joe makes a solution containing 0.100 m fluoride ions and 0.126 m hydrogen fluoride. what is the ph after the addition of 5.00 ml of 0.0100 m hcl to 25.0 ml of this solution? (ka for hf

To prepare 1.00 L solution of 0.1 M NaOH, weight 4 grams of NaOH and place in a 1 L volumetric flack and make it up with water to the mark.

Step 1: Determine the mole of NaOH in the 0.1 M NaOH solution. Details below:

Mole of NaOH = molarity × volume

= 0.1 × 1

= 0.1 mole

Step 2: Determine the mass of NaOH in the solution. Details below:

Mass of NaOH = Mole × molar mass

= 0.1 × 40

= 4 grams

Step 3: Weight 4 grams of NaOH and place in a 1 L volumetric flack and make it up with water to the mark.

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We know that:

- It is a constant-pressure calorimeter

- the temperature of 60.0 g of water increases by 4.50 °C

- Specific heat capacity of water = 4.2 J/goC

And we must find the amount of heat that is transferred to the water.

To find it, we need to use the formula for heat absorbed

Where,

Q represents the heat absorbed by the water

m represent the mass

Cp represents specific heat, in this case of the water

(T2 - T1) represents the variation of the temperature

Using the given information, we know that:

m = 60.0 g

Cp = 4.2[J/(g°C)]

(T2 - T1) = 4.50 °C

Now, replacing in the formula

Finally, we can convert J to KJ

ANSWER:

B. 1.13 KJ

Answer:303.15

Explanation:

The inlet pressure of the pipeline in (bar) is 6.7 bar. To calculate the inlet pressure of the pipeline, we can use the Darcy-Weisbach equation.

Darcy-Weisbach equation relates pressure drop, flow rate, pipe characteristics, and fluid properties. The equation is given as:

ΔP = (fLρV²) / (2D) where:

ΔP is the pressure drop

f is the Darcy friction factor

L is the length of the pipeline

ρ is the density of water

V is the velocity of water

D is the diameter of the pipeline

First, we need to convert the flow rate from m³/hr to m³/s:

Flow rate = 27 m³/hr = (27/3600) m³/s = 0.0075 m³/s

Next, we need to calculate the velocity of water:

Area of the pipeline =\(\pi \times \frac {(76/1000)^2}{4} = 0.004556 m^2\)

Velocity
= Flow rate / Area of the pipeline
= 0.0075 m³/s / 0.004556 m² = 1.646 m/s

Now, we can calculate the pressure drop using the Darcy-Weisbach equation. Since we need to calculate the inlet pressure, we assume ΔP is the difference between the outlet pressure and the inlet pressure:

ΔP = (fLρV²) / (2D)

\(\triangle P = \frac {(0.015 \times 4000 \times 1000 \times 1.646^2)}{(2 \times 0.076)} = 10.69 \times 10^5 Pa\)

= 10.7 bar (approx)

Rearranging the equation to solve for the inlet pressure:

Inlet pressure = ΔP - outlet pressure = 10.7 bar - 4 bar = 6.7 bar

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The correct answer is D) Cr is larger than Cr3+ and Se is larger than Se2-.

When an atom gains or loses electrons, its size changes. When an atom gains electrons, it becomes larger because the additional electrons repel each other and push the electron cloud outward. When an atom loses electrons, it becomes smaller because there are fewer electrons to repel each other and the electron cloud contracts.
In the case of Cr3+, the atom has lost three electrons, making it smaller than the neutral Cr atom. In the case of Se2-, the atom has gained two electrons, making it larger than the neutral Se atom. Therefore, the correct answer is D) Cr is larger than Cr3+ and Se is larger than Se2-.

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Answer:

The salts that are insoluble in water are CaS (calcium sulfide) and PbF2 (lead(II) fluoride).

Explanation:

The solubility of a salt in water depends on the balance between the forces holding the ions together in the solid state (ionic lattice) and the forces between the ions and water molecules.

In the case of calcium sulfide (CaS), it is considered insoluble in water. This is because the forces of attraction between the calcium cations (Ca2+) and sulfide anions (S2-) in the solid lattice are relatively stronger than the forces of attraction between these ions and water molecules. As a result, the solid CaS does not readily dissociate into its ions in water, leading to low solubility.

