A 0.5998 g sample of a new compound has been analyzed and found to contain the following masses of elements: carbon, 0.1565 g; hydrogen, 0.02627 g; oxygen, 0.4170 g. Calculate the empirical formula of the compound.

The filtration technique can be used by Lucan to separate the mixture.

Filtering is a physical separation procedure that uses a filter media with a complicated structure that only allows fluid to flow through it to separate solid particles and liquid from a combination. Filtrate is the term for the fluid that passes through large solid particles that cannot pass through the filter medium.

The procedure for utilizing filtration to separate the mixture is as follows:

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The reaction between solid sodium and iron III oxide is:

2 Na(s) + Fe2O3(s) → 2 NaFeO2(s) + 1/2 O2(g).

Sodium is a chemical element with the symbol Na and atomic number 11. It is a soft, silvery-white metal that belongs to the alkali metal group in the periodic table. Sodium is highly reactive, particularly in the presence of water, and is never found free in nature. It was first isolated by Sir Humphry Davy in 1807 using electrolysis. Sodium is an essential element for all living organisms, and it plays a crucial role in various physiological processes, including fluid balance, nerve impulse transmission, and muscle function. Sodium is also widely used in the production of many industrial chemicals, including sodium hydroxide (caustic soda), sodium carbonate (washing soda), and sodium bicarbonate (baking soda). Sodium compounds are also used in the manufacturing of soaps, detergents, paper, and textiles. However, excessive sodium intake can lead to health problems such as high blood pressure and cardiovascular diseases.

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H2]eq = 1.1 × 10^−4 M.

The given equilibrium reaction is:

2H₂S(g) ⇌ 2H₂(g) + S₂(g)

The equilibrium constant, Kc, for the reaction at 700 °C is 9.0 × 10^−8.

We are given the initial number of moles of H2S as 0.700 moles in a 0.500 L flask. Let's assume that x moles of H2S is consumed at equilibrium, and therefore the equilibrium concentrations of H2 and S2 are both 2x, based on the stoichiometry of the reaction.

The expression for the equilibrium constant, Kc, is:

Kc = [H2]^2[S2]/[H2S]^2

At equilibrium, the concentration of H₂S will be (0.700 - x)/0.500 M, and the concentration of H₂ and S₂ will be 2x/0.500 M. Substituting these values in the expression for Kc and solving for x gives:

9.0 × 10^−8 = (2x/0.500 M)^2 / ((0.700 - x)/0.500 M)^2

Solving for x gives x = 5.6 × 10^−5 M.

Therefore, the equilibrium concentrations of H₂ and S₂ are both 2x = 1.1 × 10^−4 M, and the equilibrium concentration of H₂S is (0.700 - x)/0.500 M = 0.699 M.

Thus, the concentration of H2 at equilibrium is 1.1 × 10^−4 M, rounded to the correct number of significant figures.

Therefore, [H2]eq = 1.1 × 10^−4 M.

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Explanation:

V(CO) = 50 cm^3 = 50 ml = 0.05 l

V(air) = 150 cm^3 = 150 ml = 0.15 l

V(O2)/V(air) = 20%

_____________________________

2 CO + O2 = 2CO2

n(CO) = V(CO)/V(M) = 0.05/22.4 = 0.0022mol

V(O2)/V(air) = x/150 x=150×0.2 =30 ml=0.03l

n(O2) = V(O2)/V(M) =0.03/22.4 = 0.0013 mol

From the reaction ratio n(CO): n(O2) = 2:1

For explosing 0.0022 mol CO need 0.0011 mol O2. We have 0.0013 mol O2, then , 0.0003 mol O2 is excess.

Answer: O2 was in excess

It is not consumed and speeds up a reaction.

What is Catalyst?

A catalyst is a substance that increases the rate of a chemical reaction without being consumed in the process. It works by lowering the activation energy required for the reaction to occur, which allows the reaction to proceed more quickly and with less energy input. Catalysts can be used in a wide range of industrial processes, from the production of chemicals and fuels to the manufacture of pharmaceuticals and food products.

A catalyst is a substance that speeds up a chemical reaction without being consumed in the process. Catalysts work by lowering the activation energy required for a reaction to occur, which allows the reaction to proceed at a faster rate.

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The pH of the solution is 3.53 when potassium fluoride is added to water at a temperature of 298 k.

