what is the percent concentration by mass if 7.24 g of barium fluoride are mixed with 51.35 g of distilled water to form a solution?

If 0.090 mole of solid NaOH is added to 1.0 liter of 0.180 m [tex]CH_3COOH[/tex]. The pH of the resulting solution is 4.74.

To solve this problem, we need to use the equation for the dissociation of acetic acid:

[tex]CH_3COOH + H_2O[/tex]⇌ [tex]CH_3COO^{-} + H_3O^+[/tex]

The addition of solid NaOH will react with the acetic acid to form sodium acetate and water:

[tex]CH_3COOH + NaOH[/tex] → [tex]CH_3COO^{-} Na^{+} + H_2O[/tex]

To calculate the pH of the resulting solution, we need to determine the new concentrations of [tex]CH_3COOH[/tex]and [tex]CH_3COO^-[/tex]. We can use the initial concentration of [tex]CH_3COOH[/tex]and the amount of NaOH added to calculate the new concentration of [tex]CH_3COOH[/tex]:

moles of [tex]CH_3COOH[/tex]= initial concentration x volume = 0.180 M x 1.0 L = 0.180 moles

moles of NaOH = 0.090 moles

moles of [tex]CH_3COOH[/tex]remaining = 0.180 moles - 0.090 moles = 0.090 moles

volume of solution = 1.0 L + 0.090 L = 1.090 L

new concentration of [tex]CH_3COOH[/tex]= moles / volume = 0.090 moles / 1.090 L = 0.0826 M

Since NaOH is a strong base, it will completely dissociate in water to form [tex]Na^+[/tex] and [tex]OH^-[/tex]. The [tex]OH^-[/tex] ions will react with the remaining [tex]CH_3COOH[/tex]to form [tex]CH_3COO^-[/tex], so the new concentration of [tex]CH_3COO^-[/tex] will be equal to the moles of NaOH added:

new concentration of [tex]CH_3COO^-[/tex] = 0.090 moles / 1.090 L = 0.0826 M

Now we can use the equilibrium expression for the dissociation of acetic acid to calculate the pH of the buffer:

[tex]Ka = [CH_3COO^{-}][H_3O^{+}] / [CH_3COOH]\\[H_3O^{+}] = Ka * [CH_3COOH] / [CH_3COO^{-}]\\[H_3O^{+}] = 1.8 * 10^{-5} * 0.0826 M / 0.0826 M\\[H_3O^{+}] = 1.8 * 10^{-5} M\\pH = -log[H_3O^{+}]\\pH = -log(1.8 * 10^{-5})\\pH = 4.74[/tex]

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Answer:

Explanation: The equation of water is H2+1/2O2=H2O

->5 moles H2O*1 mol O2/2 mole

-> H2O=2.5 moles O2

->Now, molar mass of O2= 2*16=32gms

To find mass of O2,

-> No. of moles*Molar mass of O2

->2.5*32=0 gms

So, the mass of oxygen will be 80 gms.

Ernest Rutherford's experiments in which alpha rays pass through a thin piece of metal foil demonstrated that atom contains a tiny nucleus containing most of the atom's mass. The correct answer choice is "the atom contains a tiny nucleus containing most of the atom's mass."

The alpha-particle scattering experiment was performed by Ernest Rutherford in 1911. It was conducted to discover the nature of atomic structure. The experiment demonstrated that most of the mass of an atom and all of its positive charge are contained in a small nucleus at the center of the atom.

The electrons were found to occupy almost all of the remaining space. Therefore, the correct answer is the atom contains a tiny nucleus containing most of the atom's mass.

Therefore "the atom contains a tiny nucleus containing most of the atom's mass." is the correct answer.

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The mass of aluminum that can be produced in 2.40 h is 1390 kg aluminum production by Electrolysis.

The amount of aluminum produced can be calculated using Faraday's law, which states that the amount of substance produced is directly proportional to the amount of electric charge passed through the cell, and the molar mass of the substance.

The balanced chemical equation for the electrolysis of [tex]Al_2O_3[/tex] is:

[tex]2 Al_2O_3(l)[/tex] -> [tex]4 Al(l) + 3 O_2(g)[/tex]

From the equation, we see that for every 4 moles of aluminum produced, 6 moles of electrons are needed.

