Which pair of atoms do you think have highest degree of solid solution solubility based on the information that is given a. Fe (BCC) & Al (FCC) b. Lithium (BCC) & Magnesium (HCP) c. Copper (FCC) & Aluminum (FCC) S
d. ilver (FCC) & Tungsten (BCC)

After adding the correct amount of solute (NaCl) to a large beaker to prepare 500 ml of a 1.35 M aqueous solution of NaCl,

You should follow these steps:

Add some distilled water to the beaker and dissolve the NaCl completely. Use a stirring rod to assist in the dissolution of the NaCl.

Once the NaCl is completely dissolved, add distilled water to the beaker until the volume reaches 500 ml.

Mix the solution thoroughly using a stirring rod or a magnetic stirrer to ensure that the concentration is uniform throughout the solution.

If necessary, adjust the concentration of the solution by adding more NaCl or distilled water, as required.

Transfer the solution to a clean and dry storage container, such as a plastic bottle or a glass flask, and label it with the concentration, date of preparation, and any other relevant information.

It is important to accurately measure the mass of NaCl to be added and use distilled water for the preparation to ensure that the final concentration of the solution is precise.

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The solubility of [tex]N_{2}[/tex] (mol [tex]N_{2}[/tex]/L solution) in an aqueous solution at a temperature of 37°C is approximately 0.015 g/L. Solubility refers to the maximum amount of a solute that can dissolve in a solvent to form a homogeneous solution.

The solubility of a gas in a liquid depends on several factors, including temperature, pressure, and the chemical nature of the gas and liquid. At higher temperatures, the solubility of gases typically decreases, which is known as the "temperature effect." In this case, the temperature is 37°C, which suggests that the solubility of [tex]N_{2}[/tex] in water will be relatively low. Other factors that can affect gas solubility include pressure, the presence of other solutes, and the chemical reactivity of the gas and liquid.

Thus, the solubility of [tex]N_{2}[/tex] (mol [tex]N_{2}[/tex]/L solution) in an aqueous solution at a temperature of 37°C is approximately 0.015 g/L.

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Under a rate law where [tex]C_2H_4[/tex] and [tex]Br_2[/tex] are both elevated by a factor of two. As a result, choice c is the appropriate response.

The rate law consistent with the proposed mechanism is:
rate = [tex]k[C_2H_4][Br_2]1/2[/tex]

This is because the second step is the rate-determining step, meaning it is the slowest step in the mechanism and limits the overall reaction rate. The rate law for the slow step involves both [tex]C_2H_4Br[/tex] and Br, but since Br is a reactant in the first step, we can substitute [Br] in terms of [tex][Br_2][/tex] using the fast equilibrium step. This results in a rate law with both [tex]C_2H_4[/tex] and [tex]Br_2[/tex] raised to the power of 1/2. Therefore, the correct answer is option c.

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50 g of sodium nitrate will precipitate out of the solution when it is cooled from 50°C to 20°C.

When a solution is saturated, it means that it contains the maximum amount of solute that can dissolve in the solvent at a particular temperature. If the temperature of the solution is lowered, the solubility of the solute decreases and some of it will start to precipitate out of the solution.

In this problem, we are given a solution of sodium nitrate ([tex]\mathrm{NaNO_3}[/tex]) in water at 50°C that is saturated. We are asked to determine how much sodium nitrate will precipitate out of the solution if it is cooled to 20°C.

To solve this problem, we need to know the solubility of sodium nitrate in water at both 50°C and 20°C. According to the solubility curve for sodium nitrate, the solubility of [tex]\mathrm{NaNO_3}[/tex] in water is approximately 113 g/100 g of water at 50°C and 88 g/100 g of water at 20°C.

Since we have 200 g of water in our solution, we can use these solubility values to determine how much sodium nitrate is dissolved in the solution at 50°C and how much will precipitate out at 20°C.

At 50°C, the solution is saturated with sodium nitrate, so we can assume that it contains 113 g of [tex]\mathrm{NaNO_3}[/tex] for every 100 g of water. Therefore, the total amount of [tex]\mathrm{NaNO_3}[/tex] in the solution is:

[tex]\left(\frac{113 \textrm{ g NaNO}_3}{100 \textrm{ g H}_2\textrm{O}}\right) \times (200 \textrm{ g H}_2\textrm{O}) &= 226 \textrm{ g NaNO}_3\end{align*}[/tex]

When we cool the solution to 20°C, the solubility of [tex]\mathrm{NaNO_3}[/tex] in water decreases to 88 g/100 g of water. This means that the maximum amount of [tex]\mathrm{NaNO_3}[/tex] that can remain dissolved in the solution is:

[tex]\left(\frac{88 \textrm{ g NaNO}_3}{100 \textrm{ g H}_2\textrm{O}}\right) \times (200 \textrm{ g H}_2\textrm{O}) &= 176 \textrm{ g NaNO}_3[/tex]

The difference between the amount of [tex]\mathrm{NaNO_3}[/tex] in the saturated solution at 50°C and the maximum amount that can remain dissolved at 20°C is the amount that will precipitate out of the solution:

[tex]226 \textrm{ g NaNO}_3 - 176 \textrm{ g NaNO}_3 &= 50 \textrm{ g NaNO}_3[/tex]

Therefore, 50 g of sodium nitrate will precipitate out of the solution when it is cooled from 50°C to 20°C.

