When three electrons are added to the neutral atom, an anion is formed. The charge on an ion will be negative (-3) when three electrons are added.
The correct option is a.
When four electrons are removed from the neutral atom, a cation is formed. The charge on an ion will be positive (+4) when four electrons are removed.Two neutrons are added to the nucleus, but since neutrons are electrically neutral, this does not change the charge on the ion.
Therefore, there will be no change in the charge on the ion when two neutrons are added to the nucleus.Therefore, the charges on the ions will be as follows:(a) The charge on an ion would be -3.(b) The charge on an ion would be +4.(c) The charge on an ion would not change.
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The Haber-Bosch process reacts nitrogen gas with hydrogen gas to form liquid ammonia. Write balanced chemical equation (including phases) for this reaction.
The chemical equation for the Haber-Bosch process is: N2(g) + 3H2(g) → 2NH3(g)
The Haber-Bosch process is an industrial process used to synthesize ammonia from nitrogen gas and hydrogen gas. It is named after its inventors, Fritz Haber and Carl Bosch. This process is crucial for the production of fertilizers and other chemical products.
The reaction requires high temperature (around 450 °C) and high pressure (around 200 atm) to proceed. The reaction is exothermic and releases a significant amount of heat, which must be managed to prevent the reaction from becoming uncontrollable.
The catalyst used in the Haber-Bosch process is iron, which helps to speed up the reaction rate and improve the yield of ammonia. The ammonia that is produced in the reaction is in the form of a gas, which is then cooled and compressed to form a liquid.
The liquid ammonia can then be used in the production of fertilizers and other chemical products.
The Haber-Bosch process has had a significant impact on agriculture and the chemical industry. By allowing for the production of large quantities of ammonia, it has enabled the production of fertilizers on a massive scale, which has helped to increase crop yields and improve food security around the world.
In addition, the Haber-Bosch process has made it possible to produce a wide range of chemical products, including plastics, synthetic fibers, and explosives.
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Consider an squeous 1.00 m solution of Na3PO4 a a compound with useful detergent properties (kg=0.51∗Cm−1 h. (a) Cakulate the boiling point of the solution on the assumption that it does not ionize at all in solution. ∘C (b) Do the same calculation assuming that the var't Hoff foctor for Na.8O sefiectes 100% dissociation into ions. " C
(a) Assuming no ionization, the boiling point of the 1.00 m Na3PO4 aqueous solution is elevated by 0.51°C.
(b) Assuming 100% dissociation of Na3PO4, the boiling point of the solution is elevated by 2.04°C due to the presence of 4 particles from the complete dissociation.
(a) The boiling point of the solution, assuming no ionization, can be calculated using the equation:
ΔTb = Kb * m
Where ΔTb is the boiling point elevation, Kb is the molal boiling point elevation constant for the solvent, and m is the molality of the solution. Since Na3PO4 does not ionize in this scenario, the concentration of particles is considered to be the same as the concentration of Na3PO4.
Given that the molality (m) of the solution is 1.00 m and the molal boiling point elevation constant (Kb) is 0.51°C * kg⁻¹ * mol⁻¹, we can calculate the boiling point elevation:
ΔTb = 0.51 * 1.00 = 0.51°C
Therefore, the boiling point of the solution is elevated by 0.51°C.
(b) In this scenario, we assume that Na3PO4 dissociates completely into ions. The van 't Hoff factor (i) represents the number of particles into which a compound dissociates in solution. For Na3PO4, the van 't Hoff factor would be 4 since it dissociates into 3 Na+ ions and 1 PO4³⁻ ion.
Using the same equation as before, ΔTb = Kb * m, we can calculate the boiling point elevation. However, this time we need to consider the van 't Hoff factor:
ΔTb = i * Kb * m
Substituting the values, we have:
ΔTb = 4 * 0.51 * 1.00 = 2.04°C
Therefore, considering 100% dissociation of Na3PO4, the boiling point of the solution is elevated by 2.04°C.
It's important to note that in reality, the extent of ionization of Na3PO4 in solution may not be complete, and the actual boiling point elevation would depend on the degree of ionization.
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When a rotating disk electrode (RDE) is held at a high enough potential, the rate of the reaction is governed by the diffusion rate, the rate at which the analyte diffuses through the diffusion layer to the electrode. The thickness of this diffusion layer, δ, is calculated as δ=1.61D
1/3
ω
−1/2
ν
1/6
where D is the diffusion coefficient ( m
2
/s),ν is the kinematic viscosity of the liquid ( m
2
/s ), and ω is the rotation rate (radians/s) of the electrode. Additionally, the current density, J( A/m
2
), is measured using the Levich equation, J=0.62nFD
2/3
ω
1/2
ν
−1/6
C
0
where n is the number of electrons transferred in the half-reaction, F is the Faraday constant, and C
0
is the concentration of the electroactive species (mol/m
3
). Calculate δ, in micrometers, and the current density, J, for the reduction of 0.025MTl
3
+
+
Tl
+
at a gold electrode in 1 FHCl at 0.95 V vs. SHE at 2200rpm, where D=1.4×10
−9
m
2
/s and ν=2.0×10
−6
m
2
/s.
The diffusion layer thickness (δ) is approximately 245 nm, and the current density (J) is approximately 2.53 A/m².