The charge passed through the cell can be calculated using the formula:

charge = current × time

Where current is measured in amperes (A) and time is measured in hours (h). We need to convert 2.40 h to seconds:

[tex]2.40* 3600 = 8640 s[/tex]

The charge passed through the cell is:

charge = 1.15 million A × 8640 s = [tex]9.936 * 10^9[/tex] C

The number of moles of electrons passed through the cell can be calculated as:

moles of electrons = charge / Faraday's constant

Where Faraday's constant is 96,485 C/mol.

moles of electrons = [tex]\frac{9.936 * 10^9}{96,485} = 103,068[/tex] mol

Therefore, the number of moles of aluminum produced is half of the number of moles of electrons, or:

moles of Al = 0.5 × (103,068 mol) = 51,534 mol

The mass of aluminum produced can be calculated using the molar mass of aluminum, which is 26.98 g/mol:

mass of Al = moles of Al × molar mass of Al

mass of Al = [tex]51,534 * 26.98 = 1.39 * 10^6[/tex] g or 1390 kg

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The following statements on solutions, the following are as, 1- false. 2- false. 3- false.

1. The statement "a buffer can not be destroyed by adding too much strong base. it can only be destroyed by adding too much strong acid" is false. While it is true that buffers are most commonly used to resist changes in pH caused by the addition of strong acid, adding too much strong base can also destroy a buffer.

This is because a buffer works by containing both a weak acid and its conjugate base (or a weak base and its conjugate acid). If too much strong base is added to a buffer, it can convert the weak acid to its conjugate base, removing the buffer capacity.

2. The statement "adding 0.1 mol solid naoh to a 1.0 l of 0.2 m hcl solution could prepare a buffer" is false. In order to prepare a buffer, you need to have both a weak acid and its conjugate base (or a weak base and its conjugate acid) present in the solution.

The addition of NaOH to HCl will only create a strong acid-base reaction that will result in the formation of salt and water.

3. The statement "adding a small amount of acid to a buffer increases the ph of the buffer" is also false. When a small amount of acid is added to a buffer, the buffer will resist the change in pH caused by the addition of the acid.

This is because the buffer contains both a weak acid and its conjugate base (or a weak base and its conjugate acid), which can neutralize the added acid. As a result, the pH of the buffer will remain relatively unchanged.

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The 2 processes in which the particles will lose or gain energy are:

1) Temperature Change Or

2) State Change

A particle losses or gains energy when its Temperature changes i.e. A vessel filled with boiling water (100 degrees C ) cools down at the room temperature OR

when its State changes i.e. A bucket full of ice is kept at a room temperature (state changes from Solid to Liquid)

It is because of the breaking or formation of bonds, which results in loss or gain in energy.

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The order of bonds from highest polarity to the lowest is:1. O--Cl2. H--N3. C--S4. N--Si

Polarity can be explained as the extent to which electron density is unevenly distributed between atoms within a molecule. This can result in the molecule having a partial positive or negative charge.

The polarity of the bond is decided by the electronegativity difference between the two atoms involved in the bond. If the difference is greater, the bond will be more polar.

O--Cl bond: Oxygen is more electronegative than chlorine, so the bond is polar. Hence, O--Cl has the highest polarity.

H--N bond: The difference in electronegativity between hydrogen and nitrogen is not as great as that between oxygen and chlorine, but it is still significant. Hence, H--N has the second-highest polarity.

C--S bond: The difference in electronegativity between carbon and sulfur is less significant, making the bond less polar. Hence, C--S has the third-highest polarity.

N--Si bond: The difference in electronegativity between nitrogen and silicon is the least significant of all the bonds given. Thus, N--Si has the lowest polarity.

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The arrangement of radial, symmetric, and asymmetric letters is found in the attachment.

Bilateral Letters: These are letters that have a symmetrical shape where the left and right sides are mirror images of each other. In other words, if you were to draw a vertical line down the center of the letter, both sides would be identical.

Examples of bilateral letters include B, C, D, E, G, H, K, M, O, P, Q, R, S, U, and V.