Similarly, lead(II) fluoride (PbF2) is also considered insoluble in water. The forces between the lead cations (Pb2+) and fluoride anions (F-) in the solid lattice are strong enough to prevent easy dissociation in water.

On the other hand, magnesium sulfate (MgSO4) and copper(II) chloride (CuCl2) are both soluble in water. The forces between the ions in these salts are weaker than the forces between the ions and water molecules, allowing them to dissociate into their respective ions and form a solution in water.

Here are some general rules to follow when predicting solubility.

1. Most nitrate (NO3-) salts are soluble in water.

2. Most salts of alkali metals (Group 1 elements) and ammonium (NH4+) are soluble.

3. Most chloride (Cl-), bromide (Br-), and iodide (I-) salts are soluble, except for those of silver (Ag+), lead (Pb2+), and mercury (Hg2+).

4. Most sulfate (SO4 2-) salts are soluble, except for those of barium (Ba2+), strontium (Sr2+), lead (Pb2+), and calcium (Ca2+).

5. Most hydroxide (OH-) salts are insoluble, except for those of alkali metals (Group 1 elements) and barium (Ba2+), strontium (Sr2+), and calcium (Ca2+).

6. Most carbonate (CO3 2-) and phosphate (PO4 3-) salts are insoluble, except for those of alkali metals (Group 1 elements) and ammonium (NH4+).

These are just a few examples of energy transformations associated with different processes. Energy transformations occur in various forms throughout nature and technology, and understanding them is crucial for studying energy systems and their impacts on the environment and society.

Here are some common energy transformations and their associated processes:

Combustion:

Process: Burning of fossil fuels such as coal, oil, or natural gas.

Energy Transformation: Chemical energy in the fuel is converted into thermal energy (heat) and light energy.

Photosynthesis:

Process: Occurs in plants, where they convert sunlight, carbon dioxide, and water into glucose (chemical energy) and oxygen.

Energy Transformation: chemical energy is transformed into chemical energy stored in glucose.

Cellular Respiration:

Process: Takes place in cells, breaking down glucose and other organic molecules to release energy.

Energy Transformation: Chemical energy stored in glucose is converted into ATP (adenosine triphosphate), which is a form of usable energy for cells.

Electric Power Generation:

Process: Power plants use various methods (such as coal, nuclear reactions, or renewable sources like wind or hydro) to generate electricity.

Energy Transformation: The energy source (e.g., fossil fuels, nuclear reactions, or renewable sources) is converted into electrical energy.

Solar Panels:

Process: Solar panels capture sunlight using photovoltaic cells and convert it into electricity.

Energy Transformation: Solar energy is transformed into electrical energy.

Nuclear Fission:

Process: In nuclear power plants, the process of splitting atoms (fission) is used to release a tremendous amount of energy.

Energy Transformation: Nuclear potential energy is converted into thermal energy, which is then used to generate electricity.

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Xe gas atoms do experience attraction to each other, that is called a London dispersion forces. Compared to Kr gas atoms, Xe has stronger London dispersion forces, thus Xe is more difficult to boil.

London dispersion force is an intermolecular force, the weakest force that is only temporarily active when the electrons in two adjacent atoms occupy positions that make the atoms form temporary dipoles. London dispersion forces strength is very dependent on the molecular weight. Xe has heavier molecules than Kr, so Xe also has stronger London dispersion forces. As a result, Xe is more difficult to boil than Kr.

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a. The number of grams of sodium oxide produced by 100 g Na would be 134.85 grams while the amount produced by 60 g oxygen would be 232.5 grams.

b. The mass of excess reactant left over at the conclusion of the reaction would be 24.624 grams of oxygen.

The unbalanced equation goes thus: \(Na + O_2 -- > Na_2O\)

First, we need to balance the equation of the reaction. This balanced equation is written as:

\(4 Na + O_2 -- > 2Na_2O\)

From the equation:

With 100 g Na:

Mole = 100/23

        = 4.35 moles

Equivalent mole of \(Na_2O\) = 4.35/2

                                          = 2.175 moles

Mass of 2.175 moles of \(Na_2O\) = 2.175 x 62

                                                 = 134.85 grams

Thus, 134.85 grams of \(Na_2O\) will be produced with 100g of Na.

With 60 g of oxygen:

Mole = 60/32

        = 1.875 moles

Equivalent moles of \(Na_2O\) = 1.875 x 2

                                           = 3.75 moles

Mass of 3.75 moles \(Na_2O\) = 3.75 x 62

                                           = 232.5 grams.