To calculate the diffusion layer thickness (δ) and current density (J), we can use the given equations and substitute the provided values:
Given:
D (diffusion coefficient) = 1.4 × 10⁻⁹ m²/s
ν (kinematic viscosity) = 2.0 × 10⁻⁶ m²/s
ω (rotation rate) = 2200 rpm (rotations per minute)
n (number of electrons transferred) = 1
F (Faraday constant) = 96485 C/mol
C0 (concentration of electroactive species) = 0.025 M (mol/m³)
First, let's convert the rotation rate from rpm to radians/s:
ω (rotation rate) = 2200 rpm * (2π radians/minute) * (1/60 minute/second) = 229.18 radians/s
Now, we can calculate the diffusion layer thickness (δ):
δ = 1.61 * D¹⁽³ * ω⁻¹⁽² * ν¹⁽⁶
= 1.61 * (1.4 × 10⁻⁹ m²/s)¹⁽³ * (229.18 radians/s)⁻¹⁽² * (2.0 × 10⁻⁶ m²/s)¹⁽⁶
Next, we can calculate the current density (J):
J = 0.62 * n * F * D²⁽³ * ω¹⁽² * ν⁻¹⁽⁶ * C0
= 0.62 * (1) * (96485 C/mol) * (1.4 × 10⁻⁹ m²/s)²⁽³ * (229.18 radians/s)¹⁽² * (2.0 × 10⁻⁶ m²/s)⁻¹⁽⁶ * (0.025 mol/m³)
Now, we can calculate the values of δ and J:
δ = 1.61 * (1.4 × 10⁻⁹ m²/s)¹⁽³ * (229.18 radians/s)⁻¹² * (2.0 × 10⁻⁶ m²/s)¹⁽⁶
≈ 2.45 × 10⁻⁷ meters (or 245 nm)
J = 0.62 * (1) * (96485 C/mol) * (1.4 × 10⁻⁹ m²/s)²⁽³ * (229.18 radians/s)¹⁽² * (2.0 × 10⁻⁶ m²/s)⁻¹⁽⁶ * (0.025 mol/m³)
≈ 2.53 A/m²
Therefore, the diffusion layer thickness (δ) is approximately 245 nm, and the current density (J) is approximately 2.53 A/m² for the reduction of 0.025 M Tl³⁺ at a gold electrode in 1 F HCl at 0.95 V vs. SHE with a rotation rate of 2200 rpm.
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Biphenyl, C
12
H
10
, is a nonvolatile, nonionizing solute that is soluble in benzene, C
6
H
6
. At 25
∘
C, the vapor pressure of pure benzene is 100.84Torr. What is the vapor pressure of a solution made from dissolving 11.1 g of biphenyl in 30.7 g of benzene? P
solution
=[1
Given: The molecular formula of Biphenyl is C12H10, 11.1g of Biphenyl is dissolved in 30.7 g of benzene and at 25°C, the vapor pressure of pure benzene is 100.84 Torr.To calculate: The vapor pressure of a solution made from dissolving 11.1 g of Biphenyl in 30.7 g of benzene.
First, calculate the mole fraction of Biphenyl. Mole fraction of Biphenyl = mole of Biphenyl / Total moles of the solution We have the mass of solute and solvent, using this, we can calculate moles of solute and solvent. The molecular weight of Biphenyl (C12H10) = (12 * 12.01 g/mol) + (10 * 1.008 g/mol) = 154.22 g/mol
Moles of Biphenyl = Mass
Molecular weight = 11.1 g / 154.22 g/mol
= 0.0719 mol The molecular weight of Benzene .
Total moles of solution = moles of Benzene + moles of Biphenyl= 0.3931 + 0.0719= 0.465 molesMole fraction of Biphenyl = 0.0719 / 0.465 = 0.1546Vapor pressure of the solution can be calculated using the formula:P solution = P°benzene * XbenzeneWhere, P°benzene is the vapor pressure of the pure solvent i.e. 100.84 .
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how many energy sublevels are in the 2nd energy level
The second energy level is composed of two energy sublevels. An energy level can be divided into sublevels, each of which corresponds to a specific shape and energy level.
Energy levels are a set of energy states that an atom or molecule can have. They are normally represented as electron orbits that can hold a specific number of electrons. The first energy level can hold up to two electrons, while the second energy level can hold up to eight electrons, and so on. Each energy level is made up of sublevels. These sublevels are known as s, p, d, and f. Each sublevel has a specific shape and energy, and it can hold a specific number of electrons. The s sublevel can hold a maximum of two electrons, the p sublevel can hold a maximum of six electrons, the d sublevel can hold a maximum of ten electrons, and the f sublevel can hold a maximum of fourteen electrons.
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The decomposition of HI is a first order reaction. At a particular temperature, the rate constant for the reaction is 4.2×10−5/sec. What is the percent HI decomposed at 750 secs after the reaction begins
The decomposition of HI is a first order reaction. At a particular temperature, the rate constant for the reaction is 4.2×10−5/sec, the percent HI decomposed at 750 secs after the reaction begins is 86%.
The rate equation for this first order reaction can be written as:rate = k [HI]t = (1/k) ln [HI]0/[HI], where t is the time taken for the reaction, k is the rate constant, [HI] is the concentration of HI at any given time, and [HI]0 is the initial concentration of HI.
Now, to find the percent HI decomposed at 750 secs after the reaction begins, we need to use the formula:% decomposition = (initial conc. of HI – conc. of HI after 750 sec)/initial conc. of HI x 100%
We know that the rate constant, k = 4.2×10−5/sec and the time, t = 750 sec, therefore, we can calculate the concentration of HI at any given time using the formula: ln [HI]0/[HI] = kt[HI] = [HI]0/e^(kt)
Plugging in the values, we get:[HI] = [HI]0/e^(4.2×10−5 × 750) = [HI]0/e^31.5 = 0.14 × [HI]0
Therefore, the percent HI decomposed after 750 sec is: % decomposition = (1 – 0.14) × 100% = 86%
Therefore, the percent HI decomposed at 750 sec after the reaction begins is 86%.
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on which side of a reduction half-reaction do the electrons appear?
In a reduction half-reaction, electrons appear on the left side.
This is because reduction is a process that involves the gain of electrons. In other words, the species that is being reduced is gaining electrons and thus they need to appear on the side where this species is located.
A reduction half-reaction is a chemical reaction that describes the reduction of a species by the gain of electrons. It is typically written in the form of a chemical equation that shows the species being reduced along with the electrons that are being gained.
For example, the reduction half-reaction for the reduction of copper (II) ion to copper (I) ion can be written as follows:Cu2+ + e- → Cu+In this reaction, the copper (II) ion is being reduced to copper (I) ion by the gain of one electron. The electron is represented on the left-hand side of the equation.
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how many hydrogen atoms is the carbonyl group in a ketone bonded to?
Answer:
mark it as brainlist......
Explanation:
Ketones. In ketones, the carbonyl group has two hydrocarbon groups attached.
An ion of the element oxygen (O) has an overall charge of 2−. Therefore, the number of electrons in this oxygen ion is
.
Considering the definition of atomic number, the number of electrons in this oxygen ion is 18.
Definition of atomic numberAll atoms are made up of subatomic particles: protons and neutrons, which are part of their nucleus, and electrons, which revolve around them. Protons are positively charged, neutrons are neutrally charged, and electrons are negatively charged (electrons).
The atomic number represents the number of protons present in the nucleus of an atom of any element. If the atom is electrically neutral, the number of electrons is equal to the number of protons.
