which substances are strong acids? [select all that apply] group of answer choices hf hi hcl h2so4 hno3 hbr hclo hclo2 hclo3 hclo4 h2s ch3cooh h3po4 nh3 nh4cl koh fecl3 h2n2 ca(oh)2 ch3nh2

The correct formula for the precipitate formed when aqueous solutions of [tex]FeCl_{3}[/tex] and [tex](NH_{4})2S[/tex] are mixed is [tex]Fe_{2}S_{3}[/tex].

А precipitаte is аn insoluble solid thаt forms from а chemicаl reаction in а solution. It hаppens when two solutions thаt contаin soluble sаlts аre mixed, аnd а new insoluble sаlt is formed. In this cаse, when аqueous solutions of [tex]FeCl_{3}[/tex] аnd [tex](NH_{4})2S[/tex] аre mixed, а solid precipitаte forms.

To determine the correct formulа for the precipitаte, we need to consider the reаction thаt tаkes plаce during mixing. Aqueous solutions of [tex]FeCl_{3}[/tex] and [tex](NH_{4})2S[/tex] react to form [tex]Fe_{2}S_{3}[/tex] (Iron(III) sulfide) and [tex]6NH_{4}Cl[/tex] (Ammonium chloride) as shown below:

[tex]Fe_{2}S_{3}[/tex] (aq) + 3 [tex](NH_{4})2S[/tex] (aq) → [tex](NH_{4})2S[/tex] (s) + [tex]6NH_{4}Cl[/tex] (aq)

So the correct formula for the precipitate formed is [tex]Fe_{2}S_{3}[/tex].

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Friedel crafts acylation is preferred over Friedel craft alkylation. Friedel crafts acylation reactions have many benefits as compared to Friedel crafts alkylation reactions.

Friedel-Crafts acylation and Friedel-Crafts alkylation reactions are both types of electrophilic substitution reactions that involve the formation of carbocations as intermediates. However, acylation is preferred over alkylation in certain situations.

Here are some benefits of Friedel-Crafts acylation reactions compared to Friedel-Crafts alkylation reactions:

1. Friedel-Crafts acylation reactions produce pure compounds as their major products because they do not involve any byproducts like Friedel-Crafts alkylation reactions.

2. The yields of Friedel-Crafts acylation reactions are often higher than those of Friedel-Crafts alkylation reactions.

3. Friedel-Crafts acylation reactions are more selective than Friedel-Crafts alkylation reactions because the acyl group is a better electrophile than the alkyl group.

4. The carbonyl group in the acylating agent (usually an acid chloride) can be selectively protected or modified using a variety of functional groups without affecting the aromatic ring. This is not possible in Friedel-Crafts alkylation reactions.

5. Friedel-Crafts acylation reactions can be carried out with a wider range of substrates (such as anisole or benzene) than Friedel-Crafts alkylation reactions.

6. The products of Friedel-Crafts acylation reactions are often more reactive than the starting materials, which allows for further functionalization or modification of the aromatic ring.

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The equilibrium concentration of HF would be 0.625 mol/L.

When 1.50 mol of HF is removed from the equilibrium mixture, the reaction will proceed to the left in order to regain equilibrium. The equation for this reaction is:

HF(g) ↔ H+(aq) + F−(aq)

Since the equilibrium constant, Kc, is constant, the ratio of the product and reactant concentrations remains unchanged.

Therefore, the equilibrium concentrations of HF can be determined by dividing the original equilibrium concentration by the number of moles of HF that have been removed.

The final equilibrium concentration of HF would be 0.625 mol/L (1.50 mol / 2.40 mol).

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A valid electronic configuration should be written as: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d¹⁰ 6p⁶ 7s².


An electron configuration is used to show the distribution of electrons among the orbitals of an atom in its ground state, and it is written in the order of increasing energy of the orbitals.

Let's now figure out which of the following is an incorrect electron configuration.

2d5 is not a possible electron configuration according to the rules of electron configuration.

However, it is incorrect because, in the modern periodic table, the d orbital comes after the s orbital, so it should be written as: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d¹⁰ 6p⁶ 7s².

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A wavelength of [tex]10^{-2}[/tex] meters corresponds to a frequency of about 3 × [tex]10^{14}[/tex] Hz, which places it in the microwave range of the electromagnetic spectrum.