Radial Letters: These are letters that have a symmetrical or circular shape around a central point. If you were to draw a circle around the letter, it would fit within that circle.

Examples of radial letters include A, C, D, M, and O.

Asymmetric Letters: These are letters that do not have symmetry or balance. If you were to draw a vertical line down the center of the letter, the two sides would not be mirror images of each other.

Examples of asymmetric letters include I, J, L, N, T, U, V, W, X, Y, and Z.

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Answer:

a. it is impossible to anticipate the pH profile of an acid and base without performing the experiment in the lab.

Explanation:

Ionic and covalent network solids are hard, have high melting points, and are poor electrical conductors.

Commonly, materials that are hard, have high liquefying focuses, and poor electrical conductivities are ionic solids or covalent organization solids.

Ionic solids are made out of a three-layered exhibit of emphatically and adversely charged particles kept intact by electrostatic powers. The solid ionic connections between the particles make these solids hard and high softening, while the shortfall of free electrons makes them unfortunate channels of power.

Covalent organization solids, then again, are made out of an immense organization of covalent connections between particles in a gem cross section structure. These covalent bonds are major areas of strength for incredibly, these solids high softening focuses and hardness. Once more, the shortfall of free electrons implies these materials are unfortunate conduits of power.

Instances of ionic solids incorporate NaCl (table salt) and MgO (magnesium oxide), while instances of covalent organization solids incorporate jewel and silicon dioxide (quartz).

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19.7 g of calcium carbonate must be decomposed to produce 5.00 L of carbon dioxide gas at STP.

Calcium carbonate, also known as limestone, can be heated to produce calcium oxide (lime), an industrial chemical with a wide variety of uses, and carbon dioxide. In order to produce 5.00 L of carbon dioxide gas at STP, we need to determine how many grams of calcium carbonate must be decomposed.The chemical equation for the thermal decomposition of calcium carbonate is:CaCO3 (s) → CaO (s) + CO2 (g).To find out the number of grams of calcium carbonate that must be decomposed to produce 5.00 L of carbon dioxide gas at STP, we can use the following steps:

Step 1: Write down the given information.V = 5.00 L (volume of carbon dioxide gas)T = 273 K (temperature at STP)P = 1 atm (pressure at STP)

Step 2: Use the ideal gas law to calculate the number of moles of carbon dioxide gas.n = PV/RT = (1 atm) x (5.00 L) / (0.0821 L atm/mol K x 273 K) = 0.197 mol

Step 3: Use the stoichiometry of the chemical equation to calculate the number of moles of calcium carbonate required to produce this amount of carbon dioxide gas.

The stoichiometry of the chemical equation tells us that 1 mole of CaCO3 produces 1 mole of CO2. Therefore, we need 0.197 moles of CaCO3 to produce 0.197 moles of CO2.Step 4: Calculate the mass of calcium carbonate required using its molar mass.The molar mass of CaCO3 is 100.1 g/mol. Therefore, the mass of CaCO3 required is:m = n x M = 0.197 mol x 100.1 g/mol = 19.7 g.

Therefore, 19.7g calcium carbonate must be decomposed to produce 5l of co2 gas at STP .

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In 8.00 seconds, approximately 0.84 mol of NO2 decomposes. The closest answer among the given options is 0.85 mol.

To find out how much NO2 decomposes in 8.00 s given the initial concentration and rate constant, we can use the second-order reaction formula:

1/[A]t = kt + 1/[A]₀

Where:
- [A]t is the concentration of NO2 at time t (which we want to find)
- k is the rate constant (0.255 M⁻¹s⁻¹)
- t is the time (8.00 s)
- [A]₀ is the initial concentration of NO2 (1.33 M)

Step 1: Plug the values into the equation.
1/[A]t = (0.255 M⁻¹s⁻¹)(8.00 s) + 1/(1.33 M)

Step 2: Calculate the value on the right side of the equation.
1/[A]t = 2.04 M⁻¹

Step 3: Solve for [A]t (concentration of NO2 at 8.00 s).
[A]t = 1/2.04 M⁻¹ = 0.49 M

Step 4: Calculate the change in concentration (how much NO2 decomposes).
Change in concentration = [A]₀ - [A]t = 1.33 M - 0.49 M = 0.84 M

Step 5: Convert the change in concentration to moles (since the volume is 1.00 L, the change in concentration is equal to the change in moles).
Change in moles = 0.84 mol

So, the correct option is 0.85 as it is closest to the answer.

