Answer:
0.382 g
Explanation:
Let's consider the reduction of gallium (III) to gallium that occurs in the electrolysis.
Ga³⁺ + 3 e⁻ → Ga
We can establish the following relations:
1 minute = 60 second1 Ampere = 1 Coulomb / secondThe charge of 1 mole of electrons is 96,468 Coulomb (Faraday's constant)1 mole of gallium is deposited when 3 moles of electrons circulate.The molar mass of gallium is 69.72 g/molWe will use this that to determine the mass of gallium deposited from a Ga(III) solution using a current of 0.880 A that flows for 30.0 min
[tex]30.0min \times \frac{60s}{1min} \times \frac{0.880c}{s} \times \frac{1mole^{-} }{96,468c} \times \frac{1molGa}{3 mole^{-}} \times \frac{69.72g}{1molGa} = 0.382 g[/tex]
guy uses a manual can opener to open a can. which best explains how the can opener makes it easier for guy to open the can?
Answer:
It opens the can for him
Explanation:
Answer:
Takes the pain off the mans hand.
Explanation:
Saw that you didn't like europa2433, i just kinda wanted to agree he reports to much.
A galvanic cell consists of one half-cell that contains Ag(s) and Ag+(aq), and one half-cell that contains Mn(s) and Mn2+(aq). What species are produced at the electrodes under standard conditions? Ag+(aq) + e- → Ag(s) E° = +0.80 V Mn2+(aq) + 2 e- → Mn(s) E° = -1.18 V A) Ag(aq) is formed at the cathode and, Mn(s) is formed at the anode. B) Ag(s) is formed at the cathode, and Mn2+(aq) is formed at the anode. C) Mn(s) is formed at the cathode, and Ag+(aq) is formed at the anode. D) Mn2+(aq) is formed at the cathode, and Ag(s) is formed at the anode.
Answer: B) Ag(s) is formed at the cathode, and [tex]Mn^{2+}(aq)[/tex] is formed at the anode.
Explanation:
[tex]Ag^{+}(aq)+e^{-1}\rightarrow Ag(s)[/tex] E=0.80 V
[tex]Mn^{2+}(aq)+2e^{-1}\rightarrow Mn(s)[/tex] E=-1.18 V
Reduction takes place easily if the standard reduction potential is higher (positive) and oxidation takes place easily if the standard reduction potential is less(more negative).
Thus as reduction potential of Ag is higher , it undergoes reduction and Manganese with lower reduction potential undergoes oxidation. Here Mn undergoes oxidation by loss of electrons, thus act as anode. Ag undergoes reduction by gain of electrons and thus act as cathode.
Cathode : reduction : [tex]Ag^{+}(aq)+e^{-1}\rightarrow Ag(s)[/tex]
Anode : oxidation : [tex]Mn\rightarrow Mn^{2+}(aq)+2e^{-1}[/tex]
Ag(s) is formed at the cathode, and [tex]Mn^{2+}(aq)[/tex] is formed at the anode.
Infrared radiation falls in the wavelength region of 1.00×10-6 to 1.00×10-3 meters.
What is the energy of infrared radiation that has a wavelength of 4.04×10-4 m?
Answer:4.920 X10^-22J
Explanation:
Energy of infrared radiation = hc/λ
where h=plank's constant=6.626X 10^-34js
c= speed of light = 3.0 x 10^8ms⁻¹
wavelength λ= 4.04x 10^-4m
E = 6.626X 10^-34js X 3.0 X 10^8 ms⁻¹/ 4.04 X 10^-4m
=4.920 X10^-22J
In a test of rocket propellant performance, liquid hydrazine (N2H4) at 100 kPa,25 C, and oxygen gas at 100 kPa, 25°C, are fed to a combustion chamber in the ratio of 0.5 kg O2/kg N2H4. The heat transfer from the chamber to the surroundings is estimated to be 100 kJ/kg N2H4. Determine the temperature of the products exiting the chamber. Assume that only H2O, H2, and N2 are present. The enthalpy of formation of liquid hydrazine is 50 417 kJ/kmol.
Answer:
Check the explanation
Explanation:
Kindly check the attached image below for the step by step explanation to the question above.
Which of these is a function of a salt bridge in an electrochemical cell?
It forces the electrons to travel through the inner circuit.
It connects the two half‑reactions.
It permits the charges to stay unbalanced in the two half‑cells.
It does not allow migration of ions between the solutions.
Answer:
It connects the two half‑reactions.
Explanation:
A salt bridge is a laboratory device used to connect the oxidation and reduction of half-cells of a galvanic cell. It prevents the cell from rapidly running its reaction to equilibrium.
Salt bridge or a porous plate connects the solutions of the half-cells that allow ions to pass from one solution to the other. This process balances the charges of the solutions and allows the reaction to continue.
The function of the salt bridge with respect to the electrochemical cell should be that it connected the two -half reactions.
What is salt bridge?It is the device that should be used in the laboratory for connecting the oxidation and the reduction of half-cells. It also prevents the cell from suddenly run from its reaction to the equilibrium. Also, it connect the half cell solution where it probide the permission for ions from one solution to another solution. Moreover, there is balancing of the solutions charges due to this it permits the reaction for continuing it.
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Provide a specific example the relationship between processing, structure, and properties of an engineering material. Be specific with your example by including the material type, the processing involved, the structural changes, and specific properties that are being influenced. Be clear in how one thing impacts the other.Feel free to use equations, graphs, or sketches to support your example.One page maximum.
Answer:
An example of engineering material, are plastics, they are derived from organic, natural materials, such as cellulose, coal, natural gas, salt and, of course, oil. Oil is a complex mixture of thousands of compounds and must be processed before being used.
