Aluminum reacts with excess hydrochloric acid to form aqueous aluminum chloride and 190.6 ml of hydrogen gas (collected over water at 26°C and 0.89 atm). (Vapor pressure of water at 26ºC = 25.2 mmHg.) How many grams of aluminum reacted? Enter to 4 decimal places.

Answer:

C2H4Cl2

Explanation:

Firstly, we know that the compound contains only three elements. These are carbon, hydrogen and oxygen. We have the percentage compositions of carbon and hydrogen, thus we need the one for chlorine. To get the one for chlorine, we simply subtract that of carbon and hydrogen from a total of 100%.

Hence percentage composition of chlorine = 100 - 24.27 - 4.07 = 71.66%

Now, we divide the percentage compositions by the atomic masses. The atomic masses of carbon, hydrogen and chlorine are 12, 35.5 and 1 respectively. We go on to the divisions as follows.

C = 24.27/12 = 2.0225

H = 4.07/1 = 4.07

Cl = 71.66/35.5 = 2.02

We then go on to divide each by the smallest which is 2.02

C = 2.0225/2.02 = 1

H = 4.07/2.02 = 2

Cl = 2.02/2.02 = 1

Hence the empirical formula is CH2Cl

Now, since the molecular mass is 98.95, we need to calculate the molecular formula

Hence, [CH2Cl]n = 98.95

[12 + 2(1) + 35.5]n = 98.95

[12 + 2 + 35.5]n = 98.95

49.5n = 98.95

n = 98.95/49.5 = 2

The molecular formula is thus [CH2Cl]2 = C2H4Cl2

a) ΔH of  Fermentation = - 2816 kJ/mol

ΔH of Respiration =  - 1409.2 kJ/mol

b)C₂H₅OH (l) + 3O₂ (g) → 3H₂O(g) + 2CO₂(g)

c) Combustion per mole of sugar

Let's solve for each part one  by one:

a)

C₆H₁₂O₆ (s) + 6O₂ (g) → 6O₂ (g) + 6H₂O (l)

ΔHrxn = ΔHformation (products) - ΔHformation (reactants)

[tex]= (6 mol * -393.5kJ/mol)+ (6mol * -285.8kJ/mol) - (1 mol * -1260kJ/mol + 6mol * 0)\\\\= -2 816 kJ/mol[/tex]

C₂H₅OH (l) + 3O₂ (g) → 3H₂O(g) + 2CO₂(g)

Let's assume that 1 mole of ethanol is burnt. The heat of reaction, ΔHrxn is given by this equation:

ΔHrxn = ΔHformation (products) - ΔHformation (reactants)

[tex]= (2 mol* - 393.5kJ/mol) + ( 3mol * -285.8kJ/mol) - [ (1mol * -235 kJ/mol) + ( 3mol * 0.0000kJ)]\\\\= -1644 - (-235.2)\\\\= - 1409.2 kJ/mol[/tex]

The negative sign indicates that the process of combustion is exothermic.

b) The combustion reaction for ethanol can be given as:

C₂H₅OH (l) + 3O₂ (g) → 3H₂O(g) + 2CO₂(g)

c) From the comparisons of the ΔHrxn, sugar produces more energy.

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Final answer:

The enthalpy change for fermentation and respiration (combustion) can't be calculated without additional data on standard enthalpies of formation. The combustion reaction for ethanol is balanced, and to compare the heat released from sugar or ethanol combustion, we would need their specific heats of combustion in kJ/mol.

Explanation:

The question asks us to calculate the change in enthalpy (ΔHorxn) for both fermentation and respiration (combustion) processes and write a balanced chemical equation for the combustion of ethanol. To answer these, we start by looking at each process.

Fermentation

The balanced chemical equation for the fermentation of glucose to ethanol (CH3CH2OH) and carbon dioxide (CO2) is given by:
C6H12O6 → 2 CH3CH2OH + 2 CO2
We cannot calculate ΔHorxn for fermentation without the standard enthalpy of formation for each species.

Respiration (Combustion)

The combustion of glucose can be represented as:
C6H12O6 + 6 O2 → 6 CO2 + 6 H2O
Again, to calculate ΔHorxn, we need the standard enthalpies of formation for each compound.

Combustion Reaction for Ethanol

The balanced combustion reaction of ethanol is:
C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(l)+ 29.7 kJ/g
This reaction shows that for each mole of ethanol burned, carbon dioxide and water are produced, and heat is released.

Comparison of Heat Released

To compare which releases more heat per mole of carbon, sugar or ethanol, we would need to know the specific heat of combustion for each substance in kJ/mol.

