An atom X is 9 times heavier than 1/12 the mass of a carbon-12 atom. What is the relative atomic mass of atom X ?​

Isolating and modifying cellulose from a melon seed coat involves several steps. Here's a general outline of the process:

Seed Coat Removal ; Cleaning ; Drying ; Grinding ; Cellulose Extraction ; Cellulose Modification.

Seed Coat Removal: Start by separating the melon seed coat from the seeds. This can be done by manually peeling or cutting away the outer coat.

Cleaning: Wash the seed coat thoroughly to remove any dirt, debris, or other impurities.

Drying: After cleaning, allow the seed coat to dry completely. This can be done by spreading it out on a clean surface and leaving it to air dry or using a low-temperature oven.

Grinding: Once the seed coat is dry, grind it into a fine powder. This can be achieved using a blender, food processor, or mortar and pestle.

Cellulose Extraction: Extract cellulose from the powdered seed coat using a chemical or enzymatic method. One common approach is the acid hydrolysis method:

a. Prepare a solution of dilute acid (such as hydrochloric acid) and add the powdered seed coat to it.

b. Heat the mixture under controlled conditions, typically around 60-80°C, for a specific duration.

c. Filter the mixture to separate the cellulose residue from the liquid acid solution.

d. Wash the cellulose residue with water multiple times to remove any remaining acid or impurities.

e. Dry the cellulose thoroughly, either through air drying or using a low-temperature oven.

Cellulose Modification: The isolated cellulose can be modified to enhance its properties or functionality. This step can include processes such as chemical derivatization or physical treatments. For example, cellulose can be chemically modified by esterification or etherification to introduce new functional groups.

It's important to note that the specific conditions and methods for isolation and modification may vary depending on the desired characteristics of the cellulose and the intended applications.

Additionally, proper safety precautions and adherence to applicable regulations should be followed when working with chemicals and equipment during the process.

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The partial pressures of nitrogen, oxygen, and neon in the mixture are approximately 366.13 torr, 312.55 torr, and 214.32 torr, respectively.

To find the partial pressures of the three gases, we can use Dalton's Law of Partial Pressures, which states that the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures of each gas.

Given:

Total pressure = 893 torr

Mole fractions: nitrogen (N2) = 0.41, oxygen (O2) = 0.35, neon (Ne) = 0.24

To calculate the partial pressures, we can multiply the total pressure by the mole fraction of each gas:

Partial pressure of nitrogen (P(N2)) = Mole fraction of nitrogen (X(N2)) * Total pressure

P(N2) = 0.41 * 893 torr

Partial pressure of oxygen (P(O2)) = Mole fraction of oxygen (X(O2)) * Total pressure

P(O2) = 0.35 * 893 torr

Partial pressure of neon (P(Ne)) = Mole fraction of neon (X(Ne)) * Total pressure

P(Ne) = 0.24 * 893 torr

Calculating the values:

P(N2) = 0.41 * 893 torr = 366.13 torr

P(O2) = 0.35 * 893 torr = 312.55 torr

P(Ne) = 0.24 * 893 torr = 214.32 torr

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To calculate the amount of heat released when cooling water, we can use the formula:

Q = m × c × ΔT

where:

Q is the heat energy released or absorbed (in Joules),

m is the mass of the water (in grams),

c is the specific heat capacity of water (in J/g°C),

ΔT is the change in temperature (in °C).

Given:

m = 100.0 g (mass of water)

c = 4.186 J/g°C (specific heat capacity of water)

ΔT = 10.0°C (change in temperature, from 75.0°C to 10.0°C)

Using the formula:

Q = 100.0 g × 4.186 J/g°C × (10.0°C - 75.0°C)

Q = 100.0 g × 4.186 J/g°C × (-65.0°C)

Q = -27187 J

The negative sign indicates that heat is released during the cooling process.

Therefore, the correct answer is:

d) -27,209 J

At a specific temperature, we can use the following equation to determine the root mean square (rms) speed of molecules or atoms:

rms speed = √(3RT/M)

where

R is the gas constant (8.314 J/(mol·K)),

T is the temperature in Kelvin, and

M is the molar mass in kilograms per mole.

For [tex]O_2[/tex] molecules at 427 K:

Molar mass of O2 = 2 * 16.00 g/mol = 32.00 g/mol = 0.032 kg/mol

Substituting the values into the equation:

rms speed = √((3 * 8.314 J/(mol·K) * 427 K) / 0.032 kg/mol)

rms speed ≈ √(28165.3 J/kg)

rms speed ≈ 167.9 m/s

The rms speed of [tex]O_2[/tex] molecules at 427 K is approximately 167.9 m/s.