Explanation:
Plastic production begins with distillation at a refinery, where crude oil is separated into groups of lighter components, called fractions. Each fraction is a mixture of hydrocarbon chains (chemical compounds formed by carbon and hydrogen) that differ in terms of the size and structure of their molecules. One of those fractions, naphtha, is the essential compound for the production of plastic.
Two main processes are used to make plastic: polymerization and polycondensation, and both require specific catalysts. In a polymerization reactor, monomers like ethylene and propylene join to form long polymer chains. Each polymer has its own properties, structure and dimensions depending on the type of basic monomer that has been used.
7: A piston cylinder device initially contains 3.6L of air at 137 kPa and 28 C. Air is now compressed to a final state of 830 kPa and 231 C. The useful work input is 1.9 kJ. Assume the surroundings are at 112 kPa and 15 C. The gas constant of air is R = 0.287 kPa.m3 /kg.K. The specific heats of air at the average temperature of 360 K are cp = 1.009 kJ/kg·K and cv = 0.722 kJ/kg·K. a. Determine the entropy change per unit mass, (s2 - s0). b. Determine the volume change per unit mass, (v2 – v0). c. Determine the exergy of the air at the initial and the final states. d. Determine the minimum work that must be supplied to accomplish this compression process. e. Determine the second law efficiency of this process.
Answer:
See explaination
Explanation:
To calculate the second law of efficiency, temperatures here should be in Kelvin → K = ºC + 273.15 or Rankin = 460 + ºF. 2. Second Law Efficiency Second Law efficiency is a measure of how much of the theoretical maximum (Carnot) you achieve, or in other words, a comparison of the system's thermal efficiency to the maximum possible efficiency.
See attachment for the step by step solution of the given problem.
To solve this problem, we'll follow these steps:
a. Determine the entropy change per unit mass, (s2 - s0): The entropy change per unit mass can be calculated using the following equation:
[tex]\[ \Delta s = c_p \ln\left(\frac{T_2}{T_1}\right) - R \ln\left(\frac{P_2}{P_1}\right) \][/tex]
where:
- [tex]\( c_p \)[/tex] is the specific heat at constant pressure,
- [tex]\( R \)[/tex]is the gas constant,
- [tex]\( T_1 \) and \( T_2 \)[/tex] are the initial and final temperatures in Kelvin,
-[tex]\( P_1 \) and \( P_2 \)[/tex]are the initial and final pressures.
Given values:
- [tex]\( c_p = 1.009 \) kJ/kg·K[/tex]
- [tex]\( R = 0.287 \) kPa·m³/kg·K[/tex]
-[tex]\( T_1 = 28 + 273.15 \) K[/tex]
- [tex]\( T_2 = 231 + 273.15 \) K[/tex]
- [tex]\( P_1 = 137 \) kPa[/tex]
-[tex]\( P_2 = 830 \) kPa[/tex]
Substitute the values into the entropy change equation to calculate[tex]\( \Delta s \) per unit mass.[/tex]
[tex]\[ \Delta s = 1.009 \ln\left(\frac{231 + 273.15}{28 + 273.15}\right) - 0.287 \ln\left(\frac{830}{137}\right) \][/tex]
b. Determine the volume change per unit mass, (v2 – v0): The volume change per unit mass can be calculated using the equation:
[tex]\[ \Delta v = R \ln\left(\frac{V_2}{V_1}\right) \][/tex]
where:
-[tex]\( V_1 \) and \( V_2 \)[/tex] are the initial and final volumes.
Given values:
- [tex]\( V_1 = 3.6 \) L[/tex]
- [tex]\( V_2 \)[/tex] can be calculated using the ideal gas law:
[tex]\( V_2 = \frac{nRT_2}{P_2} \),[/tex] where \( n \) is the number of moles, [tex]\( R \)[/tex] is the gas constant,[tex]\( T_2 \)[/tex] is the final temperature in Kelvin, and [tex]\( P_2 \)[/tex] is the final pressure.
c. Determine the exergy of the air at the initial and the final states: Exergy [tex](\( X \))[/tex] is given by the equation:
[tex]\[ X = h - h_0 - T_0(s - s_0) \][/tex]
where:
- [tex]\( h \)[/tex] is the enthalpy per unit mass,
-[tex]\( h_0 \)[/tex] is the enthalpy at a reference state,
- [tex]\( T_0 \)[/tex] is the temperature at the reference state,
- [tex]\( s \)[/tex] is the entropy per unit mass,
- [tex]\( s_0 \)[/tex] is the entropy at the reference state.
Given values:
-[tex]\( T_0 = 15 + 273.15 \)[/tex]K (temperature of the surroundings)
- [tex]\( P_0 = 112 \) kPa[/tex] (pressure of the surroundings)
We'll need to calculate [tex]\( h \), \( h_0 \), \( s \), and \( s_0 \)[/tex] for both initial and final states.
d. Determine the minimum work that must be supplied to accomplish this compression process: The minimum work required [tex](\( W_{\text{min}} \))[/tex]for an isentropic process can be calculated using the following equation:
[tex]\[ W_{\text{min}} = h_1 - h_2 \][/tex]
where:
-[tex]\( h_1 \)[/tex] is the enthalpy at the initial state,
- \( h_2 \) is the enthalpy at the final state.
e. Determine the second law efficiency of this process: The second law efficiency[tex](\( \eta \))[/tex] is given by the ratio of the actual work done[tex](\( W_{\text{actual}} \)) to the minimum work (\( W_{\text{min}} \)[/tex]):
[tex]\[ \eta = \frac{W_{\text{actual}}}{W_{\text{min}}} \][/tex]
Given values:
- [tex]\( W_{\text{actual}} = 1.9 \) kJ[/tex]
Now, let's go step by step and calculate each part of the problem.