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There are 0.511 moles of CO 2 in 22.5 gm of the substance

Determine the number of moles of carbon dioxide, we need to use the molecular weight of CO₂ and the given mass of 22.5 g.

The molecular weight of CO₂ is:

c = 12.01 gm/mol

o = 16.00 gm/mol

2 x o = 2 x 16.00 gm/mol = 32.00 gm/mol

So, the molecular weight of CO₂ is 12.01 gm/mol + 32.00 gm/mol = 44.01 gm/mol.

Now, we can use the formula:

no of moles = Given mass / Molar mass

where n is the number of moles, m is the given mass, and M is the molecular weight.

Plugging in the values, we get:

n = 22.5 g / 44.01 g/mol

n = 0.511 moles"

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Answer:

Check to make sure the equation is balanced

Explanation:

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Answer:

When 5.0 mL of chlorine gas is consumed, 1.3 mL of carbon tetrachloride gas is produced.

Explanation:

The balanced chemical equation for the reaction between methane gas and chlorine gas is:

CH4(g) + 4Cl2(g) → 4HCl(g) + CCl4(g)

From this equation, we can see that for every 4 moles of chlorine gas that react, 1 mole of carbon tetrachloride is produced.

To determine the volume of carbon tetrachloride produced when 5.0 mL of chlorine gas is consumed, we need to use the ideal gas law. Assuming standard temperature and pressure (STP), which is 0°C and 1 atm, the volume of one mole of any ideal gas is 22.4 L.

First, we need to calculate the number of moles of chlorine gas consumed:

n(Cl2) = V(Cl2) / Vm(Cl2)

n(Cl2) = 5.0 mL / 22.4 L/mol

n(Cl2) = 0.00022321 mol

According to the balanced equation, 4 moles of chlorine gas react to produce 1 mole of carbon tetrachloride. Therefore, the number of moles of carbon tetrachloride produced is:

n(CCl4) = n(Cl2) / 4

n(CCl4) = 0.00022321 mol / 4

n(CCl4) = 5.58025 x 10^-5 mol

Finally, we can calculate the volume of carbon tetrachloride produced using the ideal gas law:

V(CCl4) = n(CCl4) x Vm(CCl4)

V(CCl4) = 5.58025 x 10^-5 mol x 22.4 L/mol

V(CCl4) = 0.001251 L

Rounding to the correct number of significant digits, the volume of carbon tetrachloride produced is 0.0013 L or 1.3 mL.

Therefore, when 5.0 mL of chlorine gas is consumed, 1.3 mL of carbon tetrachloride gas is produced.

Answer:

Thus, to yield the enthalpy of the combustion reaction, we then sum the enthalpies of formation of the products, weighted by their stoichiometric coefficients, and subtract the enthalpy of formation of ethanol (-277.69kJ/mol). The result is ethanol's standard enthalpy of combustion of -1234.79kJ/mol.

The heat of formation of the given reaction is -56.4 kJ/mol.

What is Heat?

Heat is a form of energy that flows from a hotter object to a colder object. It is a type of energy transfer that occurs due to a temperature difference between two objects. The direction of heat flow is always from the object with higher temperature to the object with lower temperature until they reach thermal equilibrium.

The heat of formation of a compound is defined as the enthalpy change when one mole of the compound is formed from its constituent elements in their standard states at a given temperature and pressure. The standard state of an element is its most stable form at 1 atm pressure and a specified temperature.

Using the heat of formation values from standard tables, we can calculate the heat of formation of the products and reactants in the given reaction as follows:

Heat of formation of NO2 = -33.2 kJ/mol

Heat of formation of NO = +90.3 kJ/mol (since NO is an unstable gas at room temperature, we use the heat of formation of NO at 298 K instead of the standard heat of formation)

Heat of formation of O2 = 0 kJ/mol

Therefore, the heat of formation of the given reaction is:

ΔHf = Σ(heat of formation of products) - Σ(heat of formation of reactants)

ΔHf = [2(-33.2 kJ/mol)] - [2(90.3 kJ/mol) + 0 kJ/mol]

ΔHf = -56.4 kJ/mol

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According to galvanic cell video which is provided in the introduction page of this module, the statement which is true about the four main parts of the galvanic cell will be; "The cathode will be the electrode where reduction can occurs." Option D is correct.