How to determine pH ?

Fluoride ion (F-) and then convert it to the dissociation constant (Ka) of its conjugate acid (HF) using the relationship:

Kw = Ka x Kb

where Kw is the ion product constant for water (1.0 x 10⁻¹⁴ at 298 K).

The Kb of fluoride ion is 1.5 x 10⁻¹¹ at 298 K. Therefore, the Ka of HF is:

Ka = Kw / Kb = (1.0 x 10⁻¹⁴) / (1.5 x 10⁻¹¹) = 6.67 x 10⁻⁴

The dissociation of HF in water is:

HF + H2O ⇌ H3O+ + F-

The equilibrium constant expression for this reaction is:

Ka = [H3O+][F-] / [HF]

At equilibrium, the concentration of HF will be equal to the initial concentration of KF, since KF is a salt of HF. Let x be the concentration of F- and [HF] = 0.251 M.

Then,

Ka = [H3O+][F-] / [HF]

6.67 x 10⁻⁴ = x²  / 0.251

x = 1.13 x 10⁻³ M

The concentration of H3O+ can be calculated using the equilibrium constant expression:

Ka = [H3O+][F-] / [HF]

6.67 x 10⁻⁴ = [H3O+](1.13 x 10⁻³) / 0.251

[H3O+] = 2.95 x 10⁻⁴ M

Therefore, the pH of the solution is:

pH = -log[H3O+]

pH = -log(2.95 x 10⁻⁴)

pH = 3.53

Thus, the pH of the solution is 3.53.

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Propan-1-ol and ethanoate are esterified Ethanoic acid and propan-1-ol combine to generate an ester. Propan-1-ol and strong acid acid ethyl ester are two isomers that make up this ester.

Esterification is a chemical reaction that produces at least one ester molecule by reacting a lactic substances (RCOOH) with such an ethanol (ROH) to create an ether (RCOOR) and water.

Up to around 1 moldm3, redox reactions are linked to [H+] concentration. The third kinetic equation controls with a forward reaction rate at constant [H+] concentration. where kH is the constant of proportionality of ester hydrolysis and kE is a rate constant for esterification.

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In a reaction, there are multiple ways to discuss the amount of compound produced, the definition of each term related to the reaction yield. The calculated amount of product under ideal conditions is theoretical yield, the amount of product experimentally produced  is actual yield, the comparison of the experimental amount of product to the calculated amount is the percent yield

In a reaction, there are multiple ways to discuss the amount of compound produced, the definition of each term related to the reaction yield is are 1. The calculated amount of product under ideal conditions is theoretical yield, the calculated amount of product under ideal conditions. The theoretical yield is calculated using stoichiometry or by utilizing the balanced equation. 2. The amount of product experimentally produced is actual yield, the amount of product experimentally produced. Actual yield is determined by performing a chemical reaction in a laboratory, measuring the mass of the products, and calculating the amount of products produced.

3. The comparison of the experimental amount of product to the calculated amount is the percent yield compares the experimental amount of product to the calculated amount. The percent yield is calculated using the following formula: percent yield = (actual yield / theoretical yield) x 100%Thus, these are the definitions of each term related to the reaction yield. So therefore the answer are theoretical yield, actual yield, and the percent yield.

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The ultimate BOD (BODu) is the BOD remaining after a long period of incubation (usually 20 days). The BOD of a waste can be modelled using the following equation:

BODt = BODu ( 1 - e^(-kt) )

where BODt is the BOD at time t, BODu is the ultimate BOD, k is the rate constant, and e is the base of the natural logarithm.

We can use the given k values and BOD5 to solve for the ultimate BOD for each waste.

For waste 1 (k = 0.0800 d^-1):

BOD5 = BODu ( 1 - e^(-k5) )

280.0 ml/L = BODu ( 1 - e^(-0.08005) )

280.0 ml/L = BODu ( 1 - e^(-0.4) )

280.0 ml/L = BODu ( 0.3297 )

BODu = 851.4 ml/L

Therefore, the ultimate BOD for waste 1 is 851.4 ml/L.

