If 100. 0 g of Cl2 are needed, what mass of NaOCl must be reacted? NaOCl + HCl → NaOH + Cl2​

Answer:

solution

Explanation:

Alloy is solid solution as it is a mixture of two or more metals or a metal and a non-metal.

The amount of energy released would be 1917.96 kJ.

We can start by calculating the number of moles of methane used up in the reaction:

Number of moles = mass / molar mass

molar mass of CH4 = 12.01 + 4(1.01) = 16.05 g/mol

Number of moles = 38.0 g / 16.05 g/mol = 2.37 mol (rounded to two decimal places)

Next, we can calculate the total energy released by the combustion of this amount of methane:

Energy released = energy released per mole x number of moles

Energy released = 808 kJ/mol x 2.37 mol = 1917.96 kJ (rounded to two decimal places)

Therefore, if you used up 38.0 g (or 2.37 mol) of methane in 5 minutes, you would have released approximately 1917.96 kJ of energy.

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The pH of an acidic solution depends on the concentration of hydrogen ions (H+) in the solution.

However, in this case, we are given the concentration of fluoride ions (F-) in the solution, which is not directly related to the pH.

Assuming that the acid in the solution is hydrofluoric acid (HF), we can write the following equilibrium expression:

[tex]HF + H2O H3O+ + F-[/tex]

where HF is the weak acid, H2O is water, H3O+ is the hydronium ion, and F- is the fluoride ion.

The equilibrium constant expression for this reaction is:

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

At equilibrium, the concentration of HF will be equal to the initial concentration minus the concentration of F-, since one F- ion reacts with one H+ ion to form HF:

[HF] = [HF]initial - [F-]

Assuming that the initial concentration of HF is also 1 x 10^-3 M (since the problem does not give this information), we can substitute these values into the equilibrium expression and solve for the concentration of H3O+:

[tex]Ka = [H3O+][F-]/([HF]initial - [F-])1.8 x 10^-4 = [H3O+][1 x 0^-3]/(1 x 10^-3 - 1 x 10^-3)1.8 x 10^-4 = [H3O+][1 x 10^-3]/0[/tex]

Since the denominator is zero, this equation is undefined and cannot be solved. Therefore, we cannot determine the pH of the solution based on the given information.

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The gas law that applies to this scenario is the Ideal Gas Law, which states:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the universal gas constant, and T is the temperature in Kelvin.

Rearranging the Ideal Gas Law to solve for pressure (P), we get:

P = nRT/V

Substituting the given values, we get:

P = (36.0 mol)(0.0821 L•atm/K•mol)(25.0 + 273 K) / 12.4 L

P = 98.9 atm

Therefore, the hydrogen gas must be compressed to a pressure of 98.9 atm.

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The addition of 1.25 g of sodium formate increases the pH of the solution by 2.12 units.

To calculate the change in pH, we need to determine the new concentrations of the formic acid and its conjugate base after the addition of the sodium formate. We can then use the Henderson-Hasselbalch equation to calculate the new pH.

First, let's calculate the initial concentration of formic acid:

0.50 M = moles of formic acid / 0.027 L

moles of formic acid = 0.50 x 0.027 = 0.0135 mol

Now, we need to calculate the concentration of formate ion (HCOO-) after adding 1.25 g of sodium formate (HCOONa) to the solution. Since sodium formate dissociates completely in water, the number of moles of formate ion added will be the same as the number of moles of sodium formate:

moles of HCOO- = moles of NaHCOO = 1.25 g / (68.01 g/mol) = 0.0184 mol

Next, we can calculate the new concentrations of formic acid and formate ion:

[ HCOOH ] = (0.0135 mol) / (0.027 L) = 0.5 M

[ HCOO- ] = (0.0135 mol + 0.0184 mol) / (0.027 L) = 1.15 M

Now, we can use the Henderson-Hasselbalch equation to calculate the new pH:

pH = pKa + log([ HCOO- ] / [ HCOOH ])

pKa = -log(Ka) = -log(1.8 x 10^-4) = 3.74

pH = 3.74 + log(1.15 / 0.5) = 3.74 + 0.40 = 4.14

Therefore, the change in pH is:

ΔpH = 4.14 - 2.02 = 2.12

So, the addition of 1.25 g of sodium formate increases the pH of the solution by 2.12 units.