Answer: r=132pm

Explanation:

r = ⟨r⟩=5a0/Z

r=(5(5.29177*〖10〗^(-11) m) x^2)/2

r=132pm

Answer:

0.1976 atm is the total pressure in the container at equilibrium.

Explanation:

The value of [tex]K_p[/tex] for the given reaction = 0.842

Partial pressure of the hydrogen sulfide = 0.104 atm

Partial pressure of the hydrogen sulfide at equilibrium = (0.104-x) atm

Partial pressure of the hydrogen gas at equilibrium = x

Partial pressure of the sulfur gas at equilibrium = x

[tex]H_2S(g)\rightleftharpoons 2H_2(g)+S(g)[/tex]

Initially

0.104 atm                      0  0

At equilibrium

(0.104 -2x)                      2x   x

The expression of  [tex]K_p[/tex] is given by ;

[tex]K_p=\frac{x\times x}{(0.104 -x) }[/tex]

[tex]0.842 =\frac{x^2}{(0.104-x)}[/tex]

Solving for x:

x = 0.0936 atm

Partial pressure of the hydrogen sulfide at equilibrium = (0.104-x) atm

= (0.104-0.0936 ) atm = 0.0104 atm

Partial pressure of the hydrogen gas at equilibrium = x = 0.0936 atm

Partial pressure of the sulfur gas at equilibrium = x = 0.0936 atm

Total pressure in the container will be sum of all the partial pressure of the hydrogen sulfide gas, hydrogen gas and sulfur gas at the equilibrium.

P = 0.0104 atm + 0.0936 atm + 0.0936 atm = 0.1683 atm

0.1976 atm is the total pressure in the container at equilibrium.

The total pressure in the system at equilibrium is calculated using the equilibrium constant Kp and the initial pressure of the system. By setting up a equilibrium constant expression and solving the quadratic equation for 'x', the change in pressure, we find that the total pressure at equilibrium in the system is ~0.101 atm.

The decomposition reaction of H2S into H2 and S is a chemical equilibrium problem in which the equilibrium constant is given in terms of pressure, or Kp. We know that at the beginning, only H2S is present at a pressure of 0.104 atm, and at equilibrium, the pressure of H2S decreases by 'x', being 'x' the pressure of H2 and S at equilibrium. According to the expression of the equilibrium constant Kp, Kp = (pH2* pS) / p_H2S = x^2 / (0.104-x).

From here, if we solve for 'x' using the quadratic formula and the given Kp = 0.842, we find two solutions, ~0.0967 atm and ~0.0072 atm. We discard the result 0.0967 because it cannot be higher than the initial pressure of the system, so the correct 'x' is ~0.0072 atm. Therefore, the total pressure will be the sum of the individual pressures at equilibrium. Total Pressure = pH2 + pS + p_H2S = 2x + (0.104-x) = 2*0.0072 + (0.104-0.0072) = ~0.101 atm.

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Answer:

The mass of water is 18 g

Explanation:

The reaction of calcium with water can be represented in the equation below:

Ca + 2H₂O --------->Ca(OH)₂ +H₂

1 Mole of gas at STP = 22.4L

From the displacement reaction above, calculate the mass of water that will produce 22.4L of hydrogen gas at STP.

Mass of 2H₂O = 2(2x1 + 16) = 2X18 = 36 g/mol

Using proportional analysis;

36 g of 2H₂O produced 22.4 L of H₂, then

what mass of 2H₂O will produce 11.2L of  H₂ ?

Mathematically,

22.4 L ----------------------------------> 36g

11.2 L -----------------------------------> ?

Cross and multiply, to obtain the expression below

= (11.2 X 36)/22.4

= 18 g

Therefore, the mass of water is 18 g

Answer:

A suitable scale, say 1 cm: 100 km can be used.

Explanation:

Thinking process:

The best way to approach the question will be to consider the requirements. This is a simple case of scaling. In order to achieve the objective, you need to choose a scale that does not consume space and that presents more details at the same time.

For instance, a scale of 1 cm to 100 km will give me lines which are a little more then 5 cm. This can be presented as:

1: 500

This is appreciable for the paper size.

Answer:C

Explanation:

Chromatography separates compounds by taking advantage of their polarity. The stationary phase is generally very polar. The mobile phase can be pure hexane or various ratios of hexane with a polar eluent added. The more polar the compound, the more it interacts with the stationary phase and won’t move very far up the plate compared to the non-polar or less polar compounds that interact more with the non-polar hexane.