For waste 2 (k = 0.1200 d^-1):

BOD5 = BODu ( 1 - e^(-k5) )

280.0 ml/L = BODu ( 1 - e^(-0.12005) )

280.0 ml/L = BODu ( 1 - e^(-0.6) )

280.0 ml/L = BODu ( 0.4457 )

BODu = 628.4 ml/L

Therefore, the ultimate BOD for waste 2 is 628.4 ml/L.

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Answer:

oxygen should be present for burning to happen,

3906.114 grams of CaCO₃ is required to completely react with 1250 grams of SO₂.

Scrubbing mechanism is a method of removing pollutants, such as sulfur dioxide (SO₂), from industrial exhaust gases. In this mechanism, a substance, such as limestone or lime, is added to the exhaust gases, which react with the pollutants to form solid waste products that can be easily removed.

The balanced chemical equation for the reaction of SO₂ gas and CaCO₃ is:

SO₂ + CaCO₃ → CaSO₃ + CO₂

From the equation, we can see that one mole of SO₂ reacts with one mole of CaCO₃. Therefore, we need to first determine the number of moles of SO₂ in 1250 g of SO₂:

molar mass of SO₂ = 32.066 g/mol

moles of SO₂ = mass of SO₂ / molar mass of SO₂

moles of SO₂ = 1250 g / 32.066 g/mol

moles of SO₂ = 39.012 mol

Since one mole of SO₂ reacts with one mole of CaCO₃, we need 39.012 moles of CaCO₃ to react with the 39.012 moles of SO₂. The molar mass of CaCO₃ is 100.086 g/mol, so we can calculate the mass of CaCO₃ needed as:

mass of CaCO₃ = moles of CaCO3₃ × molar mass of CaCO₃

mass of CaCO₃ = 39.012 mol × 100.086 g/mol

mass of CaCO₃ = 3906.114 g

Therefore, we need 3906.114 grams of CaCO₃ to completely react with 1250 grams of SO₂.

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0.56 kPa is the final pressure.

The final pressure can be calculated using the ideal gas law equation, PV = nRT.

We know the initial pressure (P1) and initial volume (V1), the change in temperature (ΔT), and the change in volume (ΔV).

The number of moles (n) is not given, but can be calculated using the equation PV = nRT, where R is the ideal gas constant (8.314 J/molK).

[tex]n = \frac{P_1V_1 }{ RT_1}\\\\n=\frac{ (0.76 kPa)(V_1) }{ (8.314 J/molK)(19.0 + 273.15)}\\\\n = 0.039 moles[/tex]

Therefore, the final pressure (P2) can be calculated using the equation PV = nRT, where T2 is the final temperature in Kelvin and V2 is final volume.

[tex]P_2 =\frac{ nRT_2 }{ V_2}[/tex]

[tex]P_2 =\frac{ (0.039 moles)(8.314 J/molK)(-14.0 + 273.15) }{ (V_1 + \Delta V)}\\\\P_2 = 0.56 kPa[/tex]

Therefore, the final pressure is 0.56 kPa.

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An example of a secondary pollutant is option (c)- photochemical oxidants.

Secondary pollutants are formed when primary pollutants react with each other or with other substances in the atmosphere. Photochemical oxidants, such as ozone, are created through reactions involving sunlight and primary pollutants like VOCs (volatile organic compounds) and nitrogen oxides.

Hence, an example of a secondary pollutant is option (c)- photochemical oxidants.

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HF, or hydrogen fluoride, is likely to dissolve in water (H2O), a polar solvent.

This is due to the principle "like dissolves like," which means that polar solvents tend to dissolve polar solutes, and nonpolar solvents tend to dissolve nonpolar solutes.  HF is a polar molecule because of the difference in electronegativity between hydrogen (H) and fluorine (F) atoms, resulting in a polar covalent bond. This bond creates a dipole moment, causing the HF molecule to have a partial positive charge on the hydrogen end and a partial negative charge on the fluorine end.

Water, as a polar solvent, has a similar dipole moment due to the difference in electronegativity between oxygen (O) and hydrogen (H) atoms. The partial positive charges on the hydrogen atoms in water molecules can interact with the partial negative charges on the fluorine atoms in HF molecules, forming hydrogen bonds. These interactions between polar molecules make it possible for HF to dissolve in water.


In conclusion, HF is likely to dissolve in water (H2O) because both are polar substances that can form hydrogen bonds, following the "like dissolves like" principle.

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A brief explanation of bilateral, radial, and asymmetric can be found below.