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The oxygen that is released during the light reactions comes from water.

Let's know more about photosynthesis:

1. Photosynthesis is the process by which the plant can produce food. The process of photosynthesis converts light energy into chemical energy, which organisms can use. In this process, water and carbon dioxide produce oxygen and glucose. This process is carried out in the chloroplasts of plants.

2. Light Reaction: During the process of photosynthesis, two types of reactions take place, light reaction, and dark reaction. The light reaction takes place in the thylakoid membrane of the chloroplast. It needs light to occur. Here, the light energy is absorbed by pigments such as chlorophyll. This light energy is converted into chemical energy. ATP and NADPH are the two products of the light reaction.

3. Oxygen is also released during the light reaction. The oxygen that is released during the light reactions comes from water. This reaction also helps in the formation of the electron transport chain.

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An argumentative, plausible view of the literary work being evaluated is the claim in a literary analysis.

The claim in a literary analysis, which is an interpretation of a literary work, is the author's argument or viewpoint regarding the relevance or meaning of the work.

The assertion needs to be clear, debatable and backed up by textual evidence.

The claim in literary analysis is the main viewpoint or argument that the author is advancing regarding the relevance or meaning of the literary work under consideration.

The assertion should be a reasonable, disputed opinion that can be backed up by textual evidence, and it should be sufficiently detailed to be convincing and understandable to the reader.

For instance, in an interpretation of William Shakespeare's play "Hamlet," a writer can contend that, rather than a lack of courage, Hamlet's hesitation to exact revenge on his father's murderer stems from his desire for justice and his battle with indecision.

This assertion is both plausible and problematic because different readers or critics may interpret Hamlet's actions differently.

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The reactor needs to process 2,488,076 m³/hour of ethylene to remove 0.8219 metric tons/day of acetylene. The reactor design will depend on the specifics of the plant, such as the available space, pressure and temperature requirements, and the desired efficiency of the process.

To remove 1% acetylene from 1.5 million metric tons per year of ethylene, we need to calculate the flow rate of ethylene and acetylene. Let's assume that the ethylene stream contains 2% acetylene, so we need to remove 1% of 2%, which is 0.02%.

1.5 million metric tons per year of ethylene is equivalent to 4,109.59 metric tons per day. To calculate the flow rate of acetylene, we multiply the flow rate of ethylene by the concentration of acetylene:

Acetylene flow rate = 4,109.59 metric tons/day x 0.02% = 0.8219 metric tons/day

Now we can use the given information about the catalyst to design a reactor. The weight hourly space velocity (WHSV) is defined as the weight of reactants per hour per weight of catalyst. We can use the following equation to calculate the required weight of catalyst:

WHSV = (Weight of reactants per hour) / (Weight of catalyst)

Weight of catalyst = (Weight of reactants per hour) / WHSV

Since we know the WHSV and the acetylene flow rate, we can calculate the required weight of ethylene per hour:

Weight of ethylene per hour = (Weight of acetylene per hour) / (0.9 × 0.5 × WHSV)

Weight of ethylene per hour  [tex]=\frac{(0.8219)}{(0.9*0.5*800*24)} = 1,368.94[/tex] metric tons/hour

Assuming a density of 0.55 kg/L for ethylene, the volumetric flow rate of ethylene is:

Volumetric flow rate of ethylene = Weight of ethylene per hour / (Density of ethylene x 1000 L/m³)

Volumetric flow rate of ethylene [tex]= \frac{(1,368.94)}{(0.55 * 1000)} = 2,488,076[/tex] m³/hour

This indicates that in order to remove 0.8219 metric tons of acetylene per day, the reactor must process 2,488,076 m³/hour of ethylene each hour. The reactor design will be determined by the characteristics of the facility, including the amount of space available, the necessary pressure and temperature, and the desired process efficiency. Based on these variables, the catalyst weight and reactor size will need to be determined.