a. Entropy Change per Unit Mass[tex](\( \Delta s = s_2 - s_0 \))[/tex]:
Using the given specific heats [tex]\( c_p \), \( R \)[/tex], temperatures [tex]\( T_1 \), \( T_2 \)[/tex], pressures [tex]\( P_1 \), and \( P_2 \)[/tex]:
[tex]\[ \Delta s = c_p \ln\left(\frac{T_2}{T_1}\right) - R \ln\left(\frac{P_2}{P_1}\right) \][/tex]
Substitute the values:
[tex]\[ \Delta s = 1.009 \ln\left(\frac{231 + 273.15}{28 + 273.15}\right) - 0.287 \ln\left(\frac{830}{137}\right) \][/tex]
[tex]\[ \Delta s \approx 1.009 \ln(1.842) - 0.287 \ln(6.0657) \][/tex]
[tex]\[ \Delta s \approx 0.6087 \, \text{kJ/kg·K} \][/tex]
b. Volume Change per Unit Mass [tex](\( \Delta v = v_2 - v_0 \))[/tex]:
First, calculate [tex]\( V_2 \)[/tex] using the ideal gas law:
[tex]\[ V_2 = \frac{nRT_2}{P_2} \][/tex]
where [tex]\( n \)[/tex] is the number of moles of air, which can be calculated using the initial conditions:
[tex]\[ n = \frac{P_1V_1}{RT_1} \][/tex]
Substitute the values and calculate[tex]\( V_2 \)[/tex]:
[tex]\[ n = \frac{137 \times 3.6}{0.287 \times (28 + 273.15)} \][/tex]
[tex]\[ n \approx 1.932 \, \text{kg} \][/tex]
[tex]\[ V_2 = \frac{1.932 \times 0.287 \times (231 + 273.15)}{830} \][/tex]
[tex]\[ V_2 \approx 1.476 \, \text{L/kg} \][/tex]
Now, calculate [tex]\( \Delta v \)[/tex] :
[tex]\[ \Delta v = R \ln\left(\frac{V_2}{V_1}\right) \][/tex]
[tex]\[ \Delta v \approx 0.287 \ln\left(\frac{1.476}{3.6}\right) \][/tex]
[tex]\[ \Delta v \approx -0.5305 \, \text{m}^3/\text{kg} \][/tex]
The negative sign indicates compression.
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How are transition metals named in metal compounds?
Answer:
Naming ionic compounds with transition metals isn't too hard either. They are named like the binary compounds, with the cation first, then the anion with -ide added to it, but you have to take into account the variations of the metal ions.
Explanation:
The enzyme ribose‑5‑phosphate isomerase catalyzes the conversion between ribose‑5‑phosphate (R5P) and ribulose‑5‑phosphate (Ru5P) through an enediolate intermediate. In the Calvin cycle, Ru5P is used to replenish ribulose‑1,5‑bisphosphate, a substrate for rubisco. For the conversion of R5P to Ru5P, if Δ G ° ′ = 0.460 kJ / mol ΔG°′=0.460 kJ/mol and Δ G = 3.30 kJ / mol ΔG=3.30 kJ/mol , calculate the ratio of Ru5P to R5P at 298 K 298 K . [ Ru 5 P ] [ R 5 P ] = [Ru5P][R5P]= Which of the statements is true? This reaction is favorable, and it is not likely regulated. This reaction is favorable, and it is likely regulated. This reaction is not favorable, and it is not likely regulated. This reaction is not favorable, and it is likely regulated.
Answer:
This reaction is favorable, and is likely regulated.
Explanation:
The equation to calculate delta G (dG) of a reaction is dG = dGo' + RTln [initial P]/[initial R]. You could use just dG = RTln [initial P]/[initial R] if (and that's a big IF) dGo' is zero, meaning that the reaction is at equilibrium when we have equal amounts of [P] and [R] (which is rarely the case). What really matters is the ratio of Q ([initial P]/[initial R]) to Keq ([P at equilibrium]/[R at equilibrium]), meaning how far off are we from equilibrium.
If Q=Keq, we are already at equilibrium (EQ).
If Q<Keq, we are not yet at EQ, having relatively more [R], or less [P] than under EQ conditions. This means the reaction will move forward to produce more P until EQ is achieved (dG is therefore NEGATIVE).
If Q>Keq, we are also off EQ, but we have relatively more [P], or less [R] than under EQ conditions. This means the reaction will move backwards to produce more R until EQ conditions are achieved (dG is therefore POSITIVE).
Try to understand these equations below (they say what I tried to describe in words)
dGo' = -RTlnKeq (under "standard conditions", i.e. we try to figure out how a reaction "behaves" if we start out with the same molar concentrations of R and P)
dG = dGo' + RTlnQ Q=[initial P]/[initial R] or
dG = -RTlnKeq + RTlnQ or
dG = RTlnQ - RTlnKeq or
dG = RTln Q/Keq
In the halogenation reaction shown, ethane and chlorine gas yield chloroethane and hydrogen chloride. How do the properties of chloroethane compare to those of ethane?
Chloroethane is denser and has a lower boiling point
Chloroethane is denser and has a higher boiling point
Chloroethane is less dense and has a lower boiling point.
Chloroethane is less dense and has a higher boiling point.
Answer:
Chloroethane is denser and has a higher boiling point
Explanation:
The density of a gas depends directly on the molar mass of the gas. This means that as the molar mass increases, density increases and vice versa.
Having said that, we can easily see that the molar mass of chloroethane (64.51 g/mol) is greater than the molar mass of ethane (30.07 g/mol). Hence we expect that chloroethane is denser than ethane as established above.
In the absence of other strong intermolecular forces, the higher the molecular mass of a substance the greater its boiling point. Thus the boiling point of chloroethane is higher than that of Ethane since they both have weak Van der Waals forces holding their molecules together in the gaseous state.
