For the titration of 25.00 mL of 0.150 M HCl with 0.250 M NaOH, calculate:

(a) the pH when neutralization is 50% complete;

(b) the pH when 1.00 mL of NaOH is added beyond the equivalence point.

Answer:

True

Explanation:

Answer:

D.

The molecular structure has changed.

Explanation:

I got it right on Plato.

Answer: Volume decreases to half of original volume

Explanation:

Boyle's Law: This law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.

[tex]P\propto \frac{1}{V}[/tex]     (At constant temperature and number of moles)

[tex]{P_1V_1}={P_2V_2}[/tex]

where,

[tex]P_1[/tex] = initial pressure of gas = p

[tex]P_2[/tex] = final pressure of gas = 2p

[tex]V_1[/tex] = initial volume of gas = v

[tex]V_2[/tex] = final volume of gas = ?

Now put all the given values in the above equation, we get the final pressure of gas.

[tex]{p\times v}=2p\times V_2[/tex]

[tex]V_2=\frac{v}{2}[/tex]

Therefore, the final volume of the gas will become half of initial volume.

Covalent compounds can be identified by their low melting and boiling points, non-conductivity of electricity, and solubility in nonpolar solvents.

Covalent compounds, also referred to as molecular compounds, are usually identified through several physical properties such as their low melting and boiling points, non-conductivity of electricity, and their solubility in nonpolar solvents.

Low melting and boiling points: Covalent compounds don't have strong intermolecular forces, resulting in low melting and boiling points.

Non-conductivity of electricity: Since most covalent compounds don't dissociate into ions in solutions, they typically do not conduct electricity.

Solubility in nonpolar solvents: Due to the 'like soluble like' rule, covalent compounds tend to be soluble in nonpolar solvents rather than polar solvents such as water.

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Three equivalents of hydrogen can react with one equivalent of phosphorus to form compounds like phosphine, where each hydrogen atom forms a bond with one of the unpaired electrons of phosphorus.

Phosphorus typically has three unpaired electrons and hydrogen has one unpaired electron, which means that three equivalents of hydrogen can react with one equivalent of phosphorus. For instance, in the formation of phosphine, PH₃, three hydrogen atoms will combine with one phosphorus atom, each hydrogen providing one electron to form a single bond with phosphorus. Since phosphorus has three unpaired electrons available, it is able to form three single bonds with three hydrogen atoms, resulting in the phosphine compound.

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Answer : The perchloric acid and potassium hydroxide base is used to prepare the salt of potassium perchlorate, [tex](KClO_4)[/tex]

Explanation :

when the perchloric acid react with the potassium hydroxide as a base to form a salt of potassium perchlorate, [tex](KClO_4)[/tex]

The balanced chemical reaction will be,

[tex]HClO_4+KOH\rightarrow KClO_4+H_2O[/tex]

By the stoichiometry, we can say that 1 mole of perchloric acid react with the 1 mole of potassium hydroxide base to give 1 mole of potassium perchlorate and 1 mole of water as a product.

Hence, the perchloric acid and potassium hydroxide base is used to prepare the salt of potassium perchlorate, [tex](KClO_4)[/tex]

Answer:the answer is the Velocity of molecules in the body

Explanation:hope this helped have a great day

To find the Ka value for the weak acid, we can use the given pH and concentration of the acid solution. The Ka value is calculated using the equation Ka = ([H3O+][A-])/[HA].

To find the value of Ka for the weak acid, we can use the given pH and concentration of the acid solution. We know that the pH is a measure of the concentration of H3O+ ions in a solution, so we can use the pH to calculate the [H3O+] concentration. From the given pH of 3.25, we can determine that the [H3O+] concentration is 10^(-pH). So, [H3O+] = 10^(-3.25) M.

Now, the equilibrium equation for the dissociation of the weak acid is HA(aq) + H2O(l) -> H3O+(aq) + A-(aq). Since we know the [H3O+] concentration, we can assume that the [HA] concentration is equal to the [H3O+] concentration. So, [HA] = [H3O+] = 10^(-3.25) M.

The Ka value is calculated using the equation Ka = ([H3O+][A-])/[HA]. Substituting the given values, Ka = (10^(-3.25)^2)/10^(-3.25). Simplifying this expression gives us the value of Ka for the acid.