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The enthalpy change for the formation of lead chloride by the reaction of lead chloride with chlorine is -11.6 kJ/mol.

It can be calculated by using Hess's Law. We can break down the overall reaction into two steps: the first is the dissociation of lead chloride into lead ions and chloride ions, and the second is the reaction of chlorine with the chloride ions to form lead chloride.

The enthalpy change for the first step, the dissociation of lead chloride, is given as ΔH = +92.8 kJ/mol. The enthalpy change for the second step, the reaction of chlorine with the chloride ions, is given as ΔH = -104.4 kJ/mol.
To calculate the enthalpy change for the overall reaction, we add the enthalpy changes for the two steps:
ΔH = ΔH1 + ΔH2
ΔH = (+92.8 kJ/mol) + (-104.4 kJ/mol)
ΔH = -11.6 kJ/mol

Therefore, by using formula of the enthalpy change, the result for the formation of lead chloride is -11.6 kJ/mol.

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Answer:

0.251 L

Explanation:

convert grams of solute to moles of solute using molar mass

13.3 g LiCl / 42.39 g LiCl gives 0.314 moles of LiCl

Molarity = moles / liter

rework this to solve for liters, which is

moles/molarity = liter

0.314 mol/1.25M = 0.251

The cruising depth at which the proposed desalination process would work is approximately 100 meters.

The osmotic device in the Navy submarines would work at a cruising depth where the pressure is high enough to allow for the process of osmosis to occur. Osmosis is the movement of water molecules from an area of low concentration to an area of high concentration through a semi-permeable membrane. In this case, the membrane would allow water molecules to pass through but not salt molecules.
To determine the cruising depth at which the osmotic device would work, we need to calculate the osmotic pressure of seawater at different depths. The osmotic pressure is the pressure required to prevent the movement of water molecules from a solution through a semi-permeable membrane.
The osmotic pressure of seawater can be calculated using the Van't Hoff equation:
Π = iMRT
where Π is the osmotic pressure, i is the van't Hoff factor (the number of particles the solute dissociates into), M is the molarity of the solution, R is the gas constant, and T is the temperature in Kelvin.
Assuming a temperature of 25°C (298 K), the osmotic pressure of seawater at the surface would be:
Π = 2 x 0.5 M x 0.0821 L atm mol-1 K-1 x 298 K
Π = 24.44 atm
At a depth of 100 meters, the pressure would be approximately 11 atm. Using the same equation, we can calculate the osmotic pressure of seawater at that depth:
Π = 2 x 0.5 M x 0.0821 L atm mol-1 K-1 x 298 K x (11 atm/1 atm)
Π = 268.84 atm
At this pressure, the osmotic device would be able to convert seawater into pure water through osmosis.  

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Geometric isomers of 3-Chloropent-2-ene are E and Z-type isomers of the compound. The structures of the compound in E and Z configurations are shown in the figures.

The isomers are of two types in this case that are E- type and Z- type. This system is based on different priorities given to different atoms in the compound.

When the higher priority group of both sides of the double bond are on the same side this is known as a Z-type configuration. And when opposite sides have the highest priority group this type of arrangement is known as an E-type configuration.

Z comes from the German word zusammen which means together while E comes from the German word entgegen meaning opposite

The real advantage of the E-Z system is that it works in every case. In contrast, the cis-trans system is not applicable to each case such as the one given in the question.

Since Cl is higher in priority than C in the ethyl group, the compound having -Cl and [tex]CH_3[/tex]  on the same side is called Z-type while one having them on different sides is known as E-type.

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The species that will not be involved in the net ionic equation are Zn²⁺(aq) and NO₃⁻(aq).

In order to determine which species will not be involved in the net ionic equation, we first need to write out the complete ionic equation. This equation shows all of the soluble ionic species in the reaction as separate ions. The complete ionic equation for the given reaction is:

Zn²⁺(aq) + 2NO₃⁻(aq) + 2H⁺(aq) + S²⁻(aq) → ZnS(s) + 2H⁺(aq) + 2NO₃⁻(aq)

From this equation, we can see that all of the species are involved in the reaction, including NO₃⁻(aq), Zn²⁺(aq), S²⁻(aq), and ZnS(s). However, the question asks which species will not be involved in the net ionic equation.