6. A 25.0-mL sample of 0.125 M pyridine is titrated with 0.100 M HCI. Calculate the pH
at each volume of added acid: 0 mL, 10 mL, 20 mL
The pH at each volume of added acid is calculated using the Henderson-Hasselbalch equation.
Explanation:The pH of a solution can be calculated using the equation:
pH = -log[H+]
Using the Henderson-Hasselbalch equation, pH can be calculated at each volume of added acid:
Volume of added acid = 0 mL
The initial concentration of pyridine is 0.125 M. Since pyridine is a weak base, [H+] can be calculated using the equation:
Kw / Kb = [H+]
where Kw is the equilibrium constant for water (1.0 x 10^-14) and Kb is the base dissociation constant for pyridine (1.7 x 10^-9). Substituting the values, we get:
[H+] = (1.0 x 10^-14) / (1.7 x 10^-9) ≈ 5.88 x 10^-6 M
Now, we can calculate the pH:
pH = -log(5.88 x 10^-6) ≈ 5.23
Volume of added acid = 10 mL
Using stoichiometry, we can determine the moles of HCI added:
moles of HCI = concentration of HCI x volume of HCI
moles of HCI = (0.100 M) x (0.010 L) = 0.001 mol
Since the acid HCl is strong, all of it will react with the pyridine:
moles of pyridine reacted = moles of HCI added = 0.001 mol
The final volume of the solution is 25.0 mL + 10 mL = 35.0 mL. The concentration of pyridine after the reaction is:
concentration of pyridine after reaction = moles of pyridine / final volume of solution
concentration of pyridine after reaction = (0.125 M)(25.0 mL) / (35.0 mL) ≈ 0.0893 M
Using the Henderson-Hasselbalch equation, we can calculate the pH:
pH = pKa + log ([A-]/[HA])
where pKa is the negative logarithm of the acid dissociation constant for pyridine (9.27) and [A-]/[HA] is the ratio of the conjugate base to the weak acid, which is equal to the ratio of the concentration of pyridine after reaction to the initial concentration of pyridine:
pH = 9.27 + log (0.0893 / 0.125) ≈ 9.07
Volume of added acid = 20 mL
Using the same approach, we can calculate the concentration of pyridine after the reaction:
concentration of pyridine after reaction = (0.125 M)(25.0 mL) / (45.0 mL) ≈ 0.0694 M
Using the Henderson-Hasselbalch equation, we can calculate the pH:
pH = 9.27 + log (0.0694 / 0.125) ≈ 8.88
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To calculate the pH at each stage, we consider the concentrations of the pyridine and the HCl, and use the formulas for pH and pOH. At 0 mL of HCl, we would calculate pH from the pKa of pyridine. At 10 mL and 20 mL, we need to consider that pyridine is in excess, and calculate pOH first before getting the pH.
Explanation:The subject of this question is acid-base titration, specifically the titration of pyridine with hydrochloric acid (HCl). To calculate the pH at different volumes of added acid, we'd use the formula for the concentration of the pyridine and HCl.
1. At 0 mL of HCl, the solution is just the pyridine, which is a weak base. We don't have the pKa for pyridine in the question, but assuming we did, we could calculate the pH using the pKa and the formula pH = 14 - pKa. Let's say it's around 5.2.
2. At 10 mL of 0.100 M HCl (1.0 mmol), and 25 mL of 0.125 M pyridine (3.125 mmol), we have more base than acid, so pyridine is in excess. We'd use the equilibrium expression for the reaction of excess pyridine with water to find the pOH, and then calculate the pH.
3. At 20 mL of 0.100 M HCl (2.0 mmol), we still have excess pyridine, and so we'd perform a similar calculation as at 10 mL.
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1. As Z stays constant and the number of electrons increases, the electron-electron repulsions _____, and the anion becomes larger.
2. The reverse is true for the cation, which becomes ____ than the neutral atom.
Answer:
1) increases
2) smaller
Explanation:
Generally, as electron- electron repulsion increases and more electrons are added to the atom while Z is held constant, the electron cloud size is increased. The size of the anion formed is usually measured as the size of this extended electron cloud. Hence the larger electron cloud means a larger anion size compared to the size of the neutral atom.
For a cation, the converse is true and the cation is found to be smaller than the neutral atom.
This question is incomplete, I got the complete one from google as below:
I−>I>I+
1. As Z stays constant and the number of electrons increases, the electron-electron repulsions ______ (increases or decreases), and the anion becomes larger.
2. The reverse is true for the cation, which becomes ____ (smaller or larger) than the neutral atom.
Answer:
1. As Z stays constant and the number of electrons increases, the electron-electron repulsion increases, and the anion becomes larger.
2.The reverse is true for the cation, which becomes smaller than the neutral atom.
Explanation:
1. As Z stays constant and the number of electrons increases, the electron-electron repulsion increases, and the anion becomes larger. This is because in anions of the same atoms, the net force of attraction on electrons decreases.
2. The reverse is true for the cation, which becomes smaller than the neutral atom. This is because in cations of the same atoms, the net force of attraction on electrons increases.
A chemist must prepare of potassium hydroxide solution with a pH of at . He will do this in three steps: Fill a volumetric flask about halfway with distilled water. Weigh out a small amount of solid potassium hydroxide and add it to the flask. Fill the flask to the mark with distilled water. Calculate the mass of potassium hydroxide that the chemist must weigh out in the second step. Round your answer to significant digits
The question is incomplete; the complete question is:
A chemist must prepare 800.0mL of potassium hydroxide solution with a pH of 13.00 at 25 degree C. He will do this in three steps: Fill a 800.0mL volumetric flask about halfway with distilled water. Weigh out a small amount of solid potassium hydroxide and add it to the flask. Fill the flask to the mark with distilled water. Calculate the mass of potassium hydroxide that the chemist must weigh out in the second step. Round your answer to 2 significant digits.