Therefore, the type of radiation that would have a wavelength of [tex]10^{-2}[/tex]meters is a microwave radiation. Microwaves are a type of electromagnetic radiation that has a longer wavelength than visible light but shorter than radio waves. They are commonly used in communication, heating, and cooking applications. In particular, microwave radiation is used in microwave ovens to heat food by causing water molecules to vibrate, which generates heat. Additionally, microwave radiation is used in telecommunications, such as mobile phones and satellite communications.

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The combination of elements that are required for a compound to be considered organic are carbon and hydrogen. The correct answer among the given options is carbon and hydrogen.

Organic compounds are the fundamental components of life and are classified by the presence of carbon atoms, which are covalently linked to one another and to other elements such as oxygen, nitrogen, and sulfur, as well as by the lack of ionic bonding.

To summarize, an organic compound is a compound that contains carbon atoms bonded to hydrogen atoms, among other elements, in a covalent bond. The majority of organic compounds contain a carbon-carbon bond, which is the foundation of organic chemistry.

The following are some examples of organic compounds:

Methane, CH4

Ethanol, C2H5OH

Ethanoic acid, CH3COOH

Acetone, (CH3)2CO

Amino acid glycine, NH2CH2COOH

As a result, the correct combination of elements that are required for a compound to be considered organic are carbon and hydrogen.

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The percent yield of carbon dioxide is 66.90%.

To calculate the percent yield of carbon dioxide, we need to compare the actual yield of carbon dioxide with the theoretical yield of carbon dioxide that would be expected from the balanced chemical equation.

Let's say the chemical equation for the reaction that produces carbon dioxide is:

2 A + 3 B → 2 CO2 + C

Assuming that carbon dioxide is the only product, we can calculate the theoretical yield of carbon dioxide from the given amount of reactants used in the reaction.

If we know the mass of the limiting reactant that was used, we can use stoichiometry to calculate the theoretical yield of carbon dioxide.

Let's say that we used 5.0 g of reactant A, and that reactant A is the limiting reactant. If we know the molar mass of reactant A and the stoichiometric coefficients of the reactants and products in the equation, we can calculate the theoretical yield of carbon dioxide:

Calculate the number of moles of reactant A used:

moles of A = mass of A / molar mass of A

Use the stoichiometry of the equation to calculate the number of moles of carbon dioxide produced:

moles of CO2 = (moles of A) x (2 moles of CO2 / 2 moles of A)

Calculate the mass of carbon dioxide produced:

mass of CO2 = moles of CO2 x molar mass of CO2

Once we have calculated the theoretical yield of carbon dioxide, we can calculate the percent yield by dividing the actual yield by the theoretical yield and multiplying by 100:

percent yield = (actual yield / theoretical yield) x 100

Let's assume that the theoretical yield of carbon dioxide is calculated to be 7.25 g based on the amount of reactants used. If the actual yield of carbon dioxide is measured to be 4.85 g, the percent yield can be calculated as follows:

percent yield = (4.85 g / 7.25 g) x 100

percent yield = 66.90%

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Answer:

a. 3.19 m

b. 361.45 m

Explanation:

wavelength = speed of light ÷ frequency

speed of light = 3.00 x 10^8 m/s

AM is KILOhertz

830 kHz = 830,000 Hz

FM is MEGAhertz

93.9 MHz = 93,900,000 Hz

a.

wavelength = 3.00 x 10^8 m/s ÷ 830,000 Hz =

361.45 m

b.

wavelength = 3.00 x 10^8 m/s / 93,900,000 Hz = 3.19 m

Methane can be produced from various natural and human-related activities. However, one activity that is not a source of methane is photosynthesis.

Methane is a potent greenhouse gas that can contribute to climate change. It has a much greater warming effect than carbon dioxide, although it stays in the atmosphere for a shorter period of time. The effects of methane emissions can include:

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As you rise from low light intensity to higher light intensity, the rate of photosynthesis will increase because there is more light available to drive the reactions of photosynthesis.

The balanced equation is  [tex]2H_{2} O_{2}[/tex]–> [tex]2H_{2} O[/tex] +[tex]O_{2}[/tex]. This involves the spontaneous decomposition of hydrogen peroxide down into water and oxygen.