For He atoms at 427 K:

Molar mass of He = 4.00 g/mol = 0.004 kg/mol

Substituting the values into the equation:

rms speed = √((3 * 8.314 J/(mol·K) * 427 K) / 0.004 kg/mol)

rms speed ≈ √(7061.925 J/kg)

rms speed ≈ 83.9 m/s

The rms speed of He atoms at 427 K is approximately 83.9 m/s.

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Answer:

A Brønsted-Lowry acid is a substance which donates protons to other substances. (option c.)

Explanation:

For Brønsted-Lowry acids, are the ones that donates protons and bases, the ones that accepts H⁺ from others compounds.

In this reaction:

NH₃  +  HCl → NH₄Cl

The hydrocloridic acid donates the proton to ammonia to make ammonium chloride. So the HCl, is a Brønsted-Lowry acid while the ammonia is a base.

There are cases, like water that behaves as acid and base.

These compounds are called amphoteric because it can donate H⁺ or OH⁻ at the same time.

H₂O ⇄  H⁺  +  OH⁻

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The event of the fan stopping when it is switched off in a classroom provides an opportunity to discuss scientific concepts and learning. Here are the various steps involved in this event.

Observation: The students observe that when the switch of the fan is turned off, the fan stops rotating.

Questioning: Encourage students to ask questions about why the fan stops when the switch is turned off, leading to discussions about energy transfer, electrical circuits, and mechanical motion.

Explanation: Explain to the students that when the switch is turned off, it breaks the electrical circuit, interrupting the flow of current to the fan motor. Without the electrical energy from the current, the fan motor stops rotating.

Discussion: Engage students in a discussion about the conversion of electrical energy to mechanical energy in the fan motor. Discuss concepts such as electromagnetic induction, electric motors, and the role of magnetic fields in generating rotational motion.

Application: Relate this event to everyday life, highlighting the importance of understanding electrical circuits and energy conversions in various appliances.

By discussing these concepts, students can deepen their understanding of the principles behind the event and foster scientific learning.

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77.7K

Ideal gas laws allow us to calculate the characteristics of gases through their changes.

Gay-Lussac's Law

One of the ideal gas laws is Gay-Lussac's Law. This law describes the relationship between temperature and pressure. It states that temperature and pressure are directly proportional. This means that as temperature increases, pressure also increases. In equation form, this law is shown as:

Finding Temperature
To find the temperature, plug in the known information and solve for T₂.

Now, to find T₂, multiply both sides by 210.0 kPa.

Since this question is based on measured values, we should round according to significant figure rules. We should round to 3 sig figs because 208K has 3. This means the temperature of the gas tank must be 77.7 Kelvin.

If 2.2 L of an aqueous solution that contains 261 g of calcium acetate  Ca(C₂H₃O₂)₂, at 37 °C, the osmotic pressure is 57.5 atm.

According to question:

Volume of solution = 2.2 L

Mass of solute calcium acetate Ca(C₂H₃O₂)₂  = 261 g

Molar mass of solute     = 158.17 g/mol

Temperature  = 370C  

=  ( 37 + 273 ) K  

=  310 K

No.of moles of solute   = mass ÷ molar mass

= 261g ÷ 158.17 g/mol

=  1.650 mol

Molarity of solute   = no.of moles solute ÷ volume of solution (L)

=  1.650 mol ÷ 2.2 L

=  0.753 mol/L

Osmotic pressure (π) is given by

π    =  i × M × R × T

i  =  van't Hoff factor  = 3

It is the figure of ions formed from 1 formula unit of a solute in solution.

Ca(C₂H₃O₂)₂  →   Ca2+   +    2C₂H₃O₂⁻

Thus, 1 Ca2+ and 2 Ca(C₂H₃O₂)₂ is formed.

i = 1+2  = 3

M = Molarity = 0.753 mol/L

R = universal gas constant  = 0.0821 L atm mol-1 K-1

T = Temperature ( K )   = 310 K

π   =  3 × 0.753 mol/L  × 0.0821 L atm mol-1 K-1  × 310 K

=  57.5 atm

Therefore, the osmotic pressure is 57.5 atm.

