calculation:

[tex]\[ \text{Molarity (M)} = 18.0 \, \text{mol/L} \][/tex]

[tex]\[ \text{Mass of H}_2\text{SO}_4 = 18.0 \, \text{mol} \times 98.09 \, \text{g/mol} = 1,765.62 \, \text{g} \][/tex]

Calculate the total mass of the solution:

Since the solution is 98.0 wt% H₂SO₄:

[tex]\[ \text{Mass percentage} = \frac{\text{Mass of H}_2\text{SO}_4}{\text{Total mass of solution}} \times 100\% \][/tex]

[tex]\[ 98.0\% = \frac{1,765.62 \, \text{g}}{\text{Total mass}} \times 100\% \][/tex]

[tex]\[ \text{Total mass} = \frac{1,765.62 \, \text{g}}{0.980} = 1,803.70 \, \text{g} \][/tex]

Calculate the density of the solution:

[tex]\[ \text{Density} = \frac{\text{Total mass}}{\text{Volume}} = \frac{1,803.70 \, \text{g}}{1.00 \, \text{L}} = 1,803.70 \, \text{g/L} \][/tex]

Answer:

If it is tested in a controlled setting with repeated results.

Explanation:

Hello,

Every scientific foundation is preceded by a series of repetitive tests which allow the scientists to state both reliable and trustful conclusions. In such a way, by repeating the aforesaid test, the conclusion is strengthened, nevertheless, those tests must be carried out in a controlled manner in order to assure that the proposed conclusion is suitable under the specified conditions for scientific claims.

Best regards.

The question is asking for the number of moles of benzoic acid and sodium bicarbonate, and whether the given quantity of sodium bicarbonate is enough to react with all of the benzoic acid. Without additional information, a detailed calculation is not possible. However, an example has been provided for clarity.

Unfortunately, insufficient information has been provided to determine the number of moles of benzoic acid in your experiment. To make this calculation, we would need to know either the mass and molar mass of benzoic acid or, if it is in solution, the volume and molarity. Similarly, for the sodium bicarbonate, while you've given a 10% solution, we also need the density of the solution for the accurate calculation.

After the moles of benzoic acid and sodium bicarbonate are calculated, the stoichiometry of the reaction would be needed to assess whether there is sufficient sodium bicarbonate to react with all of the benzoic acid. Typically, one mole of bicarbonate reacts with one mole of acid.

As an example, say we have 2 grams of benzoic acid (C7H6O2). The molar mass of benzoic acid is about 122.12 g/mol. Therefore, there would be approximately 0.016 moles of benzoic acid (2 grams divided by 122.12 g/mol). If we have more than 0.016 moles of sodium bicarbonate, then it is sufficient.

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Final answer:

In an aqueous solution of LiCl, the compound dissociates completely into hydrated lithium ions (Li+(aq)) and chloride ions (Cl-(aq)), both surrounded by water molecules due to the polarity of water.

Explanation:

In an Aqueous Solution of Lithium Chloride (LiCl), What Ions or Molecules Are Present?

Lithium chloride (LiCl) is an ionic compound that dissociates completely in water due to the strong ion-dipole interactions between the ions and the polar water molecules. When LiCl is dissolved in water, it separates into its constituent ions, lithium (Li+) and chloride (Cl−). The process is facilitated by the polar nature of water molecules which arrange themselves around the ions to form a hydration shell. The oxygen ends of water molecules point towards lithium ions due to their positive charge, while the hydrogen ends tend to surround the chloride ions because of their negative charge. As a result, the solution contains hydrated lithium ions (Li+(aq)) and hydrated chloride ions (Cl−(aq)), indicating they are surrounded by water molecules in the solution.

Thus, in an aqueous solution of lithium chloride, the present species are Li+(aq) and Cl−(aq), fully dissociated and surrounded by water molecules, reflecting their hydrated state.

Final answer:

To determine the half-life of iron-59, the decay rate formula is applied, using the provided decay rates and the time period. After calculations, the half-life of iron-59 is found to be approximately 44.5 days.