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the complete question is:

Acetylene is formed as a by-product in the manufacture of ethylene. It is removed by selective hydrogenation using noble metal catalysts (see for example US 7453 017). A particular catalyst achieves 90% acetylene saturation with 50% hydrogen selectivity for acetylene at a weight hourly space velocity of 800 h-1 based on ethylene Design a reactor to remove 1% acetylene from ethylene in a plant that produces 15 million metric tons per year of ethylene using this catalyst.

The formula 3Mg(O3H2)3 represents a molecule composed of 3 magnesium atoms and 9 groups of hydroxide ions (O3H2), each containing 3 oxygen atoms and 6 hydrogen atoms.

What is a Molecule?

A molecule is a colle-ction of two or more than two atoms that are held toge-ther by chemical-bonds. These atoms could be of the same or different elements. For example, H2O (water) is a molecule made up of two (2) hydrogen-atoms and one oxygen atom.

Another example is carbon dioxide (CO2), which is a molecule consisting of one carbon atom and two oxygen atoms. Molecules are the fundamental building blocks of many substances, and their unique arrangement and properties play a critical role in various chemical reactions and biological processes.

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[tex]Pb(NO_{3})_{2} ,And K_{2} S, Ba(NO_{3})_{2} and K_{2}CO_{3} ,K_{2}CO_{3} and Ni(NO_{3})_{2}[/tex]can be tested in the interactive, these can form precipitate (ppt), so it can be tested in interactive.

[tex]K_{2} SO_{4} and K_{2} CO_{3} , KI and K_{2} SO_{4}, NI(NO_{3} )_{2} and BO(NO_{3} )_{2}[/tex] cannot be tested in the interactive. There is no formation of precipitate (ppt), therefore it cannot be tested in interactive. When certain cations and anions interact in an aqueous solution, precipitates are insoluble ionic solid products of the process. The variables that affect how a precipitate forms can be different. Certain reactions, like those involving buffer solutions, are temperature-dependent, whereas others just depend on the concentration of the solution.The following is an illustration of a precipitation reaction: One of the products is solid, and both reactants are watery. The reactants dissociate and are soluble since they are ionic and watery.

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1.82 g of ammonia are created before the reaction comes to an end. By computing the moles of each reactant and comparing their ratios using the balanced chemical equation, we may do this.

We determine the lithium's moles:

1 mol Li divided by 6.941 g Li to get 4.87 g Li equals 0.700 mol Li

The moles of water are then determined:

1 mol H2O divided by 18.015 g H2O in 5.80 g gives 0.322 mol H2O.

As there are fewer molecules of water (0.322 mol) than there are molecules of lithium (0.700 mol), water is the limiting reactant.

We may determine the number of moles of ammonia created by using the mole ratio from the balanced chemical equation:

1 mol NH3 / 3 mol H2O 0.322 mol H2O = 0.107 mol NH3

Finally, we may convert moles to grammes using the molar mass of NH3:

1.82 g NH3 from 0.107 mol NH3 (17.031 g NH3 / 1 mol NH3)

As a result, 1.82 g of ammonia are created before the reaction comes to an end.

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One way to prove that all the products were formed when sodium bicarbonate solution and calcium chloride solution were combined is by carrying out some simple tests for the presence of each product.

Calcium Carbonate: Calcium carbonate is a white solid that is insoluble in water. To test for its presence, add a few drops of hydrochloric acid to the solution obtained from the reaction. If calcium carbonate is present, it will react with the acid to produce carbon dioxide gas, which can be seen as bubbles in the solution. Alternatively, you can filter the solution and wash the solid with water to remove any soluble salts. The solid residue left after drying is calcium carbonate.

Sodium Chloride: Sodium chloride is a soluble salt that dissolves completely in water. To test for its presence, simply evaporate the solution obtained from the reaction to dryness. If sodium chloride is present, it will remain as a white solid after evaporation.