Spontaneous decomposition or chemical decomposition is defined as the process or effect of simplifying a single chemical entity into two or more fragments. It is is usually regarded and defined as the exact opposite of chemical synthesis.

This [tex]2H_{2} O_{2}[/tex]–> [tex]2H_{2} O[/tex] +[tex]O_{2}[/tex] reaction is of the spontaneous decomposition of hydrogen peroxide down into water and oxygen. here add 2 molecules of hydrogen peroxide and 2 molecules of water. Because of the oxygen is naturally diatomic the total number of atoms of each element is now the same on both sides of the equation so it is balanced equation. This decomposition reaction is one of the exceptions to the endothermic nature of decomposition reactions.

The chemical reaction usually need some driving force to make them spontaneous. In this reaction of hydrogen peroxide, becoming water and oxygen, the driving force is energy in the form of electricity.

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The correct question is,

How would you balance the following equation:

[tex]H_{2} O_{2}[/tex]–> [tex]H_{2} O[/tex] +[tex]O_{2}[/tex]

Answer:

claim

Explanation:

CLAIM: The newt population became more poisonous because the snakes in this environment caused poison to be an adaptive trait, and Poison Level 10 is the most common because the newts with this trait were able to live longer and reproduce more than other newts.

True, a saturated hydrocarbon has the maximum amount of hydrogens attached to the carbon skeleton.

Hydrocarbons are organic molecules that are made up of only carbon and hydrogen atoms. They may be composed of chains of various lengths, rings of various sizes, or a combination of both. The simplest hydrocarbons, such as methane (CH4), ethane (C2H6), and propane (C3H8), are gaseous at room temperature, whereas larger hydrocarbons are liquids, such as hexane (C6H14), or solids, such as hexadecane (C16H34).

Unsaturated hydrocarbons have carbon-carbon double or triple bonds in their structures, indicating that they are not completely saturated with hydrogen atoms. These hydrocarbons are commonly referred to as alkenes or alkynes, respectively. Alkenes have one double bond, whereas alkynes have one triple bond.

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Answer:

Complete combustion takes place in the presence of a sufficient amount of oxygen while an incomplete combustion reaction takes place when there is an insufficient amount of oxygen supply.

The equation for incomplete combustion of propane is: 2 C3H8 + 9 O2 → 4 CO2 + 2 CO + 8 H2O + Heat. If not enough oxygen is present for complete combustion, incomplete combustion occurs. The result of incomplete combustion is, once again, water vapour, carbon dioxide and heat.

The Ultimate BOD ( Biochemical Oxygen Demand) of the waste is 81.3 mg/L when the 5-day BOD is 20 mg/L and the BOD rate coefficient is 0.15/day.

The Ultimate Biochemical Oxygen Demand (BOD) is defined as he quantity of oxygen required to stabilize or,

eliminate biodegradable organic matter in wastewater by the action of aerobic microorganisms under controlled laboratory conditions at a specified temperature over a period of time.

The 5-day BOD is measured by calculating the amount of oxygen consumed by microorganisms over a period of five days.

The Ultimate BOD of a waste can be determined by knowing the 5-day BOD and BOD rate coefficient. The following formula is used to determine the Ultimate BOD:

Ultimate BOD = 5-day BOD × [(e^(k×t))-1] / e^(k×t)Where k is the BOD rate coefficient and t is the time required to reach the Ultimate BOD.

The Ultimate BOD of the waste as follows: 5-day BOD = 20 mg/L k = 0.15/day t = ? Ultimate BOD = 5-day BOD × [(e^(k×t))-1] / e^(k×t) Ultimate BOD = 20 × [(e^(0.15×t))-1] / e^(0.15×t)

The Ultimate BOD is reached after 20 days. Ultimate BOD = 20 × [(e^(0.15×20))-1] / e^(0.15×20) Ultimate BOD = 81.3 mg/L

Therefore, the Ultimate BOD of the waste is 81.3 mg/L when the 5-day BOD is 20 mg/L and the BOD rate coefficient is 0.15/day. The coefficient is the numerical multiplier of a variable or quantity that follows a term or a factor.

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The pH of the solution after 13.2 mL of barium hydroxide has been added is 13.69. The volume of the hydrocyanic acid solution is 26.5 mL, which is 0.0265 L.