Answer:

The lengths of the C-C bonds increases with the decrease in the resistance of said bond, for example, a triple bond has a shorter length than in the case of a single bond. Butane has the single bonds and the CC bond is hybridized with sp3 hybridization, however in the butylcyclohexane structure the CC bond is also sp3 and the angle is 120°, however the angle shown is equal to 109.5°, so there is a certain angular tension and it is very unstable with respect to butane

Explanation:

The chemical formula of the bright yellow precipitate is PbI₂ (lead iodide).

A chemical equation is the representation of a chemical reaction which consists of reactants participating, formed products, and an arrow indicating the direction of the chemical reaction.

The equation that has the number of atoms of substances equal on either side of the chemical equation is known as a balanced chemical equation.

The law of conservation of mass has to be followed by a balanced chemical equation, according to which, the total mass of the elements on the reactant side must be equal to the total mass of elements on the product side.

The chemical equation of the reaction of lead nitrate and an aqueous solution of sodium iodide:

[tex]Pb(NO_3)_2(aq) + 2NaI \longrightarrow 2NaNO_3(aq) + PbI_2 (s)[/tex]

The bright yellow precipitate formed in the above chemical reaction has the chemical formula PbI₂.

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The correct answer is water.

During hydrolysis, _____water______ must be added before the bonds can be broken.

For example, sucrose undergoes hydrolysis to break into glucose and fructose. Here sucrose is a disaccharide and glucose and fructose are monosaccharides.  

Sucrose + H₂O → glucose + fructose

Esters are formed by the combination of carboxylic acid and alcohol.

But the hydrolysis of ester causes the release of carboxylic acid and alcohol.

RCOOR'(ester) + H₂O → RCOOH(carboxylic acid) + R'OH(alcohol)

The equilibrium constant for isomerization reaction is [tex]\boxed{9.615}[/tex]

Further Explanation:

The standard Gibbs free energy change in a reaction [tex]\left( {{{\Delta {\text{G}}_{{\text{rxn}}}^{{^\circ }}} \right)[/tex] is the difference of sum of the standard free energies of formation of product molecules and sum of standard free energies of formation of reactant molecules at the standard conditions. The formula used to calculate the value of standard Gibbs free energy  change for a reaction [tex]\left( {{{\Delta {\text{G}}_{{\text{rxn}}}^{{^\circ }}} \right)[/tex] is as follows:

[tex]\Delta\text{G}_{\text{rxn}}^{\circ}=\sum\text{n}\Delta\text{G}_{\text{f}(\text{products})}^{\circ}-\sum\text{m}\Delta\text{G}_{\text{f}(\text{reactants})}^{\circ}[/tex]

Here, n is the stoichiometric coefficients of products, and m are the stoichiometric coefficients of reactants in a balanced chemical equation.

The formula to determine the relationship between change in standard Gibbs free energy [tex]\left( \Delta{\text{G}^{\circ}} \right)[/tex] and equilibrium constant [tex]\left({\text{K}}\right)[/tex] is given as follows:

[tex]{\Delta }}{{\text{G}}^{{^\circ }}} = - {\text{RTlnK}}[/tex]       ......(1)

Here,

[tex]\Delta{\text{G}^{\circ}[/tex] is the standard Gibbs free energy change.

[tex]{\text{R}[/tex] is the gas constant.

[tex]{\text{T}}[/tex] is the temperature in Kelvin.

[tex]{\text{K}}[/tex] is the equilibrium constant.

The isomerization of glucose-1-phosphate to fructose-6-phosphate occurs in 2 steps:  

The reaction of step 1 is as follows:

[tex]{\text{glucose - 1 - phosphate}} \to {\text{glucose - 6 - phosphate}}[/tex]

                                       ......(2)

[tex]\Delta{\text{G}^{\circ}_{1}[/tex] for equation (2) is  [tex]- 7.28\;{\text{kJ/mol}}[/tex]

The reaction of step 2 is as follows:

[tex]{\text{fructose - 6 - phosphate}} \to {\text{glucose - 6 - phosphate}}[/tex]

                                                   ......(3)

[tex]\Delta{\text{G}^{\circ}_{2}[/tex] for equation (3) is  [tex]- 1.67\;{\text{kJ/mol}}[/tex]

Reverse the reaction of step 2.