Water: Water is a liquid that cannot be easily separated from the other products. However, you can test for its presence by using a drying agent such as anhydrous calcium chloride or silica gel. Simply add a small amount of the drying agent to the solution and stir for a few minutes. If the drying agent absorbs water, then water is present in the solution.

Cabon Dioxide:r Carbon dioxide is a gas that is released during the reaction. To test for its presence, you can bubble the gas through lime water (calcium hydroxide solution). If carbon dioxide is present, it will react with the lime water to form a white precipitate of calcium carbonate.

By carrying out these tests, you can prove that all the products have been formed and that no reactants are left in the solution.

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The mass of the Dinitrogen tetroxide that contains a million oxygen atoms is 23 x 10⁶ .

Atoms and molecules are very small in both size and mass. One sample mole's weight is the molar mass. To get the molar mass, connect the atomic masses (atomic weights) of each atom in the molecule. Using the mass listed in the Periodic Table or Atomic Weight Table, determine the atomic mass for each element.

To calculate the molecular mass of a molecule, multiply the subscript (number of atoms) by the atomic mass of each element in the molecule and add those masses together. Typically, molar mass is stated in either grammes (g) or kilogrammes (kg) (kg).

From the molecular formula of dinitrogen tetroxide ;

molar mass of dinitrogen tetroxide = 92 g/mol

hence;

92 g of dinitrogen tetroxide contains 4 atoms of oxygen

x g of dinitrogen tetroxide will contain 1 x 10⁶ atoms of oxygen

x = 92 x 1 x 10⁶/4

x =23 x 10⁶ atoms of oxygen.  

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Complete question:

Calculate the mass of dinitrogen tetroxide (N₂O₄) that contains a million (1.000 x 10₆) oxygen atoms. Be sure your answer has a unit symbol if necessary, and round it to 4 significant digits.

The color of the new bonds that form between a solute and solvent depends on the specific solute and solvent involved, as well as the nature of the bonding between them, and it may or may not result in a visible color change.

The color of the new bonds that form between the solute and solvent depends on the specific solute and solvent involved, as well as the nature of the bonding between them. In general, the formation of new bonds between a solute and solvent may not necessarily result in a visible color change.

However, there are some cases where the formation of new bonds can result in a change in color. For example, in the case of transition metal complexes, the formation of new coordination bonds between a transition metal ion and a ligand can result in a change in color due to the absorption of light by the metal-ligand complex.

This is known as a ligand-to-metal charge transfer (LMCT) or metal-to-ligand charge transfer (MLCT) transition, and the resulting color change can be used to identify and quantify the complex.

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Answer: Segment B

Explanation:

On a graph, a line traveling upwards from left to right is considered positive. A line traveling downwards from left to right is considered negative. A line neither traveling upwards or downwards is neutral. Based on this, we can see that segment B is traveling downwards, so it would represent a negative acceleration.

a) Carbon dioxide can be dissolved in the blood on the right side of heart.

b) Carbon dioxide

c) Oxygen-rich blood flows from the lungs to the left atrium.

8)  Circular Motion

9)Valves

10. Ventricular septum

11)  Higher altitudes have fewer oxygen molecules per volume of air.

One carbon atom is covalently double linked to two oxygen atoms in each of the compounds that make up carbon dioxide, which has the molecular formula CO2. At ambient temperature, it exists as a vapour.

In the atmosphere, carbon dioxide serves as a greenhouse gas because it captures infrared energy despite being invisible to visible light. It has increased from pre-industrial values of 280 ppm to a minor gas in the Earth's atmosphere at 421 parts per million (ppm), or approximately 0.04% by volume (as of May 2022).

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Answer:

Elements placed in Group VIIIA of the periodic table are also known as the noble gases, and they are helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), and radon (Rn).

These elements do not take part in chemical reactions because they have a full outermost electron shell, also known as the valence shell. This means that their valence shell contains the maximum number of electrons that it can hold, and therefore, these elements do not have any electron available for bonding with other atoms.