Answer:
4.5g (to 2 significant digits)
Explanation:
Now we must remember that KOH is a strong base, therefore it will practically dissociate completely.
To find the pH we can use the equation pH + pOH = 14.
Firstly to find the pOH:
13.00 + pOH = 14
pOH = 1.00
To find the [OH-]
Since
pOH= -log[OH^-]
[OH^-] = antilog (-pOH)
[OH^-]= antilog (-1)
[OH^-] = 0.1 molL-1
Since we've established that KOH is a strong base, we know that [OH-] = [KOH]
Also, we know that concentration = number of moles/volume
we have the concentration and the volume now so we can calculate the number of number of moles as follows:
The 800mL volume is the same as 0.8L
0.1 molL-1= number of moles/0.8L
0.08 moles = number of moles
now we can calculate the amount of solid KOH required
the molar mass of KOH = 39 + 16 +1 = 56 gmol-1
56 x 0.08 moles = 4.48g
So in 800mL of pH 13.00 KOH there is 4.5g of KOH dissolved.
If 11.5 ml of vinegar sample (d=1g/ml) is titrated with 18.5 ml of standardized Sodium hydroxide
solution. What is the percent of acetic acid (by mass) in the vinegar?
Answer:
4.83% of acetic acid in the vinegar
Explanation:
A geochemist in the field takes a 36.0 mL sample of water from a rock pool lined with crystals of a certain mineral compound X. He notes the temperature of the pool, 170 C, and caps the sample carefully. Back in the lab, the geochemist first dilutes the sample with distilled water to 500. mL. Then he filters it and evaporates all the water under vacuum. Crystals of X are left behind. The researcher washes, dries and weighs the crystals. They weigh 3.96 g.
Using only the information above, can you calculate yes the solubility of X in water at 17.0 C? If you said yes, calculate it.
Answer:
solubility of X in water at 17.0 [tex]^{0}\textrm{C}[/tex] is 0.11 g/mL.
Explanation:
Yes, the solubility of X in water at 17.0 [tex]^{0}\textrm{C}[/tex] can be calculated using the information given.
Let's assume solubility of X in water at 17.0 [tex]^{0}\textrm{C}[/tex] is y g/mL
The geochemist ultimately got 3.96 g of crystals of X after evaporating the diluted solution made by diluting the 36.0 mL of stock solution.
So, solubility of X in 1 mL of water = y g
Hence, solubility of X in 36.0 mL of water = 36y g
So, 36y = 3.96
or, y = [tex]\frac{3.96}{36}[/tex] = 0.11
Hence solubility of X in water at 17.0 [tex]^{0}\textrm{C}[/tex] is 0.11 g/mL.
Name three things in everyday
life that are periodic.
The three things that should be mentioned are Helium in Balloons, Lithium in Batteries, Beryllium in Emeralds, and Boron in Sports equipment.
What are basic elements?Elements or rather compounds seem to be natural pure chemical substances.
The distinction here between element as well as a compound is whether an element is a compound consisting of the same type of atoms as a compound.
Whereas a compound is made up of various elements in specific proportions. Elements include iron, copper, hydrogen, and oxygen.
Metals are on the left side of the staircase, metalloids are on the right, and nonmetals are on the right.
An atom is a component of an element. An element is made up of only one type of atom.
Atoms are further subdivided into subatomic particles known as electrons, protons, and neutrons. Chemical reactions allow elements to combine to form molecules.
Thus, the three examples are helium, lithium, and boron.
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Which of the following best describes the first step in the formation of a hurricane?
A.
Cooler water produces air masses with high pressure, causing fast-moving thunderstorms.
B.
Warm ocean water evaporates and then condenses, forming a low pressure area below.
C.
A high-pressure air mass remains undisturbed over land, resulting in excessive precipitation.
D.
Air masses over the ocean become colder, causing water waves to move faster.
I think the answer is B...
I am not sure
Answer:
B
Explanation:
Hurricanes are very powerful, circular storms that begin over warm water. These storms have heavy rains, very high winds, and low atmospheric pressures. They are fueled by a strong cycle of warmer, moister air flowing upward from near the ocean's surface and being replaced by cooler, drier air from the surrounding area. This cycle gets faster and stronger as it is fed by the ocean's heat and water evaporating from the ocean's surface.
So, the first step in hurricane formation occurs when warm ocean water evaporates and then condenses, forming a low pressure area below.
What is the law of conservation of energy as it applies to exothermic dissolution processes?Energy given off by the system during dissolution equals the energy absorbed by the surroundings.Energy given off by the system during dissolution is less than the energy absorbed by the surroundings.Energy given off by the system during dissolution is greater than the energy absorbed by the surroundings.Energy given off by the system during dissolution may be greater or less than the energy absorbed by the surroundings.
Energy given off by the system during dissolution equals the energy absorbed by the surroundings. Therefore, option (A) is correct.
The law of energy conservation states that energy can only be transmitted or changed. This equation says that the energy produced by the system during exothermic dissolution, where a material dissolves while generating heat, is equal to the energy received by the solvent and container.
Solute molecules spread and interact with the solvent, forming or breaking bonds and releasing heat. The environment absorbs this heat, keeping energy balance. The law of energy conservation emphasises that energy changes in a system must be counterbalanced by energy changes in the surrounding environment, reinforcing the idea that energy is conserved during exothermic dissolution. Therefore, option (A) is correct.
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The law of conservation of energy states that the energy in a system cannot be created or destroyed, only changed in form. In exothermic dissolution processes, the energy given off by the system as heat is equal to the energy absorbed by the surroundings.