Explanation:

The half-life of a radioisotope can be determined by understanding how the decay rate changes over time. In this case, we need to find the half-life of iron-59, knowing that its initial decay rate is 16,800 disintegrations per minute, which falls to 10,860 disintegrations after 28.0 days.

To find the half-life of iron-59, we use the decay rate formula:
N(t) = N0 * (1/2)^(t/T), where N(t) is the decay rate at time t, N0 is the initial decay rate, and T is the half-life.

Plugging in the values we have, we get 10,860 = 16,800 * (1/2)^(28/T). To solve for T, we rearrange and take the natural logarithm of both sides to get T as the subject. In so doing, we find that the half-life of iron-59 is approximately 44.5 days.

Answer:  L2/L1=1.5

Explanation:

Not my work but heres the way the answer is obtained cause the other guy was totally incorrect

Write the expression to calculate the moment of inertia of the meter stick about second axis of rotation. I 2 = M L 2 12 + M ( x 1 − x 2 ) 2

Substitute the values in the above expression. I 2 = M ( 1 m ) 2 12 + M ( 0.5 m − 0.3 m ) 2 I 2 = M 12 + 0.04 M I 2 = 1.48 M 12

. . . . . . ( i i ) Divide expression (ii) by expression (i):

I 2 I 1 = 1.48 M 12 M 12 = 1.48 ≃ 1.5

The ratio of the moment of inertia through the second axis (30-cm mark) to the moment of inertia through the first axis (50-cm midpoint) is 7/3. This calculation uses the parallel-axis theorem.

The question deals with the moment of inertia of a slender uniform rod rotated around two different axes perpendicular to the rod, where one passes through the rod's midpoint, and the other through a point that is not the midpoint. To find the ratio of moments of inertia about these two axes, we can use the parallel-axis theorem, which states that the moment of inertia I about any axis parallel to and a distance d from the center of mass axis is I = Icm + Md2, where Icm is the moment of inertia about the center of mass and M is the mass of the rod.

For the first axis, which passes through the 50-cm mark, the moment of inertia is I50 = ML2/12, because it is the midpoint of the rod (center of mass). For the second axis, which passes through the 30-cm mark, we apply the parallel-axis theorem to find I30 = I50 + M(0.22), because the distance d from the center of mass (50 cm) to the 30-cm mark is 20 cm or 0.2 m. Thus, the ratio I30 / I50 is (ML2/12 + M(0.2)2) / (ML2/12), which simplifies to 1 + 4/3 = 7/3. Therefore, the moment of inertia of the rod about the axis through the 30-cm mark is 7/3 times greater than the moment of inertia about the 50-cm mark.

Answer:

[tex]2POCl_3(g) --> 2PCl_3(g) + O_2(g)[/tex]

Explanation:

Hello,

For the listed chemical reactions, one can predict the one which has a positive ΔS° is:

[tex]2POCl_3(g) --> 2PCl_3(g) + O_2(g)[/tex]

This is foreseen owing to the greater number of products (2) by contrast with the number of reagents (1). Moreover, for such reaction, there is a greater amount of moles at the products than at the reagents.

Best regards.

Answer:

true

Explanation:

The balanced chemical equation  between sodium lithium metal and diatomic nitrogen gas is given by;

6Li(s) + N2(g) → 2Li3N(s)

Keywords: Chemical equations, balancing of chemical equations

Level: high school

Subject: Chemistry

Topic: Chemical equations

Sub-topic: Balancing chemical equations  

The balanced chemical equation for the reaction between lithium metal and nitrogen gas to form lithium nitride is 6 Li (s) + N₂ (g) → 2 Li₃N (s). This reaction is exothermic.

The reaction between solid lithium metal and diatomic nitrogen gas produces lithium nitride.

The balanced chemical equation for this reaction is as follows:

This reaction is exothermic, meaning it releases heat.

Correct question is: Solid lithium metal and diatomic nitrogen gas react spontaneously to form a solid product. give the balanced chemical equation (including phases) that describes this reaction. indicate the phases using abbreviation (s), (I), (g) for solid, liquid and gas

Answer: The final volume should be [tex]\frac{V_1}{2}[/tex]

Explanation:

To calculate the new pressure, we use the equation given by Boyle's law. This law states that pressure is directly proportional to the volume of the gas at constant temperature.  