Thus, 232.5 grams of \(Na_2O\) will be produced with 60 g of oxygen.

To get the excess reactant:

Mole of 100 g Na = 4.35 moles

Mole of 60 g oxygen = 1.875 moles

Mole ratio of Na to oxygen = 4:1

Thus the excess reactant is oxygen.

Actual amount of oxygen = 4.35/4

                                           = 1.0875 moles

Excess mole of oxygen = 1.857 - 1.0875

                                       = 0.7695 moles

Mass of excess oxygen = 0.7695 x 32

                                       = 24.624 grams

Thus, the mass of the excess reactant left over after the conclusion of the reaction is 24.624 grams of oxygen.

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The correct order of increasing strength of intermolecular forces is C*H_{4} < C*H_{3}*C*H_{3} < C*H_{3}*C*H_{2}*C*H_{3}. Hence, the option (d) is correct.

The strength of intermolecular forces depends on the type of molecules present. There are three types of intermolecular forces that are generally present in molecular substances: van der Waals forces, dipole-dipole forces, and hydrogen bonding. These forces increase in strength as the polarity of the molecule increases. Vanderwaal's forces are considered the weakest intermolecular force between molecules. Van der Waals forces are also called dispersion forces, London forces, or instantaneous dipole forces. These forces arise when there is a temporary formation of dipoles between atoms. These are more pronounced in larger molecules as there is more space for temporary dipoles to occur, which implies that CH4 has the weakest Vanderwaal's forces due to the presence of multiple carbon atoms and the presence of hydrogen atoms in between them. CH3CH3 has a higher intermolecular force than CH4.

Therefore, CH3CH3 has higher intermolecular forces than CH4 and CH3CH2CH3 has the highest intermolecular force out of the three.

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Delta H°for the reaction 2Al2O3 (s) ==>4Al (s) + 3O2 (g) is  +3,340 kJ/mol.

To find Delta H° for the reaction 2Al2O3 (s) ==>4Al (s) + 3O2 (g), we can use Hess's Law which states that the enthalpy change for a reaction is the same regardless of the pathway taken.

First, we need to write the reaction as a combination of the given equation and its reverse:

2Al (s) + (3/2)O2 (g) ==> Al2O3 (s)      Delta H° = -1,670 kJ/mol

Reverse the given equation:

Al2O3 (s) ==> 2Al (s) + (3/2)O2 (g)      Delta H° = +1,670 kJ/mol

Next, we need to multiply the second equation by 2 in order to cancel out the Al and O2 in the final equation:

2Al2O3 (s) ==> 4Al (s) + 3O2 (g)       Delta H° = 2(1,670 kJ/mol) = +3,340 kJ/mol

Therefore, Delta H° for the reaction 2Al2O3 (s) ==> 4Al (s) + 3O2 (g) is +3,340 kJ/mol.

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Answer:

Kjiiii99i9oojiiiiiiiii88

Answer:

2.24g/

Explanation:

mass of water is found by getting 5he extra mass the cylinder gained after adding

100g-80g=20g - mass of water

Density of water: 1g/cm³ that means that 20g water has a volume of 20cm³

1cm³=1ml; 20cm³=20ml

volume of stone is the extra height the water gained: 45ml-20ml =

25ml

mass of stone is the extra mass now earned:

156g-100g=56g

Density of stone=mass of stone÷divide by volume of stone

56g÷25ml=

2.24g/ml

Answer:

Nuclear process involves the splitting of an atom, which is a chemical process. Nuclear reactions are usually more efficient than chemical reactions, because the reaction can be controlled and there is no need to wait for the reaction to take place.

a and e are heterogeneous (door and salsa), b, c, and d are homogeneous (air, black coffee, and pure water), and f cannot be classified as it pertains to a person rather than a substance or mixture. Option A and E

a. A door: Heterogeneous. A door is typically made up of various materials such as wood, metal, glass, etc. These materials have different properties and can be easily distinguished, making the door a heterogeneous object.

b. The air you breathe: Homogeneous. Air is a mixture of gases, primarily nitrogen, oxygen, carbon dioxide, and trace amounts of other gases. On a macroscopic scale, air appears uniform and consistent throughout, making it a homogeneous mixture.

c. A cup of coffee (black): Homogeneous. A cup of black coffee consists of water and coffee solutes that are evenly distributed throughout the liquid. It appears uniform and consistent, indicating a homogeneous mixture.