Bilateral, radial, and asymmetric are terms used in zoology to describe the symmetry of an animal's body.

Bilateral symmetry refers to an animal's body being divided into two equal and opposite halves along a central axis. This type of symmetry is common in animals with a distinct head and tail end, such as humans.

Radial symmetry refers to an animal's body being arranged in a circular or radial pattern around a central axis. This type of symmetry is common in animals such as jellyfish, where the body parts are arranged around a central mouth.

Asymmetric refers to an animal that lacks any kind of symmetry. This type of body plan is uncommon in animals, but is seen in some types of sponges and other primitive organisms.

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Ammonium hydroxide is a weak base because c. it dissociates only slightly in water.

When dissolved in water, ammonium hydroxide donates a proton to form the ammonium ion and hydroxide ion. However, this reaction is only partially reversible because the ammonium ion is not a strong enough acid to accept a proton and regenerate the ammonium hydroxide. Therefore, the hydroxide ion concentration in the solution is relatively low, making it a weak base. In chemistry, a substance is referred to as a base if it has the ability to accept protons (H+ ions).

Ammonium hydroxide is a weak base because, when dissolved in water, it donates a proton to form the ammonium ion (NH4+) and hydroxide ion (OH−). This reaction is reversible, as represented by the chemical equation below: NH4+ (aq) + OH− (aq) ⇌ NH3 (aq) + H2O (l)However, this reaction is only partially reversible because the ammonium ion is not a strong enough acid to accept a proton and regenerate the ammonium hydroxide. Therefore, the hydroxide ion concentration in the solution is relatively low, making it a weak base.

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The pH of the resulting solution is 7.00.

When answering questions on Brainly, it is important to always be factually accurate, professional, and friendly. Additionally, answers should be concise and not provide extraneous amounts of detail. Any typos or irrelevant parts of the question should be ignored.

In this case, the student question is:If 50.0 mL of 0.60 M HCl is added to 25.0 mL of 0.60 M NaOH, what will be the pH of the resulting solution? Your answer should be rounded to two decimal places.

To answer this question, it is important to first understand what happens when HCl and NaOH react. HCl is an acid, and NaOH is a base. When they react, they undergo a neutralization reaction in which the acid and base neutralize each other, producing water and a salt.

In this case, the salt produced will be sodium chloride (NaCl).The equation for this neutralization reaction is:

HCl + NaOH → NaCl + H2OWhen HCl and NaOH react, the H+ ions from HCl combine with the OH- ions from NaOH to produce water. The remaining ions (Na+ and Cl-) combine to form the salt NaCl. Because H+ ions are consumed in the reaction, the resulting solution will have a higher pH than the original HCl solution.

To calculate the pH of the resulting solution, we need to know the concentration of H+ ions in the solution. We can calculate this by using the equation for the ionization of water:

Kw = [H+][OH-]where Kw is the ion product constant for water (1.0 × 10^-14 at 25°C).At 25°C, the concentration of OH- ions in pure water is 1.0 × 10^-7 M. Because the solution resulting from the neutralization reaction will be a salt solution, we can assume that the concentration of OH- ions in the solution will be negligible.

Therefore, we can simplify the equation to:

[H+] = Kw/[OH-]

= 1.0 × 10^-14/1.0 × 10^-7

= 1.0 × 10^-7 MWe can use the concentration of H+ ions to calculate the pH of the solution:

pH = -log[H+]

= -log(1.0 × 10^-7)

= 7.00 (rounded to two decimal places)

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Answer:combustion

Explanation:combustion, a chemical reaction between substances, usually including oxygen and usually accompanied by the generation of heat and light in the form of flame.

Answer:

The scientific term used for the chemical reaction when a fuel burns is combustion.

The density of a ocean water rises as the temperature falls. As a result, it water gets colder the denser it is. As salinity grows, seawater's density rises as well.

As the salinity rises, the density of the water rises. At all temperatures just above freezing point, seawater with a salinity more than 24.7 becomes denser as the temperature drops. Pressure increases cause seawater's density to rise.

Fresh water is introduced into the sea as a result of rising surface temperatures, melting ice, and increased precipitation, which reduces salinity. Seawater with a lower salinity has a lower density so won't sink as far as water with a higher density. Ocean currents are modified by this process.