The equation given by this law is:

[tex]P_1V_1=P_2V_2[/tex]

where,

[tex]P_1[/tex] = initial pressure of gas= P atm

[tex]V_1[/tex] =  initial volume of gas = [tex]V_1[/tex]

[tex]P_2[/tex] = final pressure of gas= 2P

[tex]V_2[/tex] = final volume of gas = ?

Putting values in above equation, we get:

[tex]P\times V_1=2P\times V_2[/tex]

[tex]V_2=\frac{V_1}{2}[/tex]

Thus the final volume should be [tex]\frac{V_1}{2}[/tex]

The combustion of ethanol produces water and carbon dioxide. Given the mass of ethanol and water produced in the experiment, the percent yield of H2O can be calculated as the ratio of actual yield to theoretical yield multiplied by 100%, which is 34.1%.

The question is about calculating the percent yield of H2O from the combustion of ethanol (C2H5OH). The balanced chemical equation for the combustion of ethanol is C2H5OH + 3O2 -> 2CO2 + 3H2O. The total volume of ethanol burned is 4.61 ml, and its density is 0.789 g/ml, hence the mass is 4.61 ml x 0.789 g/ml = 3.634 g. According to the stoichiometric ratio, 1 mol of ethanol can form 3 mol of water. In the experiment, 3.72 ml of water (which is 3.72 g because the density of water is 1g/ml) was collected. Thus, the theoretical yield is 3.634 g x 3 = 10.902 g, and the percent yield of H2O is the actual yield/theoretical yield x100% = 3.72 g/10.902 g x 100% = 34.1%.

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The percent yield of water from the combustion of ethanol is calculated as approximately 87.32%. This involves determining the theoretical yield based on the balanced chemical equation and comparing it with the actual yield collected.

To determine the percent yield of water from the combustion of ethanol, follow these steps:

The balanced chemical equation for the combustion of ethanol ([tex]C_2H_5OH[/tex]) is:

[tex]C_2H_5OH (l) + 3O_2 (g) \rightarrow 2CO_2 (g) + 3H_2O (l)[/tex]

Find the moles of ethanol: Given 4.61 mL of ethanol with a density of 0.789 g/mL, mass of ethanol = 4.61 mL × 0.789 g/mL = 3.635 g.

Calculate moles of ethanol: Using the molar mass of ethanol ([tex]C_2H_5OH[/tex], which is 46.07 g/mol), moles of ethanol = 3.635 g / 46.07 g/mol ≈ 0.0789 mol.

Convert moles of ethanol to moles of water: From the balanced equation, 1 mole of ethanol produces 3 moles of water. Therefore, moles of water = 0.0789 mol × 3 = 0.2367 mol.

Convert moles of water to grams of water: Using the molar mass of water (which is 18.02 g/mol), mass of water (theoretical) = 0.2367 mol × 18.02 g/mol ≈ 4.26 g.

Convert grams of water to mL, assuming the density of water is 1.00 g/mL (since 1 g of water = 1 mL), the volume = 4.26 mL.

Actual yield of water collected is 3.72 mL.

Percent yield = (actual yield / theoretical yield) × 100 = (3.72 mL / 4.26 mL) × 100 ≈ 87.32%.

The predominant olefin formed in the dehydration of 2-methyl-2-butanol is 2-methylbutene.

The predominant olefin formed in the dehydration of 2-methyl-2-butanol is 2-methylbutene. During the dehydration process, a water molecule is eliminated from 2-methyl-2-butanol, resulting in the formation of 2-methylbutene. This is the major product because it is the most stable and thermodynamically favored product.

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The olefin that would be predominate in the product of the dehydration of 2-methyl-2-butanol is 2-methyl-2-butene, also known as isobutene. The dehydration involves the removal of a hydrogen and a hydroxyl group, resulting in the formation of a double bond, creating the said olefin. The formation of isobutene is favored due to its relatively more stable structure.

The olefin that should predominate in the product of the dehydration of 2-methyl-2-butanol is 2-methyl-2-butene, also known as isobutene. This reaction can produce either a cis-isomer or a trans-isomer.