Number of electrons in the oxygenThe atomic number of oxygen is 16. If it is a neutral atom, the number of protons and the number of electrons is 16.
Anions are negatively charged ions. They are formed as a consequence of the gain of electrons. Therefore, anions are characterized by having more electrons than protons in their composition.
In this case, an ion of the element oxygen (O) has an overall charge of -2. It means that the neutral atom has gained 2 electrons, so it will have a number of electrons of 18.
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consider two cells the first with al and ag electrodes
Let's first understand the terms AL and Ag electrodes in order to form a detailed answer.AL electrodes: AL is the symbol of aluminum, which is an excellent conductor of electricity, lightweight, and inexpensive.
The use of aluminum (AL) electrodes in lithium-ion batteries is common because it is a cost-effective and environmentally friendly way to make high-capacity batteries. Aluminum foil is the most commonly used AL electrode material in lithium-ion batteries.
Due to its thinness and uniformity, AL foil is ideal for making battery electrodes. Ag electrodes:Ag is the symbol of silver, which is a soft, white, lustrous metal that is an excellent conductor of electricity. In the field of electrochemistry, silver is used as a reference electrode for the determination of the activity of other metals.
Ag/AgCl is the most common type of reference electrode, and it is used for numerous electrochemical measurements. The difference between the two is: The primary difference between AL and Ag electrodes is their conductivity.
Aluminum (AL) is a metal that is an excellent conductor of electricity, and it is used in numerous applications. In contrast, silver (Ag) is a more conductive metal that is used in specialized applications, such as in reference electrodes.
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Liquid phase elementary rection is carried out
2A→B+C is carried out in a tubular reactor in which the volumetric flow rate, v, is constant. Derive an equation relating the reactor volume to the entering and exiting concentrations of A, the rate constant k, and the volumetric flow rate v. Determine the reactor volume necessary to reduce the exiting concentration to 35% of the entering concentration. Given, The rate law, −rₐ =kCₐ² Entering Concentration, CA0=25 mol/dm³ Volumetric flow rate is 20 dm³/min (liters /min ) and The specific reaction rate, k, is 0.15 min⁻¹.
The mass of A that enters the reactor in time dt is equal to the mass of A that exits the reactor in the same time in Liquid phase elementary rection .
The reactor volume required to reduce the concentration of A at the exit of the reactor to 35% of the entering concentration is 47.62 dm³.
Given Data:
The specific reaction rate, k, is 0.15 min⁻¹
The rate law, −rₐ =kCₐ²
Entering Concentration, CA0=25 mol/dm³
Volumetric flow rate is 20 dm³/min (liters /min )
Reaction : 2A→B+C
Derive an equation relating the reactor volume to the entering and exiting concentrations of A, the rate constant k, and the volumetric flow rate v.
The mole balance equation for A is given by:
CA = CA0 (1 - X)Here, X is the fractional conversion of A.
CA = Concentration of A at any point in the reactor
CA0 = Initial Concentration of A
Volume of Reactor = VR
Mass balance equation gives the relation between the concentration of A and that of B or C.
Based on this relation, the concentration of A can be determined in terms of that of B or C.
To derive an equation relating the reactor volume to the entering and exiting concentrations of A, the rate constant k, and the volumetric flow rate v, we apply the mole balance to the reactant A.
The volumetric flow rate, v = 20 dm³/min
The flow rate is constant.
Therefore, the mass of A that enters the reactor in time dt is equal to the mass of A that exits the reactor in the same time.
The balance equation for A is given by,FA0 - FA = -rA×VRFA0
= CA0 × v, FA
= CA × v
For a liquid-phase, elementary reaction, the rate of reaction rA is given by the rate law,-rA = kCA²
At steady-state, FA0 - FA = -rA×VRCA0 × v - CA × v
= kCA²× VR
Canceling the terms of v from both sides and rearranging gives,
The design equation for a constant-volume, liquid-phase reactor is given by,
Where VR = Reactor Volume
CA0 = Concentration of A at the entrance of the reactor,
mol/dm³CA = Concentration of A at the exit of the reactor,
mol/dm³k = specific reaction rate,
min⁻¹v = Volumetric flow rate, dm³/min
Here, the entering concentration of A, CA0 = 25 mol/dm³
Given, the exit concentration of A is to be reduced to 35% of the entering concentration of A.
The desired concentration of A at the exit of the reactor,
CA = 0.35 × CA0
= 0.35 × 25
= 8.75 mol/dm³
Substitute the given values in the equation of reactor volume and calculate the reactor volume.
VR = (CA0 - CA) / (k × CA × v)
= (25 - 8.75) / (0.15 × 8.75 × 20)
= 47.62 dm³
Therefore, the reactor volume required to reduce the concentration of A at the exit of the reactor to 35% of the entering concentration is 47.62 dm³.
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The decomposition of N2O5 can be described with the following reaction: 2 N2O5(g) → 4NO2(g) + O2(g) The reaction is second order. The rate law for the reaction is:
a) rate = k[N2O5]
b) rate = k/ [N2O5]
c) rate = k[N2O5]2
d) rate = k ([NO2]4 x [O2]) / [N2O5]
e) rate = k/ [N2O5]2
The reaction is second order, which means that the rate of the reaction is dependent on the concentration of N₂O₅ raised to the power of 2. Therefore, the rate law for the reaction is: c) rate = k[N₂O₅]²
The rate law is a mathematical expression that describes how the rate of a reaction is influenced by the concentrations of the reactants. In this case, the reaction is second order, which means that the rate is proportional to the square of the concentration of N₂O₅.
By observing the balanced equation, we can see that for every 2 moles of N₂O₅, 4 moles of NO₂ and 1 mole of O₂ are produced. This indicates that the rate of the reaction is directly proportional to the square of the concentration of N₂O₅.
Therefore, the rate law for the decomposition of N₂O₅ is c) rate = k[N₂O₅]², where k is the rate constant.
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A carload of Hg-ore containing grains of cinnabar (86% Hg by mass; density= 8.19 g/cm3) and grains of basalt (containing no Hg; density=3.11 g/cm3) is to be sampled and analyzed for mercury.
The average density of the bulk material is found to be 3.32 g/cm3.
To keep the sampling error below 0.5% relative it is determined that 1.01×106 particles are needed in the gross sample.