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Answer:

100 rbed  KJ |0| +2k

Explanation:

In the given chemical reaction, H2SO4 reacts with water to form HSO4- (hydrogen sulfate) and H3O+ (hydronium) ions. Since H2SO4 is a strong acid, it dissociates completely in water, meaning that all the H2SO4 molecules will ionize.

Given that 0.0005 moles of H2SO4 have reacted and the solution volume is 500 L, we can calculate the concentration of H3O+ ions in the solution. Since 1 mole of H2SO4 produces 1 mole of H3O+ ions, the concentration of H3O+ ions is:Concentration of H3O+ = (0.0005 moles) / (500 L) = 1.0 x 10^-6 M

The pH of a solution is defined as the negative logarithm (base 10) of the H3O+ concentration. Therefore:pH = -log10(1.0 x 10^-6)

Using logarithmic properties, the pH can be calculated:

pH ≈ -(-6) = 6

Therefore, the pH of the solution is approximately 6, indicating that it is slightly acidic.

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Answer:

the answer is a

Explanation:

Approximately 7.26 grams of sodium propanoate must be dissolved in 0.25 dm^3 of 1 mol dm^-3 propanoic acid to create a buffer of pH 4.87.

To calculate the mass of sodium propanoate (C2H5COONa) that must be dissolved in 0.25 dm^3 of 1 mol dm^-3 propanoic acid (C2H5COOH) to create a buffer at pH 4.87, we can use the Henderson-Hasselbalch equation:

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

where:

pH is the desired pH of the buffer (4.87 in this case)

pKa is the dissociation constant of the weak acid (4.87 in this case)

[A-] is the concentration of the conjugate base (sodium propanoate)

[HA] is the concentration of the weak acid (propanoic acid)

In a buffer solution, the ratio of the concentrations of the conjugate base and the weak acid is important. Let's assume that x moles of sodium propanoate are dissolved in 0.25 dm^3 of 1 mol dm^-3 propanoic acid. This means the concentration of propanoic acid will be (1 - x) mol dm^-3, and the concentration of sodium propanoate will be x mol dm^-3.

According to the equation, we have:

4.87 = 4.87 + log (x / (1 - x))

Simplifying the equation, we get:

0 = log (x / (1 - x))

Now, let's solve for x:

x / (1 - x) = 1

x = 1 - x

2x = 1

x = 1/2

So, the concentration of sodium propanoate (C2H5COONa) in the buffer solution should be 0.5 mol dm^-3.

To calculate the mass of sodium propanoate needed, we need to convert the volume of the solution to moles. Since the concentration is 0.5 mol dm^-3 and the volume is 0.25 dm^3, the number of moles of sodium propanoate required is:

moles = concentration × volume = 0.5 mol dm^-3 × 0.25 dm^3 = 0.125 moles

The molar mass of sodium propanoate is:

(2 × atomic mass of carbon) + (5 × atomic mass of hydrogen) + atomic mass of oxygen + atomic mass of sodium

= (2 × 12.01 g/mol) + (5 × 1.008 g/mol) + 16.00 g/mol + 22.99 g/mol

= 58.08 g/mol

Finally, we can calculate the mass of sodium propanoate using the number of moles:

mass = moles × molar mass = 0.125 moles × 58.08 g/mol = 7.26 grams

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Answer:

The total number of carbon atoms on the left-hand side of the equation is 6 x 2 = 12.

Answer:

Copper Copper Copper Copper

The atomic mass of an atom tell us How much the atom weighs

The atomic mass represents the total mass of an atom, including its protons, neutrons, and electrons. It is measured in atomic mass units (amu) and gives us an indication of the mass of the atom.

Moreover, the atomic mass helps in calculating other properties of atoms, such as the number of neutrons. By subtracting the number of protons (which corresponds to the atomic number) from the atomic mass, we can determine the approximate number of neutrons in an atom. Neutrons play a vital role in stabilizing the nucleus and influencing the atom's overall behavior.

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Pressure and volume can be regarded as the entity that is inversely proportional.

It should be noted that this explanatin an be done using the law in chemistry which is the Boyle's law which states that, for a given amount of gas and constant temperature, the volume is inversely proportional to the pressure.

However the Equal quantities of all gases can be seen to have same number of molecules when subjected to the same temperature and pressure (Avogadro's law).