The process of dehydration involves the removal of a hydrogen and a hydroxyl group (OH) from the 2-methyl-2-butanol molecule under the influence of a strong acid like sulfuric acid. This results in the formation of a double bond, creating the said olefin - 2-methyl-2-butene. The formation of isobutene is favored due to its relatively more stable structure, which is attributed to increased steric hindrance in the molecule.

Finally, it's worth noting that the overall intermolecular forces (IMFs), particularly hydrogen bonding, are important factors affecting the vapor pressure of the resultant olefin.

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Since its structure is depicted by dots, which cannot describe it,

Tetrahedral molecules like methane have four comparable C-H bonds.

An organic compound's skeletal structure is made up of a group of atoms that are joined together to form the compound's fundamental structure. Chains, branches, and/or rings of bound atoms can make up the skeleton. Heteroatoms are skeletal atoms that are not carbon or hydrogen.

How can we know that methane has four spheres?

We start with what we already know: the geometry of the methane molecule has the carbon at the center and the four C-H bonds pointing towards the corners. In methane, the carbon atoms form four bonds, one to each of the four hydrogen atoms.

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Answer:

s atomic orbital

Explanation:

Lets us consider 4s, 4p, 4d and 4f. Here 4 is the energy level or the principal quantum number. For all the atomic orbital s, p, d and f the energy level is the same. But the energy level of each  

Atomic orbital is different. s atomic orbital has the lowest energy followed by p, d and f.  

Arranging the atomic orbitals in order of increasing energy:

s < p < d < f

Answer: The ions present in a given solution of silver nitrate are [tex]Ag^+,NO_3^-[/tex]

Explanation:

Silver nitrate is a strong electrolyte because it displaces into its ions in aqueous solution.

The chemical reaction for the ionization of silver nitrate solution follows the equation:

[tex]AgNO_3(aq.)\rightarrow Ag^+(aq.)+NO_3^-(aq.)[/tex]

By Stoichiometry of the reaction:

1 mole of silver nitrate solution ionizes into 1 mole of silver ions and 1 mole of nitrate ions.

Hence, the ions present in a given solution of silver nitrate are [tex]Ag^+,NO_3^-[/tex]

The chemical formula is[tex]\boxed{{\text{A}}{{\text{g}}^ + }{\text{,N}}{{\text{O}}_{\text{3}}}^ - }[/tex].

The phases of ions are[tex]\boxed{{\text{aqeuous phase}}}[/tex].

Further explanation:

Ionic compound:

Ionic compounds are those compounds formed from the ions of the species. Ions are the species that are formed due to the loss or gain of electrons. Cation forms by the loss of electrons and anion forms by the gain of electrons in a neutral atom.

Some of the properties of ionic compounds are as follows:

1. These are hard solids.

2. They have high melting and boiling points.

3. They are considered as good conductors of heat and electricity.

Ionic reaction:

Ionic reaction is a type of chemical reaction in which molecules in aqueous solution dissociate to form ions. In ionic reaction, the net charge is same on both sides.

[tex]{\text{AgN}}{{\text{O}}_{\text{3}}}[/tex]is a strong electrolyte and dissociates into ions in the aqueous solution. The reaction for dissociation of [tex]{\text{AgN}}{{\text{O}}_{\text{3}}}[/tex] solution is:

[tex]{\text{AgN}}{{\text{O}}_{\text{3}}}\left({aq}\right)\to{\text{A}}{{\text{g}}^+}\left({aq}\right)+{\text{N}}{{\text{O}}_{\text{3}}}^-\left({aq}\right)[/tex]

In this reaction, 1 mole of [tex]{\text{AgN}}{{\text{O}}_{\text{3}}}[/tex] dissociates to form 1 mole of [tex]{\text{A}}{{\text{g}}^+}[/tex] and 1 mole of [tex]{\text{N}}{{\text{O}}_{\text{3}}}^-[/tex].

Therefore, ions present in the solution of silver nitrate are [tex]{\text{A}}{{\text{g}}^+}[/tex], [tex]{\text{N}}{{\text{O}}_{\text{3}}}^-[/tex]. As ions are formed in solution only, so the phase of both the ions is aqueous phase.