In other words, their outermost electron shell is already stable, which means they have very low reactivity with other elements. Because of their unreactive nature, noble gases are often referred to as inert gases.

The 0.68 M HNO₂ solution has a pH of 1.96. This demonstrates how acidic the solution is.

The equation for the dissociation of nitrous acid is as follows:

HNO₂ (aq) ⇌ H+ (aq) + NO₂- (aq)

The equilibrium constant expression for this reaction is:

K = [H+][NO₂-]/[HNO₂]

We are given the value of K and the initial concentration of HNO₂. We can assume that x is the concentration of H+ and NO₂- produced when the HNO₂ dissociates.

Using the equilibrium constant expression and the given values, we can write:

4.0 x [tex]10^{-4}[/tex] = x² / (0.68 - x)

Because x is small compared to 0.68, we can approximate (0.68 - x) as 0.68:

4.0 x [tex]10^{-4}[/tex] = x² / 0.68

Solving for x gives us:

x = √(4.0 x [tex]10^{-4}[/tex] x 0.68) = 0.011 M

The pH of the solution can be calculated using the formula:

pH = -log[H+]

Therefore, the pH of the 0.68 M HNO2 solution is:

pH = -log(0.011) = 1.96

Therefore, the pH of the 0.68 M HNO2 solution is 1.96. This indicates that the solution is acidic.

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The temperature of the xenon gas in the balloon is approximately 276 K.

We can use the Ideal Gas Law to solve for the moles of xenon gas present in the balloon:

PV = nRT

where P is pressure, V is volume, n is number of moles, R is gas constant, and T will be the temperature in Kelvin.

Converting the given values to the correct units:

P = 954 Torr = 954/760 atm

= 1.2553 atm

V = 2.37 L

n = 0.115 mol (given)

R = 0.08206 L·atm/mol·K (gas constant)

T = ? (unknown)

Substituting the values and solving for T:

T = PV/nR = (1.2553 atm)(2.37 L)/(0.115 mol)(0.08206 L·atm/mol·K)

≈ 276 K

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There is a production of 59.14 grammes of water vapour and 65.85 grammes of nitrogen monoxide.

A. The balanced chemical equation for the combustion of ammonia with excess oxygen to form nitrogen monoxide and water gas is:

4 NH₃ (g) + 5 O₂ (g) → 4 NO (g) + 6 H₂O (g)

According to the equation, for every 4 moles of ammonia, 4 moles of nitrogen monoxide and 6 moles of water gas are produced. To calculate the number of grams of each product produced, we need to find the number of moles of ammonia in the sample:

Molar mass of NH₃ = 14.01 g/mol + 3(1.01 g/mol) = 17.04 g/mol

Number of moles of NH₃ = 37.4 g / 17.04 g/mol = 2.19 mol

Using the mole ratio from the balanced equation, we can calculate the number of moles of each product:

Number of moles of NO = 4/4 x 2.19 mol = 2.19 mol

Number of moles of H₂O = 6/4 x 2.19 mol = 3.28 mol

To find the mass of each product, we can use their molar masses:

Molar mass of NO = 30.01 g/mol

Molar mass of H₂O = 18.02 g/mol

Mass of NO produced = 2.19 mol x 30.01 g/mol = 65.85 g

Mass of H₂O produced = 3.28 mol x 18.02 g/mol = 59.14 g

Therefore, 65.85 grams of nitrogen monoxide and 59.14 grams of water gas are produced.

B. The amount of heat transferred during this reaction can be calculated using the formula:

q = CΔT

where q is the amount of heat transferred, C is the heat capacity of the calorimeter, and ΔT is the change in temperature of the calorimeter.

From the problem, we are given:

C = 952 J/°C

ΔT = 1750°C - 22.9°C = 1727.1°C

Substituting the values, we get:

q = (952 J/°C) x (1727.1°C) = 1.64 x [tex]10^{6}[/tex] J

Therefore, the amount of heat transferred during this reaction is 1.64 x [tex]10^{6}[/tex] J.

