The precipitation of an ionic substance from solution occurs when the ionic product exceeds the value of its solubility product at that temperature. Here the moles of Na₂S₂O₃ needed is
The solubility product of a sparingly soluble salt is defined as the product of the molar concentrations of its ions in a saturated solution of it at a given temperature.
Here the concentration of Ag⁺ ions = √Ksp = √3.3 × 10⁻¹³ = 1.81 × 10⁻¹³.
Moles of Ag⁺ ions: (1.82 x 10⁻¹³ M) x 1.0 L = 1.82 x 10⁻¹³ mol Ag⁺
Use the stoichiometry of the reaction to find the moles of Na₂S₂O₃ needed: 1 mol Na₂S₂O₃ / 2 mol Ag⁺ = 0.5 mol Na₂S₂O₃/mol Ag⁺
Moles of Na₂S₂O₃ required: 0.5 mol Na₂S₂O₃/mol Ag⁺ x 1.82 x 10⁻¹³ mol Ag⁺ = 9.1 x 10⁻¹⁴ mol Na₂S₂O₃
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A solution of thickness 3cm transmits 30%. calculate the concentration of the solution. E= 400dm/mol/cm
The concentration of the solution is 0.000435 mol/dm³.
What is the concentration of the solution?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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Iron pyrite (FeS2) is the form in which much of
the sulfur exists in coal. In the combustion of
coal, oxygen reacts with iron pyrite to produce
iron(III) oxide and sulfur dioxide, which is a
major source of air pollution and a substantial
contributor to acid rain. What mass of Fe2O3
is produced from 74 L of oxygen at 2.97 atm
and 161◦C with an excess of iron pyrite?
Answer in units of g
The mass of Fe₂O₃ produced is 101.9 g.
How to calculate mass ?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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We wish to determine the mass of Mg required to react completely with 250mL of 1.0 M HCI. HCI reacts with Mg according to the equation below.
2HCl(aq) + Mg(s) → MgCl₂(aq) + H₂(g)
How many moles of HCI are present in 250. mL of 1.0 M HCl?
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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Does anyone have Personal Care services on E2020
Here are some questions on Personal Care services on E2020 are:
A client with a new ileostomy has been home for four days. The HHA is giving the client a bath and notices that the pouch is full. The HHA should say to the client: D. "I'll empty the pouch for you."A client is bedridden at home and has an infected draining sacral wound. The infection control supplies that should be kept in the home is: Gloves.What is infection?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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Silver chloride, AgCl, is a sparingly soluble solid. Answer the following questions about a saturated solution prepared by placing solid silver chloride in a 2.45 10-5 M NaCl(aq) solution. At some temperature, the silver ion concentration, [Ag+], was found to be 5.36 10-6 M.
(a) What is the concentration of chloride ions, [Cl − ], in the resulting solution?
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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The following first-order reaction occurs in CCL4(l) at 45°C: N2O5》N2O4+1÷2O2. The rate consast is k=6.2×10^-4 s^-1 an 80.0 g sample of N2O5 in CCL4 is allowed to decompose at 45°C
a) how long does it take for the quantity of N2O5 to be reduced yo 2.5 g ?
b) how many liters of O2 measured at 745 mmHg and 45°C, are produced up to this point ?
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.
How to find quantity?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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An argon ion laser emits visible radiation with photons of energy 4.071 x 10-19 J. What is the
wavelength of the radiation?
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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Calculate the cell potential for the galvanic cell in which the given reaction occurs at 25 °C, given that [Ti2+]=0.00140 M and [Au3+]=0.887 M .
3Ti(s)+2Au3+(aq)↽−−⇀3Ti2+(aq)+2Au(s)
Under the specified conditions, the cell potential of the given galvanic cell is 3.11 V.
How to determine cell potential?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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How many grams of NiNO can be produced if 35.1 g of ammonium nitrate and 552 g of sodium phosphate react?
Answer:its b
Explanation:
jusut bcss
The enthalpy combustion of ethanol is -1430 kJ/mol. Determine heat given off from the combustion of 1 dm³ of ethanol. Given density of ethanol is 0.79 gcm³. (molar mass ethanol = 46 g/mol)
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.
A 7.95 L
container holds a mixture of two gases at 25 °C.
The partial pressures of gas A and gas B, respectively, are 0.352 atm
and 0.715 atm.