d. The water you drink: Homogeneous. Pure water, without any dissolved substances or impurities, is a homogeneous substance. It is composed of H2O molecules that are uniformly distributed throughout the liquid.

e. Salsa: Heterogeneous. Salsa is a mixture of various ingredients such as tomatoes, onions, peppers, and spices. These ingredients have different textures, colors, and sizes. The different components can be visually distinguished, making salsa a heterogeneous mixture.

f. Your lab partner: Heterogeneous. A lab partner refers to a person, and individuals are not considered homogeneous or heterogeneous in the same sense as substances or mixtures. They are complex beings with different physical characteristics, thoughts, and behaviors. Thus, categorizing a lab partner as homogeneous or heterogeneous is not applicable in this context. Option A and E

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According to the Freundlich adsorption isotherm, the minimum dosage of activated carbon required to reduce the odor to a residual value of 0.20 is approximately 0.77 mg/L.

The Freundlich adsorption isotherm describes the relationship between the concentration of a solute in a liquid phase and the amount of adsorbent required to remove the solute. It is often expressed as an equation:

\(\[ q = K \cdot C^{1/n} \]\)

where q is the amount of solute adsorbed per unit mass of adsorbent (mg/g), C is the concentration of the solute in the liquid phase (mg/L), K is a constant, and n is the Freundlich exponent.

In this case, the odor number is used as a measure of the concentration of the odor-causing compounds in the wastewater. The initial odor number is 10, and the desired residual odor number is 0.20.

By plotting the carbon added (mg/L) against the odor number, we can observe the data points. The Freundlich equation can be rearranged as follows to determine the values of K and n:

\(\[ \log(q) = \log(K) + \frac{1}{n} \log(C) \]\)

By taking the logarithm of both sides of the equation and plotting log(q) against log(C), a straight line can be obtained. The slope of the line corresponds to 1/n, and the intercept corresponds to log(K).

Using the given data points, a plot of log(q) against log(C) can be constructed. The slope of the resulting line is approximately -0.857, which corresponds to \(\(1/n\)\). Therefore, \(\(n \approx -1.166\)\).

To find the value of K, we can substitute the data points into the equation and solve for K. By taking the average of the calculated K values, we find that \(\(K \approx 10.1\)\).

Now, to determine the minimum dosage of activated carbon required to reduce the odor to a residual value of 0.20, we can use the Freundlich equation. Rearranging the equation, we have:

\(\[ q = K \cdot C^{1/n} \]\)

Substituting the desired residual odor number (0.20) for q and the calculated values of K and n, we can solve for C:

\(\[ 0.20 = 10.1 \cdot C^{-1.166} \]\)

Simplifying the equation, we find that \(\(C \approx 0.77\)\) mg/L. Therefore, the minimum dosage of activated carbon required to achieve the desired residual odor number of 0.20 is approximately 0.77 mg/L.

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Based on the data provided, 0.2375 L of water and 81.225 g of Kool-aid are required to make 237.5 mL of 1 M solution of Kool-aid.

The molarity of a solution is the amount in moes of a solute dissolved in a given volume of solution in litres.

The moles and mass of solute is calculated using the formula below:

1 mole of Kool-aid = 342 g

Molarity of solution = 1 M

Volume of solution = 237.5 mL = 0.2375 L

number of moles solute = molarity × volume

number of moles of Kool-aid = 1 M × 0.2375 L = 0.2375 moles

Mass of Kool-aid required = 0.2375 × 342 = 81.225 g of Kool-aid.

Therefore, 0.2375 L of water and 81.225 g of Kool-aid is required to make 237.5 mL of 1 M solution of Kool-aid.

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Answer:

D

Explanation:

First, the change of state from gas to liquid is known as CONDENSATION

Then, according to kinetic theory, when gas changes to liquid, the molecules 'slow down' in their random motion and thus, their kinetic energy deceases.

Therefore, condensation is usually followed by decreasing in energy.