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Answer:

Salt Method of Preparation

(i) Sodium nitrate (b) Neutralisation

(ii) Iron (III) chloride (e) Direct synthesis

(iii) Lead chloride (d) Double decomposition

(iv) Zinc sulphate (a) Simple displacement

(v) Sodium hydrogen sulphate (c) Decomposition by acid

Explanation:

Understanding the behavior of gases has numerous practical applications in real life. Here are some ways:

HVAC Systems ; Chemical Reactions ; Gas Storage ; Weather Forecasting ; Environmental Studies etc.

HVAC Systems: Heating, ventilation, and air conditioning (HVAC) systems rely on the principles of gas behavior to regulate temperature, humidity, and air quality in buildings. Knowledge of gas properties helps in designing efficient HVAC systems for optimal comfort and energy efficiency.

Chemical Reactions: Understanding gas behavior is crucial in chemical reactions, especially those involving gases. Knowledge of factors like pressure, volume, and temperature helps determine reaction rates, equilibrium conditions, and optimize industrial processes.

Gas Storage and Transportation: The behavior of gases is essential in designing storage tanks and transportation systems for various gases. Understanding pressure-volume relationships helps ensure safe handling and storage of gases, such as natural gas, propane, and compressed air.

Medical Applications: Medical professionals rely on the behavior of gases in several applications. For instance, respiratory therapies involve understanding the exchange of gases in the lungs, and anesthesiology utilizes gas behavior to administer and monitor the effects of anesthesia during surgeries.

Weather Forecasting: The study of gas behavior is vital in meteorology and weather forecasting. Understanding how gases behave in the atmosphere helps predict weather patterns, including temperature changes, wind patterns, and the formation of clouds and precipitation.

Environmental Studies: Gases play a significant role in environmental studies. Knowledge of gas behavior aids in studying air pollution, atmospheric chemistry, and climate change. It helps scientists model and understand the movement and distribution of pollutants in the atmosphere.

Aerospace Engineering: Understanding gas behavior is crucial in aerospace engineering for designing aircraft, rockets, and spacecraft. Factors like air resistance, aerodynamics, and propulsion systems rely on gas behavior principles for efficient and safe flight.

These are just a few examples of how understanding gas behavior has practical applications in various fields and contributes to technological advancements and our overall understanding of the natural world.

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The Microscope part and their right descriptions are as follows

Iris Diaphragm: A. Increases or decreases the light intensity

Objective Lens System: B. After light passes through the specimen, it next enters this lens system

Stage: C. Platform that supports a microscope slide

Adjustment Knob: D. Causes stage (or objective lens) to move upward or downward

Condenser: E. Concentrates light onto the specimen

Other parts of a microscope and their description that you should know about includes;

Eyepiece - The lens that you look through to see the image of the specimen.

Body tube - The tube that connects the eyepiece to the objective lenses.

Arm - The part of the microscope that supports the body tube and connects it to the base.

Base - The part of the microscope that supports the arm and provides stability.

Illuminator - The light source that provides light for the microscope.

Stage clips - The clips that hold the microscope slide in place on the stage.

Revolving nosepiece - The part of the microscope that holds the objective lenses and allows them to be rotated into place.

The above answer is in response to the full question below;

Match the names of the microscope parts in column A with the descriptions in column B. Place the letter of your choice in the space provided.

1. Iris diaphram

2. Objective lens system

3. Stage

4. Adjustment knob

5. Condenser

Increases or decreases the light intensity

2. After light passes through the specimen, it next enters this lens system

3. Platform that supports a microscope slide

4. Causes stage (or objective lens) to move upward or downward

5. Concentrates light onto the specimen

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The type of energy transfer between the frozen ice cream and the tongue is convection. The frozen ice cream will gradually melt because of the temperature difference.

Convection is the method of heat transfer that involves the transmission of heat in a fluid (liquid or gas) by the circulation of currents.

According to this question, a spoonful of frozen ice cream is placed on the tongue. The heat transfer will occur between the tongue with warmer temperature and the ice cream with colder temperature.

The heat energy will flow from the tongue to the ice cream, hence, making it defreeze or melt with time. This type of heat transfer is convection because it involves a fluid.

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The freezing point of the solution is 6.213 °C. This is less than the freezing point of ether.

Freezing point of the solution can be determined by using the concept of freezing point depression, which is a colligative property.

The freezing point depression is given by the equation:

ΔTf = Kf * m

Where:

ΔTf = Freezing point depression

Kf = Cryoscopic constant (a constant specific to the solvent)

m = Molality of the solution

First, we need to calculate the molality (m) of the solution. Molality is defined as the number of moles of solute per kilogram of solvent.

We have 1.727 mol of solute and 483 g of ether (solvent).