Learn more:

1. Draw Lewis structure of ionic compound https://brainly.com/question/6786947.

2. Identify the neutral element https://brainly.com/question/9616334

Answer details:

Grade: High School

Subject: Chemistry

Chapter: Chemical compound

Keywords: Ionic compound, aqueous solution, good conductor, high enthalpy, ions, dissociates, AgNO3, Ag+ and NO3-.

The width (y) of the NFL field is 53 1/3 yards, and the area (a) is 6399 1/3 square yards, using a length (x) of 120 yards.

The regulation NFL playing field is a rectangle. If the length is x and the width is y, the perimeter of the rectangle is represented as 2(x+y) which equals 346 2/3 or 1040/3 yards according to your question.

Given the standard dimensions of an NFL field, the length x is 360ft (120 yards) and the width y is 160ft (53 1/3 yards). Now we approach this problem using this knowledge.

To find the width y, we can rearrange the formula: y = 1040/6 - x/2. Substituting x = 120 yards into this equation, we find that y = 53 1/3 yards.

The area of the rectangle (field), represented by a, is found by multiplying the length by the width (x*y). Therefore, the area is a = 120 * 53 1/3 = 6399 1/3 square yards.

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Answer : The volume in gallons will be [tex]1.3\times 10^3\text{ gallons}[/tex]

Explanation :

The conversion used volume from liters to gallons is :

1 gallon = 3.78541 L

or,

[tex]1\text{ liter}=\frac{1}{3.78541}\text{ gallon}[/tex]

As we are given that the lung capacity of the blue whale is 5100 L. Now we have to convert the volume into gallons.

As, [tex]1\text{ liter}=\frac{1}{3.78541}\text{ gallon}[/tex]

So, [tex]5100\text{ liter}=\frac{5100\text{ liter}}{1\text{ liter}}\frac{1}{3.78541}\text{ gallon}=1.3\times 10^3\text{ gallons}[/tex]

Therefore, the volume in gallons will be [tex]1.3\times 10^3\text{ gallons}[/tex]

Harry started with [tex]\boxed{{\text{5}}{\text{.5 grams}}}[/tex] of  [tex]{\text{KAl}}{\left( {{\text{S}}{{\text{O}}_{\text{4}}}} \right)_{\text{2}}} \cdot{\text{12}}{{\text{H}}_{\text{2}}}{\text{O}}[/tex] hydrated salt.

Further explanation:

Mole is the measure of the amount of substance. Mole is the relation between the mass of the substance and molar mass of a substance. It is defined as the mass of a substance in grams divided by its molar mass (g/mol).

The expression to calculate the number of moles is as follows:

[tex]{\text{Number of moles}}\;=\;\dfrac{{{\text{Given mass}}\left({\text{g}}\right)}}{{{\text{molar mass}}\left({{\text{g/mol}}}\right)}}[/tex]

Therefore, the formula to calculate the mass of the given compound is,

[tex]{\text{Mass}}\left({\text{g}} \right)=\left({{\text{Number of moles}}}\right)\left( {{\text{molar mass}}\left({{\text{g/mol}}}\right)}\right)[/tex]

The molar mass of  [tex]{\text{KAl}}{\left( {{\text{S}}{{\text{O}}_{\text{4}}}} \right)_{\text{2}}} \cdot{\text{12}}{{\text{H}}_{\text{2}}}{\text{O}}[/tex] is 474.3884g/mol.

The molar mass of  [tex]{\text{KAl}}{\left({{\text{S}}{{\text{O}}_{\text{4}}}}\right)_{\text{2}}}[/tex] is 258.2050 g/mol.

