what is the freezing point of an aqueous solution that boils at 106.5 c?

1. The volume of water in the tank at time t is given by the equation

Volume(t) = 30 + t.

2.The appropriate initial condition for the ODE is S(0) = 0, as there is no salt initially in the tank.

3. It is not separable because the variables S(t) and t are not separable on opposite sides of the equation.

4. The solution can be expressed in terms of the integral as:
\(S(t) = (70 * \int e^{(t^{2/2} + 30t)} dt) / e^{(t^{2/2} + 30t)})\)

5.  the salt concentration in the tank as t→infinity is zero.

1. To determine the volume of water in the tank at time t, we need to consider the rate at which water is being added and drained. The tank is being filled at a rate of 2 liters per minute and drained at a rate of 1 liter per minute.

Since the tank starts with an initial volume of 30 liters, the volume of water in the tank at time t can be calculated using the equation:
Volume(t) = Initial volume + (Rate of filling - Rate of draining) * t

Volume(t) = 30 + (2 - 1) * t

So, the volume of water in the tank at time t is given by the equation

Volume(t) = 30 + t.

2. Let S(t) denote the amount of salt in the fish tank at time t in grams.

To show that S(t) satisfies the ODE S'(t) = 70 - (t+30)S(t),

we need to take the derivative of S(t) with respect to t and substitute it into the given ODE.

Taking the derivative of S(t), we have:

S'(t) = 0 - (1+0)S(t) + 0

S'(t) = -S(t)

Substituting this into the given ODE, we get:

-S(t) = 70 - (t+30)S(t)

Simplifying the equation, we have:

S'(t) = 70 - (t+30)S(t)

Therefore, S(t) satisfies the ODE S'(t) = 70 - (t+30)S(t).

The appropriate initial condition for the ODE is S(0) = 0,

as there is no salt initially in the tank.

3. This ODE is a first-order linear ordinary differential equation. It is not separable because the variables S(t) and t are not separable on opposite sides of the equation.

4. To solve the initial value problem for S(t) using the method of integrating factors, we first rewrite the ODE in standard form:

S'(t) + (t+30)S(t) = 70

The integrating factor is given by:
\(\mu(t) = e^{(\int (t+30) dt)} = e^{(t^2/2 + 30t)\)

Multiplying both sides of the equation by μ(t), we have:
\(e^{(t^2/2 + 30t)} * S'(t) + e^{(t^2/2 + 30t)} * (t+30)S(t) = 70 * e^{(t^2/2 + 30t)\)

Applying the product rule to the left side of the equation, we get:
\((e^{(t^{2/2} + 30t) * S(t))' = 70 * e^{(t^{2/2} + 30t)})\)

Integrating both sides of the equation with respect to t, we have:
\(\int (e^{(t^2/2 + 30t)} * S(t))' dt = \int (70 * e^{(t^2/2 + 30t))} dt\)

Using the fundamental theorem of calculus, the left side becomes:
\(e^{(t^2/2 + 30t)} * S(t) = \int (70 * e^{(t^2/2 + 30t))} dt\)

Simplifying the right side by integrating, we get:
\(e^{(t^2/2 + 30t)} * S(t) = 70 * \int e^{(t^2/2 + 30t)} dt\)

At this point, the integration of \(e^{(t^2/2 + 30t)\) becomes difficult to express in terms of elementary functions.

Hence, the solution can be expressed in terms of the integral as:
\(S(t) = (70 * \int e^{(t^2/2 + 30t)} dt) / e^{(t^2/2 + 30t)\)

5. As t approaches infinity, the exponential term \(e^{(t^2/2 + 30t)\) becomes very large, causing the salt concentration S(t) to approach zero. Therefore, the salt concentration in the tank as t→infinity is zero.

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The salt concentration in the tank as t approaches infinity is 70/3.

1. To determine the volume of water in the tank at time t, we need to consider the rate at which water is being poured into the tank and the rate at which water is being drained from the bottom.