C. The change in enthalpy for the reaction can be calculated using the equation:

ΔH = q/n

where q is the amount of heat transferred, and n is the number of moles of ammonia that reacted.

From part A, we know that 2.19 moles of ammonia reacted. Substituting the values, we get:

ΔH = (1.64 x [tex]10^{6}[/tex] J) / 2.19 mol = 748,858 J/mol = 748.9 kJ/mol

Therefore, the change in enthalpy for the reaction is 748.9 kJ/mol ammonia.

D. The change in enthalpy is not equal to the change in internal energy for this reaction because the bomb calorimeter is a constant volume calorimeter, which means that no work is done during the reaction (Δw = 0). Therefore, the change in enthalpy (ΔH) equals the change in internal energy (ΔU) plus the work done (Δw) by the system:

ΔH = ΔU + Δw

Since Δw = 0 for the bomb calorimeter, ΔH = ΔU.

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Answer:

To solve the problem, we need to convert the given quantity of AlBr to grams of Br₂ using the given molar ratios:

70.6 g AlBr * (1 mole AlBr / 266.68 g AlBr) * (3 moles Br₂ / 2 moles AlBr) * (159.80 g Br₂ / 1 mole Br₂) = 94.3 g Br₂

Therefore, 94.3 g of Br₂ are required.

Answer:

a

Explanation:

The half-life of sodium-24 is 15 hours, which means that after 15 hours, half of the original sample will decay, leaving half of the sample remaining. After another 15 hours (30 hours total), half of that remaining sample will decay, leaving a quarter of the original sample remaining. After another 15 hours (45 hours total), half of that remaining sample will decay, leaving an eighth of the original sample remaining. Finally, after another 15 hours (60 hours total), half of that remaining sample will decay, leaving a sixteenth of the original sample remaining.

To calculate the amount remaining after 60 hours, we can use the formula:

Amount remaining = (initial amount) x (1/2)^(t/h)

where "t" is the time elapsed and "h" is the half-life.

Plugging in the values we have, we get:

Amount remaining = (10.0 g) x (1/2)^(60/15)

= (10.0 g) x (1/2)^4

= (10.0 g) x 0.0625

= 0.625 g

Therefore, after 60 hours, approximately 0.63 g of the original 10.0 g sample of sodium-24 will remain. Rounded to the nearest hundredth place, this is 0.63 g.

Answer: 85.73 cm^3

Explanation:

Combined Gas law (temperature must be in Kelvin)

V2= P1V1 T2/(T1 x P2)

P1 = 320 mmHg

V1= 220 cm^3

T1= 295.15 K

T2= 273.15 K

P2 = 760 mmHg

320 x 220 X 273.15/(295.15 X 760) = 85.72697201 cm^3

Answer:

Fossil fuels, such as coal, oil, or natural gas, release carbon back into the atmosphere.

The processes would be decomposition, diffusion, erosion, respiration, and combustion.

Explanation:

Hope this helped?

Answer:

 The molar mass of copper (II) nitrate is 187.5 g/mol.

Explanation:

the amount of zinc needed to react with 8.45 moles of hydrochloric acid is 4.225 moles.

To solve this problem, we will use the given balanced equation and the molar ratio of reactants and products:

Zn + 2HCl → ZnCl2 + H2

Molar ratio: 1:2 (zinc to hydrochloric acid)

Therefore, the amount of zinc needed to react with 8.45 moles of hydrochloric acid is 4.225 moles.

Since the molar mass of zinc is 65.38 g/mol, the mass of zinc needed to react with 8.45 moles of hydrochloric acid is 4.225 moles x 65.38 g/mol = 276.21 g of zinc.

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The mass fraction is 0.122  for the sample that has been solved here for the solution of the CH3COOH in 3 liters of its 2M.

The term "mass fraction" refers to the proportion of a component's mass of the substance that we have in the system and it is common that we expresses this value as a percentage accordingly.