Supposing the grains in the sample had an approximate spherical radius of 3.0 mm, then the gross sample should weigh about 3.79×105 g (roughly half a ton). This is an impracticably large amount for lab analysis.
Therefore, the carload of ore is repeatedly crushed or ground, mixed and divided until the grains are of a sufficiently smaller radius such that only a 140-gram gross sample will be required in the lab to obtain the same level of sampling uncertainty.
In the final ground and mixed sample, what should be the average particle mass?
In the final ground and mixed sample, what should be the approximate particle radius?
The average particle mass in the final ground and mixed sample should be approximately 0.375 grams. The approximate particle radius in the final ground and mixed sample should be approximately 2.9999 mm.
To determine the average particle mass and approximate particle radius in the final ground and mixed sample, we can use the given information and apply some calculations.
Given: Initial gross sample weight: 3.79 × 10^5 g
Desired gross sample weight for lab analysis: 140 g
Number of particles in the gross sample: 1.01 × 10^6
Average density of the bulk material: 3.32 g/cm^3
Approximate spherical radius of the grains in the sample (initially): 3.0 mm
To find the average particle mass:
1. Calculate the reduction in sample weight:
Weight reduction = Initial gross sample weight - Desired gross sample weight
= 3.79 × 10^5 g - 140 g
= 3.7886 × 10^5 g
2. Calculate the reduction factor in weight:
Reduction factor = Weight reduction / Initial gross sample weight
= (3.7886 × 10^5 g) / (3.79 × 10^5 g)
≈ 0.99976
3. Calculate the reduction factor in the number of particles:
Reduction factor in particle count = Reduction factor^(1/3)
= 0.99976^(1/3)
≈ 0.99992
4. Calculate the average particle mass:
Average particle mass = Initial gross sample weight / Number of particles
= 3.79 × 10^5 g / (1.01 × 10^6 × Reduction factor in particle count)
≈ 0.375 g
Therefore, the average particle mass in the final ground and mixed sample should be approximately 0.375 grams.
To find the approximate particle radius:
1. Calculate the reduction in particle radius:
Reduction factor in radius = Reduction factor in particle count^(1/3)
= 0.99992^(1/3)
≈ 0.99997
2. Calculate the approximate particle radius:
Approximate particle radius = Initial particle radius × Reduction factor in radius
= 3.0 mm × 0.99997
≈ 2.9999 mm
Therefore, the approximate particle radius in the final ground and mixed sample should be approximately 2.9999 mm.
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Indicate the concentration of each ion present in the solution formed by mixing the following.
(a) 14 mL of 0.100 M HCl and 10.0 mL of 0.220 M HCl
H+ :
Cl- :
(b) 15.0 mL of 0.300 M Na2SO4 and 13.2 mL of 0.100 M NaCl
Na+ :
SO42- :
Cl- :
(c) 3.50 g of KCl in 60.0 mL of 0.458 M CaCl2 solution (assume that the volumes are additive)
K+ :
Ca2+ :
Cl- :
The concentration of H+ ions remains 0.100 M, while the concentration of Cl- ions is 0.0522 M in the mixture of HCl solutions.
For the mixture of 15.0 mL of 0.300 M Na2SO4 and 13.2 mL of 0.100 M NaCl, the concentration of Na+ ions is determined by the total moles of Na+ ions, resulting in a concentration of 0.177 M. The concentration of SO42- ions remains unchanged at 0.300 M, while the concentration of Cl- ions also remains unchanged at 0.100 M. In the mixture of 3.50 g of KCl in 60.0 mL of 0.458 M CaCl2 solution, the concentration of K+ ions can be calculated by converting the mass of KCl to moles and dividing by the total volume, resulting in a concentration of 0.672 M. The concentration of Ca2+ ions remains unchanged at 0.458 M, and the concentration of Cl- ions is determined by the total moles in the solution.
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One mole of an ideal gas contained in a piston-cyclinder assemebly is compressed from 100kPa and 27C∘till its volume is reduced to 1/15 of the original volume. The process of compression is polytropic with n=1.2. Derermine a) The final temperature and pressure of the gas b) the work done on the gas and c) the heat of interation (5 Marks)
a) The final temperature of the gas is 746.76 K, and the final pressure is 248.92 kPa. b) The work done on the gas is approximately -8.27 kJ. And c) The heat interaction is 0.
To solve this problem, we can use the ideal gas equation and the polytropic process equation: The ideal gas equation is given by: PV = nRT.The polytropic process equation is given by: PV^n = constant. Given: Initial pressure (P1) = 100 kPa Initial temperature (T1) = 27°C = 273 + 27 K = 300 K Initial volume (V1) = V Final volume (V2) = (1/15)V Polytropic exponent (n) = 1.2
a) Final temperature and pressure of the gas: We can use the ideal gas equation to find the final temperature (T2) and pressure (P2). Using the polytropic process equation, we have: P1V1n = P2V2n. Substituting the given values, we get: (100 kPa)(V)1.2 = P2[(1/15)V]1.2
Simplifying, we have: P2 = (100 kPa)(15)^1.2 = 248.92 kPa. Now, using the ideal gas equation: P1V1 / T1 = P2V2 / T2. Substituting the given values and the calculated P2, we get: (100 kPa)(V) / 300 K = (248.92 kPa)[(1/15)V] / T2. Simplifying, we find: T2 = (300 K)(248.92 kPa) / (100 kPa) = 746.76 K
Therefore, the final temperature of the gas is 746.76 K, and the final pressure is 248.92 kPa. b) Work done on the gas: The work done on the gas during the polytropic process can be calculated using the formula: W = (P2V2 - P1V1) / (1 - n)
Substituting the given values and the calculated P2, we get: W = (248.92 kPa)[(1/15)V] - (100 kPa)(V) / (1 - 1.2), Simplifying, we find: W ≈ -8.27 kJ,The negative sign indicates work done on the gas. c) Heat interaction: Since the process is adiabatic, there is no heat interaction with the surroundings. Therefore, the heat interaction is zero (Q = 0).
To summarize: a) The final temperature of the gas is 746.76 K, and the final pressure is 248.92 kPa. b) The work done on the gas is approximately -8.27 kJ. c) The heat interaction is 0.Please note that rounding errors may occur during the calculations, so the values provided are approximate.
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What effect would the presence of an insoluble impurity such as sodium sulfate have on the observed melting point of a sample? explain.