The net ionic equation is obtained by canceling out any spectator ions, which are ions that are present on both sides of the equation and do not participate in the reaction. In this case, the spectator ions are Zn²⁺(aq) and NO₃⁻(aq), which are present on both sides of the equation. The net ionic equation is therefore:

S²⁻(aq) + 2H⁺(aq) → H₂S(aq)

From this equation, we can see that the species that will not be involved in the net ionic equation are Zn²⁺(aq) and NO₃⁻(aq). The species that will be involved in the net ionic equation are S²⁻(aq), H⁺(aq), and H₂S(aq).

In summary, the species that will not be involved in the net ionic equation are Zn²⁺(aq) and NO₃⁻(aq), while the species that will be involved in the net ionic equation are S²⁻(aq), H⁺(aq), and H₂S(aq).

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In this chemical reaction, aluminum hydroxide is the base and nitric acid is the acid.

What is chemical reaction?

The reactants in this equation are not given, so it is not possible to predict the parent acid and base with certainty. However, based on the products, we can make some educated guesses.

[tex]Al(NO_{3})_{3}[/tex] is an ionic compound composed of aluminum ions ([tex]Al_{3}^{+}[/tex]) and nitrate ions ([tex]NO_{3}^{-}[/tex]). When this compound is dissolved in water, it dissociates into its constituent ions. The presence of  [tex]3H_{2}O[/tex] molecules on the product side of the equation suggests that the reaction involves a hydrated metal ion.

One possible reactant that could produce these products is aluminum hydroxide ([tex]Al(OH)_{3}[/tex]), which is a common base. When [tex]Al(OH)_{3}[/tex]  reacts with nitric acid (HNO3), it forms aluminum nitrate ([tex]Al(NO_{3})_{3}[/tex]) and water ([tex]H_{2}O[/tex]) as products.

The balanced chemical equation for this reaction would be:

[tex]Al(OH)_{3}[/tex] + [tex]3HNO_{3}[/tex] → [tex]Al(NO_{3})_{3}[/tex] + [tex]3H_{2}O[/tex]

In this equation, aluminum hydroxide is the base and nitric acid is the acid.

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Complete question is: Based on the following products, the parent acid and base are nitric acid and aluminum hydroxide.

The statement that ethanol cannot be added to the test tube too quickly because it will break up the DNA precipitate is true. Ethanol is commonly used in DNA extraction protocols to precipitate DNA from solution.

When added slowly, the ethanol gradually changes the conditions in the test tube, causing the DNA to come out of solution and form a visible clump. However, if too much ethanol is added too quickly, it can cause the DNA to break apart and become less visible or even invisible. This is because the high concentration of ethanol can disrupt the hydrogen bonds that hold the DNA strands together, causing them to unravel and lose their structure. Therefore, it is important to add ethanol slowly and carefully to avoid breaking up the DNA precipitate.

In summary, it is important to use the appropriate volume of ethanol based on the volume of the DNA solution, as too much or too little ethanol can also affect the precipitation of the DNA.

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The excited state of the electron was n= 6.

When the electron drops from its excited state to a lower level n=4, it releases a photon of light with a wavelength of 651 nm. The energy difference between the excited state and the lower level n=4 can be calculated using the Rydberg formula:

1/λ = R(1/n1^2 - 1/n2^2)

where λ is the wavelength of the photon, R is the Rydberg constant, and n1 and n2 are the initial and final energy levels of the electron, respectively.

Plugging in the given values, we get:

1/651 nm = R(1/n^2 - 1/4^2)

Solving for n gives us n=6. Therefore, the excited state of the electron was n=6 before it dropped to the lower level n=4 and released a photon of light with a wavelength of 651 nm.

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1)5 moles O2 x 2 moles H2O / 1 mole O2 = 10 moles H2O

10 moles H2O

Based on the balanced equation, for every 1 mole of oxygen used, 2 moles of water are produced. Therefore, 5 moles of oxygen would produce 10 moles of water.

2)3.00 moles H2O x 1 mole O2 / 2 moles H2O = 1.50 moles O2

1.50 moles O2

Based on the balanced equation, for every 2 moles of water produced, 1 mole of oxygen is used. Therefore, to produce 3.00 moles of water, 1.50 moles of oxygen must be used.

3)2.5 moles H2O x 2 moles H2 / 2 moles H2O = 2.5 moles H2

2.5 moles H2

Based on the balanced equation, for every 2 moles of water produced, 2 moles of hydrogen are used. Therefore, to produce 2.5 moles of water, 2.5 moles of hydrogen must be used.

4)4.00 grams H2 x 1 mole H2 / 2.016 grams H2 x 2 moles H2O / 2 moles H2 = 3.97 grams H2O

3.97 grams H2O

First, the number of moles of H2 used is calculated using the molar mass of H2. Then, the number of moles of H2O produced is calculated based on the balanced equation. Finally, the mass of H2O produced is calculated using the molar mass of H2O.