First, we convert the mass of ether to kilograms:

Mass of ether = 483 g = 483/1000 kg = 0.483 kg

Now we can calculate the molality:

m = moles of solute / mass of solvent (in kg)

m = 1.727 / 0.483 ≈ 3.575 mol/kg

Next, we need the cryoscopic constant (Kf) for ether. The cryoscopic constant is specific to the solvent. For ether, the cryoscopic constant is approximately 1.74 °C kg/mol.

Finally, we can calculate the freezing point depression (ΔTf):

ΔTf = Kf * m

ΔTf = 1.74 °C kg/mol * 3.575 mol/kg

= 6.213 °C

Therefore, the freezing point of the solution is decreased by approximately 6.213 °C compared to the freezing point of pure ether.

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Answer:

The volume of Component 1 is 36 cubic inches.

Explanation:

To calculate the volume of a composite solid, we need to determine the individual volumes of the different components and then add them together.

In this case, the composite solid consists of multiple components with the following dimensions:

Component 1:

Length: 4 inches

Width: 3 inches

Height: 3 inches

To find the volume of Component 1, we multiply the length, width, and height together:

Volume of Component 1 = Length x Width x Height = 4 in x 3 in x 3 in = 36 cubic inches

Therefore, the volume of Component 1 is 36 cubic inches.

Please provide the dimensions of the remaining components of the composite solid, and I will calculate the total volume by summing up the individual volumes.

The action that would most likely reduce water pollution is A) initiating efforts to remove trash from the ocean.

Trash and plastic waste in the ocean pose a significant threat to marine life and ecosystems. Initiating efforts to remove trash from the ocean can help mitigate this problem and reduce water pollution. When plastic waste accumulates in oceans, it can break down into microplastics, which are ingested by marine organisms and can enter the food chain, causing harm to both marine life and humans.

By actively removing trash from the ocean, we can prevent it from further degrading and releasing pollutants. This can help protect marine habitats, reduce the risk of entanglement or ingestion by marine animals, and preserve the overall health of ocean ecosystems.

On the other hand, options B, C, and D would likely contribute to increased water pollution:

B) Eliminating laws and regulations on industrial waste would remove important safeguards and oversight, potentially allowing industries to discharge pollutants directly into water bodies, leading to increased water pollution.

C) Promoting the use of oil and gas can lead to oil spills, leakage, and contamination of water sources. This can have severe consequences for aquatic ecosystems and human health.

D) Removing the ban on chlorofluorocarbons (CFCs) would result in the release of these harmful substances, which deplete the ozone layer and can contaminate water sources when disposed of improperly.

In conclusion, initiating efforts to remove trash from the ocean is the most effective action to reduce water pollution, while the other options would likely exacerbate the problem. Therefore, Option A is correct.

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The differences and similarities between the Digestive system and the Excretory system, including their individual responsibilities: Like Ingestion, Digestion, Absorption, Elimination, etc.

Digestive System:

The Digestive system is responsible for the breakdown and absorption of food and nutrients. Its main functions include:

Ingestion: The process of taking in food through the mouth.

Digestion: Breaking down food into smaller molecules through mechanical and chemical processes (such as chewing and enzyme action).

Absorption: The uptake of nutrients and water by the cells lining the digestive tract into the bloodstream.

Elimination: The removal of indigestible waste material (feces) from the body through the rectum and anus.

The digestive system includes organs such as the mouth, esophagus, stomach, small intestine, large intestine, liver, gallbladder, and pancreas.

Excretory System:

The Excretory system is responsible for the removal of metabolic waste products and maintaining fluid and electrolyte balance in the body. Its main functions include:

Filtration: The process of filtering waste products and excess substances from the blood.

Reabsorption: The reabsorption of essential substances (such as water, ions, and nutrients) back into the bloodstream from the filtrate.

Secretion: The active transport of waste products, excess ions, and toxins from the blood into the filtrate.

Excretion: The elimination of the filtrate containing waste products and excess substances from the body as urine.

The excretory system primarily consists of the kidneys, ureters, bladder, and urethra. It also involves other organs like the skin, lungs, and liver, which contribute to the elimination of specific waste products.

Differences between Digestive and Excretory Systems:

Function: The digestive system is responsible for breaking down food and absorbing nutrients, while the excretory system focuses on eliminating waste products from the body.

Organs involved: The digestive system involves organs such as the mouth, stomach, and intestines, while the excretory system primarily includes the kidneys, bladder, and urethra.