[tex]{\text{glucose - 6 - phosphate}} \to {\text{frutcose - 6 - phosphate}}[/tex]

                                                ......(4)

[tex]\Delta{\text{G}^{\circ}_{3}[/tex] for equation (4) is [tex]+ 1.67\;{\text{kJ/mol}}[/tex]

Add equation (1) and (3) to get the final equation.

[tex]{\text{glucose - 1 - phosphate}} \to {\text{frutcose - 6 - phosphate}}[/tex]

To calculate [tex]\Delta {\text{G}}_{{\text{rxn}}}^{^\circ }}[/tex], add [tex]\Delta{\text{G}^{\circ}_{1}[/tex] and [tex]\Delta{\text{G}^{\circ}_{3}[/tex] as follows:  

[tex]\Delta{\text{G}^{\circ}_{\text{rxn}}=\Delta{\text{G}^{\circ}_{1}+\Delta{\text{G}^{\circ}_{3}[/tex]                       ......(5)

Substitute [tex]- 7.28\;{\text{kJ/mol}}[/tex] for [tex]\Delta{\text{G}^{\circ}_{1}[/tex] and [tex]+ 1.67\;{\text{kJ/mol}}[/tex] for [tex]\Delta{\text{G}^{\circ}_{3}[/tex] in equation (5).

[tex]\begin{aligned}\Delta {\text{G}}_{{\text{rxn}}}^{{^\circ }} &=  - 7.28\;{\text{kJ/mol + }} + 1.67\;{\text{kJ/mol}}\\{\text{}}&= - 5.61\;{\text{kJ/mol}}\\\end{aligned}[/tex]

For equilibrium constant (K), rearrange equation (1)

[tex]{\text{K}}={\text{e}}\frac{-\Delta{\text{G}}^{\circ}}{\text{RT}}[/tex]    ......(6)

Substitute [tex]- 5.61\;{\text{kJ/mol}}[/tex] for [tex]\Delta{\text{G}^{\circ}[/tex],[tex]8.314\;{\text{J/mol}} \cdot {\text{K}}[/tex] for R and [tex]298\;{\text{K}}[/tex] for T in equation (6)

[tex]\begin{aligned} {\text{K}}&= {{\text{e}}^{\frac{{ - \left( { - 5.61\;{\text{kJ/mol}}} \right)}}{{\left( {8.314\;{\text{J/mol}} \cdot {\text{K}}} \right)\left( {\frac{{{\text{1J}}}}{{1000{\text{kJ}}}}} \right)\left( {298\;{\text{K}}} \right)}}}}\\&= {{\text{e}}^{2.2634}}\\&= 9.615\\\end{aligned}[/tex]

The equilibrium constant for the reaction is 9.615.

Learn more:

1. The change in standard gibbs free is for a reaction: https://brainly.com/question/10838453

2. Determination of the equilibrium constant for pure water: https://brainly.com/question/3467841

Answer details:

Grade: Senior Secondary School

Subject: Chemistry

Chapter: Chemical Equilibrium

Keywords: Standard Gibbs free energy, equilibrium, constant, glucose-1-phosphate and fructose-6-phosphate.

ANSWER: There is only one path to get a career in chemistry.

EXPLANATION: There are many paths of career in Chemistry and not only one. It would not be justified to say that there is only one path to get a career in chemistry. Careers in chemistry would include Biochemistry, Forensic Scientist, Research Scientist, Chemical Engineer, Chemical Plant Operator and even a Science teacher or Chemistry professor in a school or university.

Answer:

answer in picture    It's B

Explanation:

Final answer:

The energy released from converting 1.0 g of mass into energy is 9 × 10¹³ joules (J), using the equation E = mc², where c is the speed of light.

Explanation:

According to Einstein's famous equation E = mc², where E represents energy, m is mass, and c is the speed of light in a vacuum, the energy released from completely converting 1.0 g of mass into energy is tremendously large. Since the speed of light, c, is approximately 3 × 10⁸ meters per second, and the mass m is 1.0 g (which is 1/1000 of a kilogram), the calculation is E = (1.0 g / 1000) × (3 × 10⁸ m/s)². This results in an energy release of 9 × 10¹³ joules (J), which is equivalent to about twice the energy released by the atomic bomb dropped on Hiroshima.

