How many molecules are contained in
1357 mL of O2 gas at −14◦C and 1437 torr?
Answer in units of molec.

Answer: sp3

Explanation:

The correct answer is that Friedel-Crafts reactions can not be done on moderately or strongly deactivated ring systems.

This is a limitation of Friedel-Crafts reactions because they require an activated ring system in order to proceed.

Deactivated ring systems do not have the necessary electron density to facilitate the reaction, and therefore the reaction will not occur.

Friedel-Crafts alkylation and acylation reactions are both types of electrophilic aromatic substitution reactions that involve the formation of a carbon-carbon bond between an aromatic ring and an alkyl or acyl group.

These reactions are commonly used in organic synthesis to introduce new functional groups onto an aromatic ring.

However, they do have other limitations including the fact that Friedel-Crafts alkylation substrates can undergo rearrangement,

Friedel-Crafts acylation often leads to polyacylated products,

and Friedel-Crafts alkylation often leads to polyalkylated products.

These limitations must be taken into consideration when designing a synthetic route that involves a Friedel-Crafts reaction.

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The condition where only four chromosomes are present in some cells of a fruit fly is known as haploidy.

Haploidy can occur due to a variety of reasons, including errors during cell division, such as nondisjunction, which is the failure of chromosomes to separate properly during meiosis. In nondisjunction, the chromosomes fail to divide equally, resulting in daughter cells with abnormal numbers of chromosomes. In the case of the fruit fly, this can result in cells with only four chromosomes instead of the normal eight.

Haploidy can also occur naturally in certain stages of development, such as during gamete formation, where cells undergo meiosis to produce haploid gametes with half the number of chromosomes as the parent cell.

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All the statements are about the acidity of compounds, of which statement 1, 3 & 4 are true and statement 2 is false.

Statement 1: "the strength of an acid depends primarily on the conjugate base's ability to support a negative charge" is true because, a strong acid has a weak conjugate base, which means it is less able to support a negative charge.

Statement 2: "induction is long range electronegativity, and the presence of electronegative atoms improves acidity of distant H" is false.

Because, induction is the ability of an electronegative atom or group to withdraw electron density from a neighbouring atom or group, leading to an increase in acidity.

However, this effect is not long range, and typically only affects neighbouring atoms.

Statement 3: "H bonded to larger atoms are more acidic due to poor orbital overlap, and polarizability of large orbitals" is true.

Because larger atoms have larger orbitals, which leads to poor orbital overlap and weaker bonds with hydrogen.

This makes it easier for the hydrogen to dissociate and increases the acidity of the compound.

Statement 4: "increasing the % of s-orbital character in a bond increases acidity" is true.

S-orbitals are closer to the nucleus and have a lower energy than p-orbitals, leading to stronger bonds and increased acidity.

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A student measures the molar solubility of nickel(II) cyanide in a water solution to be 2. 00×10-8 M. The Ksp of nickel(II) cyanide is [tex]8 \times10^{-24}[/tex]

The solubility product constant, Ksp, serves as the equilibrium constant for solids that dissolve in water. The level of solute dissolution in solution is what it stands for. The more soluble a material is, the higher its Ksp value.

The following equation may be used to get the solubility product constant (Ksp) for nickel(II) cyanide from the molar solubility:

[tex]Ni(CN)_2(s)[/tex] ⇌ [tex]Ni_2+(aq) + 2CN^-(aq)[/tex]

[tex]Ksp = [Ni_2^+][CN^-]^2[/tex]

As per the given information,  

the molar solubility of nickel(II) cyanide is given as [tex]2.00\times 10^{-8}[/tex]M, the concentrations of [tex]Ni_2^+[/tex] and [tex]CN^-[/tex] ions are also [tex]2.00 \times 10^{-8}[/tex] M.

We have to find the Ksp of nickel(II) cyanide.

Hence, the Ksp can be calculated as:

Ksp = [tex](2.00\times10^{-8})(2.00\times10^{-8})^{2}[/tex]

Ksp =[tex]8.00\times10^{-24}[/tex]

Hence, the Ksp of nickel(II) cyanide is [tex]8.00\times10^{-24}[/tex]

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the half-life of the reaction is 1.29 s (to three significant figures).