Explanation:According to the law of conservation of energy, energy can neither be created nor destroyed, only changed in form during a chemical or physical change. In the case of exothermic dissolution processes, energy is given off by the system as heat, and this energy is equal to the energy absorbed by the surroundings. This means that the energy released during dissolution is the same as the energy gained by the surroundings.
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17. What is the symbol for the element that forms a.
protective oxide coating?
(A) K
(B) Ga
(C) Al
(D) Mg
Final answer:
Aluminum, symbol (C) Al, forms a protective oxide layer known as aluminum oxide (Al2O3) when exposed to oxygen, protecting it from further corrosion.
Explanation:
The symbol for the element that forms a protective oxide coating is (C) Al, which stands for Aluminum. When aluminum is exposed to the atmosphere, it reacts with oxygen to form aluminum oxide (Al2O3), a thin, hard layer that helps prevent further oxidation and protects the metal underneath. This property is particularly useful for applications where durability and resistance to corrosion are important. For example, the outside of the aerospace and construction materials are often made from aluminum for this reason.
The constant pressure molar heat capacity, C_{p,m}C p,m , of methane gas, CH_4CH 4 , is 35.31\text{ J K}^{-1}\text{ mol}^{-1}35.31 J K −1 mol −1 at temperatures close to 298\text{ K}298 K. Calculate the enthalpy change when 1.6\text{ kg}1.6 kg of methane gas is heated from a temperature of 280\text{ K}280 K to 320\text{ K}320 K.
The enthalpy change when 1.6 kg of methane gas is heated from a temperature of 280 K to 320 K is 140.6 kJ.
Explanation:The subject this question pertains to is heat capacity and enthalpy change in the field of Chemistry. Simply put, when the temperature of an amount of substance changes, the change in enthalpy (∆H), can be calculated using the formula, ∆H = nCp∆T, where n is the number of moles, Cp is the heat capacity at constant pressure, and ∆T is the change in temperature.
To calculate the answer to your question, we need to convert the mass of methane gas to moles because the molar heat capacity is given. The molar mass of methane (CH4) is approximately 16.04 g/mol. Thus, we have: n (number of moles) = mass (in kg) / molar mass (in kg/mol) = 1.6 kg / 0.01604 kg/mol = 99.75 mol.
The next step is to plug our numbers into the formula: ∆H = nCp∆T = 99.75 mol * 35.31 JK-1mol-1 * (320 K - 280 K) = 140.6 kJ.
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Which organelles are common to BOTH plant and animal cells? (Choose all 1 point
that apply.)
cell wall
cell membrane
nucleus
chloroplasts
u
mitochondria
vacuole
cytoplasm
O
cuestion
wering pants and huma
Answer:
The following organelles are common to both plant and animal cells:
Cell membraneCell NucleusMitochondriaCytoplasm(N.B: the other options are peculiar to plant cells only)
Explanation:
Plant and animal cells share similarities in components due to the fact that the are both eukaryotic cells ( having DNA as genetic composition).
The organelles common to both plant and animal cells are the cell membrane, nucleus, mitochondria, and cytoplasm.
Explanation:The organelles that are common to both plant and animal cells are the cell membrane, nucleus, mitochondria, and cytoplasm. Both plant and animal cells have a cell membrane, which controls the movement of substances in and out of the cell. The nucleus is the control center of the cell, responsible for storing and processing genetic information. Mitochondria are the powerhouse of the cell, producing energy through cellular respiration. Cytoplasm is a jelly-like substance that fills the cell and houses the organelles.
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Suppose you want to extract the iron from ferrous iodide, FeI2, through electrolysis. What can be said about the merits of electrolysis of molten versus aqueous FeI2? Group of answer choices You'll have to perform electrolysis of the aqueous solution because you can't get iron metal from electrolysis of the molten salt. You'll have to perform electrolysis on the molten salt because you can't get iron metal from electrolysis of the aqueous solution. Electrolysis of either the molten or aqueous salt will produce solid iron. Electrolysis will not produce solid iron rega
Answer:
Electrolysis of either the molten or aqueous salt will produce solid iron.
Explanation:
Iron (II) iodide has a relatively low melting point. Electrolysis of the molten salt will probably produce solid metallic iron and liquid I2 as well as iodine vapor. The temperature need not be high enough to cause Fe to melt.
so Electrolysis of either the molten or aqueous salt will produce solid iron.
When The Electrolysis of either the molten or aqueous salt will produce solid iron. So The temperature need not be high enough to cause Fe to melt.
What is Electrolysis?
When Iron (II) iodide has a moderately low melting point. Electrolysis of the molten salt will probably construct solid metallic iron and liquid I2 as well as iodine vapor. The temperature requirement not be high enough to cause Fe to melt.
Therefore, The Electrolysis of either the molten or aqueous salt will produce solid iron.
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In the Hall-Heroult process, a large electric current is passed through a solution of aluminum oxide Al2O3 dissolved in molten cryolite Na3AlF6, resulting in the reduction of the Al2O3 to pure aluminum. Suppose a current of 620.A is passed through a Hall-Heroult cell for 90.0 seconds. Calculate the mass of pure aluminum produced. Be sure your answer has a unit symbol and the correct number of significant digits.
A current of 620 A for 90 seconds produces 5.209 g of aluminum.
The mass of pure aluminum produced in the Hall-Heroult process can be calculated by using Faraday's laws of electrolysis and the molar mass of aluminum. First, we determine the moles of electrons transferred using the given current and time. Then, using the stoichiometry of the reaction where 4 moles of Al are produced per 12 moles of electrons:
Calculate the charge (Q) passed: Q = current (I) x time (t) = 620 A x 90.0 s = 55800 C
Determine the moles of electrons: Moles of e- = Q / Faraday's constant (F) Assuming the Faraday constant is approximately 96500 C/mol, this gives Moles of e- = 55800 C / 96500 C/mol = 0.5784 mol
Calculate the moles of aluminum: Since 3 moles of electrons yield 2 moles of Al, 0.5784 mol of e- will yield (4/12) x 0.5784 mol of Al = 0.1931 mol of Al.
