Question
How many moles of Na₂S₂O3 are needed to dissolve 0.65 mol of AgBr in a solution volume of
1.0 L, if Ksp for AgBris 3.3 x 10-13 and K for the complex ion [Ag(S₂03)2] is 4.7 × 10¹3?
Remember to use correct significant figures in your answer (round your answer to the nearest
tenth). Do not include units in your response.

The concentration of the solution is  0.000435 mol/dm³.

The concentration of a solution is calculated as follows;

Concentration = (Absorbance) / (Molar absorptivity x path length)

the path length =  3cm

the molar absorptivity (E) = 400 dm/mol/cm.

if the solution transmits 30% of the light, it absorbs 70% of the incident light.

Absorbance = log (1/Transmittance)

Absorbance  = log (1/0.3)

Absorbance  = 0.523

Concentration = (0.523) / (400 dm/mol/cm x 3 cm)

= 0.000435 mol/dm³

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The Lewis structure of the compound[tex]CH_{3} Br[/tex] is shown in the image attached.

The Lewis structure of a molecule or ion is produced by arranging the atoms in a manner that lessens the attraction between their valence electron pairs and then distributes the valence electrons among the atoms to form covalent bonds.

The octet rule, which states that atoms normally gain or lose electrons to obtain a stable configuration with eight valence electrons, frequently serves as a guidance when arranging electrons in the Lewis structure.

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Answer:

39.21 psi

Explanation:

According to Dalton's Law, the total pressure in the system is the sum of all the partial pressures, so all you need to do is add the partial pressures :)

The solubility of silver chloride (AgCl) can be represented by the following equilibrium equation:

AgCl(s) ⇌ Ag+(aq) + Cl-(aq)

In a saturated solution of AgCl, the concentration of Ag+ ions is equal to the solubility product constant (Ksp) for AgCl at that temperature. Since the concentration of Ag+ ions in the solution is given as 5.36 x 10^-6 M, we can write:

[Ag+] = 5.36 x 10^-6 M

According to the stoichiometry of the equilibrium equation, the concentration of chloride ions ([Cl-]) is also equal to the concentration of Ag+ ions, as one mole of AgCl dissociates to yield one mole of Ag+ ions and one mole of Cl- ions. Therefore:

[Cl-] = 5.36 x 10^-6 M

So, the concentration of chloride ions in the resulting solution is 5.36 x 10^-6 M.

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a) The amount of N₂O₅ is lowered to 2.5 g during the course of around 4.41 × 10⁴  seconds or 12.25 hours.

b) 9.71 L of O₂ are generated at 745 mmHg and 45 °C.

a) To solve for the time required for the quantity of N₂O₅ to be reduced to 2.5 g, use the first-order integrated rate law:

ln[N₂O₅]t/[N₂O₅]0 = -kt

where [N₂O₅]t = concentration of N₂O₅ at time t, [N₂O₅]0 = initial concentration of N₂O₅, k = rate constant, and t = time.

Find the initial concentration of N₂O₅:

n(N₂O₅) = m/M = 80.0 g / 108.01 g/mol = 0.7413 mol

[N₂O₅]0 = n/V = 0.7413 mol / 0.153 L = 4.846 M

where M = molar mass of N₂O₅ and V = volume of the solution.

Substituting the given values into the equation:

ln([N₂O₅]t / 4.846 M) = -6.2×10⁻⁴ s⁻¹ × t

When the quantity of N₂O₅ is reduced to 2.5 g, the concentration is:

n(N₂O₅) = m/M = 2.5 g / 108.01 g/mol = 0.02314 mol

[N₂O₅]t = n/V = 0.02314 mol / 0.153 L = 0.151 M

Substituting this concentration into the equation and solving for t:

ln(0.151 M / 4.846 M) = -6.2×10⁻⁴ s⁻¹ × t

t = 4.41 × 10⁴ s

Therefore, it takes approximately 4.41 × 10⁴ seconds or 12.25 hours for the quantity of N₂O₅ to be reduced to 2.5 g.

b) The balanced equation for the reaction shows that 1 mole of N₂O₅ produces 1/2 mole of O₂:

N₂O₅ → N₂O₄ + 1/2 O2

Therefore, the number of moles of O₂ produced can be calculated using the stoichiometry:

n(O₂) = 1/2 × n(N₂O₅) = 1/2 × 0.7413 mol = 0.3707 mol

The ideal gas law can be used to calculate the volume of O₂ produced at 745 mmHg and 45°C:

PV = nRT

where P = pressure, V = volume, n = number of moles, R = gas constant, and T = temperature in Kelvin.