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We need to add 10.90 g of NaOH to the 250 ml sample of 1.09 molarity acetic acid to make the best buffer.

To make the best buffer from a 250 ml sample of 1.09 molarity acetic acid, we need to find out how much NaOH must be added. The Henderson-Hasselbalch equation, which relates the pH of a buffer to its pKa and the ratio of its weak acid to weak base forms.pH = pKa + log([A-]/[HA])In this case, we are dealing with an acetic acid buffer, which has a pKa of 4.76.

We are given a 250 ml sample of 1.09 molarity acetic acid. The molar mass of acetic acid is 60.05 g/mol. Therefore, the number of moles of acetic acid in the sample is:n(acetic acid) = M(acetic acid) x V(acetic acid)n(acetic acid) = 1.09 mol/L x 0.250 Ln(acetic acid) = 0.2725 molNext, we need to find out the amount of NaOH we need to add to make the buffer.

This is a bit tricky because we are given the volume of the acetic acid solution, but not the volume of the final buffer. We also don't know the concentration of the NaOH solution. However, we can use the fact that the buffer is prepared by mixing a weak acid with its conjugate base. Therefore, we can assume that the final buffer will have approximately equal concentrations of acetic acid and acetate ions, which means we need to add an equal amount of NaOH to convert half of the acetic acid to acetate ions.

The balanced chemical equation for the neutralization reaction is:CH3COOH + NaOH → CH3COONa + H2OThe stoichiometry of the reaction is 1:1, which means we need to add an equal number of moles of NaOH to neutralize the acetic acid.n(NaOH) = n(acetic acid) = 0.2725 molNow we need to find out how much NaOH that is in grams.

We can use the molar mass of NaOH to convert from moles to grams.m(NaOH) = n(NaOH) x M(NaOH)m(NaOH) = 0.2725 mol x 40.00 g/molm(NaOH) = 10.90 gTherefore, we need to add 10.90 g of NaOH to the 250 ml sample of 1.09 molarity acetic acid to make the best buffer.

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Answer:

The balanced chemical equation for the formation of ammonia gas by the reaction between nitrogen gas an hydrogen gas is given. N2+3H→2NH3.

Explanation:

I hope it's correct

Answer:

The balanced equation for the reaction between nitrogen gas and hydrogen gas to produce ammonia gas is: N2(g) + 3H2(g) -> 2NH3(g) This equation shows that one molecule of nitrogen gas reacts with three molecules of hydrogen gas to produce two molecules of ammonia gas. The coefficients in the balanced equation indicate that the number of atoms of each element is conserved in the reaction.

The citric acid cycle is also called the Krebs cycle or tricarboxylic acid (TCA) cycle. In this cycle, a series of chemical reactions that take place in living organisms' cells generate energy through oxidation of acetate derivatives.

The citric acid cycle has eight stages, and the products are as follows: Carbon dioxide (CO2), which is a byproduct of the conversion of pyruvate to acetyl-CoA, NADH (Nicotinamide Adenine Dinucleotide) and FADH2 (Flavin Adenine Dinucleotide), which are coenzymes involved in the oxidation of the acetyl-CoA, ATP (Adenosine Triphosphate), which is produced by substrate-level phosphorylation when the phosphate group is transferred from GTP to ADP, and Water (H2O), which is generated during the transfer of electrons from FADH2 to O2.

The citric acid cycle is an energy-producing cycle in cells that operates through oxidation of acetate derivatives. In this cycle, a series of chemical reactions that take place in living organisms' cells generate energy. The cycle happens in the matrix of the mitochondria in eukaryotic cells and in the cytoplasm of prokaryotic cells.

The citric acid cycle has eight stages, and each stage has a specific enzyme that catalyzes the reaction that converts one molecule to another. The cycle is a central metabolic pathway of aerobic organisms, and it is involved in various metabolic pathways.

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Clearing forests.

According to the text, humans impact the population growth of other ecosystems by clearing forests.

This is because forests are important habitats for many different species, and when they are destroyed or degraded, those species are forced to relocate or die off.

Additionally, forests are also important carbon sinks, which means they help to absorb and store carbon from the atmosphere. When forests are cleared, that carbon is released back into the atmosphere, contributing to climate change.