5. One of the most common errors in determining the melting point is heating the system too quickly (Melt-temp). Under these circumstances, how would you compare this observed melting point to the actual value (higher or lower)? explain
The presence of an insoluble impurity, such as sodium sulfate, in a sample will depress its melting point. This happens because the impurity alters the crystal lattice of the sample, causing the bonds between the molecules to become weaker and more disordered, and therefore less stable.
As a result, the sample will require less energy to melt and, consequently, its observed melting point will be lower than it would be without the impurity. This phenomenon is known as depression of the melting point and it can be used to determine the amount of impurity present in a sample by measuring the difference between the observed and expected melting points of the pure substance.
The observed melting point of a sample heated too quickly will be lower than the actual value. This is because when a substance is heated too quickly, its temperature will rise unevenly, causing different parts of the sample to melt at different rates.
As a result, the melting point will be observed at a lower temperature than it would be if the sample was heated more slowly and uniformly. Therefore, in order to obtain an accurate melting point, it is important to heat the sample slowly and uniformly, allowing the temperature to equilibrate before taking a reading.
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the equilibrium constant k of a certain first order reaction
In chemistry, the Equilibrium constant (K) is a mathematical relationship that relates to the concentrations of reactants and products at equilibrium in a reversible chemical reaction.
It is a dimensionless constant, and its value depends on the stoichiometry of the chemical reaction. For a certain first-order reaction, the equilibrium constant K can be calculated using the concentration of the reactant and product in the equation.K can be defined as the ratio of the concentration of products to the concentration of reactants at equilibrium for a given reaction. Mathematically, K = [products]/[reactants]where the square brackets denote the molar concentration of the product and reactant respectively.
The value of K determines the position of the equilibrium. If K > 1, then the concentration of products is higher than the concentration of reactants at equilibrium. Conversely, if K < 1, then the concentration of reactants is higher than the concentration of products at equilibrium. If K = 1, then the concentrations of products and reactants are equal at equilibrium.
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SO
2
is absorbed from air into water in a packed absorption tower, at room temperature. At a location in the tower, SO
2
is absorbed at the rate of 0.02 mol/m
2
h, with its liquid-phase mole fraction of 0.0025 at the gas-liquid interface, and 0.0003 in the bulk liquid. The diffusivity of SO
2
in water is 1.5×10
−6
cm
2
/s. a) If film theory is applicable, calculate the mass transfer coefficient and the film thickness. b) If penetration theory is applicable, calculate the contact time. c) If surface renewal theory is applicable, calculate the fractional surface renewal rate.
a) The mass transfer coefficient is 0.74 cm/h and the film thickness is 2.03μm.
b) The contact time is 3.7 s.
c) The fractional surface renewal rate is 0.09 h⁻¹ .
a) If film theory is applicable, the mass transfer coefficient and the film thickness are given by:
Mass transfer coefficient, K = (0.02 mol/m²h)/[(0.0025-0.0003)(1000/18)] = 0.74 cm/h
Film thickness, δ = (1.5 × 10⁻⁶ cm²/s) / (0.74 cm/h) = 2.03 × 10⁻⁶ cm or 2.03 μm
b) If penetration theory is applicable, the contact time is given by:δ = (1.11 × 10⁻⁵ cm²/s)tContact time, t = δ² / (1.11 × 10⁻⁵ cm²/s) = (2.03 × 10⁻⁶ cm)² / (1.11 × 10⁻⁵ cm²/s) = 3.7 s or 0.001 h.
c) If surface renewal theory is applicable, the fractional surface renewal rate is given by:f = (0.74 cm/h) / (2 × 2.03 μm) = 0.09 h⁻¹The fractional surface renewal rate is 0.09 h⁻¹.
The problem can be solved by applying three different theories of mass transfer. The first theory, film theory, is applicable when the mass transfer resistance is in the liquid film surrounding the gas bubble.
The second theory, penetration theory, is applicable when the gas bubbles penetrate into the liquid and the mass transfer resistance is mainly in the liquid phase. The third theory, surface renewal theory, is applicable when the liquid phase around the gas bubbles is constantly being renewed due to turbulent mixing.
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how many grams are in 2.00 mol of sodium chloride (nacl)?
The molar mass of NaCl can be calculated from the atomic masses of the constituent elements using the formula: Molar mass of NaCl = atomic mass of Na + atomic mass of Cl.
The atomic masses of sodium (Na) and chlorine (Cl) are 22.99 g/mol and 35.45 g/mol, respectively. Thus, the molar mass of NaCl is:
Molar mass of NaCl = 22.99 g/mol + 35.45 g/mol
= 58.44 g/mol
Therefore, 1 mol of NaCl weighs 58.44 g.
To find the mass of 2.00 mol of NaCl, we can simply multiply the molar mass by the number of moles:
Mass of 2.00 mol of NaCl = 2.00 mol × 58.44 g/mol
= 116.88 g
So there are 116.88 grams in 2.00 mol of sodium chloride (NaCl).I hope this answer is more detailed and contains 100 words.
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low ph means that the hydrogen-ion concentration is
A low pH indicates high hydrogen ion concentration, affecting enzymatic activity, protein function, and cell division. The pH scale measures acidity or basicity of a substance, with 0 being the most acidic and 14 being the most basic. Monitoring pH levels is crucial for optimal functioning, as it can damage cells and cause diseases.
When pH is low, the concentration of hydrogen ions increases, and acidity rises. The pH scale ranges from 0 to 14, with 0 being the most acidic and 14 being the most basic. Water is neutral, with a pH of 7, and anything above 7 is considered basic. A low pH means that the concentration of hydrogen ions is high. The pH scale is used to measure the concentration of hydrogen ions in a solution. It measures the acidity or basicity of a substance. The pH scale ranges from 0 to 14, with 0 being the most acidic and 14 being the most basic. pH is calculated based on the number of hydrogen ions in a solution.
Knowing the pH of a solution is critical since it can impact various factors such as enzymatic activity, protein function, and even cell division. Therefore, a slight change in the pH of a solution can result in a big change in the entire system. If the pH of a solution is too high or too low, it can damage the cells and cause diseases. Thus, it is crucial to monitor pH levels and maintain them at a level that is conducive for the system’s optimal functioning.
Conclusion A low pH means that the hydrogen-ion concentration is high. The pH scale is used to measure the concentration of hydrogen ions in a solution. It measures the acidity or basicity of a substance. Knowing the pH of a solution is critical since it can impact various factors such as enzymatic activity, protein function, and even cell division.