Explanation:

According to the the ideal gas law, PV= nRT.

This means that pressure is directly proportional to temperature.

So, when HCl and zinc are combined in a flask then it will lead to the formation of zinc chloride and hydrogen gas.

The reaction equation will be as follows.

             [tex]HCl + Zn \rightarrow ZnCl_{2} + H_{2}[/tex]

Since, the flask is sealed hence, hydrogen gas will not be able to move out of the flask.

So, when it will behave ideally then due to directly proportional relation between pressure and temperature there will occur a decrease in temperature with decrease in pressure.

The pressure in the vessel after the temperature increase is approximately 33.65 MPa.

We are given a scenario where liquefied natural gas (LNG) stored in a rigid, sealed vessel experiences a temperature increase beyond its critical point, causing it to transition from a liquid to a gas phase. We need to find the final pressure in the vessel using the ideal gas law.

2. Modeling the system:

We treat the LNG as pure methane ([tex]CH_4[/tex]) for simplification.

We assume the system behaves like an ideal gas, meaning it follows the ideal gas law.

3. Setting up the equation:

The ideal gas law relates pressure (P), volume (V), number of moles (n), gas constant (R), and temperature (T) through the equation:

PV = nRT

4. Identifying known and unknown values:

V: 0.17 m³ (volume of the vessel)

R: 8.314 J/(mol·K) (universal gas constant)

T: 200 K (final temperature)

P: Unknown (pressure we need to solve for)

5. Converting mass of LNG to moles:

Molar mass of methane ([tex]CH_4[/tex]): 16.04 g/mol

Mass of LNG (m): 55 kg = 55,000 g

Number of moles (n):

n = m / molar mass

n = 55,000 g / 16.04 g/mol

n ≈ 3433 mol

6. Solving for pressure:

Plug the known values into the ideal gas law and solve for P:

P = (n * R * T) / V

P = (3433 mol * 8.314 J/(mol·K) * 200 K) / 0.17 m³

P ≈ 33,647,247 Pa

7. Converting units and expressing final answer:

Convert pressure from Pascal (Pa) to Megapascal (MPa):

P = 33,647,247 Pa * (1 MPa / 1,000,000 Pa)

P ≈ 33.65 MPa

The question probable may be:

55 kg of liquefied natural gas (lng) are stored in a rigid, sealed 0.17 m3 vessel. in this problem, model lng as 100% methane. due to a failure in the cooling/insulation system, the temperature increases to 200 k, which is above the critical temperature; thus, the natural gas will no longer be in the liquid phase. What would be  pressure in the vessel after the temperature increase

The molar mass of carbon dioxide is 44.0 g/mol. a mass of 150.0 grams of carbon dioxide is equivalent to 3.41 mol of carbon dioxide.

In the International System of Units, the mole is the unit of substance quantity (SI). How so many elementary units of a certain substance are present in an item or sample is determined by the quantity of that material. There are precisely 6.022×10²³ elementary entities in a mole.

For instance, although having differing volumes and weights, 10 moles containing water along with ten moles of mercury both contain the same quantity of material, and the mercury includes exactly 1 atom for every molecule of water.

M(CO₂)=44.0 g/mol

m(CO₂)=150.0 g

n(CO₂)=m(CO₂)/M(CO₂)

n(CO₂)=150.0/44.0=3.41 mol

Therefore, 3.41 mol of carbon dioxide is there.

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Final answer:

To find the vapor pressure of the glycerin solution, calculate the moles of glycerin and water, determine the mole fraction of water, and apply Raoult's law using the vapor pressure of pure water at the specified temperature.

Explanation:

To calculate the vapor pressure of the solution containing glycerin in water, we will use Raoult's law, which states that the vapor pressure of a solution is directly proportional to the mole fraction of the solvent. The first step is to calculate the number of moles of glycerin (C3H8O3) by using its molar mass (92.09 g/mol), and then calculate the number of moles of water using its given density (1.00 g/mL) to convert the volume to mass and then to moles with its molar mass (18.015 g/mol).

Once we have both amounts in moles, we can calculate the mole fraction of water and apply Raoult's law to find the new vapor pressure of the solution, knowing the vapor pressure of pure water at the given temperature (30.0 °C) is 30.6 Torr.