The balanced equation for the reaction between hydrocyanic acid and barium hydroxide is:

2 HCN + Ba(OH)2 → Ba(CN)2 + 2 H2O

This reaction is a neutralization reaction, which means that the number of moles of acid is equal to the number of moles of the base at the equivalence point. We can use this information to calculate the number of moles of barium hydroxide that have reacted with the hydrocyanic acid.

n(Ba(OH)2) = M(Ba(OH)2) x V(Ba(OH)2)

where M(Ba(OH)2) is the molarity of the barium hydroxide solution and V(Ba(OH)2) is the volume of barium hydroxide solution added.

Using the given values, we have:

n(Ba(OH)2) = 0.489 mol/L x 0.0132 L

= 0.00646 mol

Since the stoichiometry of the reaction is 2:1 for HCN to Ba(OH)2, the number of moles of HCN that have reacted is half the number of moles of Ba(OH)2:

n(HCN) = 0.5 x n(Ba(OH)2)

= 0.5 x 0.00646 mol

= 0.00323 mol

The volume of the hydrocyanic acid solution is 26.5 mL, which is 0.0265 L. Thus, the initial concentration of hydrocyanic acid is:

M(HCN) = n(HCN) / V(HCN)

= 0.00323 mol / 0.0265 L

= 0.122 M

At the equivalence point, all of the hydrocyanic acids have reacted, so the concentration of hydroxide ions (OH-) in the solution is equal to the concentration of barium hydroxide:

[OH-] = M(Ba(OH)2) = 0.489 M

The hydrocyanic acid dissociates in water to form hydrogen cyanide and hydronium ions (H3O+):

HCN + H2O ⇌ CN- + H3O+

The equilibrium constant expression for this reaction is:

Ka = [H3O+][CN-] / [HCN]

The value of Ka for hydrocyanic acid is 4.9 x 10^-10.

At the equivalence point, all of the hydrocyanic acids have reacted, so the concentration of hydrogen cyanide and hydronium ions is zero. The concentration of hydroxide ions can be used to calculate the concentration of hydronium ions using the equation:

Kw = [H3O+][OH-]

where Kw is the ion product constant for water, which has a value of 1.0 x 10^-14 at 25°C.

Rearranging the equation gives:

[H3O+] = Kw / [OH-]

= 1.0 x 10^-14 / 0.489

= 2.04 x 10^-14 M

Taking the negative logarithm of this value gives:

pH = -log[H3O+]

= -log(2.04 x 10^-14)

= 13.69

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The reaction of propane and oxygen is C3H8 + 5O2 --> 3CO2 + 4H2O. If 15.0 g of C3H8 and 50.0g of O2 are allowed to react, then oxygen is the limiting reactant because there is only enough C3H8 to consume 45.0 g of oxygen, while there is enough oxygen to consume 30.0 g of C3H8.

Explanation : Propane combusts with oxygen according to the reaction C3H8+5O2-->3CO2+4H2O.If 15.0 g of C3H8 and 50.0 g of O2 are allowed to react, then oxygen is the limiting reactant.Steps to solve this problem:Calculate the number of moles of C3H8 and O2.Using the coefficients in the balanced equation, calculate the number of moles of CO2 and H2O that should be produced by the reaction.Using the mole ratio of C3H8 and O2, calculate the number of moles of O2 that are needed to react with 15.0 g of C3H8.Using the mole ratio of O2 and CO2, calculate the number of moles of CO2 that should be produced by reacting with the moles of O2 calculated in step 3.If the calculated number of moles of CO2 from step 4 is greater than the calculated number of moles of CO2 using the amount of O2 given in the problem (50.0 g), then oxygen is the limiting reactant. If the calculated number of moles of CO2 from step 4 is less than the calculated number of moles of CO2 using the amount of C3H8 given in the problem (15.0 g), then C3H8 is the limiting reactant. If the calculated number of moles of CO2 from step 4 is equal to the calculated number of moles of CO2 using either the amount of C3H8 or O2 given in the problem, then neither reactant is limiting.

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The new volume of the gas is approximately 29.5 L when the temperature is raised to 350K and the pressure is lowered to 1.5 atm.