A galvanic cell, also termed as a voltaic cell, in an electrochemical cell which  uses a spontaneous redox reaction to generate an electric current. The cell consists of two half-cells, each containing an electrode and an electrolyte solution. The two half-cells are connected by a wire as well as a salt bridge.

The other statements are not correct; The anode is the electrode where oxidation occurs, not where it receives electrons.

An external circuit with a voltmeter (not multimeter) is used to measure the potential difference (voltage) between the two electrodes.

There are two electrodes, but they are immersed in different solutions (not the same metal solutions).

The salt bridge is placed in both cells to allow the flow of ions to maintain electrical neutrality, not just in one cell.

Hence, D. is the correct option.

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--The given question is incomplete, the complete question is

"According to the galvanic cell video provided in the introduction page of this module, which statement is true about the four main parts of the galvanic cell? group of answer choices A) the anode is the electrode that receives electrons. B) an external circuit with a multimeter will be used to transfer electrons. C) there are two electrodes and they will be immersed in the same metal solutions. D) the cathode is the electrode where oxidation occurs. E) salt bridge will only be placed in one of the cells."--

Answer:

32

Explanation:

The pH at the equivalence point for a weak acid-strong base titration is affected by the dissociation constant of the weak acid and the hydrolysis of the resulting salt, leading to a pH above 7 at the equivalence point.

The pH at the equivalence point for a weak acid-strong base titration is affected by two main factors: the dissociation constant (Ka) of the weak acid and the hydrolysis of the resulting salt.
1. Dissociation constant (Ka) of the weak acid: The Ka represents the extent to which the weak acid ionizes in water. A smaller Ka value indicates a weaker acid, which means it ionizes less in solution. Since the equivalence point occurs when equal amounts of H+ and OH- ions have reacted, the degree of ionization of the weak acid will influence the pH of the solution.
2. Hydrolysis of the resulting salt: When a weak acid reacts with a strong base, a salt and water are formed. The anion of the salt, which is a conjugate base of the weak acid, can undergo hydrolysis, reacting with water to form a small amount of the weak acid and hydroxide ions (OH-). This hydrolysis increases the concentration of OH- ions in the solution, leading to a pH above 7 at the equivalence point.

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The molecular geometry of a given element is octahedral and it has no dipole moment. Therefore octahedral, No would be the correct answer.

The SF6 molecule has no dipole moment because each S−F bond dipole is balanced by one of equal magnitude pointing in the opposite direction of the other side of the molecule.

The three-dimensional configuration of the atoms that make up a molecule is known as molecular geometry. In addition to providing details about the molecule's overall shape, it also provides data on the bond lengths, bond angles, torsional angles, and any other geometrical factors that affect each atom's position.

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The concentration of the CH₃COOH solution is 0.0399 M.

The balanced chemical equation for the reaction between CH₃COOH and NaOH is:

CH₃COOH + NaOH → CH₃COONa + H₂O

From the equation, it can be seen that one mole of NaOH reacts with one mole of CH₃COOH. Therefore, the number of moles of NaOH used in the titration is:

0.880 mol/L × 0.00680 L = 0.00598 mol

Since the dilution did not affect the number of moles of CH₃COOH, the number of moles of CH₃COOH in the original solution is also 0.00598 mol.

The volume of the original solution used in the titration is:

15.0 mL/100.0 mL = 0.15

Therefore, the concentration of the CH3COOH solution is:

0.00598 mol/0.15 L = 0.0399 M

As a result, the CH₃COOH solution has a concentration of 0.0399 M.