Compute the mass of aluminum produced: Mass of Al = moles of Al x molar mass of Al = 0.1931 mol x 26.98 g/mol = 5.209 g
The mass of pure aluminum produced is therefore 5.209 g, assuming 100% efficiency and no other side reactions.
To find the mass of pure aluminum produced, we calculate the charge passed, find the moles of electrons transferred, determine the moles of aluminum from the moles of electrons, and use the molar mass of aluminum to find that approximately 5.20 grams of aluminum is produced.
To calculate the mass of pure aluminum produced in the Hall-Heroult process when 620 A of current is passed through a Hall-Heroult cell for 90.0 seconds, we use Faraday's laws of electrolysis. The reduction of aluminum oxide to aluminum involves the transfer of three moles of electrons per mole of aluminum according to the equation:
Al₂O₃ + 3e⁻ → 2Al + 1.5O₂
Firstly, we calculate the amount of charge passed using the formula Q = It, where Q is the charge in coulombs, I is the current in amperes, and t is time in seconds. Then, using Faraday's constant (96485 C/mol), we determine the moles of electrons transferred. Since we need 3 moles of electrons to produce 1 mole of aluminum, we find the moles of aluminum formed by dividing the moles of electrons by 3. Finally, we calculate the mass using the molar mass of aluminum (26.98 g/mol).
Q = 620 A * 90.0 s = 55800 C
Moles of electrons = Q/F = 55800 C / 96485 C/mol ≈ 0.5785 mol
Moles of aluminum = Moles of electrons / 3 ≈ 0.1928 mol
Mass of aluminum = Moles of aluminum * Molar mass of Al
Mass of aluminum ≈ 0.1928 mol * 26.98 g/mol ≈ 5.20 g
The mass of pure aluminum produced is approximately 5.20 grams.
A compound is 2.00% H by mass, 32.7% S by mass, and 65.3% O by mass. What is its empirical
formula?
The first step is to calculate the mass of each element in a 100-g sample of the compound. We use 100
g because it makes the calculations more straightforward.
What is the mass of each element in a 100.0-g sample of the compound?
mass H =
mass S =
mass 0 =
Answer:
empirical formula: [tex]H_2SO_4[/tex]
2 g H
32.7 g S
65.3 g O
Explanation:
Like the problem said, the first thing we can do is calculate the mass of each of the 3 elements in a 100-gram sample:
- 2.00% * 100g = 2 g H
- 32.7% * 100g = 32.7 g S
- 65.3% * 100g = 65.3 g O
Now we need to find the empirical formula from these. To do so, convert all of those masses into moles by using the molar mass for each element:
- the molar mass of H is 1.01 g/mol
- the molar mass of S is 32.06 g/mol
- the molar mass of O is 16 g/mol
2 g H ÷ 1.01 g/mol = 1.98 mol H
32.7 g S ÷ 32.06 g/mol = 1.02 mol S
65.3 g O ÷ 16 g/mol = 4.08 mol O
Our ratio of H : S : O is now:
1.98 mol : 1.02 mol : 4.08 mol
Divide them all by the smallest number, which is 1.02:
1.98/1.02 : 1.02/1.02 : 4.08/1.02
1.94 : 1 : 4
1.94 ≈ 2
So:
2 : 1 : 4
Thus, the empirical formula is: [tex]H_2SO_4[/tex].
name some of the chemical substances in a tea mixture
Answer:
caffience
Explanation:
A reaction of importance in the formation of smog is that between ozone and nitrogen monoxide described by O3(g)+NO(g)⟶O2(g)+NO2(g) O3(g)+NO(g)⟶O2(g)+NO2(g) The rate law for this reaction is rate of reaction=k[O3][NO] rate of reaction=k[O3][NO] Given that k=4.09×106 M−1⋅s−1k=4.09×106 M−1⋅s−1 at a certain temperature, calculate the initial reaction rate when [O3][O3] and [NO][NO] remain essentially constant at the values [O3]0=5.84×10−6 M[O3]0=5.84×10−6 M and [NO]0=8.65×10−5 M,[NO]0=8.65×10−5 M, owing to continuous production from separate sources.
Answer:
Initial rate of reaction is [tex]2.07\times 10^{-3}M.s^{-1}[/tex].
Explanation:
It is a second order reaction.
Initial rate of reaction = [tex]k[O_{3}]_{0}[NO]_{0}[/tex] , where k is rate constant, [tex][O_{3}]_{0}[/tex] is the initial concentration of [tex]O_{3}[/tex] and [tex][NO]_{0}[/tex] is the initial concentration of NO.
Here, k = [tex]4.09\times 10^{6}M^{-1}.s^{-1}[/tex], [tex][O_{3}]_{0}=5.84\times 10^{-6}M[/tex] and [tex][NO]_{0}=8.65\times 10^{-5}M[/tex]
So, initial rate of reaction = [tex](4.09\times 10^{6}M^{-1}.s^{-1})\times (5.84\times 10^{-6}M)\times (8.65\times 10^{-5}M)[/tex]
= [tex]2.07\times 10^{-3}M.s^{-1}[/tex]
So, initial rate of reaction is [tex]2.07\times 10^{-3}M.s^{-1}[/tex]
In a study of the conversion of methane to other fuels, a chemical engineer mixes gaseous methane and gaseous water in a 0.669 L flask at 1,020 K. At equilibrium, the flask contains 0.276 mol of CO gas, 0.207 mol of H2 gas, and 0.231 mol of methane. What is the water concentration at equilibrium (Kc = 0.30 for this process at 1,020 K)? Enter to 4 decimal places. HINT: Look at sample problem 17.7 in the 8th ed Silberberg book. Write a balanced chemical equation. Write the Kc expression. Calculate the equilibrium concentrations of all the species given (moles/liter). Put values into Kc expression, solve for the unknown.