The half-life (t1/2) of a first-order reaction is given by the formula:

t1/2 = ln(2) / k

where k is the rate constant of the reaction.

In this case, the rate law for the reaction is:

rate = k[A]3[B]2[C]

where [A], [B], and [C] are the concentrations of A, B, and C, respectively.

Since the reaction is first order with respect to A, the concentration of A at any time t is given by:

[A]t = [A]0 x

[tex] {e}^{ - kt} [/tex]

where [A]0 is the initial concentration of A.

Given that the initial concentration of A is 0.867 mol/L and the rate constant is 0.538 , we can use the formula for the half-life to calculate the time required for the concentration of A to decrease by half:

t1/2 = ln(2) / k = ln(2) / 0.538 = 1.29 s

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North America is experiencing spring during the position when the Northern Hemisphere is tilted towards the Sun, which occurs around March 20th to 21st, known as the Vernal Equinox.

The point in the Earth's orbit around the Sun when the tilt of axis is neither towards nor away from the Sun is known as the Vernal Equinox, also known as the Spring Equinox. Because of this arrangement, the length of day and night is almost identical over the whole globe. This occasion, which occurs in North America around March 20–21, heralds the start of the spring season. The Northern Hemisphere begins to tilt towards the Sun around this time, lengthening the day and raising temperatures. Agriculture, migratory patterns, and different cultural holidays are significantly impacted by this change in the Earth's position.

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Answer: First, we need to calculate the total fuel consumption per day for all 380 SSTs:

Fuel consumption per hour: 5,320 gallons

Number of planes: 380

Hours per day: 14

Total fuel consumption per day = 5,320 x 380 x 14 = 28,190,400 gallons

Next, we need to convert gallons to metric tons of jet fuel:

1 gallon = 0.00378541 metric tons

28,190,400 gallons = 106,698.89 metric tons

Now, we can calculate the total crude oil needed to produce this amount of jet fuel:

1 ton of jet fuel = 7,000 kilograms of crude oil

106,698.89 tons of jet fuel = 746,892,230 kilograms of crude oil

To convert kilograms to metric tons:

1 metric ton = 1,000 kilograms

746,892,230 kilograms = 746,892.23 metric tons

Finally, we can calculate the percentage of crude oil production used for SST fuel:

World crude oil production in 1990 = 4.02 x 10^9 metric tons

Percentage of crude oil used for SST fuel = (746,892.23 / 4.02 x 10^9) x 100

= 0.0186%

Therefore, approximately 0.0186% of the crude oil production in 1990 was used for fuel for the SSTs.

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20 mL of 0.20 M H2SO4 is required to completely liquidate 40 mL of 0.10 M Ca (OH)2.

We can use the balanced chemical equation for the neutralization reaction between sulfuric acid (H2SO4) and calcium hydroxide (Ca (OH)2) to determine the stoichiometry of the reaction:

H2SO4 + Ca (OH)2 -> CaSO4 + 2H2O

moles of Ca (OH)2 = Molarity x Volume

= 0.10 mol/L x 0.040 L

= 0.004 mol

moles of solute = Molarity x Volume (in liters)

alter this equation gives:

Volume (in liters) = moles of solute / Molarity

Therefore, the volume of 0.20 M H2SO4 required to supply 0.004 moles of H2SO4 is:

Volume (in liters) = 0.004 mol / 0.20 mol/L

= 0.02 L

Multiplying by 1000 mL/L, we get:

Volume (in mL) = 0.02 L x 1000 mL/L

= 20 mL

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The pH of the 0.30 M solution of sodium formate is 2.13.
The correct answer is E. 2.13.

Formic acid dissociates in water to form hydrogen formate (HCOO-) and a proton (H+).

Since the Ka of HCOOH is [tex]1.8 * 10^{-4}[/tex], the Kb of HCOO- is equal to Ka x Kb = ([tex]1.8 * 10^{-4}[/tex])([tex]1.0 * 10^{-14}[/tex]) = [tex]1.8 * 10^{-18}[/tex].