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(i) True. If f(n) E O(n) and g(n) E O(n), then it implies that there exist constants c1 and c2 such that f(n) ≤ c1n and g(n) ≤ c2n for all n > N, where N is some positive integer. Thus, f(n) + g(n) ≤ c1n + c2n = (c1 + c2)n for all n > N. This means f(n) + g(n) is bounded above by a constant multiple of n, i.e., f(n) + g(n) E O(n).

(ii) True. If f(n) E O(n), then it implies that there exists a constant c such that f(n) ≤ cn for all n > N, where N is some positive integer. Thus, n² x f(n) ≤ cn x n² = cn³ for all n > N. This means n² x f(n) is bounded above by a constant multiple of n³, i.e., n² x f(n) E O(n³).

(iii) True. If f(n) E O(n) and g(n) E O(n²), then it implies that there exist constants c1 and c2 such that f(n) ≤ c1n and g(n) ≤ c2n² for all n > N, where N is some positive integer. Thus, f(n) x g(n) ≤ c1n x c2n² = c1c2n³ for all n > N. This means f(n) x g(n) is bounded above by a constant multiple of n³, i.e., f(n) x g(n) E O(n³).

(iv) True. If f(n) € Θ(n log n), then it implies that there exist constants c1, c2, and N such that c1n log n ≤ f(n) ≤ c2n log n for all n > N. Thus, f(n) x n ≤ c2n² log n for all n > N. Since n^2 log n → ∞ as n → ∞, it follows that f(n) x n e o(n² log n).

(v) False. Consider f(n) = n log n and g(n) = n log n. Both f(n) and g(n) € Θ(n log n), but their product f(n) x g(n) = n² (log n)² € Θ(n² (log n)²), which is not constant.

(vi) False. Consider f(n) = n and g(n) = n. Both f(n) E Ω(n) and g(n) E Ω(n), but limn→∞ f(n) / g(n) = 1, which is not equal to any positive constant.

Overall, it is important to understand and apply the properties of asymptotic notation correctly to analyze the behavior of functions and algorithms as the input size grows.

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The ΔrH∘ for the given reaction at 25∘C is -1,484.0 kJ/mol.

To calculate ΔrH∘ for the given reaction, we need to use the standard enthalpy of formation values for the reactants and products.
2CaC2O4(s)⟶4CO(g)+O2(g)+2CaO(s)
The standard enthalpy of formation values can be found in the CRC_Std_Thermodyn_Substances and CRC_Std_Thermodyn_Aqueous-Ions tables.
Reactants:
2CaC2O4(s) -> ΔfH∘ = -2,239.0 kJ/mol

Products:
4CO(g) -> ΔfH∘ = -1,103.0 kJ/mol
O2(g) -> ΔfH∘ = 0 kJ/mol
2CaO(s) -> ΔfH∘ = -1,264.0 kJ/mol

ΔrH∘ = ΣΔfH∘(products) - ΣΔfH∘(reactants)
ΔrH∘ = [4(-1,103.0 kJ/mol) + 0 kJ/mol + 2(-1,264.0 kJ/mol)] - [2(-2,239.0 kJ/mol)]
ΔrH∘ = -1,484.0 kJ/mol

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Answer:

Explanation:

Explanation:

Solute = solution - solvent

Solute = 400 - 300

Solute = 100ml

Percentage of Solute = volume of Solute / volume of solution

= 25 %

Hope it helps you.

The δg∘rxn (free energy of reaction) for the reaction 2 H₂S (g) + 3 O₂ (g) → 2 SO₂(g) + 2 H₂O (g) is -990.6 kJ/mol

To find the δg∘rxn (free energy of reaction) for the reaction 2 H₂S (g) + 3 O₂ (g) → 2 SO₂ (g) + 2 H₂O (g), you'll need to use the following formula:

δg∘rxn = Σ (δgf products) - Σ (δgf reactants)

where, δgf is the standard Gibbs free energy of formation for each compound involved in the reaction.

The given values for δgf are:
- H₂S: -33.4 kJ/mol
- O₂: 0 kJ/mol (since it's an element in its standard state)
- SO₂: -300.1 kJ/mol
- H₂O: -228.6 kJ/mol

Now, apply the formula:

δg∘rxn = [(2 × -300.1) + (2 × -228.6)] - [(2 × -33.4) + (3 × 0)]

δg∘rxn = (-600.2 - 457.2) - (-66.8)

δg∘rxn = -1057.4 + 66.8

δg∘rxn = -990.6 kJ/mol

So, the δg∘rxn for the given reaction is -990.6 kJ/mol.

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The pH of the weak acid HA in a 1.4 M solution is around 2.64.