Final answer:

All methods of chromatography operate based on the principle that components of a mixture will distribute unequally between the mobile and stationary phase, due to differences in intermolecular forces and affinities.

Explanation:

The question asks about the basic principle on which all methods of chromatography operate. The correct answer is c. The components of the mixture will distribute unequally between the mobile and stationary phases. In chromatography, the separation of components is achieved because different components have different affinities for the stationary and mobile phases. A component with a higher affinity for the stationary phase will move more slowly through the chromatography system compared to a component with a higher affinity for the mobile phase. This differing affinity is often due to differences in the strength of intermolecular forces between the components and the phases. For example, in liquid chromatography, if a compound has strong intermolecular forces with the stationary phase (e.g., through hydrogen bonding or van der Waals forces), it will tend to be 'adsorbed' onto the stationary phase and thus move through the system more slowly compared to compounds that have weaker interactions and therefore travel with the mobile phase.

Answer:

a. in supernovae and star collisions

Explanation:

The periodical table contains some heavier elements, which are formed as neutron stars pairs hit eachother and erupt cataclysmically.

The star emitts very large quantities of energy and neutrons during supernova, which allow for the production of heavier elements than iron, such as uranium and gold. All these elements are ejected into space during the supernova explosion.

The gas laws involve Charles' Law, showing a direct proportionality between volume and temperature for a gas. The new volume of the gas after being cooled from 20°C to -60°C (converted to Kelvin), which is under constant pressure, is calculated to be approximately 30.6 mL.

This question involves the concept of the gas laws in Physics, specifically, Charles' Law which states that the volume of a given mass of an ideal gas is directly proportional to its temperature on an absolute scale if pressure and the amount of gas remain constant.

First, let's convert the Celsius temperatures to Kelvin by adding 273.15. This gives us initial temperature T1 = 20°C + 273.15 = 293.15 K and final temperature T2 = -60°C + 273.15 = 213.15 K. The initial volume V1 is 42 mL.

According to Charles' Law, V1/T1 = V2/T2. Substituting the given values, we find the new volume V2 = V1*(T2/T1) = 42 mL * (213.15 K / 293.15 K) = 30.6 mL.

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Answer: that one

Explanation:

Answer:

-1.9 KJ/mol

Explanation:

In order to solve the problem, we have to rearrange the equations in a way in which all molecules of O₂ and CO₂ are eliminated:

2C(diamond) + 2O₂(g) → 2CO₂(g)     ΔH₁= 2 x (-395.4 KJ) ------> we multiply by 2 both reactants and products

2 CO₂(g) → 2CO(g) + O₂(g)         ΔH₂= 566.0 KJ

CO₂(g) → C(graphite) + O₂(g)     ΔH₃= -1 x (-393.5 KJ) ------> we use reverse rxn

2CO(g) → C(graphite) + CO₂(g)   ΔH₄= -172.5 KJ

When we cancel the molecules that appear both in reactants and products, the total reaction is the following:

2C(diamond) → 2C(graphite)

ΔHt= ΔH₁ + ΔH₂ + ΔH₃ + ΔH₄ = 2 x (-395.4 KJ) + 566.0 KJ + (-1 x (-393.5 KJ)) - 172.5 KJ

ΔHt= 347.2 KJ

This is for 2 mol of C(diamond) which are converted in 2 mol of C(graphite). To obtain ΔH for the reaction of 1 mol C(diamond) to 1 mol (graphite) we have to divide into 2:

ΔH= -3.8 KJ/2mol= -1.9 KJ/mol

To determine ΔHrxn for the reaction C(diamond) → C(graphite), we can use Hess's law and the given equations. By canceling out common compounds/products and summing the remaining equations, we can calculate ΔHrxn. The enthalpy change can be determined by summing the enthalpy changes of the canceled equations.

This type of calculation usually involves the use of Hess's law, which states: If a process can be written as the sum of several stepwise processes, the enthalpy change of the total process equals the sum of the enthalpy changes of the various steps. For this specific reaction, we can use the given equations to determine the enthalpy change ΔHrxn for C(diamond) → C(graphite).