Therefore, it is crucial to monitor pH levels and maintain them at a level that is conducive for the system’s optimal functioning.
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LiFeSiO4 and LiAlSiO4 are both silicate materials with potential applications in lithium-ion batteries. Identify the type of silicate structure and the repeating silicate anion unit for each case. (a) What type of silicate structure is found in LiFeSiO4 ? [5 pts] (b) What type of silicate structure is found in LiAlSiO4 ? [5 pts] The possible answer choices for (a) - (b): Island silicate (orthosilicate), Single chain (pyroxene) silicate, double chain (amphibole) silicate, sheet silicate, aluminosilicate, 3D network and compare/contrast the structures. Cite all external sources used. [10 pts]
LiAlSiO4 has potential applications in the fields of microelectronics, nuclear waste immobilization, and chemical sensors because of its unique structural and physical properties.
(a) The type of silicate structure found in LiFeSiO4 is Olivine Silicate Structure.
Olivine silicates are of great interest in solid-state materials, as they are composed of tetrahedra (SiO4) that form a three-dimensional network held together by divalent metal cations.
The silicate anion unit is (SiO4)4-.(b) The type of silicate structure found in LiAlSiO4 is the Aluminosilicate structure.
Aluminosilicates are materials that have a three-dimensional network of tetrahedrally coordinated SiO4 and AlO4, where each oxygen atom is shared by two tetrahedra.
The silicate anion unit is (SiO4)4-. These structures can be compared as follows: In Olivine, SiO4 tetrahedra are bonded in chains of corner-sharing tetrahedra, while in Aluminosilicate, SiO4 tetrahedra are arranged in a three-dimensional framework.
In Olivine, metals are held in octahedral coordination with oxygen, while in Aluminosilicate, AlO4 tetrahedra are present in the structure.
LiFeSiO4 has potential applications in lithium-ion batteries because of its high theoretical capacity and superior electrochemical performance over other materials.
LiAlSiO4 has potential applications in the fields of microelectronics, nuclear waste immobilization, and chemical sensors because of its unique structural and physical properties.
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At 370 C and 1 atm fully saturated inspired air has PH20 of 47mmHg and the capacity
of water vapor amount(max.) is 43.8mg/L. If the water vapor content is content is 28.5mg/L,
calculate the %Body Humidity (%BH). 1pt
10. The total content of a gas is 40mg/L. The capacity is 44mg/L. What is the %Relative
Humidity (%RH)
The % Relative Humidity (%RH) is 90.91%.
At 370°C and 1 atm fully saturated inspired air has PH20 of 47 mmHg and the capacity of water vapor amount (max.) is 43.8 mg/L.
Water vapor content is 28.5 mg/L.
The total content of gas is 40 mg/L. The capacity is 44 mg/L.
To find:% Body Humidity (%BH) = ?%
Relative Humidity (%RH) = ?
% Body Humidity (%BH)H
ere, capacity of water vapor = 43.8 mg/LWater vapor content = 28.5 mg/L% Body Humidity (%BH) = (28.5/43.8) × 100 = 65.07%
Therefore, the %Body Humidity (%BH) is 65.07%.% Relative Humidity (%RH)
Here, capacity of gas = 44 mg/LTotal content of gas = 40 mg/L% Relative Humidity (%RH) = (Total content of gas / Capacity of gas) × 100= (40/44) × 100= 90.91%
Relative humidity is the amount of water vapor present in the air expressed as a percentage of the amount needed for saturation at the same temperature.
A high percentage of relative humidity makes hot weather more unbearable, while low humidity makes cold weather feel more severe. Both factors may cause people to seek refuge in climate-controlled environments.
The relative humidity can be calculated as:%RH = (actual water vapor content / maximum water vapor content) × 100Hence, the maximum water vapor capacity can be calculated by Dalton’s law of partial pressure and it can be calculated by using the following formula:
PH2O = (760 - PB) (XH2O)
Here, PB = partial pressure of dry gasXH2O = mole fraction of water vapor
The unit of PH2O will be in mmHg.
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What is the area in square meters of 392 g of a piece of tin foil (tin atomic symbol: Sn ) that has a thickness of 21747 nm ? The density of Sn is 5.75 g/cm
3
. Useful information: 1 nm=10
−7
cm, Volume = length x width x height, area= length x width
Weight of the piece of tin foil = 392 g Thickness of the tin foil = 21747 nmDensity of the Sn = 5.75 g/cm³To find:Area of the tin foil in square meters.
As we know, Density = Mass/VolumeSo, Volume = Mass/Density=> Volume of tin foil = 392 g/5.75 g/cm³=> Volume of tin foil = 68 cm³Now, we have to find out the area of the tin foil.To calculate the area of tin foil, we need to determine its dimensions or measurements, i.e., length and width.
The given tin foil has thickness, and we are not given its length and width, which makes it impossible to determine the area.However, we have volume of tin foil. So, Area = Volume / Thickness = 68 cm³ / (21747 nm × 10^-7 cm/nm) = 0.3135 m²Therefore, the area of tin foil is 0.3135 square meters.
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make a table to distinguish between polymer blend and composite polymer based on :
1- ingredient
2- bonding nature
3- homogeneity
Polymer blends consist of physically mixed polymers, exhibit weak intermolecular bonding, and tend to have limited homogeneity.
Composite polymers, on the other hand, involve the incorporation of a reinforcing material, exhibit strong bonding between the matrix and the reinforcement, and can be engineered to have a high degree of homogeneity. Polymer blends and composite polymers can be distinguished based on their ingredients, bonding nature, and homogeneity.
In terms of ingredients, polymer blends consist of two or more different polymers that are physically mixed together. Each polymer retains its individual chemical identity within the blend. On the other hand, composite polymers involve the incorporation of a reinforcing material, such as fibers or particles, into a polymer matrix. This reinforcing material can be made of various substances, such as glass, carbon, or ceramics.
Regarding bonding nature, polymer blends rely on weak intermolecular forces or physical interactions between the different polymers in the blend. These interactions can include van der Waals forces, hydrogen bonding, or entanglements. In contrast, composite polymers exhibit strong bonding between the polymer matrix and the reinforcing material. This bonding can be achieved through chemical reactions, such as covalent bonding, or through physical interactions, such as interlocking or adhesion.