To find the pH of a 1.4 M solution of the weak acid HA with a Ka of 4.5×10⁻⁶, we can use the given equilibrium expression:

HA(aq) + H2O(l) ⇋ [tex]H_3O^+[/tex](aq) + A−(aq)

Since HA is a weak acid, we'll assume that only a small amount of it will ionize, and we can represent that amount as x:

HA ⇋ [tex]H_3O^+[/tex] + A−
Initial: 1.4 M     0      0
Change: -x        +x     +x
Equilibrium: 1.4-x M     x      x

Now, we can plug these equilibrium concentrations into the Ka expression:

Ka =[tex][H3O^+][/tex][tex][A^-][/tex]/[HA]
4.5×10⁻⁶ = (x)(x)/(1.4-x)

Since x is small compared to 1.4, we can approximate and assume that (1.4-x) ≈ 1.4:

4.5×10⁻⁶ = x^2/1.4

Next, we can solve for x:

x^2 = 4.5×10⁻⁶ * 1.4
x ≈ 2.3×10⁻³

Since x represents the [H3O+], we can now calculate the pH:

pH = -log[H3O+]
pH = -log(2.3×10⁻³)
pH ≈ 2.64

So, the pH of the 1.4 M solution of the weak acid HA is approximately 2.64.

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You should start timing the reflux for the procedure when the reflux ring stabilizes in the condenser. This indicates that the condenser is maintaining a consistent flow of coolant, and the reaction mixture is being heated and condensed properly.

Option A is correct

Coolant is a substance, usually a liquid or gas, that is used to cool down a system or device by absorbing and dissipating heat. In automotive engines, coolant is used to prevent the engine from overheating by absorbing heat and transferring it away from the engine.

Coolant is typically a mixture of water and ethylene glycol or propylene glycol, along with additives such as corrosion inhibitors, lubricants, and antifreeze agents.

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The activity that is most likely to be fueled primarily by lactic acid is the 400-meter run.

Lactic acid is produced in the body when there is not enough oxygen available to produce energy aerobically, which is the case during high-intensity exercises such as sprinting. The 400-meter run is a middle-distance sprint that requires a high level of anaerobic energy production.

This means that the body relies more on lactic acid to fuel the muscles during the race. On the other hand, the 5-mile run, 2-hour hike, and tennis serve are all endurance activities that rely more on aerobic energy production, which uses oxygen to produce energy. These activities do not require as much lactic acid production as the 400-meter run because they are less intense and require longer periods of sustained energy.

The 100-meter run, while still a sprint, is shorter and less intense than the 400-meter run, so it also relies less on lactic acid as a fuel source.

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The enthalpy change (ΔHrxn) for the reaction Ca(s) + 1/2 O2(g) + CO2(g) → CaCO3(s) is -813.4 kJ/mol.

To calculate δhrxn for the reaction:

Ca(s) + 1/2 O2(g) + CO2(g) → CaCO3(s)

We need to first determine the standard enthalpy of formation (ΔHf°) for each of the species involved.

ΔHf° for CaCO3(s) = -1206.9 kJ/mol
ΔHf° for Ca(s) = 0 kJ/mol
ΔHf° for O2(g) = 0 kJ/mol
ΔHf° for CO2(g) = -393.5 kJ/mol

Using Hess's law, we can then calculate the overall enthalpy change for the reaction:

ΔHrxn = ΣnΔHf°(products) - ΣmΔHf°(reactants)

ΔHrxn = [1(-1206.9 kJ/mol)] - [1(0 kJ/mol) + 1/2(0 kJ/mol) + 1(-393.5 kJ/mol)]

ΔHrxn = -1206.9 + 393.5 = -813.4 kJ/mol

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Ka for the acid HA is 0.115.

The Ka value is the equilibrium constant for the dissociation of an acid in water. It is a measure of the strength of an acid and is defined as the ratio of the concentrations of the products of the reaction to the concentration of the acid in its undissociated form.

To determine the Ka value for the acid HA, we need to write the chemical equation for the dissociation of HA in water. The equation is as follows:
HA + H2O ⇌ H3O+ + A-

From the equilibrium concentrations given, we can write the expression for the equilibrium constant, Ka, as:
Ka = [H3O+][A-]/[HA]

Substituting the given concentrations into the equation, we get:
Ka = (0.522 M)2 / 2.35 M = 0.115

Therefore, the Ka value for the acid HA is 0.115 with three significant figures.

This Ka value suggests that the acid HA is a weak acid as it has a low dissociation constant. It means that only a small fraction of the acid molecules dissociate into H3O+ and A- ions in water. Hence, the solution of HA would be acidic but not very strong.

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(a) The pH after 0.020 mol NaOH (strong base) is added to 1.00 L of 0.114 M HONH₂,  is 6.8.

(b) When 0.020 mol of NaOH is added to 1.00 L of 0.114 M HONH₃Cl, the resulting pH will be 3.99.