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Explanation:

To solve this problem, we first need to calculate the number of moles of NH3 and HCl using the ideal gas law:

n = PV / RT

where:

P = pressure in atmospheres

V = volume in liters

T = temperature in kelvins

R = 0.08206 L atm/mol K (the ideal gas constant)

For NH3:

n(NH3) = (1.02 atm) (4.21 L) / (0.08206 L atm/mol K) (300 K)

n(NH3) = 0.177 mol

For HCl:

n(HCl) = (0.998 atm) (5.35 L) / (0.08206 L atm/mol K) (299 K)

n(HCl) = 0.204 mol

From the balanced equation, we can see that the stoichiometry of NH3 to HCl is 1:1, so NH3 is the limiting reactant since it has fewer moles. The balanced equation also tells us that the ratio of NH3 to NH4Cl is 1:1, so the number of moles of NH4Cl produced is also 0.177 mol.

To calculate the mass of NH4Cl produced, we need to use the molar mass of NH4Cl:

M(NH4Cl) = 14.01 g/mol (N) + 4(1.01 g/mol) (H) + 35.45 g/mol (Cl)

M(NH4Cl) = 53.49 g/mol

mass(NH4Cl) = n(NH4Cl) × M(NH4Cl)

mass(NH4Cl) = 0.177 mol × 53.49 g/mol

mass(NH4Cl) = 9.48 g

Therefore, the mass of NH4Cl produced is 9.48 g.

To determine the excess reactant, we can calculate how much of each reactant is consumed based on the stoichiometry of the reaction. Since NH3 and HCl react in a 1:1 ratio, we can assume that all of the NH3 is consumed, so we need to calculate the amount of HCl that is consumed:

n(HCl) consumed = n(NH3) = 0.177 mol

n(HCl) initially = (0.998 atm) (5.35 L) / (0.08206 L atm/mol K) (299 K)

n(HCl) initially = 0.204 mol

n(HCl) excess = n(HCl) initially - n(HCl) consumed

n(HCl) excess = 0.204 mol - 0.177 mol

n(HCl) excess = 0.027 mol

To convert this to a volume, we can use the ideal gas law:

V(HCl) excess = n(HCl) excess × RT / P

V(HCl) excess = 0.027 mol × (0.08206 L atm/mol K) (299 K) / (0.998 atm)

V(HCl) excess = 0.686 L

Therefore, the excess reactant is HCl, and the limiting reactant is NH3.

Answer: The surroundings will lower in temperature.

Explanation:

Endothermic reactions draw heat from the surroundings in order to occur. So the surroundings will feel cold, as the heat is being used as energy in the reaction.

The temperature of 1.2 moles of Helium gas at 1950 mm Hg if it occupies 15,500 ml of volume is 131°C.

We know we want to use the ideal gas law because we have the moles, pressure, and volume and we want to get the pressure: PV = nRT, where P = pressure, V = volume, n = number of moles, R = gas constant, and T = Kelvin temperature.

Here, the following values are given: 1950 mmHg of pressure, 15,500 mL (15.5 L), 1.2 mol of material, and R = 62.3638 (according to a table of gas constants). To the equation, include these:

Solve for temperature as follows: (1950) x (15.5 L) = (1.2) x (62.3638) x Temperature Temperature = 403.88 K.

We can convert this to degrees Celsius by subtracting 273 from the total:

3 sig figs = 403.88 - 273 = 130.88 - 131 (3 sig figs)

This causes the temperature to be 131°C.

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The rates are constant for this reaction 0.0782 M⁻¹s⁻¹. Second-order reactions are those in which the total of the exponents in the appropriate rate law of the chemical reaction equals two.

Second-order reactions are chemical processes that depend on either the concentrations of two first-order reactants or the concentration of one second-order reactant, according to the rate law equations provided below. A chemical reaction's half-life is the length of time it takes for half of the reactant to move through the reaction.

Second-order kinetics can be used to explain a variety of significant biological processes, including the creation of double-stranded DNA from two complementary strands. The total of the exponents in the rate law is equal to two in a second-order reaction. In this section, the two most typical types of second-order reactions will be thoroughly covered.

The differential (derivative) rate equation and the integrated rate equation are used to explain how the rate of a second-order reaction varies with the concentration of reactants or products. The integrated rate equation demonstrates how the concentration of species varies over time, whereas the differential rate law demonstrates how the reaction's rate changes over time.