If 0.240 mol
of a third gas is added with no change in volume or temperature, what will the total pressure become?
Why are leaves green
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.
Describe the intermolecular forces that must be overcome to convert each of the following from a liquid or solid to a gas.
P2O5 and HI
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.
What are the intermolecular forces that must be overcome?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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What volume of oxygen gas can be collected
at 1.05 atm pressure and 44.0◦C when 42.5 g
of KClO3 decompose by heating, according to
the following equation?
2 KClO3(s) ∆
−−−−→
MnO2
2 KCl(s) + 3 O2(g)
Answer in units of L.
005 1.0 points
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
How do i determine the volume of oxygen gas collected?We shall begin by obtaining the mole in 42.5 g of KClO₃. Details below:
Mass of KClO₃ = 42.5 g Molar mass of KClO₃ = 122.5 g/mol Mole of KClO₃ =?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:
Pressure (P) = 1.05 atmTemperature (T) = 44 °C = 44 + 273 = 317 KGas constant (R) = 0.0821 atm.L/mol KNumber of mole (n) = 0.525 moleVolume of gas (V) =?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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6. A 90.0 gram sample of an unknown solid is heated to 80 °C and placed into a calorimeter containing 120 grams of water at 24 °C. If the final temperature of the solid sample and the water is 32 °C, what is the specific heat of solid?
PLEASE SHOW WORK
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.
a student mixed 20 grams of salt into a beaker with 200 milliliters of warm water. then, the student set the cup of saltwater on a windowsill undisturbed for one week. what changes did the student observe? include what happened when salt was mixed with warm water and what most likely happened to the saltwater after one week.
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:
A sample of gas is contained in a 245 mL flask at a temperature of 23.5°C. The gas pressure is 37.8 mm Hg. The gas is moved to a new flask, which is then immersed in ice water, and which has a volume of 54 mL. What is the pressure of the gas in the smaller flask at the new temperature?
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.
What is the electron configuration for magnesium (Mg)?
O A. 1s²2s²2p²356
B. 15²25²3s23p6
C. 3s²3p 3d
D. 1s²2s²2p63s²
Answer:
D is correct.(1s22s22p63s2)
If a balloon containing a mixture of gases where the partial pressure of oxygen is 12.8 psi, the partial pressure of hydrogen is 4.31 psi, and the partial pressure of nitrogen is 22.1 psi, then the actual pressure inside the balloon is…
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 :)
How does the frequency of a string affect its wavelength?
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:
What will happen when pressure on a reactant mixture at equilibrium and with fewer moles on the reactant side is increased
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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Find the concentration of all ions present in a 0.223 M solution of PbCl2.
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:
Explain how your model is different from the model in the picture.
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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For the reaction:
S8(s) + 8 O2(g)⟶8 SO2(g) ΔH = –2368 kJ
How much heat is evolved when 25.0 moles of sulfur is burned in excess oxygen?
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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What is the volume of a 1.0 M solution that has 4.0 moles of solute?
The volume of a 1.0 M solution that has 4.0 moles of solute is 4.0 liters.
What is mole ?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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A titration setup was used to determine the unknown molar concentration of a solution of NaOH. A1.2 M HCl solution was used as the
titration standard. The following data were collected.
Trial 1
Amount of HCI
Standard Used 10.0 mL
0.0 mL
Initial NaOH
Buret Reading
Final NaOH
Buret Reading 12.2 mL
Trial 2
10.0 mL
12.2 mL
23.2 mL
Trial 3 Trial 4
10.0 mL 10.0 mL
23.2 mL 35.2 mL
35.2 mL 47.7 mL
79) Calculate the volume of NaOH solution used to neutralize 10.0 ml. of the standard HCl solution in trial 3 in the given diagram.
[Show your work.]
diagram of reaction of water,oxygen,acids
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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A solution has [H+] = 1.39x10^-6 M. What is the pH?
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.
draw the lewis structure of ch3br
The Lewis structure of the compound[tex]CH_{3} Br[/tex] is shown in the image attached.
What is the Lewis structure?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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The change in enthalpy (AH, ) for a reaction is -25.8 kJ mol.
The equilibrium constant for the reaction is 1.4 × 103 at 298 K.
What is the equilibrium constant for the reaction at 655 K?
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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