To obtain the ΔHrxn for C(diamond) → C(graphite), we need to cancel out common compounds/products in these equations. From the given equations, we can see that CO2(g) is a common compound in steps 1, 2, and 3, and thus we can cancel it out. Likewise, O2(g) is present in steps 1, 2, and 3, so we can also cancel it out. By summing the canceled equations, we get the final equation:

C(diamond) → C(graphite)

The enthalpy change for this reaction is the sum of the enthalpy changes of the canceled equations:

ΔHrxn = ΔH1 + ΔH2 + ΔH3 + ΔH4 = -395.4 kJ + 566.0 kJ + (-393.5 kJ) + (-172.5 kJ) = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + (-393.5 kJ) + 566.0 kJ + (-172.5 kJ) = -395.4 kJ - 393.5 kJ + 566.0 kJ - 172.5 kJ = -395.4 kJ - 393.5 kJ + 566.0 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ

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The density of the original piece of aluminum is 2.7 grams per cubic centimeter. If one piece is cut in half, the density of each piece would remain the same.

In order to find the density of an object, you divide the mass of the object by its volume. The mass of the aluminum piece is 270 grams and its volume is 100.0 cubic centimeters. So, the density would be 270 grams divided by 100.0 cubic centimeters, which equals 2.7 grams per cubic centimeter.

If the piece of aluminum is cut in half, the mass of each piece would be half of the original mass, so each piece would have a mass of 135 grams. Since the volume of each piece remains the same, 100.0 cubic centimeters, the density of one of the pieces would still be 2.7 grams per cubic centimeter.

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Answer:

Photosystems PSI (P700) and PSII (P680) and photolysis of water releases energy

In the form of movement of electrons which is later extracted as ATP

Explanation:

On absorbing light energy electrons obtained from photolysis of water and those present in and those present in the reaction centres PSI and PSII move to the higher energy level and electron acceptors accept them. They then move down through the electron transport chain to the lower energy level during which ATP is produced which is a chemical form of energy.

Final answer:

The water bath, filled to within 2.57 inches of the top, contains approximately 1014.48 liters of water. This calculation involves converting inches to meters, calculating the volume in cubic meters, and then converting that volume to liters.

Explanation:

To calculate the volume of water in the water bath, we need to take into account the dimensions given and the fact that it is filled to within 2.57 inches from the top. First, we need to convert the depth from which it's filled into meters, as the other dimensions are given in meters.

2.57 inches = 0.06533 meters (since 1 inch = 0.0254 meters)

The actual depth filled with water will be:

0.740 m (total depth) - 0.06533 m = 0.67467 m

Next, we multiply the length, the width, and the newly calculated depth to get the volume in cubic meters:

Volume = 1.85 m × 0.810 m × 0.67467 m = 1.01448 cubic meters

To convert the volume from cubic meters to liters, we use the conversion factor 1 cubic meter = 1000 liters:

Volume = 1.01448 m3 × 1000 L/m3 = 1014.48 liters

Therefore, there are approximately 1014.48 liters of water in the water bath.

Answer:

0

Explanation:

Δk = (kf - Ki). Kf is 0 because your stopping it [m(v_i)^2 ]/2q = V

The potential difference required to bring a proton to rest can be calculated by knowing the initial kinetic energy of the proton. This energy, when applied in the opposite direction, would act to decelerate the proton and bring it to rest.

The potential difference required to stop a moving proton can be calculated using the equation PEelec = qV, where q is the proton's charge and V is the potential difference (also known as voltage). Given that the proton's charge, q = qe = 1.6×10−¹⁹ Coulombs, we can rearrange the equation to solve for V: V = PEelec/q. If we know the initial kinetic energy of the proton (its energy due to motion, which can be thought of as the electrical energy it gained from accelerating through the potential difference), then the same voltage (potential difference) applied in the opposite direction will bring the proton to rest.

In other words, if the proton gained a certain amount of energy (in electron volts, eV) while accelerating through a potential difference, it will require the same amount of energy in the opposite direction to decelerate and come to rest.

For example, if the proton gained 1 electron Volt (1 eV) of energy from the potential difference, it would require a potential difference of -1 V to bring it to rest. This is because 1 eV = (1 V)(1.6×10−¹⁹ C).

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Answer:

b.open flame because it is fundamental end of the alcohol mixes in with the flame then it will become a bigger fire

The safest practice is to never use flammable materials like alcohol near an open flame as it can easily ignite and cause a fire.

Flammable materials such as alcohol should never be dispensed or used near an open flame (option B). These substances are highly reactive and can easily ignite, potentially causing a fire. It's crucial for safety purposes to avoid using flammable materials around direct sources of heat or open flames. This rule is applicable not only in chemistry laboratories, but also in any other scenario where flammable substances are present.