Homogeneity refers to the uniformity of the composition throughout the material. Polymer blends generally exhibit limited homogeneity, as the different polymers tend to phase separate to some extent, resulting in distinct regions or domains of each polymer within the blend.
Composite polymers, however, can be designed to have a high degree of homogeneity by dispersing the reinforcing material uniformly throughout the polymer matrix. This uniform dispersion enhances the mechanical properties and performance of the composite.
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Use your knowledge of carbocation stability to help you predict the major product of the following reaction.
The major product of the reaction can be predicted by considering the stability of carbocations.
How does carbocation stability affect the outcome of the reaction?In this reaction, the stability of the carbocation intermediate plays a crucial role in determining the major product. Carbocations are positively charged carbon species with three substituents and an empty p orbital. The stability of carbocations increases with the presence of electron-donating groups and the ability to delocalize the positive charge.
Carbocations can undergo rearrangements to form more stable intermediates. The most stable carbocation is the tertiary carbocation, followed by the secondary carbocation and then the primary carbocation. This stability trend is based on the concept of hyperconjugation, where the adjacent σ bonds can donate electron density to stabilize the positive charge.
The major product of the reaction will be formed through the most stable carbocation intermediate. If there are multiple possible carbocations, rearrangements may occur to form the most stable one before the product is generated. By considering the stability of carbocations, we can predict the favored pathway and major product in this reaction.
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Balance the following chemical equations by indicating a number for cach cocfficient (including
′
' if applicable). a. Fe(s)+Cl
2
(g)→FeCl
3
(s) b. c. ⟶C
1
H
6
O
2
(3)+…O
2
( g)→
CO
2
(Q)+…
+H
2
O(Q) d. Fe
2
(SO
4
)
3
(aq)+BaCl
2
(aq)→ FeCl
3
(aq)+ BaSO
4
(s) 5. Consider the below reaction for the synthesis of hydrazine, N
2
H
4
, a substance used as rocket fuel. a. Balance the chemical reaction by indicating a number for each coefficient. b. Provide the correct chemical name for each of the products. N
2
H
4
NH
4
Cl: Write balanced chemical equations (including phases) for the following reactions. a. Solid zinc sulfide reacts with oxygen gas to form solid zinc oxide and sulfur dioxide gas. b. An aqueous solution of nitric acid reacts with solid copper(II) carbonate to form liquid water, carbon dioxide gas, and aqueous copper(II) nitrate.
Chemical name for [tex]NH4Cl[/tex] is ammonium chloride.
a. [tex]ZnS(s) + O2(g) → ZnO(s) + SO2(g)[/tex]
b. [tex]2HNO3(aq) + CuCO3(s) → H2O(l) + CO2(g) + Cu(NO3)2(aq)[/tex]
a. [tex]Fe(s) + Cl2(g) → 2FeCl3(s)[/tex]
To balance this equation, we need two moles of [tex]FeCl3[/tex] for every one mole of Fe.
Therefore, we put a coefficient of 2 in front of [tex]FeCl3.[/tex]
b. [tex]C6H6O2(l) + O2(g) → 3CO2(g) + 3H2O(g)[/tex]
To balance this equation, we need 3 moles of [tex]CO2[/tex] and 3 moles of [tex]H2O[/tex]for every one mole of [tex]C6H6O2[/tex].
Therefore, we put a coefficient of 3 in front of [tex]CO2[/tex] and [tex]H2O.[/tex]
c. [tex]Fe2(SO4)3(aq) + 3BaCl2(aq) → 6FeCl3(aq) + Ba2(SO4)3(s)[/tex]
To balance this equation, we need 6 moles of [tex]FeCl3[/tex] and one mole of [tex]Ba2(SO4)3[/tex] for every one mole of [tex]Fe2(SO4)3.[/tex]
Therefore, we put a coefficient of 6 in front of [tex]FeCl3[/tex] and 1 in front of [tex]Ba2(SO4)3.[/tex]
[tex]N2H4 + 2NH4Cl → 4NH3 + N2[/tex]
To balance this equation, we need 4 moles of [tex]NH3[/tex] and one mole of N2 for every one mole of [tex]N2H4.[/tex]
Therefore, we put a coefficient of 4 in front of [tex]NH3[/tex] and 1 in front of N2.
The correct chemical name for [tex]NH4Cl[/tex] is ammonium chloride.
a. [tex]ZnS(s) + O2(g) → ZnO(s) + SO2(g)[/tex]
b. [tex]2HNO3(aq) + CuCO3(s) → H2O(l) + CO2(g) + Cu(NO3)2(aq)[/tex]
In both equations, the given reactants and products are balanced with appropriate coefficients to ensure the conservation of mass.
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In column chromatography using active alumina, which one will elute first
1.alcohol
2.hydrocarbon
3.carbo. acid
4.Clorobenzene
please explain why
In column chromatography using active alumina, alcohol (Option 1) will elute first because it will have the highest affinity for the stationary phase.
Column chromatography is a laboratory technique for separating and purifying chemical compounds based on their molecular weight. In column chromatography, the sample mixture is added to the top of a column packed with a stationary phase, and a mobile phase is run through the column, causing different compounds in the sample mixture to elute at different times based on their affinity for the stationary phase.
Active alumina, one of the common stationary phases, is polar. Alcohols, being polar, will have the highest affinity for the stationary phase and will therefore be retained longer in the column, causing them to elute last. Thus, among the given compounds, alcohol will elute first in column chromatography using active alumina.
Hence, the correct answer is Option 1.
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A box of volume 38.2 dm^3 initially contains 30 mols of H2 and 1 mol of N2 at 273.15 K. All of the N2 reacts with enough hydrogen to form NH3 (N2 + 3 H2 -> 2 NH3). Find the total pressure of the box and the partial pressure of each component in the finak mixture at 273.15 K. (Answer should be in atm).
The total pressure of the box is P_total = (35 moles * 0.0821 L.atm/mol.K * 273.15 K) / 38.2 L.
The partial pressure of NH₃ is (20 moles * 0.0821 L.atm/mol.K * 273.15 K) / 38.2 L.
The partial pressure of H₂ is (15 moles * 0.0821 L.atm/mol.K * 273.15 K) / 38.2 L.
The partial pressure of N₂ is 0 atm since all of it reacts and none remains in the mixture.