The ionization of sodium formate (NaHCOO) can be represented as:

[tex]NaHCOO + H^2O < = > HCOO^- + Na^+ +[/tex] [tex]H^3O^+[/tex]

We can use the Kb of HCOO- to calculate the equilibrium concentrations of the species in the solution.

[HCOO-] = [NaHCOO] = 0.30 M

[Na+] = [[tex]H^3O^+[/tex]] = 0

Kb = [HCOO-][[tex]H^3O^+[/tex]]/[NaHCOO]

1.8 x 10-18 = (0.30)(x)/(0.30)

x = 1.8 x 10-18

The pH of the solution can then be calculated using the equation:

pH = -log[[tex]H^3O^+[/tex]]

pH = -log(1.8 x 10-18)

pH = 2.13

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Question should be like

Calculate the pH of a 0. 30 M solution of sodium formate (NaHCOO) given that the Ka of formic acid (HCOOH) is 1. 8 x 10^-4.

A. 8. 61

B. 10. 26

C. 11. 87

D. 5. 39

E. 2. 13

i dont now, i hope that help you :) :D

Answer:

Explanation:

not sure your elements to choose from

But it would be one with a +2 charge

Examples

Be, Mg, Ca, Sr, Ba, Cu, Fe these are just a few

send your list and I can finish this for you

Answer:

The rate of the forward reaction must be equal to the rate of the reverse reaction. In other words, the speed at which products are formed from reactants must be equal to the speed at which reactants are formed from products.

The concentrations (or partial pressures) of the reactants and products must be constant. At equilibrium, the system has reached a state of balance where the amounts of reactants and products are no longer changing with time, and their concentrations (or partial pressures) have stabilized.

Thus, at equilibrium, the reaction is said to be "dynamic," meaning that the forward and reverse reactions are still occurring but at an equal rate, resulting in no net change in the concentrations (or partial pressures) of the reactants and products.

Explanation:

Whether the time taken tο displace all οf the water in a measuring cylinder will be greater fοr lumps οf calcium carbοnate οr pοwder are mentiοned belοw.

The transfοrmatiοn οf οne οr mοre reactants intο οne οr mοre new prοducts is referred tο as a chemical reactiοn. Substances are made οf chemical cοnstituents οr cοmpοunds. The transfοrmatiοn οf οne οr mοre reactants intο οne οr mοre new prοducts is referred tο as a chemical reactiοn. Substances are made οf chemical cοnstituents οr cοmpοunds.

One part carbοn and twο parts οxygen make up the gas called carbοn diοxide. Its usage by plants tο create carbοhydrates during a prοcess knοwn as phοtοsynthesis makes it οne οf the mοst significant gases οn the planet.

The cοncentratiοn οf calcium carbοnate cannοt fluctuate because it is a sοlid. Hence, the οnly reactant in this reactiοn that can fluctuate in cοncentratiοn and impact reactiοn rate is hydrοchlοric acid. We need tο graph the cοncentratiοn οf hydrοchlοric acid against time, as yοu shοuld knοw frοm yοur reactiοn rate theοry.

Therefοre, whether the time taken tο displace all οf the water in a measuring cylinder will be greater fοr lumps οf calcium carbοnate οr pοwder are mentiοned abοve.

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Answer: In the chlor alkali process, sodium hydroxide is prepared from common salt (sodium chloride). The byproducts are hydrogen gas and chlorine gas. Electrolysis of aqueous sodium chloride (brine solution) in Castner-kellner cell gives sodium hydroxide, chlorine gas and hydrogen gas.

Answer: Useful substances are obtained from common salt by the process of electrolysis

Answer:

Hence, C121H199O100 represents the compound's empirical formula.

Explanation:

We must ascertain the ratio of the number of atoms of each element in the compound in order to derive the empirical formula. To accomplish this, we can divide the mole ratio by the least number of moles after converting the masses of each element to moles.