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Answer:

[tex]\rho=7.15\ g/cm^3[/tex]

Explanation:

The expression for density is:

[tex]\rho=\frac {Z\times M}{N_a\times {{(Edge\ length)}^3}}[/tex]

[tex]N_a=6.023\times 10^{23}\ {mol}^{-1}[/tex]

M is molar mass of Chromium = 51.9961 g/mol

For body-centered cubic unit cell , Z= 2

[tex]\rho[/tex] is the density  

Radius = 125 pm = [tex]1.25\times 10^{-8}\ cm[/tex]

Also, for BCC, [tex]Edge\ length=\frac{4}{\sqrt{3}}\times radius=\frac{4}{\sqrt{3}}\times 1.25\times 10^{-8}\ cm=2.89\times 10^{-8}\ cm[/tex]

Thus,  

[tex]\rho=\frac{2\times \:51.9961}{6.023\times \:10^{23}\times \left(2.89\times 10^{-8}\right)^3}\ g/cm^3[/tex]

[tex]\rho=7.15\ g/cm^3[/tex]

The density of solid crystalline chromium will be "7.15 g/cm³".

According to the question,

Radius,

Molar mass of Chromium,

For body-centered,

For BCC,

The edge length will be:

= [tex]\frac{4}{\sqrt{3} }\times radius[/tex]

= [tex]\frac{4}{\sqrt{3} }\times 1.25\times 10^{-8}[/tex]

= [tex]2.89\times 10^{-8} \ cm[/tex]

hence

The expression for density will be:

→  [tex]\rho = \frac{Z\times M}{N_a(Edge \ length)^3}[/tex]

By substituting the values, we get

      [tex]= \frac{2\times 51.9961}{6.023\times 10^{23}\times (2.89\times 10^{-8})}[/tex]

      [tex]= 7.15 \ g/cm^3[/tex]

Thus the above approach is right.

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Answer:

The terms used to describe the given process in Chemical weathering.

Explanation:

Weathering of rock is defined as breaking down of the rock into small pieces.

There are two types pf weathering :

Mechanical weathering : This weathering is due to change in physical parameters : temperature change, pressure change etc.

For example : When water soaked up in cracks or crevices of rocks freezes it expands and physically breaks the rock.

Chemical weathering : This weathering is decomposition of rocks due to action of chemicals.This type of weathering also changes chemical composition of rocks.

For example : when gases like carbon dioxide, sulfur oxide get dissolves in water present in rock to form weak acid which ease up the dissolving of rock in that weak acid.

Answer:

1- bromopentane          1- bromo-3-methylbutane   1-bromo2-methylbutane   2-bromo-2-methylbutane

Explanation:  

SN square reaction is a concentrated reaction. All the bond making and bond braking occur in one step

The nucleophile attack the back side of the carbon that bears the halide and replace it

Answer:  D.  There must be as equal number of atoms of each element on both sides of the equation.

Explanation:  I just took the test on Plato and this is correct

Answer:

A

Explanation:

To label an element correctly using a combination of the symbol, mass number and atomic number furnishes some important information about the element.

We can obtain these information from the element provided that correct labeling of the element is presented. Firstly, after writing the symbol of the element, the atomic number is placed as a subscript on the left while the mass number of the atomic mass is placed as a superscript on the same left.

Looking at the question asked, we have the element symbol in the correct position as Ca, with 42 also in the correct position which is the mass number. The third number which is 20 is thus the atomic number of the element.

Answer:

Unstable Nuclei. ... Too many neutrons or protons upset this balance disrupting the binding energy from the strong nuclear forces making the nucleus unstable. An unstable nucleus tries to achieve a balanced state by given off a neutron or proton and this is done via radioactive decay.

Explanation:

Explanation:

GAAP is a generally accepted accounting principle in U.S. it refers to common sets of accepted accounting principle, standards, procedures that the companies and its accountants must follow in order to compile their financial statement.

IFRS are sets of international accounting standards That specify how the financial statements will disclose different types of transactions and other activities. The International Accounting Standards Board (IASB) issues IFRS which defines precisely how accountants are required to maintain and record their accounts. In an attempt to have an universal accounting system, IFRS was developed so that business and accounts can be interpreted from industry to industry, and country to country.

Answer:FeS

Explanation:

Mass of sulphur=5.95-3.78=2.17g

Fe S

3.78/56. 2.17/32

0.0675/0.0675. 0.0678/0.0675

1:1

Empirical formula=Fe1S1=FeS

An empirical formula is a compound's chemical formula that only specifies the ratios of the elements it contains and not the precise number or arrangement of atoms. The empirical formula is FeS.

The formula of a material expressed with the smallest integer subscript is referred to as an empirical formula for a compound. The empirical formula provides details regarding the ratio of atom counts in the molecule. A compound's empirical formula is directly related to its % content.