(c) When 0.020 mol of NaOH is added to 1.00 L of pure H₂O, the resulting pH will be 12.30.

(d) When 0.020 mol of NaOH is added to 1.00 L of a mixture containing 0.114 M of both HONH₂ and HONH₃Cl, the resulting pH will be 8.63.

(a) HONH₂ is a weak base and NaOH is a strong base. When NaOH is added, it reacts completely with the weak base to form water and the conjugate base ONH₂⁻.

HONH₂ + NaOH → ONH₂⁻ + H₂O

The initial concentration of HONH₂ was 0.114 M, and 0.020 mol of NaOH is added, which is equivalent to 0.020 M. The final concentration of ONH₂⁻ is also 0.020 M.

To find the pH, we need to find the pOH first. The Kb value of HONH₂ is given as 1.1×10⁻⁸.

Kb = [OH⁻][ONH₂⁻]/[HONH₂]

1.1×10⁻⁸ = [OH-][0.020]/[0.114]

[OH⁻] = 6.27×10⁻⁸ M

pOH = -log[OH⁻] = -log(6.27×10⁻⁸ M) = 7.2

pH = 14 - pOH = 6.8

(b) HONH₃Cl is an ammonium salt and NaOH is a strong base. When NaOH is added, it reacts completely with the ammonium cation to form water and the conjugate base NH₂⁻.

HONH₃Cl + NaOH → NH₂⁻ + H₂O + NaCl

The initial concentration of HONH₃Cl was 0.114 M, and 0.020 mol of NaOH is added, which is equivalent to 0.020 M. The final concentration of NH₂⁻ is also 0.020 M.

To find the pH, we need to find the pOH first. The Kb value of NH₃ is given as 1.8×10⁻⁵, and we need to use the Kw value to account for the H⁺ ion from water.

Kw = [H⁺][OH⁻] = 1.0×10⁻¹⁴

Kb = [NH₂⁻][H⁺]/[NH₃]

1.8×10⁻⁵ = [0.020][H⁺]/[0.114]

[H⁺] = 1.02×10⁻⁴ M

[OH⁻] = Kw/[H⁺] = 9.7×10⁻¹¹ M

pOH = -log[OH⁻] = -log(9.7×10⁻¹¹) = 10.01

pH = 14 - pOH = 3.99

(c) Pure H₂O is neutral with a pH of 7. Adding NaOH to pure water will increase the OH⁻ concentration and decrease the H⁺ concentration, resulting in a higher pH.

The initial concentration of H⁺ and OH⁻ is 1.0×10⁻⁷ M. The addition of 0.020 mol of NaOH is equivalent to 0.020 M.

[OH⁻] = 0.020 M

[H⁺] = Kw/[OH⁻] = 5.0×10⁻¹³ M

pH = -log[H⁺] = -log(5.0×10⁻¹³) = 12.30

(d) This is a buffer solution containing both HONH₂ and HONH₃Cl. The buffer capacity will resist changes in pH upon addition of a small amount of strong acid or base.

When NaOH is added, it will react with both HONH₂ and HONH₃Cl to form the conjugate bases ONH₂⁻ and NH₂⁻.

HONH2 + NaOH → ONH₂⁻ + H2O

HONH3Cl + NaOH → NH₂⁻ + H2O + NaCl

The initial concentration of each weak base was 0.114 M, and 0.020 mol of NaOH is added, which is equivalent to 0.020 M. The final concentration of ONH₂⁻ and NH₂⁻ will be 0.114 - 0.020 = 0.094 M.

To find the pH, we need to calculate the buffer capacity and determine which acid or base is present in excess.

pKa = -log(Ka) = -log(1.1×10⁻⁸) = 7.96

pH = pKa + log([A⁻]/[HA])

For the HONH₂/ONH₂⁻ buffer:

[HA] = 0.020 M (initial concentration of NaOH)

[A-] = 0.094 M (final concentration of ONH₂⁻)

pH = 7.96 + log(0.094/0.020) = 8.63

For the HONH₃Cl/NH₂⁻ buffer:

[HA] = 0.114 M (initial concentration of HONH₃Cl)

[A-] = 0.094 M (final concentration of NH₂⁻

pH = 7.96 + log(0.094/0.114) = 7.87

Since the pH of the HONH₂/ONH₂⁻ buffer is higher than the pH of the HONH₃Cl/NH₂⁻ buffer, it is in excess and will determine the final pH.

Therefore the pH = 8.63.

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Your monthly payment for a 30-year mortgage with a principal of $160,000 and an APR of 6.00% based on monthly compounding will be $959.60.

To calculate the monthly payment, we can use the formula for mortgage payments:

P = L[c(1 + c)^n]/[(1 + c)^n - 1]

where P is the monthly payment, L is the principal (loan amount), c is the monthly interest rate (APR/12), and n is the total number of payments (30 years x 12 months = 360).