Second order reaction for calculating rate constant is

t1/2 = 1/KCAO

Half-Life period = 18 s

Initial Concentration of reactant = 0.71M

18s = 1/K(0.71M)

K = 0.0782 M⁻¹s⁻¹

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The mass of 11.2 L of H2 gas at STP is 1.008 grams.

To find the mass of hydrogen gas in 11.2 L, we need to use the ideal gas law, which relates the pressure (P), volume (V), number of moles (n), and temperature (T) of a gas:

PV = nRT

where R is the gas constant.

To solve for the mass of H2, we need to know the pressure, temperature, and number of moles of the gas. Let's assume that the H2 is at standard temperature and pressure (STP), which is 0 degrees Celsius (273.15 K) and 1 atmosphere of pressure (101.3 kPa).

At STP, 1 mole of any gas occupies 22.4 L of volume. Therefore, the number of moles of H2 in 11.2 L can be calculated as:

n = (11.2 L) / (22.4 L/mol) = 0.5 mol

Now we can use the molar mass of hydrogen (2.016 g/mol) to convert the number of moles to mass:

mass = n x molar mass = 0.5 mol x 2.016 g/mol = 1.008 g

Therefore, the mass of 11.2 L of H2 gas at STP is 1.008 grams.

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The average of the three titrations is 0.104 m.

The standard deviation is 0.0055

The average of the three titrations is as follows:

(0.105 + 0.0990 + 0.110) / 3  = 0.104 m.

To calculate the standard deviation, we first need to calculate the variance. The variance is the sum of the squared differences from the mean, divided by the number of measurements minus one. Using the formula for variance, we get:

Variance = [(0.105 - 0.104)² + (0.0990 - 0.104)² + (0.110 - 0.104)²] / (3 - 1)

               = 3.05 x 10⁻⁵

The standard deviation is the square root of the variance, which is:

standard deviation = √(3.05 x 10⁻⁵) = 0.0055

The standard deviation is relatively small compared to the average, indicating that the results are precise. However, we cannot determine if the results are accurate without knowing the true value of the NaOH solution concentration. Therefore, we need to compare the average to the expected value or to a certified reference material.

If the average is within an acceptable range of the expected value or certified reference material, then the results are acceptable. Otherwise, more titrations should be performed to increase the precision and accuracy of the measurements. It is recommended to consult with a teacher or a supervisor to determine the appropriate number of titrations needed for the specific experiment.

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When 350 L of nitrogen gas and 900 L of hydrogen gas are combined at STP using the Haber process, 329 L of ammonia gas is produced.

Assuming the reaction goes to completion, the Haber process synthesizes ammonia at elevated temperatures and pressures from nitrogen gas and hydrogen gas. As a result, balanced chemical equation can be given as:N2(g) + 3H2(g) → 2NH3(g)The balanced equation establishes that one mole of nitrogen reacts with three moles of hydrogen to produce two moles of ammonia.

This implies that the stoichiometric ratios of the reactants are 1:3. Thus, if 350 L of nitrogen and 900 L of hydrogen are mixed in a reaction vessel and the reaction proceeds to completion, the limiting reactant will be nitrogen. Hence, the number of moles of nitrogen present is calculated using the ideal gas equation n = PV/RT as follows:n(N2) = PV/RT = (1 atm) x (350 L) / (0.08206 L atm/mol K) x (273 K) = 14.07 mol

Similarly, the number of moles of hydrogen is calculated as:n(H2) = PV/RT = (1 atm) x (900 L) / (0.08206 L atm/mol K) x (273 K) = 36.25 molSince nitrogen is the limiting reactant, it will completely react with 1/3 of the amount of hydrogen present to produce ammonia. As a result, the amount of ammonia generated will be equivalent to the quantity of nitrogen that reacted. Therefore, the volume of ammonia is calculated as follows:n(NH3) = n(N2) = 14.07 molV(NH3) = n(NH3) x (RT/P) = 14.07 x (0.08206 x 273 / 1) / (1) = 329 L

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Answer:

What is the heat in joules required to raise the temperature of 25 grams of water from 0 degrees C to 100 degrees C? What is the heat in calories? Answer: 10450 J or 2500 calories of heat energy are required to raise the temperature of 25 grams of water from 0 degrees C to 100 degrees C.