Processes: The digestive system involves processes like ingestion, digestion, and absorption, while the excretory system involves filtration, reabsorption, secretion, and excretion.

Types of waste: The digestive system eliminates undigested food material as feces, while the excretory system removes metabolic waste products, such as urea and excess ions, in the form of urine.

Similarities between Digestive and Excretory Systems:

Both systems contribute to maintaining homeostasis in the body by regulating the internal environment.

Both systems involve the transport of substances across cell membranes, either for absorption (digestive system) or excretion (excretory system).

Both systems interact with other bodily systems to ensure overall functioning and balance.

While the digestive and excretory systems have distinct functions and processes, they are interconnected in maintaining overall health by facilitating the intake of essential nutrients and removing waste products from the body.

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The statement that correctly describes residual radiation from the patient's body after this type of treatment is the patient's body will be radioactive for a very short period after the treatment.

The correct option is A.

Residual radiation refers to the lingering radiation that remains after a nuclear or radioactive event has occurred.

It is the radiation that persists in the environment or materials even after the primary source of radiation has been removed or decayed.

Residual radiation can arise from various sources, including nuclear accidents, radioactive waste disposal, and from radiotherapy.

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The equations that show ionization are:

NaCl(s) → Na+(aq) + Cl-(aq)

H3O+(aq) + Cl-(aq)

CaCl2(s) → Ca2+(aq) + 2Cl-(aq)

The equations that show ionization are:

NaCl(s) → Na+(aq) + Cl-(aq)

This equation represents the dissociation of solid sodium chloride (NaCl) into sodium ions (Na+) and chloride ions (Cl-) in aqueous solution.

H3O+(aq) + Cl-(aq)

This equation represents the ionization of hydronium ion (H3O+) and chloride ion (Cl-) in an aqueous solution.

CaCl2(s) → Ca2+(aq) + 2Cl-(aq)

This equation represents the dissociation of solid calcium chloride (CaCl2) into calcium ions (Ca2+) and chloride ions (Cl-) in aqueous solution.

The other equations listed do not specifically show ionization. The equation O2(g) → O2(aq) represents a change of state from a gas (g) to an aqueous solution (aq), but it does not involve the formation of ions. The equation CO2(g) + 2H2O(l) → HCO3-(aq) + H3O+(aq) represents a chemical reaction involving the formation of ions (bicarbonate ion and hydronium ion), but it does not show ionization from a solid or gas. The equation HCI(g) + H2O(l) does not show ionization.

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Answer:

K= ( [h2]² [O2] )/[H2O]²

Explanation:

characters arent in my keyboard

For a solution of sodium hydroxide:

1. To make a buffer with pH = 5.00, have a ratio of

[tex]\frac{[A-]}{[HA]} = 10^{-5.50}[/tex] = 0.316. The concentration of acetic acid is 0.200 M, so add enough sodium hydroxide to make the concentration of acetate 0.316 M.

The volume of sodium hydroxide needed is calculated as follows:

V(NaOH) = (0.316 M - 0.200 M) / 4.50 M = 0.0316 L = 31.6 mL

Therefore, 31.6 mL of 4.50 M sodium hydroxide must be added to 250.0 mL of 0.200 M acetic acid solution to make a buffer with pH = 5.00.

2. The pH of the buffer is calculated as follows:

pH = pKa + log([tex]\frac{[A-]}{[HA]}[/tex])

= 4.76 + log(0.2/0.1)

= 4.86

Therefore, the pH of the buffer is 4.86.

3. The mass of salt that must be dissolved in 0.25 dm³ of 1 mol dm³ propanoic acid to give a buffer of pH 4.87 is calculated as follows:

[tex]\frac{[A-]}{[HA]} = 10^{-4.87}[/tex] = 0.0114

The concentration of acetate is 0.0114 M, and the concentration of propanoic acid is 1 mol dm³. Therefore, the mass of acetate that must be dissolved is calculated as follows:

Mass of acetate = (0.0114 mol dm³)(0.25 dm³) = 0.00285 g

Therefore, 0.00285 g of sodium propanoate must be dissolved in 0.25 dm³ of 1 mol dm³ propanoic acid to give a buffer of pH 4.87.

4. The pH of the buffer is calculated as follows:

pH = pKa + log([tex]\frac{[A-]}{[HA]}[/tex])

= 4.74 + log(0.1/0.1)

= 4.74

Therefore, the pH of the buffer is 4.74.

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