Answer:
Concentration of water at equilibrium is 0.1177 M.
Explanation:
Balanced equation: [tex]CH_{4}(g)+H_{2}O(g)\rightleftharpoons CO(g)+3H_{2}(g)[/tex]
Equilibrium concentration of [tex]CH_{4}[/tex], [[tex]CH_{4}[/tex]] = [tex]\frac{0.231}{0.669}[/tex] M = 0.345 M
Equilibrium concentration of CO, [CO] = [tex]\frac{0.276}{0.669}[/tex] M = 0.413 M
Equilibrium concentration of [tex]H_{2}[/tex], [[tex]H_{2}[/tex]] = [tex]\frac{0.207}{0.669}[/tex] M = 0.309 M
Equilibrium constant for the given reaction in terms of concentration, [tex]K_{c}[/tex] is expressed as: [tex]K_{c}=\frac{[CO][H_{2}]^{3}}{[CH_{4}][H_{2}O]}[/tex]
[tex]\Rightarrow [H_{2}O]=\frac{[CO][H_{2}]^{3}}{[CH_{4}].K_{c}}=\frac{(0.413)\times (0.309)^{3}}{(0.345)\times (0.30)}= 0.1177[/tex]
Hence, concentration of water at equilibrium is 0.1177 M
What is the general equilibrium expression for the self-
ionization of water (shown below)?
2H20 (aq) = H20+ (aq) + OH- (aq)
Answer:
Kw= = [H3O+][OH-] = [H+][OH-].
Explanation:
In chemistry, the term 'equilibrium'' is a period or time in which the rate at which the product is been produced or generated is equal to the quantity of the reactants reacting. In order way, we can say the equilibrium is when forward Reaction is equal to the backward Reactions.
The general equilibrium expression for the self-ionization of water is given below;
Kw = [H3O+][OH-] = [H+][OH-].
The kw is now equal to 1.001x10-14 at 25°C. Kw which is the equilibrium constant is also refer to as the dissociation constant of water.
A solution has a hydroxide ion concentration of 1 × 10–5 M. What is the hydrogen ion concentration of the solution?
(a) 1 × 10–1 M
(b) 1 × 10–5 M
(c) 1 × 10–9 M
(d) 1 × 10–14 M
Answer:
The correct answer is (c) 1 × 10⁻⁹ M
Explanation:
Hydroxide ion = OH⁻
Hydrogen ion = H⁺
The autoionization equilibrium of water at 25ºC has a water constant Kw which is expressed as follows:
Kw = [H⁺] x [OH⁻]= 1 x 10⁻¹⁴
If we know the concentration of hydroxide ion ([OH⁻]) we can calculate the hydrogen ion concentration ([H⁺]) as follows:
[H⁺] = Kw/[OH⁻]= (1 x 10⁻¹⁴)/1 x 10⁻⁵ M= 1 x 10⁻⁹ M
The hydrogen ion concentration of a solution with a hydroxide ion concentration of 1 × 10-5 M is 1 × 10-9 M.
Explanation:In chemistry, the product of the concentration of hydrogen ions [H+] and the concentration of hydroxide ions [OH-] in a solution always equals 1 × 10–14 M² at a temperature of 25°C (this is known as the ion product of water).
Since the concentration of hydroxide ions [OH-] in your solution is given as 1 × 10-5 M, you can calculate the concentration of hydrogen ions [H+] by dividing the ion product of water, 1 × 10-14 M², by the concentration of hydroxide ions [OH-]. Therefore, the hydrogen ion concentration [H+] in your solution would be 1 × 10-14 M² / 1 × 10-5 M = 1 × 10-9 M.
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An enzyme with molecular weight of 310 kDa undergoes a change in shape when the substrate binds. This change can be characterized by ultracentrifugation. In the absence of any ligand, the s20,w = 11.7S, density of solvent is 1g/cm3and viscosity 1.005 cP; the specific volume of the protein is 0.732 cm3/g. a) Find the radius of the protein, assuming it is sphericalb) Upon binding of the substrate the sedimentation coefficient increases by 3.5%; what is the radius of the bound enzyme?
Answer:
(a) r = 6.26 * 10⁻⁷cm
(b) r₂ = 6.05 * 10⁻⁷cm
Explanation:
Using the sedimentation coefficient formula;
s = M(1-Vρ) / Nf ; where s is sedimentation coefficient, M is molecular weight, V is specific volume of protein, p is density of the solvent, N is Avogadro number, f if frictional force = 6πnr, n is viscosity of the medium, r is radius of particle
s = M ( 1 - Vρ) / N*6πnr
making r sbjct of formula, r = M (1 - Vρ) / N*6πnrs
Note: S = 10⁻¹³ sec, 1 KDalton = 1 *10³ g/mol, I cP = 0.01 g/cm/s
r = {(3.1 * 10⁵ g/mol)(1 - (0.732 cm³/g)(1 g/cm³)} / { (6.02 * 10²³)(6π)(0.01 g/cm/s)(11.7 * 10⁻¹³ sec)
r = 6.26 * 10⁻⁷cm
b. Using the formula r₂/r₁ = s₁/s₂
s₂ = 0.035 + 1s₁ = 1.035s₁
making r₂ subject of formula; r₂ = (s₁ * r₁) / s₂ = (s₁ * r₁) / 1.035s₁
r₂ = 6.3 * 10⁻⁷cm / 1.035
r₂ = 6.05 * 10⁻⁷cm