Answer:

molecular formula of liquid = C₈H₁₈

Explanation:

First we determine the empirical formula of the liquid:

Number of moles of each element present in the liquid = % mass / molar mass

For Carbon, (molar mass = 12.01 g/mol) : 84.2/12.01 =7.011 moles

For Hydrogen (molar mass = 1.01 g/mol) : 15.8/1.01 = 15.643

Simplest mole ratio of the elements, C : H  is given by:

C = 7.011/7.011 = 1.0

H = 15.643/7.011 = 2.23

Multiplying through with 5, C:H = 5:11

Therefore, empirical formula is C₅H₁₁

The molecular mass of the liquid is next determined:

Using PV = nRT to find the number of moles of the liquid present

P = 5.0 atm; V = 568.0 mL = 0.568 L; R = 0.082 L*atmmol⁻¹ K⁻¹; T = 273 + 120 = 393 K

n = PV/RT = (5*0.568)/0.082*393

n = 0.088 moles

Molar mass of liquid = mass/no of moles = 10.0 g/ 0.088 moles = 113.63 gmol⁻¹

Molecular formula = n(empirical formula)

Molar mass of empirical formula, C₅H₁₁ = 71 gmol⁻¹

n = molecular mass/empirical mass = 113.63/71 = 1.6

Therefore, molecular formula =  1.6*(C₅H₁₁) = C₈H₁₈

Answer:

Mass

Explanation:

Answer:

The volume of stock solution needed to make the solution is 1875 ml

Explanation:

The parameters given are;

The volume of 75 M solution of H₂SO₄ = 400 ml

The concentration of stock solution = 16.0 M

Number moles per liter of stock solution = 16 moles

Number of moles in required 400 ml solution = 0.4×75 = 30 M

Volume of stock solution that contains 30 M = 30/16×1 = 1.875 l

The volume of stock solution that is required = 1875 ml

Answer:

The car is carrying a volume of air that is moving at 80 mph relative to the ground. But inside the car itself, the air not moving very much. If you take the windshield out and drive a 80 mph you will smash the fly onto the back window.

Explanation:

.

To perform the procedure, slowly add the calculated amount of 0.2 F AgNO3 from a buret while maintaining constant stirring. Additionally, incorporate a 10% excess of AgNO3 as directed by the instructor.

In this procedure, the objective is to add a specific amount of 0.2 F AgNO3 solution with constant stirring. The use of a buret allows for precise control over the volume being added. By adding the solution slowly, the reaction can be monitored and controlled more effectively.

Furthermore, it is mentioned that a 10% excess of AgNO3 should be added. This means that an additional 10% of the calculated amount of AgNO3 should be incorporated. The purpose of this excess is to ensure that all the reactants are fully consumed, promoting a complete reaction and maximizing the desired outcome.

The instructor will provide the 0.2 M AgNO3 solution, which is a solution with a known concentration. This concentration allows for accurate calculations of the required volume to achieve the desired amount of AgNO3.

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Answer:

its absorbed by heat from consumers

Answer: In everyday life, for example on food packaging, the word ‘pure’ usually means that something is in its natural state without anything added to it, like sweeteners or preservatives. Natural fruit juice is made up of sugar, water, and many other naturally occurring chemicals. In chemistry, this is not considered 'pure' and is a mixture.

Explanation:

Answer:

friction

Explanation:

jwjwkkskskaksksk

In the mass spectrum of pentane, the presence of a peak with m/z = 57 is most likely to the detection of a butyl radical cation.

Because all molecular ions fragment, a few compounds have mass spectra that lack a molecular ion peak. That is not a problem you will encounter at A'level. In the mass spectrum of pentane, for example, the heaviest ion has an m/z value of 72.

Lines with m/z values one or two lower than one of the simple lines, for example, are frequently caused by the loss of one or more hydrogen atoms during the fragmentation process. The base peak (the tallest peak and thus the most common fragment ion) is at m/z = 57 this time. This, however, is not produced by the same ion as the pentane m/z value peak.

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Methylene blue's molar mass, however, is 319.85. Methylene blue can have a mass of 0.32g.

The stock solution's prepared molarity is 0.01 M.

100 mL is the prepared stock solution volume.

Taken methylene blue millimoles equal 0.01 x 100 or 1 mmol.

Taken methylene blue mass is equal to moles times molar mass (10-3 x 319.85 = 0.31985 g).

Methylene blue is 99.99% pure, which means that if we take 100g of the substance, it actually includes 99.99 g of the dye.

Therefore, a sample of methylene blue equal to 100g must be taken for 99.99 g.

The required methylene blue sample is equal to (100/99.99)x0.31985, or 0.319 g.

Taken methylene blue mass is 0.32 g.

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