At a rate of 2 litres per minute, water is being poured into the tank. So after t minutes, the amount of water poured into the tank is 2t litres.

At a rate of 1 litre per minute, water is being drained from the tank. So after t minutes, the amount of water drained from the tank is t litres.

Since the tank was initially filled with 30 litres of fresh water, the volume of water in the tank at time t is given by:
Volume(t) = 30 + 2t - t
Volume(t) = 30 + t

2. Let S(t) denote the amount of salt in the fish tank at time t. To determine the ODE for S(t), we need to consider the salt being poured into the tank and the salt being drained from the tank.

The salt concentration in the water being poured into the tank is 35 grams per litre. So the amount of salt being poured into the tank per minute is 35 * 2 = 70 grams.

The amount of salt being drained from the tank per minute is S(t)/Volume(t) * 1.

Therefore, the ODE for S(t) is:
S'(t) = 70 - S(t)/Volume(t)

The initial condition for this ODE is S(0) = 0, since there was no salt in the tank initially.

3. The ODE S'(t) = 70 - S(t)/Volume(t) is a first-order linear ODE. It is not separable since the variables S(t) and Volume(t) are mixed together.

4. To solve the initial value problem for S(t), we can rewrite the ODE as:
Volume(t) * S'(t) + S(t) = 70 * Volume(t)

This is a linear ODE of the form y'(t) + p(t)y(t) = g(t), where p(t) = 1/Volume(t) and g(t) = 70 * Volume(t).

To solve this type of ODE, we can multiply both sides by an integrating factor, which is the exponential of the integral of p(t).

The integrating factor is exp(integral of 1/Volume(t) dt) = exp(ln(Volume(t))) = Volume(t).

Multiplying both sides of the ODE by the integrating factor, we get:
Volume(t) * S'(t) + S(t) = 70 * Volume(t)
Volume(t) * S'(t) + Volume(t) * S(t) = 70 * Volume(t)^2
( Volume(t) * S(t) )' = 70 * Volume(t)^2

Integrating both sides with respect to t, we get:
Volume(t) * S(t) = 70/3 * Volume(t)^3 + C

S(t) = 70/3 * Volume(t)^2 + C/Volume(t)

Using the initial condition S(0) = 0, we can solve for C:
0 = 70/3 * 30^2 + C/30
C = -70000

Therefore, the solution for S(t) is:
S(t) = 70/3 * Volume(t)^2 - 70000/Volume(t)

5. As t approaches infinity, the volume of water in the tank becomes very large. In this case, we can approximate the volume of the tank as t, since the rate at which water is being poured in is 2 litres per minute. So the salt concentration in the tank as t approaches infinity is given by:
S(t)/Volume(t) = (70/3 * t^2 - 70000/t) / t
As t approaches infinity, the second term (-70000/t) approaches 0, so the salt concentration in the tank as t approaches infinity is:
S(t)/Volume(t) = 70/3 * t^2 / t = 70/3 * t

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Answer:

Carbon monoxide (CO) is not a greenhouse gas because it does not trap heat in the Earth's atmosphere.

Explanation:

Answer:

Carbon monoxide (CO) is not a greenhouse gas.

Carbon monoxide is a colorless, odorless, poisonous gas, CO, that burns with a pale-blue flame, produced when carbon burns with insufficient air: used chiefly in organic synthesis, metallurgy, and in preparation of metal carbonyls, as nickel carbonyl.

Water vapor is a dispersion, in air, of molecules of water, especially as produced by evaporation at ambient temperatures rather than by boiling.

The chemical compound with the formula CH4 is methane, a hydrocarbon and primary component of natural gas. It is used in the manufacture of plastics and other commercially important organic chemicals. Methane is also a greenhouse gas that affects the earth's temperature and climate system.

Any of the gases whose absorption of solar radiation is responsible for the greenhouse effect, including carbon dioxide, methane, ozone, and the fluorocarbons.