To solve this problem, we can use the combined gas law, which states that,

(P1 × V1) / T1 = (P2 × V2) / T2

where P1, V1, and T1 are the initial pressure, volume, and temperature, respectively, and P2, V2, and T2 are the final pressure, volume, and temperature, respectively.

We can plug in the given values to get,

(2.3 atm × 17 L) / 299 K = (1.5 atm × V2) / 350 K

Solving for V2,

V2 = (2.3 atm × 17 L × 350 K) / (1.5 atm × 299 K)

V2 = 29.5 L

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The one IV bag of potassium chloride 10 meq in d5w 50 ml IV should last 24 hours  and is because the rate is set at 50 ml/hr, so after 24 hours, the full 50 ml of the IV bag will have been infused.

To calculate the duration of the IV bag, you need to divide the total volume (50 ml) by the rate (50 ml/hr). This gives you a duration of 1 hour.

To convert this to 24 hours, you need to multiply the result by 24, giving you a total of 24 hours.

Therefore, the one IV bag of potassium chloride 10 meq in d5w 50 ml IV should last for 24 hours when given at a rate of 50 ml/hr.

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The concentration of nitrate ions after mixing equal volumes of 1M NaNO3 and 1M KCl is 0.5M.

When equal volumes of 1M NaNO3 and 1M KCl are mixed, the nitrate ions (NO3-) and potassium ions (K+) will undergo a cation-anion exchange reaction to form potassium nitrate (KNO3) and sodium chloride (NaCl) as follows:

NaNO3 + KCl -> KNO3 + NaCl

The concentrations of Na+ and Cl- ions will remain unchanged after the reaction because they are spectator ions. However, the concentrations of NO3- and K+ ions will change.

Since the initial concentration of both NaNO3 and KCl is 1M, the initial concentration of NO3- is also 1M.

After the reaction, the moles of NO3- will be equal to the moles of K+ ions formed, which is 1/2 the initial concentration of KCl or 0.5M.

Therefore, the concentration of nitrate ions after mixing equal volumes of 1M NaNO3 and 1M KCl is 0.5M.

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The osmotic pressure of a solution made by dissolving 7.19 g of glucose in 18.9 ml of solution at 17.4°C can be calculated using the formula: Osmotic Pressure (atm) = Molarity (M) × Gas Constant (R) × Temperature (T).

Molarity = (Mass of Solute/ Molar Mass of Solute) / Volume of Solution
= (7.19 g / 180.2 g/mol) / 18.9 ml
= 0.3999 M

Gas Constant (R) = 0.08206 liter atm/mol K
Temperature (T) = 17.4°C + 273.15 = 290.55 K

Therefore, Osmotic Pressure (atm) = 0.3999 M × 0.08206 liter atm/mol K × 290.55 K
= 0.983 atm

The osmotic pressure of a solution is the hydrostatic pressure required to balance the osmotic pressure of a solution. This is determined by the concentration of the solute molecules, temperature, and the properties of the solvent. The osmotic pressure of a solution can be used to determine the boiling point, vapor pressure, and vapor pressure of a solution. Additionally, it is important for the transport of substances across biological membranes, as well as for the stability of colloidal suspensions.

In summary, the osmotic pressure (in atm) of a solution made by dissolving 7.19 g of glucose in 18.9 ml of solution at 17.4°C can be calculated using the formula: Osmotic Pressure (atm) = Molarity (M) × Gas Constant (R) × Temperature (T), and is equal to 0.983 atm.

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Answer: The compound of bromine and fluorine used to make UF6 has an empirical formula of BrF8, which contains 1 atom of bromine and 8 atoms of fluorine. This compound is composed of 58.37 mass percent bromine and 41.63 mass percent fluorine.

The compound of bromine and fluorine used to make UF6 is composed of 58.37 mass percent bromine. To determine its empirical formula, we can use the following equation:

Molecular Mass = Mass Percent Bromine/Atomic Mass Bromine * Number of Bromine Atoms + Mass Percent Fluorine/Atomic Mass Fluorine * Number of Fluorine Atoms

Using this equation, we can determine the empirical formula by rearranging the equation and making it easier to calculate. To do this, we can make all terms on the right side of the equation be a multiple of the smallest mass percent of the elements in the compound. In this case, the smallest mass percent is bromine, so we must make the fluorine mass percent be a multiple of 58.37.