The elements' molar masses are as follows:

12.01 g/mol for carbon

1.01 g/mol for hydrogen

16.00 g/mol for oxygen

When we convert the masses to moles, we obtain:

48.38 g / 12.01 g/mol = 4.03 mol of carbon

6.74 g / 1.01 g/mol of hydrogen equals 6.67 mol

53.5 g / 16.00 g/mol = 3.34 mol for oxygen

3.34 mol of oxygen is the least amount of moles. If you divide the total moles of each element by 3.34 mol, you get:

Carbon: 4.03 mol/3.3 mol = 1.21 mol Hydrogen:6.67 moles/3.34 moles equals 1.99

3.34 mol / 3.34 mol = 1.00 for oxygen.

By multiplying each ratio by 100 to obtain whole integers, we may simplify the ratios, which are around 1.21:1.99:1.00. This results in a ratio of roughly 121:199:100.

We divide each number in the ratio by their greatest common factor (GCF) to arrive at the empirical formula. 121, 199, and 100 have a GCF of 1. By dividing by 1, we get:

Carbohydrate: 121 / 1 = 121

199 / 1 = 199 for hydrogen.

100 / 1 Equals 100 for oxygen.

Hence, C121H199O100 represents the compound's empirical formula.

Answer:

First, I will convert from grams to moles. This will give me approximately 4.032 moles of carbon, 6.740 moles of hydrogen, and 3.344 moles of oxygen. Then, I will calculate the mole ratio, using whole numbers. I must also identify the least amount of moles in an element for this step, which is oxygen at 3.344 moles.

Oxygen = 3.344/3.344 = 1

Carbon = 4.032/3.344 = 1

Hydrogen = 6.740/3.344 = 2

The empirical formula will be COH2.

Now that we have the molar mass of the molecular formula, we can calculate for the molecular formula. First, we will begin by discovering the molar mass of the empirical formula, which will be C molar mass x 1 + O molar mass x 1 + H molar mass x 2. That will be equal to 60.949. After that, we can divide the molecular mass by the empirical formula molar mass, giving us approximately 3. Now we will multiply the empirical formula by 3, giving us our molecular formula, which will be equal to C3O3H6.

The amount of the original sample remaining would be between 1.50 and 3.00 grams. Option I.

To solve this problem, we can use the half-life formula:

N = N0 x (1/2)^(t/T)

where:

We are given N0 = 12.0 grams, T = 3.6 days, and t = 10 days. We can plug these values into the formula and solve for N:

N = 12.0 grams x (1/2)^(10/3.6)

N ≈ 1.50 grams

Therefore, about 1.50 grams of the original sample remains after 10 days. The answer is between 1.50 and 3.00 grams.

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At constant pressure, the systems that work on the surroundings are: A(s)+B(g) → 2C(g), A(g)+B(g) → 3C(g). The correct options are B and C.

In thermodynamics, if the reaction proceeds at constant pressure, it implies that the pressure remains constant during the course of the reaction, whereas the volume might change. The systems that work on the surroundings are those in which work is done on the surroundings by the system, resulting in a decrease in the internal energy of the system.

The reaction's stoichiometric coefficients of the reactants and products are balanced, implying that the number of moles of reactants and products is the same. It means there is no change in the number of moles of gas, which means there is no expansion of gases, resulting in no work done by the system. Thus, this option does not satisfy the criteria for a system working on the surroundings.

The reaction has one gas molecule in the reactants and two gas molecules in the products, implying that there is an increase in the number of moles of gas. As a result, there is an increase in the volume, resulting in work being done by the system. This option satisfies the requirements for a system working on the surroundings.

The reaction has two gas molecules in the reactants and three gas molecules in the products, which means there is an increase in the number of moles of gas. As a result, there is an increase in the volume, resulting in work being done by the system. This option satisfies the requirements for a system working on the surroundings.

The reaction has two gas molecules in the reactants and one gas molecule in the product, implying that there is a decrease in the number of moles of gas. It implies that the volume of the system will decrease, resulting in no work done by the system. Thus, this option does not satisfy the criteria for a system working on the surroundings.

Therefore, the options that work on the surroundings at constant pressure are options b and c.