To find the total pressure and partial pressure of each component in the final mixture, we can use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant (0.0821 L.atm/mol.K), and T is the temperature.
First, let's calculate the total moles of gas in the box after the reaction:
Since N₂ reacts with 3 H₂ to form 2 NH₃, and we have 30 moles of H₂, we will need 1/3 of that amount of N₂ (1/3 * 30 = 10 moles of N₂).
Now, let's find the partial pressure of each component in the final mixture:
The total pressure is the sum of the partial pressures. Therefore, we need to calculate the partial pressure of NH₃, H₂, and N₂.
1. NH₃:
Since 2 moles of NH₃ are formed for every 1 mole of N₂, the number of moles of NH₃ after the reaction will be (2/1 * 10 moles of N₂ = 20 moles of NH₃).
2. H₂:
Since there are initially 30 moles of H₂ and 3 moles of H₂ react with 1 mole of N₂ to form 2 moles of NH₃, the number of moles of H₂ remaining after the reaction will be (30 moles - 3/2 * 10 moles of N₂ = 15 moles of H₂).
3. N₂:
All of the N₂ reacts, so there will be no N₂ remaining after the reaction.
Now, let's calculate the partial pressures:
First, convert the volume from dm³ to liters by multiplying by 1 dm³/1 L.
The total pressure is given by the equation PV = nRT:
P_total * V = (20 moles of NH₃ + 15 moles of H₂) * 0.0821 L.atm/mol.K * 273.15 K.
Now, let's solve for P_total:
P_total = (35 moles of gas * 0.0821 L.atm/mol.K * 273.15 K) / V.
Next, calculate the partial pressures:
Partial pressure of NH₃ = (20 moles of NH₃ * 0.0821 L.atm/mol.K * 273.15 K) / V.
Partial pressure of H₂ = (15 moles of H₂ * 0.0821 L.atm/mol.K * 273.15 K) / V.
Finally, plug in the given volume value of 38.2 dm³ (which is equal to 38.2 L) into the equations to find the total pressure and partial pressures.
Therefore, the total pressure of the box is P_total = (35 moles * 0.0821 L.atm/mol.K * 273.15 K) / 38.2 L.
The partial pressure of NH₃ is (20 moles * 0.0821 L.atm/mol.K * 273.15 K) / 38.2 L.
The partial pressure of H₂ is (15 moles * 0.0821 L.atm/mol.K * 273.15 K) / 38.2 L.
The partial pressure of N₂ is 0 atm since all of it reacts and none remains in the mixture.
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Which of the following is the correctly charged balanced equation for the reaction below? Cr(s)+Fe3+(aq)→Fe(s)+Cr2+(aq) 3Cr(s)+Fe3+(aq)→Fe(s)+3Cr2+(aq)3Cr(s)+2Fe3+(aq)→2Fe(s)+3Cr2+(aq)Cr(s)+2Fe3+(aq)→Fe(s)+3Cr2+(aq)Cr(s)+2Fe3+(aq)→2Fe(s)+Cr2+(aq)
The correct balanced equation for the given reaction is: Cr(s) + 2Fe[tex]_{3}[/tex]+(aq) → 2Fe(s) + Cr[tex]_{2}[/tex]+(aq). The answer is D).
To balance the equation, we need to ensure that the number of atoms of each element is the same on both sides of the reaction.
In the given equation, we have one chromium atom (Cr) on the left side and two chromium atoms (Cr[tex]_{2}[/tex]+) on the right side. To balance the chromium atoms, we need to multiply the Cr(s) by 2 and the Cr[tex]_{2}[/tex]+ by 1.
Next, we have one iron atom (Fe) on the left side and two iron atoms (Fe) on the right side. To balance the iron atoms, we need to multiply the Fe(s) by 2 and the Fe[tex]_{3}[/tex]+ by 1.
Finally, we need to balance the charges. On the left side, the total charge is 0 (since Cr is neutral). On the right side, we have a total charge of +6 (from 2Fe[tex]_{3}[/tex]+ and 3Cr[tex]_{2}[/tex]+). To balance the charges, we need to multiply the Fe(s) by 3 and the Cr[tex]_{2}[/tex]+ by 2.
After balancing the equation, we get: Cr(s) + 2Fe3+(aq) → 2Fe(s) + Cr2+(aq).
Therefore, the answer is D) Cr(s) + 2Fe3+(aq) → 2Fe(s) + Cr2+(aq).
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1.2 grams of hydrochloric acid in 25 mL of solution 6) 98 grams of sodium hydroxide in 2.2 liters of solution 7) 45 grams of ammonia in 0.75t of solution Calculate the molarities of the following solutions:
The molarities of the given solutions are as follows:
1) Hydrochloric acid: 2.24 M
2) Sodium hydroxide: 1.84 M
3) Ammonia: 7.47 M
To calculate the molarity of a solution, we need to use the formula:
Molarity (M) = (moles of solute) / (volume of solution in liters)
1) Hydrochloric acid:
Given that 1.2 grams of hydrochloric acid is present in 25 mL of solution, we first need to convert grams to moles. The molar mass of HCl is approximately 36.5 g/mol. Therefore, the moles of HCl can be calculated as (1.2 g) / (36.5 g/mol) = 0.0329 mol.
Next, we convert the volume of the solution from milliliters to liters: 25 mL = 0.025 L.
Finally, we can calculate the molarity using the formula: Molarity = 0.0329 mol / 0.025 L ≈ 2.24 M.
2) Sodium hydroxide:
Given that 98 grams of sodium hydroxide is present in 2.2 liters of solution, we need to calculate the moles of NaOH. The molar mass of NaOH is approximately 40 g/mol. So, the moles of NaOH are (98 g) / (40 g/mol) = 2.45 mol.
The volume of the solution is already given in liters as 2.2 L.
Using the formula, Molarity = 2.45 mol / 2.2 L ≈ 1.84 M.
3) Ammonia:
Given that 45 grams of ammonia is present in 0.75 liters of solution, we calculate the moles of NH3. The molar mass of NH3 is approximately 17 g/mol. So, the moles of NH3 are (45 g) / (17 g/mol) ≈ 2.65 mol.
The volume of the solution is given as 0.75 L.
Using the formula, Molarity = 2.65 mol / 0.75 L ≈ 7.47 M.
In summary, the molarities of the given solutions are 2.24 M for hydrochloric acid, 1.84 M for sodium hydroxide, and 7.47 M for ammonia.
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