The Greenhouse Effect is an atmospheric heating phenomenon, caused by short-wave solar radiation being readily transmitted inward through the earth's atmosphere but longer-wavelength heat radiation less readily transmitted outward, owing to its absorption by atmospheric carbon dioxide, water vapor, methane, and other gases; thus, the rising level of carbon dioxide is viewed with concern.

Answer:

5867ml is the answer to 5.867L to ml

Answer:

s orbital and p orbital

Explanation:

Answer:

s orbital and p orbital

Explanation:

Answer:

If I remember correctly, 0.0149991895187944

Explanation:

hope this helps:)

Answer:

volume is the amount of space in a certain object

The product of the reduction of ethyl 4-oxobutanoate with sodium borohydride in ethanol at room temperature for 30 minutes is Ethyl 4-hydroxybutanoate (Option A).

What is ethyl 4-oxobutanoate and what does it look like?

Ethyl 4-oxobutanoate, also known as ethyl acetoacetate, is an organic compound with the chemical formula CH₃C(O)CH₂CO₂C₂H₅. It is derived from acetoacetic acid and is an ester of acetoacetic acid.

Ethyl 4-oxobutanoate is a colorless liquid with a fruity odor.

During the reduction reaction, sodium borohydride (NaBH₄) acts as a reducing agent, donating hydride ions (H-) to the carbonyl group of the ethyl 4-oxobutanoate. This results in the conversion of the carbonyl group (C=O) to a hydroxyl group (OH) and the formation of the corresponding alcohol.

Therefore, the correct product of the reduction of ethyl 4-oxobutanoate with sodium borohydride is Ethyl 4-hydroxybutanoate (Option A).

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Answer:

Explanation:

Decomposition reaction, also known as analysis or dissociation, is a type of chemical reaction in which a compound breaks down into simpler substances or elements. In this reaction, a single reactant undergoes a chemical change and produces two or more products.

The decomposition reaction can be represented by the general equation:

AB → A + B

Where AB is the reactant, and A and B are the products. The reactant AB is usually a compound, and it breaks down into its constituent elements or simpler compounds.

There are different types of decomposition reactions, including:

Thermal decomposition: It occurs when a compound is heated, resulting in its decomposition into simpler substances. For example, the thermal decomposition of calcium carbonate (CaCO3) produces calcium oxide (CaO) and carbon dioxide (CO2):

CaCO3 → CaO + CO2

Electrolytic decomposition: It takes place when an electric current is passed through an electrolyte, causing it to break down into its component ions. For instance, the electrolysis of water (H2O) leads to the decomposition into hydrogen gas (H2) and oxygen gas (O2):

2H2O → 2H2 + O2

Photochemical decomposition: It occurs when a compound undergoes decomposition due to exposure to light energy. Chlorine gas (Cl2) can decompose into chlorine atoms (Cl) under the influence of light:

Cl2 → 2Cl

These are just a few examples of decomposition reactions. They are important in various chemical processes and are used in industries, laboratory experiments, and natural phenomena. By understanding and controlling decomposition reactions, scientists can gain insights into the behavior of different compounds and develop practical applications in fields such as chemistry, materials science, and environmental science.

Answer:

Explanation:

reaction in which a compound breaks down into simpler substances or elements

Answer:

4.73 mol NH

Explanation:

What we're doing here is calculating basic mole-mole relationships, something that you'll be doing quite a bit!