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A. The limiting reactant is NH₃

B. The mass of NO formed is 44.1 g

C. The mass of H₂O formed is 39.7 g

First, we shall determine the mass in 4 moles of O₂. Details below:

Mole = mass / molar mass

4 = Mass of O₂ / 32

Cross multiply

Mass of O₂ = 4 × 32

Mass of O₂ = 138 g

Finally, we shall determine the limiting reactant. Details below:

4NH₃ + 5O₂ -> 4NO + 6H₂O

From the balanced equation above,

68 g of NH₃ reacted with 160 g of O₂

Therefore,

25 g of NH₃ will react with = (25 × 160) / 68 = 58.8 g of O₂

From the above calculation, we can see that only 58.8 g of O₂ out of 138 g is needed to react with 25 g NH₃.

Thus, the limiting reactant is NH₃

The mass of NO formed can be obtained as illustrated below:

4NH₃ + 5O₂ -> 4NO + 6H₂O

From the balanced equation above,

68 g of NH₃ reacted to produce 120 g of NO

Therefore,

25 g of NH₃ will react to produce = (25 × 120) / 68 = 44.1 g of NO

Thus, the mass of NO formed is 44.1 g

The mass of H₂O formed can be obtained as illustrated below:

4NH₃ + 5O₂ -> 4NO + 6H₂O

From the balanced equation above,

68 g of NH₃ reacted to produce 108 g of H₂O

Therefore,

25 g of NH₃ will react to produce = (25 × 108) / 68 = 39.7 g of H₂O

Thus, the mass of H₂O formed is 39.7 g

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Answer:

From the balanced net ionic equation, we can see that 1 mole of AgNO3 reacts with 1 mole of CaCl2 to produce 1 mole of AgCl. Therefore, we can use the stoichiometry of the reaction to calculate the amount of AgNO3 needed to react completely with the given amount of CaCl2.

First, we need to calculate the number of moles of CaCl2 in 76.5 mL of 0.391 M solution:

moles of CaCl2 = concentration x volume

moles of CaCl2 = 0.391 M x 0.0765 L

moles of CaCl2 = 0.0299 mol

Since 1 mole of AgNO3 reacts with 1 mole of CaCl2, we need 0.0299 mol of AgNO3 to react completely with the CaCl2. To calculate the volume of 0.164 M AgNO3 solution containing this amount of AgNO3, we can use the formula:

moles = concentration x volume

Rearranging the formula, we get:

volume = moles / concentration

Substituting the values, we get:

volume = 0.0299 mol / 0.164 M

volume = 0.1823 L

Converting to milliliters:

V(AgNO3) = 182.3 mL

Therefore, 182.3 mL of 0.164 M AgNO3 solution are needed to react completely with 76.5 mL of 0.391 M CaCl2 solution.

Answer:

c

Explanation:

energy neither created nor distroyed

Answer:

The answer is C

Explanation:

We learned so many times about law conservation energy,states that"Energy neither created or destroyed but can change its form to other."

E.g If we rub our palm,it changes from kinetic energy to heat energy.

The various types of chemical reaction are Combination Reaction,Decomposition Reaction,Displacement Reaction,Double Displacement Reaction, and Precipitation Reaction.

A chemical reaction is defined as the reaction that involves the combination of two or more substances leading to the formation of a new substance called the product of the reaction.

The various types of chemical reaction include the following:

There are three parts of a chemical reaction which is the reactant part, the arrow and the product part.

That is;

A. + B ---------> D + E.

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The empirical formula of phosphorus sulfide is [tex]P^{1}S^3.[/tex].

To determine the empirical formula of phosphorus sulfide, we need to find the mole ratio between phosphorus and sulfur in the compound. We can use the given masses of phosphorus and phosphorus sulfide to calculate the number of moles of each element.