Plugging in the values, we get:

P = 160000[(0.06/12)(1 + 0.06/12)^360]/[(1 + 0.06/12)^360 - 1]

P = $959.60

Therefore, the monthly payment for the given mortgage is $959.60.

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To convert from an absolute VO2 value to a relative VO2 value, you would divide the absolute VO2 (measured in L/min) by the individual's body weight (in kg). This will give you the relative VO2 value, which is expressed in ml/kg/min. This adjustment accounts for differences in body size and allows for more accurate comparisons between individuals.

To convert from an absolute to a relative VO2 value, you would divide the absolute VO2 value by the individual's body weight. This will give you the relative VO2 value, which takes into account the individual's size and allows for comparisons across different body types. The relative VO2 value is often used in exercise physiology and fitness testing to evaluate an individual's aerobic fitness level.

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Using the value of 1.27 g/cm^3 as silk density, calculate a fiber equivalent diameter in μm if such fiber linear density was 1 den: the fiber equivalent diameter in μm for a linear density of 1 den is 15.56 μm.

The first step is to understand the terms "value", "density", and "equivalent" in the context of the question.

Value refers to a numerical quantity that represents the worth or usefulness of something. In this case, the value is the density of silk, which is 1.27 g/cm^3.

Density refers to the mass of a substance per unit volume. In this case, it is the mass of silk per unit volume, which is expressed in grams per cubic centimeter (g/cm^3).

Equivalent refers to something that is equal or interchangeable with something else. In this case, it is the fiber equivalent diameter that is equivalent to a linear density of 1 den.

Now, to calculate the fiber equivalent diameter in μm, we need to use the formula:

Linear density (den) = mass (grams) / length (meters)

We know that the linear density is 1 den and the density of silk is 1.27 g/cm^3. We also know that the length is 9000 meters for 1 den (this is a standard conversion factor for denier).

So, we can rearrange the formula to solve for the mass:

Mass (grams) = linear density (den) x length (meters)

Substituting the values we know, we get:

Mass (grams) = 1 den x 9000 meters = 9000 grams

Now, we can use the density formula to calculate the volume of the silk:

Volume (cm^3) = mass (grams) / density (g/cm^3)

Substituting the values we know, we get:

Volume (cm^3) = 9000 grams / 1.27 g/cm^3 = 7086.61417 cm^3

Finally, we can use the formula for the diameter of a cylinder to solve for the fiber equivalent diameter:

Diameter (μm) = (4 x volume (cm^3) / π)^(1/3) x 10^4

Substituting the values we know, we get:

Diameter (μm) = (4 x 7086.61417 cm^3 / π)^(1/3) x 10^4 = 15.56 μm

Therefore, the fiber equivalent diameter in μm for a linear density of 1 den is 15.56 μm.

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During the dissolving process, 3,080 J, 146.4 J, and -0.01464 kJ/g of heat are supplied to the water.

The formula may be used to determine how much heat is absorbed by water.

Q = m * c * ΔT

where Q represents the heat absorbed, m represents the mass of the water and salt, c represents the heat capacity of the water, and T represents the temperature change.

We are aware that T is 3.50 °C and that the mass of water and salt is 200.0 g + 10.0 g = 210.0 g. When these values are added to the formula, we obtain:

Q=210.0 g*4.184 J/g°C *3.50 °C = 3,080 J

B) The following formula may be used to determine how much heat the salt loses:

Q = m * c * ΔT

where Q is heat loss, m is salt mass, c is water's specific heat capacity, and T is temperature change.

We are aware that the mass of the salt is 10.0 g, and that T is still 3.50 °C. When these values are added to the formula, we obtain:

Q = 10 g * 4.184 J/g°C * 3.50 °C = 146.4 J

C) By dividing the heat lost by the mass of salt and then converting to kilojoules, it is possible to compute the heat lost by the chemicals on a kilojoules per gramme basis:

Heat loss per gramme is equal to (-146.4 J) / (10 g) or -14.64 J/g.

Heat loss per gramme in kJ is calculated as (-14.64 J/g) / 1000 J/kJ = -0.01464 kJ/g.

As a result, the compounds lose -0.01464 kilojoules of heat per gramme of substance.

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The pH of a saturated solution of Ba(OH)2 tells us that there is an excess of OH- ions present in the solution. This means that the concentration of Ba2+ ions in the solution is equal to the solubility of Ba(OH)2.

The solubility product (Ksp) expression for Ba(OH)2 is:

Ksp = [Ba2+][OH-]^2

Since the pH of the saturated solution is 12, we know that the concentration of OH- ions is 10^-2 M. Therefore, the concentration of Ba2+ ions in the solution is also 10^-2 M.