Therefore, mass of 12[tex]{{\text{H}}_2}{\text{O}}[/tex]  molecules is,

[tex]\begin{aligned}{\text{Mass}}\left( {\text{g}} \right)&={\text{Moles}}\times{\text{molar mass of }{{\text{H}}_2}{\text{O}}\\&=12{\text{ mol }}\times 18{\text{ g/mol }}\\&=216{\text{ g}}\\\end{aligned}[/tex]

Since initially in [tex]{\text{KAl}}{\left( {{\text{S}}{{\text{O}}_{\text{4}}}} \right)_{\text{2}}} \cdot{\text{12}}{{\text{H}}_{\text{2}}}{\text{O}}[/tex] , 1 mole of   [tex]{\text{KAl}}{\left({{\text{S}}{{\text{O}}_{\text{4}}}}\right)_{\text{2}}}[/tex] contained 12 moles of [tex]{{\text{H}}_2}{\text{O}}[/tex] . Thus 258.2050 g of [tex]{\text{KAl}}{\left({{\text{S}}{{\text{O}}_{\text{4}}}}\right)_{\text{2}}}[/tex] should contains 216 g of [tex]{{\text{H}}_2}{\text{O}}[/tex] before evaporation.

Therefore, the mass of [tex]{{\text{H}}_2}{\text{O}}[/tex] contained by 1 g of   [tex]{\text{KAl}}{\left({{\text{S}}{{\text{O}}_{\text{4}}}}\right)_{\text{2}}}[/tex] is calculated as,

[tex]{\text{Mass}}\left({\text{g}}\right){\text{ of }}{{\text{H}}_2}{\text{O in 1 g of KAl}}{\left( {{\text{S}}{{\text{O}}_{\text{4}}}} \right)_{\text{2}}}{\text{ }}=\dfrac{{{\text{216 g of }}{{\text{H}}_2}{\text{O}}}}{{258.2050{\text{ g of KAl}}{{\left({{\text{S}}{{\text{O}}_{\text{4}}}} \right)}_{\text{2}}}}}[/tex]

Therefore, the mass of [tex]{{\text{H}}_2}{\text{O}}[/tex] contained by 3.0 g of   [tex]{\text{KAl}}{\left({{\text{S}}{{\text{O}}_{\text{4}}}}\right)_{\text{2}}}[/tex] is calculated as follows:

[tex]\begin{aligned}{\text{Mass}}\left({\text{g}} \right){\text{ of }}{{\text{H}}_2}{\text{O in 3 g of KAl}}{\left({{\text{S}}{{\text{O}}_{\text{4}}}}\right)_{\text{2}}}&=\dfrac{{{\text{216 g of }}{{\text{H}}_2}{\text{O}}}}{{258.2050{\text{ g of KAl}}{{\left( {{\text{S}}{{\text{O}}_{\text{4}}}}\right)}_{\text{2}}}}}\times 3{\text{ g of KAl}}{\left({{\text{S}}{{\text{O}}_{\text{4}}}}\right)_{\text{2}}}\\&=2.50{\text{ g }}\\\end{aligned}[/tex]

This 2.5 g is the mass of water that is evaporated from hydrated salt.

The total mass of hydrated of [tex]{\text{KAl}}{\left( {{\text{S}}{{\text{O}}_{\text{4}}}} \right)_{\text{2}}} \cdot{\text{12}}{{\text{H}}_{\text{2}}}{\text{O}}[/tex] taken initially by harry is the sum of the mass of water that is evaporated and mass of dehydrated salt that is left after evaporation process.

[tex]\begin{aligned}{\text{Total mass of hydrated salt}}&={{\text{m}}_{{\text{KAl}}{{\left( {{\text{S}}{{\text{O}}_{\text{4}}}}\right)}_{\text{2}}}}}+{{\text{m}}_{{{\text{H}}_2}{\text{O evaporated}}}}\\&=3.0{\text{ g}}+{\text{2}}{\text{.5 g}}\\&=5.5{\text{ g }}\\\end{aligned}[/tex]

Learn more:

1. Determine the number grams of solute in 500 ml of 0.189 M KOH.: https://brainly.com/question/2847466

2. Rank the gases in decreasing order of effusion:https://brainly.com/question/1946297

Answer details:

Grade: Senior school

Subject: Chemistry

Chapter: Mole concept

Keywords: harry, hydrate salt, KAl(SO4)2•12H2O, dehydration process, hydration process, anhydrated salt, KAl(SO4)2, 3.0 gram KAl(SO4)2, grams of start with, 5.5 g.