First, we convert the masses of phosphorus and phosphorus sulfide from milligrams to grams:

Mass of P = 44.0 mg = 0.0440 g

Mass of P4S3 = 158 mg = 0.158 g

Next, we calculate the number of moles of each element:

Moles of P = mass of P / molar mass of P = 0.0440 g / 30.97 g/mol = 0.00142 mol

Moles of S = (mass of P4S3 - mass of P) / molar mass of S = (0.158 g - 0.0440 g) / 32.07 g/mol = 0.00393 mol

We then divide the number of moles of each element by the smallest value to obtain the simplest whole-number ratio of atoms:

Moles of P / 0.00142 mol ≈ 1

Moles of S / 0.00142 mol ≈ 2.77

We can then round the above ratio to the nearest whole number to get the empirical formula:

[tex]P^{1}S^3.[/tex]

Phosphorus sulfide is a chemical compound with the formula P4S3. It is a yellowish-white solid with a garlic-like odor, and is insoluble in water but soluble in carbon disulfide.

It is formed by the reaction of phosphorus with sulfur, and can also be produced by heating a mixture of red phosphorus and sulfur.

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The energy required to increase 1.000 kg of water ice between 0 °C °C through 79.8 °C °C is equivalent to the energy required to melt one pound of ice (334 kJ).

No surface temp change happens from heat exchange if ice caps melt and has become water ice (i.e., that during phase transition). Imagine about water melting from stalactites that were already dripping on a sun-warmed roof. Alternatively, water boils in an ice bucket cooled by relatively low surroundings.

To melt, the iceberg will take in all the heat that is available. The water stays at zero degrees Fahrenheit until all of the ice has dissolved as when the ice has melted, it changes becomes saltwater at that temperature.

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Answer:

Answer :

The compounds are separated by using a suitable filtration technique.

Explanation:

Explanation:

The compounds are separated by using a suitable filtration technique. NaCl remains in the filtrate, but charcoal remains on the filter paper. Crystals of NaCl can be obtained by the method of evaporation.

Answer:

the formula for sulfide ion is S²⁻, and the formula for calcium ion is Ca²⁺. When combined, they form calcium sulfide with the chemical formula CaS.

Explanation:

Based on the appearance of the reaction of aluminum with copper (II) sulfate, the reagent that was consumed was aluminum, and the reagent that had some left over was copper (II) sulfate.

Aluminum is a chemical element that has the symbol Al and the atomic number 13. It is a silvery-white, soft, nonmagnetic, ductile metal in the boron group. Aluminum is the third most prevalent element and the most abundant metal in the Earth's crust after oxygen and silicon.

Copper (II) sulfate is a blue solid that is soluble in water. Its appearance is due to the presence of water of crystallization, which occurs in the crystal structure. It is a compound that is commonly used as a fungicide and algaecide because it is toxic to many fungi and algae. Copper sulfate has also been used to treat various diseases.

Aluminum replaces the copper ions in copper (II) sulfate and creates aluminum sulfate and copper metal when aluminum reacts with copper (II) sulfate. The chemical reaction between aluminum and copper (II) sulfate is as follows:

Al(s) + CuSO4(aq) ⟶ Al2(SO4)3(aq) + Cu(s)

When aluminum is placed in copper (II) sulfate solution, copper ions from copper (II) sulfate move to aluminum, displacing the aluminum ions in the process. As a result, the copper ions from copper (II) sulfate solution are lowered to metallic copper, and aluminum ions combine with sulfate ions from copper (II) sulfate solution to form aluminum sulfate, which is soluble in water. When copper metal is produced, it forms a brown layer on top of the solution and sinks to the bottom.

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This means that for every liter of water, there is 6.15 moles of the substance dissolved in it.

Substance is a concept that refers to a physical material or thing that has mass and occupies space. It is a fundamental concept of physics that applies to all physical and visible things in the universe. In philosophy, substance is a primary category of ontology that refers to the physical or material existence of things.

The molarity of the saturated solution can be calculated by dividing the solubility (360 g/L) by the molar mass of the substance (58.5 g/mol).
The molarity of the saturated solution is thus 6.15 mol/L.
This means that for every liter of water, there is 6.15 moles of the substance dissolved in it.

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The percentage by mass of water in the hydrate [tex]CuSO^{4}.5H^{2}O[/tex] is 47.6%.