Plugging these values into the Ksp expression, we get:

Ksp = (10^-2)(10^-2)^2 = 10^-6

Therefore, the value of Ksp for Ba(OH)2 is 1.0 x 10^-6, which is closest to option a. 4.0 x 10^-6.

A saturated solution is one in which the maximum amount of solute has dissolved in the solvent, and any additional solute will not dissolve. Solubility refers to the maximum amount of solute that can dissolve in a solvent under specific conditions.

For a saturated solution of Ba(OH)₂ with a pH of 12, the concentration of OH⁻ ions can be calculated using the relationship:

pOH = 14 - pH

pOH = 14 - 12 = 2

The concentration of OH⁻ ions is 10⁻² M. Since Ba(OH)₂ dissociates into one Ba²⁺ ion and two OH⁻ ions, the concentration of Ba²⁺ ions will be half that of OH⁻ ions (i.e., 10⁻²/2 M).

The solubility product (Ksp) for Ba(OH)₂ can be calculated using the expression:

Ksp = [Ba²⁺] * [OH⁻]²

Ksp = (10⁻²/2) * (10⁻²)² = 5.0 x 10⁻⁷

Therefore, the value of the solubility product (Ksp) of Ba(OH)₂ is 5.0 x 10⁻⁷, which corresponds to option d.

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The equilibrium concentration of Ammonia is 0.79 M at the specified temperature.

There are no limiting reagents because the reactants are present in stoichiometric ratios. A. The reaction between 4 moles of Hydrogen and 2 moles of Oxygen results in the formation of 4 moles of water.

H(g) + I(g) ⇌ 2HI(g)

Keq = [HI]² / [H][I]

At equilibrium, the concentrations are:

[H] = 4 moles / 1 L = 4 M

[I] = 4 moles / 1 L = 4 M

[HI] = 1 mole / 1 L = 1 M

These values are entered into the expression for Keq and result in: Keq = (1)² / (4)(4) = 0.0625

Therefore, at 250°C, the equilibrium constant for the reaction H(g) + I2(g) ⇌ 2HI(g) is 0.0625.

The balanced chemical equation for the reaction is:

N(g) + 2O(g) ⇌ 2NO(g)

Keq = [NO]² / [N][O]²

Substituting these values into the expression for Keq gives:

42 = (1.5)² / (1)([O]²)

Solving for [O] gives:

[O] = √(1.5² / 42) = 0.128 M

Therefore, at the given temperature, the equilibrium concentration of Nitrogen mono oxide is 1.5 M and the equilibrium concentration of Oxygen is 0.128 M.

3H(g) + N(g) ⇌ 2NH₃(g)

Keq = [NH₃]² / [H]³[N]

At equilibrium, the concentrations are:

[H] = 0.40 M

[N] = 0.25 M

[NH₃] = ? (to be calculated)

20 = [NH₃]² / (0.40)³(0.25)

Solving for [NH₃] gives:

[NH₃] = √(20 × (0.40)³ × (0.25)) = 0.79 M

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The molar solubility of manganese hydroxide is 0.00157 M.

a. To find the molar concentration of OH- in the saturated solution, we need to first determine the moles of HCl used in the titration:

moles HCl = Molarity x Volume (in L)
moles HCl = 0.0045 M x 0.00872 L
moles HCl = 3.93 x 10^-5 mol

Since Mn(OH)2 reacts with 2 moles of HCl for every 1 mole of Mn(OH)2, we can calculate the moles of Mn(OH)2 in the saturated solution:

moles Mn(OH)2 = 0.5 x moles HCl
moles Mn(OH)2 = 0.5 x 3.93 x 10^-5 mol
moles Mn(OH)2 = 1.97 x 10^-5 mol

Now we can calculate the molar concentration of OH-:

Molarity of OH- = moles OH- / Volume (in L)
Molarity of OH- = 1.97 x 10^-5 mol / 0.025 L
Molarity of OH- = 0.000788 M

b. The balanced equation for the dissolution of manganese hydroxide is:

Mn(OH)2 (s) ⇌ Mn2+ (aq) + 2OH- (aq)

The Ksp expression for this reaction is:

Ksp = [Mn2+][OH-]^2

At the saturation point, the concentration of Mn2+ is equal to the solubility product, so we can use the molar concentration of OH- found in part a to calculate Ksp:

Ksp = [Mn2+][OH-]^2
Ksp = (0.000788 M)(2 x 0.000788 M)^2
Ksp = 2.48 x 10^-8

Therefore, the Ksp for manganese hydroxide is 2.48 x 10^-8.

c. The molar solubility of manganese hydroxide is the molar concentration of Mn(OH)2 in a saturated solution. From part b, we know the solubility product of manganese hydroxide, so we can use that value to find the molar solubility:

Ksp = [Mn2+][OH-]^2
2.48 x 10^-8 = x(2x)^2
x = 0.00157 M

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