Convert the pressure to atm and the temperature to Kelvin:

P = 745 mmHg / 760 mmHg/atm = 0.980 atm

T = 45°C + 273.15 = 318.15 K

Substituting the values and solving for V:

V = nRT/P = (0.3707 mol) × (0.08206 L·atm/mol·K) × (318.15 K) / (0.980 atm) = 9.71 L

Therefore, the volume of O₂ produced at 745 mmHg and 45°C is 9.71 L.

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Here are some questions on Personal Care services on E2020 are:

An infection is the entrance and growth of dangerous microorganisms in the body that harm the host, such as bacteria, viruses, fungus, or parasites.

Infections can be systemic (affecting the entire body) or localized (affecting a particular area of the body), and they can be moderate to severe.

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The amount of heat evolved when 25 moles of Sulfur is burned in excess oxygen is -74000 kJ.

The balanced reaction is given that is:

[tex]S_8(s) + 8 O_2(g) \rightarrow 8 SO_2(g)[/tex]

We can see that 1 mole of [tex]S_8[/tex] reacts with 8 moles of [tex]O_2[/tex] to produce 8 moles of [tex]So_2[/tex].

If 25.0 moles of [tex]S_8[/tex] reacts with excess Oxygen, then the amount of [tex]O_2[/tex] which is required in the reaction will be:

8 moles [tex]O_2[/tex] / 1 mole S8 × 25.0 moles S8 = 200 moles [tex]O_2[/tex]

We can use the enthalpy change and calculate the amount of heat evolved:

[tex]\Delta H[/tex] = -2368 kJ/ 8 moles [tex]SO_2[/tex]

The heat evolved = [tex]\Delta H[/tex] × moles of [tex]SO_2[/tex] produced

Moles of [tex]SO_2[/tex] produced =  8 moles [tex]SO_2[/tex] / 1 mole [tex]S_8[/tex] × 25.0 moles [tex]S_8[/tex]

= 200 moles [tex]SO_2[/tex].

Therefore, Heat evolved= -2368 kJ/ 8 moles [tex]SO_2[/tex] × 200 moles [tex]SO_2[/tex]

= -74000 kJ

The amount of heat evolved when 25 moles of Sulfur is burned in excess oxygen is -74000 kJ, the negative sign here indicates that the reaction is exothermic.

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The mass of Fe₂O₃ produced is 101.9 g.

The balanced chemical equation for the combustion of iron pyrite is:

4FeS₂(s) + 11O₂(g) → 2Fe₂O3(s) + 8SO₂(g)

From the equation, 11 moles of oxygen are required to produce 2 moles of Fe₂O₃. Convert the given volume of oxygen to moles:

n(O2) = PV/RT = (2.97 atm)(74 L)/(0.0821 L·atm/mol·K)(161 + 273 K) = 3.51 mol

Since the reaction requires 11 moles of O₂ for every 2 moles of Fe₂O₃, calculate the moles of Fe₂O₃ produced:

n(Fe₂O₃) = (2/11) × n(O₂) = (2/11) × 3.51 mol = 0.638 mol

Finally, use the molar mass of Fe₂O₃ to convert moles to grams:

m(Fe₂O₃) = n(Fe₂O₃) × M(Fe₂O₃) = 0.638 mol × 159.69 g/mol = 101.9 g

Therefore, the mass of Fe₂O₃ produced is 101.9 g.

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when pressure of the reactant mixture at the equilibrium and with the fewer moles in reactant side will be increased and the equilibrium will be shift to the side in the reaction where the fewer moles of the gas.

According to the Le Chartelier, when the reaction is in the equilibrium phase and the one of the constraints which will affect the rate of the reactions, and the equilibrium will be shift to the cancel out  this effect that the constraint had.

Therefore, If the pressure of the system or the reaction is in the equilibrium is change, the equilibrium of the reaction will be change that is depending on the side of the reaction with the highest number of the molecules.

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Answer:its b

Explanation:

jusut bcss

Answer:

Water molecules pull the sodium and chloride ions apart, breaking the ionic bond that held them together. After the salt compounds are pulled apart, the sodium and chloride atoms are surrounded by water molecules, as this diagram shows. Once this happens, the salt is dissolved, resulting in a homogeneous solution.

Explanation:

For P₂O₅ the intermolecular forces such as van der Waals forces, dipole-dipole interactions, and hydrogen bonding must be overcome.

For HI the intermolecular forces that must be overcome are as van der Waals forces, dipole-dipole interactions, and hydrogen bonding.