The steps to solving mole-mole problems like this are

write the balanced chemical equation for the reaction (this is given)

divide the number of moles of the given known substance (3.55) by that substance's coefficient in the chemical equation (3)

multiply that number by the coefficient of the substance you're trying to find (4)

Using simple dimensional analysis, it looks like this:

Answer:

4.68 H2O is needed to make 3.12 moles of NH3

Explanation:

Using stoichiometry, we can convert Moles of NH3 into Moles of H20

since we're converting moles to moles, we take the given mole of NH3

3.12, and multiply it by the Mole ratio, which in this case is (H20/NH3) or (6/4)

so then: 3.12 * 6/4 = 4.68 mols of H2O

Answer:

True. The limiting reagent is the reactant that is completely consumed in a chemical reaction and limits the amount of product that can be formed. The theoretical yield is the maximum amount of product that can be obtained from the limiting reagent, assuming that the reaction goes to completion and no side reactions occur. However, in practice, it is common for side reactions to occur, which can reduce the actual yield of the product. Therefore, while the limiting reagent does control the theoretical yield of a reaction, the actual yield may vary due to the presence of side reactions or other factors that can affect the efficiency of the reaction.

Explanation:

D. Able to answer for one's conduct and obligations

The amygdala is a pair of almond-shaped structures located in the brain's temporal lobes. It is part of the limbic system, which is involved in the processing of emotions and memory. The amygdala is involved in the formation of memories associated with emotional events, as well as the reaction to those memories.

The amygdala also plays a role in the regulation of the body's fight-or-flight response, which is the autonomic nervous system's response to danger or perceived danger. It is thought to be involved in the formation of fears and phobias, as well as the development of aggressive behaviors. In addition, the amygdala is believed to be involved in the processing of social cues, such as facial expressions and body language.

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The molecular equation for the reaction between hydrochloric acid and calcium carbonate can be written as:

HCl + CaCO3 → CaCl2 + CO2 + H2O

Hard water is water that contains a high concentration of dissolved minerals, specifically calcium and magnesium ions. These minerals are picked up by the water as it flows through soil and rocks, such as limestone, chalk, or gypsum.

The degree of water hardness is usually measured in parts per million (ppm) of calcium carbonate. Water with a concentration of calcium carbonate between 60-120 ppm is considered moderately hard, while water with a concentration above 120 ppm is considered hard.

The molecular equation for the reaction between hydrochloric acid and calcium carbonate can be written as:

HCl + CaCO3 → CaCl2 + CO2 + H2O

In this reaction, the hydrochloric acid (HCl) reacts with the calcium carbonate (CaCO3) to produce calcium chloride (CaCl2), carbon dioxide (CO2) gas, and water (H2O). The bubbling and fizzing observed during the reaction is due to the release of carbon dioxide gas.

Therefore, base on the equation HCl + CaCO3 → CaCl2 + CO2 + H2O  calcium chloride and water produced are both soluble in water and do not contribute to the hard water deposit.

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Answer: C

Explanation:

We can research about the growth rate of plant that the water has been given, since we already did 10 mL. It would be best if we research more to see if 20 mL is enough water for the plant or will it make the plant go turgid and flaccid!

The pH of a buffer solution prepared from 0.0500 M hydrofluoric acid (HF) and 0.00500 M sodium fluoride (NaF) can be calculated using the Ka value of HF.

To determine the pH of the buffer solution, we need to consider the equilibrium reaction between hydrofluoric acid (HF) and its conjugate base, fluoride ion (F-):

HF + H2O ⇌ H3O+ + F-

The Ka expression for this equilibrium can be written as:

Ka = [H3O+][F-] / [HF]

Given that the Ka value of HF is 6.8 x 10^-4, we can set up an ICE table to determine the concentrations of the species involved:

Initial concentrations:

[HF] = 0.0500 M

[F-] = 0.00500 M

[H3O+] = 0 (since the initial pH is not given)

Change in concentrations:

Let's assume that x mol/L of HF dissociates. Therefore, the change in concentrations is:

[HF] = 0.0500 M - x

[F-] = 0.00500 M + x

[H3O+] = x

At equilibrium, the concentrations become the equilibrium concentrations.