Mass of sulphur = 5.95 - 3.78 = 2.17g

Moles of Fe and S are:

Fe = 3.78/56

n = 0.0675

S = 2.17/32

n = 0.0678

Dividing with the smallest number:

0.0675/0.0675 = 1

0.0678/0.0675 = 11

The ratio is:

1:1

Empirical formula = FeS

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Answer: The molar composition of nitrogen gas is 0.6 and that of oxygen gas is 0.4

Explanation:

To calculate the molarity of solution, we use the equation:

[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}[/tex]

Given mass of nitrogen gas = 51.3 mg = 0.0513 g     (Conversion factor:  1 g = 1000 mg)

Molar mass of nitrogen gas = 28 g/mol

Volume of solution = 2 L

Putting values in above equation, we get:

[tex]\text{Molarity of nitrogen gas}=\frac{0.0513g}{28g/mol\times 2L}\\\\\text{Molarity of nitrogen gas}=9.16\times 10^{-4}mol/L[/tex]

To calculate the partial pressure, we use the equation given by Henry's law, which is:

[tex]C_{N_2}=K_H\times p_{N_2}[/tex]

where,

[tex]K_H[/tex] = Henry's constant = [tex]6.1\times 10^{-4}mol/L.atm[/tex]

[tex]C_{N_2}[/tex] = molar solubility of nitrogen gas = [tex]9.16\times 10^{-4}mol/L[/tex]

Putting values in above equation, we get:

[tex]9.16\times 10^{-4}mol/L=6.1\times 10^{-4}mol/L.atm\times p_{N_2}\\\\p_{N_2}=\frac{9.16\times 10^{-4}mol/L}{6.1\times 10^{-4}mol/L.atm}=1.50atm[/tex]

We are given:

Total pressure of the mixture = 2.50 atm

Partial pressure of oxygen gas = 2.50 - 1.50 = 1.00 atm

To calculate the mole fraction of a substance at 25°C, we use the equation given by Raoult's law, which is:

[tex]p_{A}=p_T\times \chi_{A}[/tex]       ......(1)

where,

[tex]p_A[/tex] = partial pressure

[tex]p_T[/tex] = total pressure

[tex]\chi_A[/tex] = mole fraction

We are given:

[tex]p_{N_2}=1.50atm\\p_T=2.50atm[/tex]

Putting values in equation 1, we get:

[tex]1.50atm=2.50\times \chi_{N_2}\\\\\chi_{N_2}=\frac{1.50}{2.50}=0.6[/tex]

We are given:

[tex]p_{O_2}=1.00atm\\p_T=2.50atm[/tex]

Putting values in equation 1, we get:

[tex]1.00atm=2.50\times \chi_{O_2}\\\\\chi_{O_2}=\frac{1.00}{2.50}=0.4[/tex]

Hence, the molar composition of nitrogen gas is 0.6 and that of oxygen gas is 0.4

The mole fraction of nitrogen and oxygen is 0.6 and 0.4 respectively.

Data Given;

The law states that the mass of a dissolved gas in a given volume of solvent at equilibrium will be proportional to the partial pressure of the gas.

Mathematically;

C = KP

The number of moles Nitrogen dissolved is

[tex]n = mass/ molar mass\\ n = 0.0513/28\\ n = 0.00183127 moles[/tex]

The concentration of Nitrogen in water is

[tex]\frac{0.00183127}{2} *1 = 0.0009156M[/tex]

Applying Henry's law,

[tex]0.0009156 = 6.1*10^-^4 * P\\ P = 1.5atm[/tex]

The partial pressure of nitrogen in the mixture is 1.5atm

The total pressure of the gas is 2.50atm

Partial Pressure of oxygen = total pressure - partial pressure of nitrogen

partial pressure of oxygen = 2.50 - 1.50 = 1

The pressure fraction of the gas is the ratio between the partial pressure to the total pressure

pressure fraction of nitrogen = 1.5/2.5 = 0.6

pressure fraction of oxygen = 1/2.5 = 0.4

But partial pressure is equal to molar fraction.

This makes the mole fraction of nitrogen equals 0.6 and mole fraction of oxygen equals 0.4.

Learn more about Henry's law here;

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Answer:

called theoretic yield

In most chemical reactions, the amount of product obtained is typically less than the theoretical yield due to various reasons such as incomplete reactions, loss of product during isolation, or the reversible nature of some reactions.