P₂O₅ is a covalent compound and it is solid. To convert P₂O₅ from a solid to a gas, intermolecular forces such as van der Waals forces, dipole-dipole interactions, and hydrogen bonding must be overcome.

HI is a covalent compound that is a gas at room temperature and pressure. To convert HI from a liquid to a gas, intermolecular forces such as van der Waals forces, dipole-dipole interactions, and hydrogen bonding must be overcome.

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Answer:

as the frequency goes down, the speed goes down by the same factor, and so the wavelength doesn't change.

Explanation:

Answer: as the frequency goes down, the speed goes down by the same factor, and so the wavelength doesn't change.

Explanation:

The wavelength of the radiation emitted by the argon ion laser is [tex]4.854 * 10^-7 m[/tex].

Wavelength is a property of any type of wave that refers to the distance between two adjacent points on the wave that is in phase, i.e., at the same point in their respective cycles. It is usually denoted by the Greek letter lambda (λ) and is measured in units of length, such as meters or nanometers.

The energy carried by the photon (E) is related to the wavelength ([tex]\lambda[/tex]) through the following equation:

[tex]E=hc/\lambda[/tex]; where 'h' is the Plank's Constant and 'c' is the speed of light which is [tex]3* 10^{-7} m/s[/tex].

We can say that

[tex]\lambda - hc/E[/tex]

Now after substituting the given values, we get:

[tex]\lambda = (6.626 * 10^{-34} J.s * 3.00 * 10^8 m/s) / (4.071 * 10^{-19} J)\\\lambda = 4.854 * 10^-7 m[/tex]

Therefore the wavelength of the radiation emitted by the argon ion laser is [tex]4.854 * 10^-7 m[/tex].

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When water and oxygen react in the presence of an acid, the oxygen can oxidize the acid to produce a compound and release hydrogen ions.

Reaction:

The reaction is as follows and it's diagram mentioned below.

Acid + Oxygen + Water → Compound + Hydrogen ions

if we take the acid hydrochloric acid (HCl), the reaction with oxygen and water can produce the compound chlorine dioxide ([tex]ClO_{2}[/tex]) and hydrogen ions ([tex]H^{+}[/tex]):

2 HCl + [tex]O_{2}[/tex] + [tex]H_{2}O[/tex] → 2 [tex]ClO_{2}[/tex] + 4 [tex]H^{+}[/tex]

This type of reaction is known as an oxidation-reduction reaction or a redox reaction, where one species is oxidized (loses electrons) while the other is reduced (gains electrons).

What is redox reaction?

A redox reaction, also known as an oxidation-reduction reaction, is a type of chemical reaction in which there is a transfer of electrons between two species. One species undergoes oxidation, meaning it loses electrons, while the other species undergoes reduction, meaning it gains electrons.

Redox reactions are fundamental to many processes in nature and technology, including photosynthesis, respiration, corrosion, and energy production in batteries and fuel cells. They are also important in many industrial processes, such as the production of metals, chemicals, and pharmaceuticals.

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Under the specified conditions, the cell potential of the given galvanic cell is 3.11 V.

The standard reduction potentials for the half-reactions involved in the cell reaction are:

Ti²⁺(aq) + 2e- ⇆ Ti(s) E° = -1.63 V

Au³⁺(aq) + 3e- ⇆ Au(s) E° = +1.50 V

The cell potential (Ecell) is given by:

Ecell = E°cathode - E°anode

where E°cathode = standard reduction potential of the cathode (reduction half-reaction) and E°anode = standard reduction potential of the anode (oxidation half-reaction).

In this case, Ti²⁺ is oxidized (anode) and Au³⁺ is reduced (cathode). Therefore:

E°anode = -1.63 V

E°cathode = +1.50 V

So, Ecell = +1.50 V - (-1.63 V) = +3.13 V

The Nernst equation can be used to calculate the cell potential (Ecell) under non-standard conditions:

Ecell = E°cell - (RT/nF) ln(Q)

where R = gas constant (8.314 J/(molK)), T = temperature in Kelvin (25 °C = 298 K), n = number of electrons transferred in the balanced equation (3 in this case), F = Faraday constant (96,485 C/mol), and Q = reaction quotient.

For the given concentrations:

[Ti²⁺] = 0.00140 M

[Au³⁺] = 0.887 M

The reaction quotient Q can be written as:

Q = ([Ti²⁺]³/[Au³⁺]²)

Substituting the values into the Nernst equation:

Ecell = E°cell - (RT/nF) ln([Ti²⁺]³/[Au³⁺]²)

Ecell = 3.13 V - (8.314 J/(molK) × 298 K / (3 × 96,485 C/mol)) ln(0.00140³/0.887²)

Ecell = 3.13 V - 0.0217 V

Ecell = 3.11 V

Therefore, the cell potential for the given galvanic cell under the given conditions is 3.11 V.