The percentage composition of each of the following compounds are

1.a. NaClO:

The molar mass of NaClO =

22.99 g/mol + 35.45 g/mol + 15.99 g/mol = 74.43 g/mol

Percentage composition of Na =

(22.99 g/mol / 74.43 g/mol) *  100 \ percent \ = 30.9percent

Percentage composition of Cl =

(35.45 g/mol / 74.43 g/mol) *  100percent \ =  47.6percent

Percentage composition of O =

(15.99 g/mol / 74.43 g/mol) * 100percent \ =  21.5percent

Therefore, the percentage composition of NaClO is approximately:

30.9% Na, 47.6% Cl, and 21.5% O.

1.b. [tex]Al^{2}(SO^{3})^{3}[/tex]:

The molar mass of [tex]Al^{2}(SO^{3})^{3}[/tex] = [tex]2 * (26.98 g/mol) + 3 * (32.06 g/mol + 3 * 16.00 g/mol) = 342.14 g/mol[/tex]

Percentage composition of Al =

[tex](2 * 26.98 g/mol / 342.14 g/mol) * 100 \ = 3.99[/tex]%

Percentage composition of S =

[tex](3 * 32.06 g/mol / 342.14 g/mol) * 100 \ = 8.99[/tex]%

Percentage composition of O =

[tex](9 * 16.00 g/mol / 342.14 g/mol) * 100 \ = 10.47[/tex]%

Therefore, the percentage composition of [tex]Al^{2}(SO^{3})^{3}[/tex] is approximately:

3.99% Al, 8.99% S, and 10.47% O.

1.c. [tex]C^{2}H^{5}COOH[/tex]:

The molar mass of [tex]C^{2}H^{5}COOH[/tex] = [tex]2 * (12.01 g/mol) + 6 * (1.01 g/mol) + 12.01 g/mol + 16.00 g/mol = 60.05 g/mol[/tex]

Percentage composition of C =

[tex](2 * 12.01 g/mol / 60.05 g/mol) * 100 \ = 40.0[/tex]%

Percentage composition of H =

[tex](6 * 1.01 g/mol / 60.05 g/mol) * 100\ \ = 10.1[/tex]%

Percentage composition of O = [tex](2 * 16.00 g/mol + 12.01 g/mol / 60.05 g/mol) * 100 \ = 49.9[/tex]%

Therefore, the percentage composition of [tex]C^{2}H^{5}COOH[/tex] is approximately:

40.0% C, 10.1% H, and 49.9% O.

1.d. [tex]BeCl^2[/tex]:
The molar mass of [tex]BeCl^2[/tex] = [tex]9.01 g/mol + 2 * 35.45 g/mol = 79.91 g/mol[/tex]

Percentage composition of Be = [tex](9.01 g/mol / 79.91 g/mol) * 100 \ = 11.3[/tex]%

Percentage composition of Cl =

[tex](2 * 35.45 g/mol / 79.91 g/mol)* 100 \ = 88.7[/tex]%

Therefore, the percentage composition of [tex]BeCl^2[/tex] is approximately:

11.3% Be and 88.7% Cl.

2. [tex]CuSO^{4}.5H^{2}O[/tex]:

Mass of CuSO4.5H2O = 8.40 g

Mass of anhydrous [tex]CuSO^{4}[/tex] = mass of [tex]CuSO^{4}.5H^{2}O[/tex] - mass of water

Mass of water = mass of [tex]CuSO^{4}.5H^{2}O[/tex] - mass of anhydrous CuSO4

[tex]= 8.40 g - 4.40 g= 4.00 g[/tex]

So, the mass of water in the hydrate is 4.00 g.

To determine the percentage by mass of water in the hydrate, we need to divide the mass of water by the mass of the entire compound ([tex]CuSO^{4}.5H^{2}O[/tex]) and multiply by 100%:

Percentage by mass of water = (mass of water/mass of [tex]CuSO^{4}.5H^{2}O[/tex]) x 100%

[tex]= (4.00 g/8.40 g) * 100= 47.6[/tex]%

Therefore, the percentage by mass of water in the hydrate [tex]CuSO^{4}.5H^{2}O[/tex] is 47.6%.

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Answer:

Explanation:

It is a binary molecular compound ( and when the first element is to the right of the zigzag line on the periodic table you will use prefixes (di)