In most chemical reactions, the amount of product obtained is usually less than the theoretical yield which is the amount predicted by a stoichiometric calculation based on the number of moles of all reactants present. This is because some reactants may not react to form the product, some products may be lost during the isolation step, or the reaction may not go to completion because it is reversible. For instance, if you mix 10 grams of hydrogen with 10 grams of oxygen in a sealed container, you won't get 20 grams of water, but less, because not all the hydrogen and oxygen will react.

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Answer:

Option B.

Explanation:

Let's replace the data given in the formula of Ideal Gases Law.

P. V = n . R . T

3L . 1000kPa = n . 8.31 L.kPa/mol.K . 298K

(3L . 1000 kPa) / ( 8.31 L.kPa/mol.K . 298K) = n

1.21 moles

The number of moles of oxygen by using ideal gas equation is 1.21 moles. Hence, the correct option is B.

[tex]PV=nRT[/tex]

Here, P is the pressure, V is the volume, n is the number of moles, R is the gas constant, T is the temperature.

We can calculate the number of moles by using the above equation as follows:-

[tex]PV = n R T\\ 1000kPa\times 3L = n \times 8.31 L.kPa/mol.K \times298K\\(3L . 1000 kPa) / ( 8.31 L.kPa/mol.K . 298K) = n\\n=1.21 moles[/tex]

So, the number of moles of oxygen is 1.21 moles.

Learn more about the ideal gas equation here

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Answer:

9.3 mL of 6 M acetic acid needs to be added to 500 mL of a solution of 0.20 M sodium acetate to a achieve a pH of 5.0

Explanation:

This problem can be solved by the Henderson-Hasselbalch equation. It's formula is:

[tex]pH=pKa+log(\frac{[CH_3COO^-]}{[CH_3COOH]})[/tex]

The molar concentration can be replaced by the moles of the solute as the volume of the buffer will be the same for both species

[tex]pH=pKa+log(\frac{n_{CH_3COO^-}}{n_{CH_3COOH}})[/tex]

Placing the given data:

[tex]5.0=4.75+log(\frac{0.1}{n_{CH_3COOH}})[/tex]

[tex]n_{CH_3COOH}=(\frac{0.1}{10^{5.0-4.75}})\\\\n_{CH_3COOH}=(\frac{0.10}{1.778})\\\\ n_{CH_3COOH}=0.056moles[/tex]

The volume required to obtain the above-calculated moles can be determined by the molarity of acetic acid

[tex]M_{CH_3COOH}=\frac{n_{CH_3COOH}}{V_{CH_3COOH}(L)}\\\\V_{CH_3COOH}=\frac{n_{CH_3COOH}}{M_{CH_3COOH}}\\\\V_{CH_3COOH}=\frac{0.056}{6.0}\\\\V_{CH_3COOH}=0.0093L\\\\or\\\\V_{CH_3COOH}=9.3mL[/tex]

The above volume of acetic acid can be added to 500 mL of acetate to form 5.0 pH buffer

The volume of 6 M acetic acid needed is (0.356/6)x(500+V(A-)).

To calculate the volume of 6 M acetic acid needed to achieve a pH of 5.0, we need to use the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])

In this case, the ratio of sodium acetate ([A-]) to acetic acid ([HA]) should be 1:1 to achieve a pH of 5.0.

Using the equation, we can rearrange it to find the concentration of acetic acid:

pH = pKa + log([A-]/[HA])

[HA] = [A-] x 10^(pH - pKa)

Plugging in the values, we get:

[HA] = 0.20 M x 10^(5.0 - 4.75) = 0.20 M x 10^0.25 = 0.20 M x 1.78 = 0.356 M

Now, we can calculate the volume of 6 M acetic acid needed:

[HA] x V(HA) = [HA] x V(A-)

0.356 M x V(HA) = 0.356 M x (500 mL + V(A-))

V(HA) = 0.356 M x (500 mL + V(A-)) / 6 M

V(HA) = (0.356/6)x(500+V(A-)) mL

Since the ratio of sodium acetate to acetic acid is 1:1, the volume of 6 M acetic acid needed is equal to the volume of 0.20 M sodium acetate added to the solution. Therefore, the volume of 6 M acetic acid needed is (0.356/6)x(500+V(A-)).

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Answer:

The following must be true for diffusion of a solute to occur across a partition that separates two compartments

The partition is semi permeable

There is  concentration gradient across the partition

The process is known as Osmosis

Explanation:

Diffusion is the migration of particles of a substance, liquid or gas, from a region of high concentration to a region of lower concentration and it only occurs where there is a concentration gradient for example spraying a perfume at a corner of a room, the spray sent spreads across the room as there is a concentration gradient

Osmosis is the movement or diffusion of solvent molecules across a semi permeable membrane from a region of low solute concentration to a region of high solute concentration to create a balanced concentration of solute on both sides of the compartment.