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Answer:

The enthalpy of combustion of ethanol is -1430 kJ/mol, which means that for every mole of ethanol that is burned, 1430 kJ of heat is released.

To determine the amount of heat given off from the combustion of 1 dm³ of ethanol, we need to first calculate the number of moles of ethanol in 1 dm³.

1 dm³ is equivalent to 1000 cm³. Since the density of ethanol is 0.79 g/cm³, the mass of 1 dm³ of ethanol can be calculated as:

mass = density x volume

mass = 0.79 g/cm³ x 1000 cm³

mass = 790 g

To convert this mass to moles, we need to divide by the molar mass of ethanol:

moles = mass / molar mass

moles = 790 g / 46 g/mol

moles = 17.17 mol

Therefore, 1 dm³ of ethanol contains 17.17 moles of ethanol.

To calculate the heat given off from the combustion of 1 dm³ of ethanol, we can use the following equation:

heat = enthalpy of combustion x moles of ethanol

heat = -1430 kJ/mol x 17.17 mol

heat = -24,551 kJ

Therefore, the heat given off from the combustion of 1 dm³ of ethanol is -24,551 kJ, or approximately 24,551 kJ of heat is released.

There are 0.25 moles of HCl present in 250 mL of 1.0 M HCl.

We have to calculate the number of moles of HCl present in some mL of 1.0M HCl.  A mole is defined as the amount of substance in a system that contains as many elementary entities as there are atoms in 0.012 kg of carbon 12. We represent mole by the symbol 'mol'. Now, we will see how to calculate the number of moles.

We can calculate the number of moles of a substance using the following expression;

Molarity = no of moles of an element/volume

According to this question, we were given 250. mL of 1.0 M HCl. The number of moles will be calculated by the formula as follows;

no of moles of HCl = 0.250L × 1.0M

no of moles of HCl = 0.250 moles.

Therefore, 0.25 moles are present in 250 mL of 1.0 M HCl.

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The volume of a 1.0 M solution that has 4.0 moles of solute is 4.0 liters.

A mole is a unit of measurement used in chemistry to represent the quantity of a chemical.

We can use the following formula to get the volume of a 1.0 M solution containing 4.0 moles of solute:

moles of solute = molarity x volume of solution

To determine the volume of the solution, we can rearrange this formula as follows:

Volume of solution = molarity / moles of solute.

By entering the specified values, we obtain:

4.0 moles / 1.0 M is the solution's volume.

Solution volume = 4.0 L

Therefore, the volume of a 1.0 M solution that has 4.0 moles of solute is 4.0 liters.

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Answer:

Since that means that we have 0.223 moles of PbCl2 in 1000mL of solution.

Also since mole ratio of the ions Pb2+:Cl- is 1:2

Thus, moles of Pb2+ = 0.223moles

concentration of Pb2+= 0.223M

Moles of Cl- = 2x0.223 moles

Concentration of Cl- = 0.446M

Explanation:

The equilibrium constant for the reaction at 655 K is [tex]e^{6.96}[/tex] ≈ 1.05 × 10^3.

The equilibrium constant (K) for a reaction is related to the change in Gibbs free energy (ΔG) through the equation:

ΔG = -RTlnK

where R is the gas constant, T is the temperature in kelvin, and ln is the natural logarithm. Since ΔG and ΔH (the change in enthalpy) are related by the equation:

ΔG = ΔH - TΔS

where ΔS is the change in entropy, we can rearrange the first equation to get:

lnK = -ΔH ÷ RT + ΔS ÷ R

At 298 K, we can use the given values of ΔH and K to solve for ΔS:

lnK = -ΔH ÷ RT + ΔS ÷ R

ln(1.4 × 10³) = (-(-25.8 × 10³ J/mol) ÷ (8.314 J/mol K × 298 K)) + ΔS ÷ 8.314 J/mol K

ΔS = 78.2 J/mol K

Now we can use the equation above to solve for lnK at 655 K, using the same value of ΔH and the newly calculated value of ΔS:

lnK = -ΔH ÷ RT + ΔS ÷ R

lnK = -(-25.8 × 10³ J/mol) ÷ (8.314 J/mol K × 655 K) + (78.2 J/mol K) ÷ 8.314 J/mol K

lnK = 6.96

e ≈ 1.05 × 10³

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Answer:

1.02 J/g°C.

Explanation:

We can use the equation:

q = m * c * ΔT

where q is the heat absorbed or released, m is the mass of the substance (in grams), c is the specific heat, and ΔT is the change in temperature (in Celsius).