Therefore, forest clearing by humans can have significant impacts on the population growth of other ecosystems

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When sodium phosphate is added to a solution containing 0.0076 M aluminium nitrate and 0.047 M calcium chloride, the concentration of the second ion begins to precipitate is 6.87 x 10⁻⁸ M.

The concentration of this ion when the second ion begins to precipitate is determined by comparing their solubility products (Ksp) and using the ion product (IP) concept.

First, determine which ion will precipitate first by comparing the ion product (IP) to the solubility product (Ksp) for each compound. The ion with the higher IP/Ksp ratio will precipitate first.

IP(AlPO₄) / Ksp(AlPO₄) = [Al₃+][PO4³⁻] / (9.84 x 10⁻²¹)

IP(Ca₃(PO₄)₂) / Ksp(Ca₃(PO₄)₂) = [Ca²⁺]³[PO₄³⁻]² / (2.07 x 10⁻³³)

Since the IP/Ksp ratio for aluminium phosphate is larger than that for calcium phosphate, Al³⁺ will precipitate first as AlPO₄.

Next, find the concentration of Al³⁺ when Ca²⁺ begins to precipitate. At this point, the IP for Ca₃(PO₄)₂ will equal its Ksp.

Ksp(Ca₃(PO₄)₂) = = [Ca²⁺]³[PO₄³⁻]²
2.07 x 10⁻³³ = (0.047)³[PO₄³⁻]²

Solve for PO₄³⁻, then use the IP equation for AlPO₄:

9.84 x 10⁻²¹ = [Al³⁺][PO₄³⁻]

Finally, solve for the concentration of Al³⁺ when Ca²⁺ begins to precipitate. The concentration of Al³⁺ when the second ion (Ca²⁺) begins to precipitate is approximately 6.87 x 10⁻⁸ M.

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B. N2 is the limiting reactant. The reactant that has the smaller number of moles in the balanced equation is the limiting reagent.

What is Limiting Reagent?

The limiting reagent is the reactant that is entirely consumed when the reaction goes to completion, and thus limits the amount of product that can be formed. The other reactants that are present in excess are not entirely consumed and do not limit the amount of product that can be formed.

To determine the limiting reactant, we need to calculate the amount of product that can be produced from each reactant and then choose the reactant that produces the smallest amount of product.

The balanced equation for the reaction is:

N2 + 3H2 → 2NH3

The molar mass of N2 is 28 g/mol, and the molar mass of H2 is 2 g/mol. Using these values, we can calculate the number of moles of each reactant:

Number of moles of N2 = 3.5 g / 28 g/mol = 0.125 mol

Number of moles of H2 = 3.5 g / 2 g/mol = 1.75 mol

According to the balanced equation, 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3. So, the number of moles of NH3 that can be produced from each reactant is:

Moles of NH3 produced from N2 = 0.125 mol × (2 mol NH3 / 1 mol N2) = 0.25 mol NH3

Moles of NH3 produced from H2 = 1.75 mol × (2 mol NH3 / 3 mol H2) = 1.17 mol NH3

Therefore, the limiting reactant is N2, since it produces a smaller amount of NH3 compared to H2.

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A piece of magnesium with 8.63 mol of magnesium will react with 17.26 mol of HCl to form 820.9 g of magnesium chloride, given the reaction Mg + 2HCl → MgCl2 + H2.

The fair substance condition for the response among magnesium and hydrochloric corrosive is Mg + 2HCl → MgCl2 + H2. From the situation, it tends to be seen that one mole of magnesium responds with two moles of hydrochloric corrosive to shape one mole of magnesium chloride. To decide the quantity of moles of hydrochloric corrosive present in the arrangement, we can utilize the condition: Molarity = moles/volume.

Thusly, moles of HCl = Molarity × volume = 2.88 mol/L × 5.98 L = 17.26 mol. Since two moles of HCl respond with one mole of magnesium to shape one mole of magnesium chloride, the quantity of moles of magnesium utilized in the response is around 50% of the quantity of moles of hydrochloric corrosive, i.e., 17.26/2 = 8.63 mol. At long last, we can utilize the molar mass of magnesium chloride (95.21 g/mol) to work out the mass of magnesium chloride shaped: Mass of MgCl2 = 8.63 mol × 95.21 g/mol = 820.9 g. Accordingly, 820.9 g of magnesium chloride will be shaped.