Mass of the solution = DV

Mass = 0.981 * 3000 mL

= 2943 g

We know that;

Number of moles = CV

Moles = 2M * 3 L

= 6 moles

Mass = Number moles * molar mass

= 6 * 60 g/mol

= 360 g

Mass fraction = 360 g/2943 g

= 0.122

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Exercise 1. To balance the number of electrons, we multiply the oxidation half-reaction by 5 and the reduction half-reaction by 1:

5Fe₂+ → 5Fe₃⁺ + 5e⁻

MnO₄⁻ + 8H⁺ + 5e⁻ → Mn₂⁺ + 4H₂O

We can combine the two half-reactions and cancel out the electrons. Finally, we balance the equation by adding water molecules and hydrogen ions (H+) as needed to balance the oxygen and hydrogen atoms:

The balanced equation is:

MnO₄⁻ + 8H⁺ +5Fe₂ → Mn₂⁺ + 4H₂O + 5Fe₃⁺

To balance the number of electrons, we multiply the oxidation half-reaction by 6 and the reduction half-reaction by 1:

6H₂S → 6S + 12H⁺ + 12e⁻

Cr₂O₇²⁻ + 14H⁺ +  6e⁻ → 2Cr₃⁺ + 7H₂O

We can combine the two half-reactions and cancel out the electrons.

Finally, we balance the equation by adding water molecules and hydrogen ions (H+) as needed to balance the oxygen and hydrogen atoms:

The balanced equation is:

Cr₂O₇²⁻ + 14H⁺ + 6H₂S → 2Cr₃⁺ + 7H₂O + 6S

Exercise2.

1. The anode is where oxidation occurs, and the cathode is where reduction occurs. In this case, Al is a stronger reducing agent than Cu, so Al will undergo oxidation and Cu will undergo reduction. Therefore, the anode is the Al electrode, and the cathode is the Cu electrode.

2. Electrons flow from the anode to the cathode in a galvanic cell.

3. The symbolic representation of the galvanic cell is:

Al(s) | Al₃⁺(aq) || Cu₂⁺(aq) | Cu(s)

4. The two half equations and the balanced equation are:

Oxidation half-reaction (anode): Al(s) → Al₃⁺(aq) + 3e-

Reduction half-reaction (cathode): Cu₂⁺(aq) + 2e- → Cu(s)

Overall balanced equation: 2Al(s) + 3Cu₂⁺(aq) → 3Cu(s) + 2Al₃⁺(aq)

5. The salt bridge is used to maintain charge neutrality in the two half-cells by allowing the flow of ions between them. As the electrons flow from the anode to the cathode, positively charged Al₃⁺ ions build up in the anode compartment, and negatively charged SO₄²⁻ ions build up in the cathode compartment. The salt bridge allows the movement of SO₄²⁻ ions from the cathode compartment to the anode compartment to balance the charges.

6.The EMF of the cell can be calculated using the formula:

EMF = E0(cathode) - E0(anode) = 0.34V - (-1.67V) = 2.01V

7.  Q = It = (1 A)(3600 s) = 3600 C

The number of moles of electrons transferred can be calculated from the stoichiometry of the balanced equation:

2Al(s) + 3Cu₂⁺(aq) → 3Cu(s) + 2Al₃⁺ (aq)

2 moles of Al will react for every 3 moles of Cu2+ reduced.

From the mass change of the cathode, we can calculate the number of moles of Cu reduced using its molar mass:

molar mass of Cu = 63.55 g/mol

change in mass of Cu = 0.192 g

moles of Cu reduced = (0.192 g)/(63.55 g/mol) = 0.0030 mol

Therefore, the number of moles of Al oxidized is:

moles of Al oxidized = (2/3)(0.0030 mol) = 0.0020 mol

Finally, the change in mass of the anode can be calculated using the molar mass of Al:

molar mass of Al = 26.98 g/mol

change in mass of Al = (0.0020 mol)(26.98 g/mol) = 0.054 g

Therefore, the variation in the anode's mass is 54 mg (rounded to the nearest mg).

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