Diffusion requires a concentration gradient and a membrane permeable to the solute; in osmosis, water diffuses to balance solute concentrations.

For the diffusion of a solute to occur across a partition that separates two compartments, there must be a concentration gradient present, where the solute molecules move from an area of higher concentration to an area of lower concentration. Additionally, permeability of the membrane to the solute is crucial, as only materials capable of passing through will diffuse.

In certain situations, such as osmosis, water will move across the membrane to balance the concentration gradient of solutes, when the solutes themselves cannot pass through the membrane. In living systems, osmosis is a dynamic and continuous process that maintains the balance of fluids.

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Average C-H bond energy: Ethane = 764.93 kJ/mol, Ethene = 627.79 kJ/mol, Ethyne = 528.26 kJ/mol.

To find the average C-H bond energy in ethane (C₂H₆), ethene (C₂H₄), and ethyne (C₂H₂), we can use Hess's law and the given reactions along with the average C-H bond energy in methane (CH₄), which is 415 kJ/mol.

1. **Calculate the average C-H bond energy in ethane (C₂H₆):**

  Given reaction:

[tex]\[C2H6 (g) + H2 (g) \rightarrow 2CH4 (g) \quad \Delta H_{\text{rxn}} = -65.07 \, \text{kJ/mol}\][/tex]

  This reaction breaks one C-C bond and adds two C-H bonds.

  Change in bond energy = Total energy of bonds broken - Total energy of bonds formed

[tex]\[= 1 \times (\text{C-C bond energy}) - 2 \times (\text{C-H bond energy})\][/tex]

  From the given reaction, the change in bond energy is -65.07 kJ/mol.

  Substituting the known values:

 [tex]\[-65.07 \, \text{kJ/mol} = 1 \times (\text{C-C bond energy}) - 2 \times (415 \, \text{kJ/mol})\][/tex]

  Solving for the C-C bond energy:

[tex]\[\text{C-C bond energy} = 2 \times 415 - 65.07 \, \text{kJ/mol} = 764.93 \, \text{kJ/mol}\][/tex]

2. **Calculate the average C-H bond energy in ethene (C₂H₄):**

  Given reaction:

[tex]\[C2H4 (g) + H2 (g) \rightarrow 2CH4 (g) \quad \Delta H = -202.21 \, \text{kJ/mol}\][/tex]

  This reaction breaks one C=C bond and adds two C-H bonds.

  Change in bond energy = Total energy of bonds broken - Total energy of bonds formed

[tex]\[= 1 \times (\text{C=C bond energy}) - 2 \times (\text{C-H bond energy})\][/tex]

  From the given reaction, the change in bond energy is -202.21 kJ/mol.

  Substituting the known values:

[tex]\[-202.21 \, \text{kJ/mol} = 1 \times (\text{C=C bond energy}) - 2 \times (415 \, \text{kJ/mol})\][/tex]

  Solving for the C=C bond energy:

 [tex]\[\text{C=C bond energy} = 2 \times 415 - 202.21 \, \text{kJ/mol} = 627.79 \, \text{kJ/mol}\][/tex]

3. **Calculate the average C-H bond energy in ethyne (C₂H₂):**

  Given reaction:

  [tex]\[C2H2 (g) + 3H2 (g) \rightarrow 2CH4 (g) \quad \Delta H = -376.74 \, \text{kJ/mol}\][/tex]

  This reaction breaks one C≡C bond and adds four C-H bonds.

  Change in bond energy = Total energy of bonds broken - Total energy of bonds formed

[tex]\[= 1 \times (\text{C≡C bond energy}) - 4 \times (\text{C-H bond energy})\][/tex]

  From the given reaction, the change in bond energy is -376.74 kJ/mol.

  Substituting the known values:

[tex]\[-376.74 \, \text{kJ/mol} = 1 \times (\text{C≡C bond energy}) - 4 \times (415 \, \text{kJ/mol})\][/tex]

  Solving for the C≡C bond energy:

 [tex]\[\text{C = C bond energy} = 4 \times 415 - 376.74 \, \text{kJ/mol} = 528.26 \, \text{kJ/mol}\][/tex]

So, the average C-H bond energy in ethane [tex](C2H6) is \(764.93 \, \text{kJ/mol}\), in ethene (C2H4) is \(627.79 \, \text{kJ/mol}\), and in ethyne (C2H2) is \(528.26 \, \text{kJ/mol}\).[/tex]