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When heating under reflux, you will need the following glassware equipment to set up the apparatus: Round bottom flask, condenser, hot plate, thermometer, adapter, clamps and stands, water supply.

When heating under reflux, you will need the following glassware equipment to set up the apparatus:

1. Round-bottom flask: This is the flask that contains the reaction mixture that needs to be heated.

2. Condenser: This is a glass tube that is attached to the round-bottom flask and has a water jacket around it. The water is circulated through the jacket to cool down the hot vapours and condense them back into liquid form.

3. Heating mantle or hot plate: This is the equipment that provides heat to the round-bottom flask.

4. Thermometer: This is an instrument used to measure the temperature of the reaction mixture.

5. Adapter: This is a glassware piece that connects the condenser to the round-bottom flask.

6. Clamps and stands: These are used to hold the glassware in place during the heating process.

7. Water supply: This is required to run water through the condenser's jacket to cool down the vapours.

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we can use the ideal gas law: PV = nRT where P is the pressure of the gas in pounds per square inch (psi), V is the volume of the gas in cubic inches (in^3), n is the number of moles of gas,

R is the gas constant (which is 10.73 psi·in^3/mol·K for argon), and T is the temperature of the gas in Kelvin (K).

Since we are only interested in the change in the amount of gas, we can assume that the volume of the cylinder remains constant. Therefore, we can write:

n1 = (P1 V) / (R T1)

where n1 is the initial number of moles of gas, P1 is the initial pressure of the gas, V is the volume of the gas (which we assume remains constant), R is the gas constant for argon, and T1 is the initial temperature of the gas.

To find the final number of moles of gas, we can use the same equation with the final pressure and temperature:

n2 = (P2 V) / (R T2)

where n2 is the final number of moles of gas, P2 is the final pressure of the gas, and T2 is the final temperature of the gas.

To find the change in the amount of gas, we can subtract n2 from n1:

Δn = n2 - n1 = (P2 V) / (R T2) - (P1 V) / (R T1)

Substituting the given values, we get:

Δn = [(14 psi) × V] / [(10.73 psi·in^3/mol·K) × (273 K)] - [(10 psi) × V] / [(10.73 psi·in^3/mol·K) × (300 K)]

Simplifying this expression, we get:

Δn = (4 psi × V) / (3483.9 psi·in^3/mol)

where we have converted the temperatures to Kelvin.

To find the mass of argon that remains in the cylinder, we can use the molar mass of argon, which is 39.95 g/mol. The mass of argon in the cylinder is:

m = n1 × 39.95 g/mol

Substituting n1 from the first equation, we get:

m = [(P1 V) / (R T1)] × 39.95 g/mol

To find the final mass of argon, we can use the same equation with the final number of moles:

m' = [(P2 V) / (R T2)] × 39.95 g/mol

To find the change in mass, we can subtract m' from m:

Δm = m' - m = [(P2 V) / (R T2)] × 39.95 g/mol - [(P1 V) / (R T1)] × 39.95 g/mol

Substituting the given values and simplifying, we get:

Δm = (4 psi × V) / 87.67 g

Therefore, the change in mass of argon in the cylinder is:

Δm = (4 psi × V) / 87.67 g

Note that the result depends on the volume of the cylinder, which is not given in the problem.

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D) is reaction 1 exothermic under standard conditions?  "no, because the standard enthalpy change is negative. "

When a reaction is exothermic, it releases heat as it proceeds. In this case, the given question is asking if Reaction 1 is exothermic under standard conditions.

The answer is no, because the standard enthalpy change is negative.A negative standard enthalpy change means that the reaction is endothermic, not exothermic. Endothermic reactions absorb heat from the surroundings as they proceed. Therefore, heat is not released and added to the system,

which rules out option a). The percent yield of NH₃ being greatest at high pressure, as stated in option c), is not relevant to whether the reaction is exothermic or endothermic.

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The reaction of cyclohexanol with phosphoric acid catalyst typically results in the formation of cyclohexene and water as the two products.

This is a dehydration reaction in which a molecule of water is eliminated from cyclohexanol to form cyclohexene. The chemical equation for this reaction can be written as:

C6H11OH + H3PO4 → C6H10 + H2O

where C6H11OH is cyclohexanol, H3PO4 is phosphoric acid, C6H10 is cyclohexene, and H2O is water.

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