58.37/Atomic Mass Bromine * Number of Bromine Atoms = Mass Percent Fluorine/Atomic Mass Fluorine * Number of Fluorine Atoms

Using this equation, we can calculate the number of bromine atoms and fluorine atoms. The atomic mass of bromine is 79.9 and the atomic mass of fluorine is 19. In this equation, the number of bromine atoms is 1, and the number of fluorine atoms is 8. This results in an empirical formula of BrF8.

In conclusion, the compound of bromine and fluorine used to make UF6 has an empirical formula of BrF8, which contains 1 atom of bromine and 8 atoms of fluorine. This compound is composed of 58.37 mass percent bromine and 41.63 mass percent fluorine.


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The pH of a 0.20 M acetic acid solution is 2.72.

The pH of a 0.20 M acetic acid solution can be calculated using the Ka of acetic acid, CH3COOH, which is 1.8 x 10-5.

We will use the equation for the dissociation of acetic acid to calculate the pH of the solution.

CH3COOH(aq) + H2O(l) ⇌ H3O+(aq) + CH3COO-(aq)

The equilibrium constant expression for the dissociation of acetic acid is given by

Ka = [H3O+][CH3COO-] / [CH3COOH].

Since we know the value of Ka and the initial concentration of acetic acid, we can solve for

the concentration of H3O+.Ka = [H3O+][CH3COO-] / [CH3COOH]

1.8 x 10-5 = [H3O+]2 / 0.20[H3O+]2 = 3.6 x 10-6[H3O+] = 1.9 x 10-3 M

The pH of the solution can then be calculated as:

pH = -log[H3O+]pH = -log(1.9 x 10-3)

pH = 2.72

Therefore, the pH of a 0.20 M acetic acid solution is 2.72.

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The volume of NaOH required to reach the equivalence point in the titration of a strong acid is typically smaller than the volume of NaOH required to reach the equivalence point in the titration of a weak acid.

This is because the strong acid is more reactive and therefore requires less base to neutralize it.

In a titration, the volume of a base (such as NaOH) required to reach the equivalence point is determined by the strength of the acid being titrated.

Generally speaking, a stronger acid will require a smaller volume of base than a weaker acid to reach the equivalence point.

This is because the stronger acid is more reactive, and it therefore requires less base to neutralize it.

When titrating a strong acid with a base such as NaOH, the equivalence point is reached when the number of moles of the acid is equal to the number of moles of the base.

In this situation, a relatively small volume of base will be required to completely neutralize the acid.

On the other hand, when titrating a weak acid with NaOH, the equivalence point is reached when the pH of the solution reaches the pKa of the acid.

This requires a much larger volume of NaOH than is required for titrating a strong acid, as the weak acid is much less reactive and therefore requires a larger volume of base to neutralize it.

In summary, the volume of NaOH required to reach the equivalence point in the titration of a strong acid is typically smaller than the volume of NaOH required to reach the equivalence point in the titration of a weak acid.

This is because the strong acid is more reactive and therefore requires less base to neutralize it.

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The stoichiometric factor is 6:1 that is 6 moles of [tex]Na_2S_2O_3[/tex] reacts with one mole of [tex]KIO_3[/tex]

The stoichiometric factor is a factor that shows the number of moles of a reactant or product that takes part in the chemical reaction. The balanced chemical equation provides the ratio of the reactants and products involved in a chemical reaction.

It is used to determine the stoichiometric factor which is the number of moles of a compound in a balanced equation.

The balanced equation for the given reaction is:

[tex]Na_2S_2O_3 + 2KIO_3 + H_2O \rightarrow I_2 + 2NaHSO_4 + 2KHSO_4[/tex]

First, write the balanced equation of the reaction between

[tex]Na_2S_2O_3 \times 2H_2O\ and\ KIO_3.KIO_3 + 6Na_2S_2O_3 + 9H_2O \rightarrow 3I_2 + 6Na_2SO_4 + 9H_2SO_4[/tex]

So, the stoichiometric factor, that is the number of moles, of [tex]Na_2S_2O_3\times 2H_2O[/tex] reacting with one mole of [tex]KIO_3[/tex] is 6 moles.