First, we can calculate the heat gained by the water:

q_water = m_water * c_water * ΔT_water

where m_water is the mass of the water (in grams), c_water is the specific heat of water (4.184 J/g°C), and ΔT_water is the change in temperature of the water.

m_water = 120 g

c_water = 4.184 J/g°C

ΔT_water = (32°C - 24°C) = 8°C

q_water = (120 g) * (4.184 J/g°C) * (8°C) = 4009 J

This means that the heat lost by the unknown solid is equal to the heat gained by the water:

q_solid = -q_water

q_solid = -4009 J

Next, we can calculate the change in temperature of the solid:

ΔT_solid = (32°C - 80°C) = -48°C

Now, we can solve for the specific heat of the solid:

q_solid = m_solid * c_solid * ΔT_solid

-4009 J = (90.0 g) * c_solid * (-48°C)

c_solid = -4009 J / (90.0 g * -48°C)

c_solid = 1.02 J/g°C

Therefore, the specific heat of the unknown solid is 1.02 J/g°C.

The volume of oxygen gas, O₂ collected at 1.05 atm pressure and 44.0 °C when 42.5 g of KClO₃ decomposed is 13.01 L

We shall begin by obtaining the mole in 42.5 g of KClO₃. Details below:

Mole = mass / molar mass

Mole of CaC₂ = 42.5 / 122.5

Mole of CaC₂ = 0.35 mole

Next, we shall determine the mole of oxygen gas, O₂. produced. Details below:

2KClO₃ -> 2KCl + 3O₂

From the balanced equation above,

2 moles of KClO₃ decomposed to produced 3 mole of O₂

Therefore,

0.35 mole of KClO₃ will decompose to produce = (0.35 × 3) / 2 = 0.525 mole O₂

Finally, we shall determine the volume of oxygen gas, O₂ collected. Details below:

PV = nRT

1.05 × V = 0.525 × 0.0821 × 317

Divide both sides by 1.05

V = (0.525 × 0.0821 × 317) / 1.05

Volume of oxygen gas = 13.01 L

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Answer:

the pH of the solution is approximately 5.857.

Explanation:

The pH of a solution can be calculated using the formula:

pH = -log[H+]

where [H+] is the concentration of hydrogen ions in moles per liter (M).

In this case, [H+] = 1.39x10^-6 M, so:

pH = -log(1.39x10^-6)

= 5.857

Therefore, the pH of the solution is approximately 5.857.

Answer:

D is correct.(1s22s22p63s2)

We can use the combined gas law to solve this problem:

(P1V1/T1) = (P2V2/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 are given that the initial pressure is P1 = 37.8 mm Hg and the initial volume is V1 = 245 mL. The initial temperature is T1 = 23.5°C, which we need to convert to Kelvin by adding 273.15:

T1 = 23.5°C + 273.15 = 296.65 K

We are also given that the final volume is V2 = 54 mL, and the final temperature is the temperature of the ice water, which is 0°C or 273.15 K.

Now we can solve for the final pressure, P2:

(P1V1/T1) = (P2V2/T2)

P2 = (P1V1T2) / (V2T1)

P2 = (37.8 mm Hg * 245 mL * 273.15 K) / (54 mL * 296.65 K)

P2 = 24.4 mm Hg

Therefore, the pressure of the gas in the smaller flask at the new temperature is 24.4 mm Hg.

My model is distinct from the model in the image in that it takes a more thorough and all-encompassing approach to comprehending the fundamental parts of a system.

It considers the interactions between various system elements as well as the connections between those elements and their surroundings. It also looks at how the system changes over time, and how different components interact with each other.

As a result, the system may be understood more precisely, and management choices can be made with more knowledge. In order to offer a more precise and current picture of the system, my model also integrates the most recent research and technological advancements.

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Answer:

Leaves are green due to the presence of an organelle chloroplast (in abundance) which contains the pigment chlorophyll

Explanation:

Now saying chlorophyll pigment  is a green pigment might be slightly incorrect. The two famous types (Chlorophyll a, Chlorophyll b) only absorb red and blue light from the atmosphere and reflect green light hence giving the pigment a green appearance and lastly giving the leaves a green color too

Answer:

Chlorophyll

Explanation:

Plants are often seen as green to the human eye due to the presence of chlorophyll, which is the primary pigment used in photosynthesis. Chlorophyll absorbs light in the red and blue-violet parts of the spectrum, but reflects or transmits green light, resulting in the characteristic green color of leaves.