Therefore, 6 moles of  [tex]Na_2S_2O_3\times 2H_2O[/tex] are needed to react with one mole of [tex]KIO_3[/tex].

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At standard pressure, which substance becomes less soluble in water as temperature increases from 10.°C to 80.°C is : d. NH4Cl

Solubility of substance increases with temperature, as higher temperatures allow more particles to dissolve in solvent. However, there are some exceptions, where solubility decreases with increasing temperature.

In this case, we are looking for substance that becomes less soluble in water as temperature increases from 10°C to 80°C at standard pressure.

The correct answer is d. NH4Cl.

At standard pressure, solubility of NH4Cl decreases with increasing temperature due to its endothermic dissolution process. As temperature increases, heat absorbed by the solution also increases, which makes dissolution process less favorable. Therefore, solubility of NH4Cl decreases with increasing temperature.

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Answer: A popular classroom demonstration involves placing a paper cup with water in it on a burner and boiling the water in the cup. Although part of the cup may burn, the part containing the water does not. This is because of the phenomenon of surface tension.

Surface tension is the force that causes the molecules at the surface of a liquid to be attracted to one another, creating a film of molecules across the surface of the liquid. This causes the water molecules to stick together and form a barrier against the heat of the flame, thus protecting the water from the heat.

The water molecules at the surface of the cup create a protective film, allowing the heat of the flame to be distributed evenly throughout the cup. This prevents the water in the cup from boiling and keeps it from burning.

The surface tension phenomenon can also be seen in other forms of liquids such as soaps and detergents. When these liquids are placed in a container and agitated, the molecules form a protective film over the surface of the liquid and prevent it from evaporating.

Surface tension is a fascinating phenomenon that can be seen in everyday life, and it can be used to explain why the paper cup does not burn when placed on a burner.

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The temperature should be raised by 28.15°C to run 100 times faster than it does at room temperature with the catalyst.

To determine the temperature increase needed to make the catalyzed reaction run 100 times faster, we can use the Arrhenius equation:

[tex]k_{2}[/tex]/[tex]k_{1}[/tex] = e^(-Ea/R * (1/[tex]T_{2}[/tex] - 1/[tex]T_{1}[/tex])

Where [tex]k_{1}[/tex] and [tex]k_{2}[/tex] are the rate constants at temperatures [tex]T_{1}[/tex] and [tex]T_{2}[/tex], Ea is the activation energy (98.4 kJ mol-1), and R is the gas constant (8.314 J [tex]K^{-1}[/tex] [tex]mol^{-1}[/tex]).

Since we want the reaction to be 100 times faster, k2/k1 = 100. Now we can rearrange the equation and solve for [tex]T_{2}[/tex]:

1/[tex]T_{2}[/tex] - 1/[tex]T_{1}[/tex] = -R * ln(100)/Ea

Assuming room temperature ([tex]T_{1}[/tex]) is 298 K (25°C), we can plug in the values:

1/[tex]T_{2}[/tex] - 1/298 = -8.314 * ln(100)/98,400

1/[tex]T_{2}[/tex] = 1/298 + (8.314 * ln(100)/98,400)

[tex]T_{2}[/tex] = 1 / (1/298 + (8.314 * ln(100)/98,400))

Now, calculate the value of [tex]T_{2}[/tex]:

[tex]T_{2}[/tex] ≈ 326.3 K

To convert [tex]T_{2}[/tex] to °C, subtract 273.15:

[tex]T_{2}[/tex] = 326.3 - 273.15 ≈ 53.15°C

Therefore, you would need to raise the temperature by approximately 28.15°C (53.15 - 25) to make the catalyzed reaction run 100 times faster.

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The variation in phase observed at room temperature can be explained by the presence of polar bonds in one structure as opposed to nonpolar bonds in the other structure.

Due to the formation of hydrogen bonds between highly electronegative atoms (F, O, and N) and hydrogen, they are stronger than dipole-dipole interactions. As compared to any polar bond that has dipole-dipole interactions, the dipole is stronger because of the greater electronegativity differential.

Dipole-dipole interactions, London dispersion interactions (sometimes referred to as Van der Waals interactions), hydrogen bonds, and ionic bonds are the four basic intermolecular interaction types in charge of a compound's physical characteristics.

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