Using the Ka expression, we can substitute the equilibrium concentrations and solve for x:

Ka = [H3O+][F-] / [HF]

6.8 x 10^-4 = x * (0.00500 + x) / (0.0500 - x)

Since the value of x is expected to be small compared to 0.0500 M, we can approximate 0.0500 - x to be approximately 0.0500. Simplifying the equation:

6.8 x 10^-4 = x * (0.00500 + x) / 0.0500

Next, we can solve this equation to find the value of x using algebraic methods, such as factoring or quadratic formula. After solving for x, we can determine the equilibrium concentrations of [HF], [F-], and [H3O+].

Finally, we can calculate the pH using the equation:

pH = -log[H3O+]

Note: The detailed calculations to solve for x and determine the equilibrium concentrations are not included in the response due to space constraints.

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Answer:

84.313 g/mol

Explanation:

Magnesium Carbonate molecular weight. Molar mass of MgCO3 = 84.313 g/mol.

The ion channel associated with the NMDA receptor is permeable to calcium (Ca²⁺), while the ion channel associated with the AMPA receptor is permeable to sodium (Na⁺).

The NMDA receptor is a type of glutamate receptor found in the central nervous system (CNS) that plays a crucial role in synaptic plasticity and learning.

When glutamate, the primary excitatory neurotransmitter, binds to the NMDA receptor, it allows the influx of calcium ions (Ca²⁺) into the postsynaptic neuron.

Calcium entry through the NMDA receptor is important for long-term potentiation (LTP), a process involved in strengthening synaptic connections and facilitating learning and memory.

On the other hand, the AMPA receptor is also a type of glutamate receptor that mediates fast synaptic transmission in the CNS.

When glutamate binds to the AMPA receptor, it opens an ion channel that is permeable to sodium ions (Na⁺), leading to depolarization of the postsynaptic membrane and generation of an excitatory postsynaptic potential (EPSP).

This EPSP can trigger the firing of an action potential in the postsynaptic neuron, allowing for the transmission of signals between neurons.

Therefore, the NMDA receptor's ion channel allows the passage of calcium (Ca²⁺), while the AMPA receptor's ion channel permits the flow of sodium (Na⁺).

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Complete question here:

The ion channel associated with the NMDA receptor is permeable to ____ while the ion channel associated with the AMPA receptor is permeable to a. iron; sodium b. glutamate; potassium C. calcium, sodium and potassium; sodium d. calcium and selenium, potassium and sodium e. calmodulin; glutamate

(a) sp2- hybrid orbital of C1 and p-orbital of O1 , because, central atom i.e. carbon has sp2-hybridization and in surrounded oxygen atom last electrons enters in p-orbital. (b) 1200 bond angle, because shape is trigonal planar.

Atomic orbitals are generally specific through a mixture of numerals and letters that constitute precise residences of the electrons related to the orbitals—for example, 1s, 2p, 3d, 4f. The numerals, referred to as primary quantum numbers, imply strength degrees in addition to relative distance from the nucleus. The simple names s orbital, p orbital, d orbital, and f orbital refer to orbitals with angular momentum quantum number ℓ = 0, 1, 2, and 3 respectively.

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Answer:

neutral

Explanation:

The Tollens test is positive in the presence of an aldehyde because aldehydes can be converted to silver ions (Ag+) when they interact with the Tollens reagent. This reaction forms a silver mirror on the inside of the test tube, indicating the presence of an aldehyde.

The Tollens reagent is made up of silver nitrate (AgNO3) and ammonia (NH3), and the reaction involves the reduction of Ag+ ions to metallic silver (Ag) by the aldehyde. This reaction is used to distinguish aldehydes from ketones, as ketones do not react with the Tollens reagent.

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Answer:

potassium is likely to be harder

Explanation:

because whenan atom loss electron they can form particle ion acquare stable a arrangement that loos electron can transfer to another which that also to acquaire stable

To make 1 liter of a 10% NaCl solution from a solid stock, you will require the following materials and containers.MaterialsSolid NaClDistilled water1-Liter volumetric flask250-mL volumetric flask 2-beakersProcedureTo prepare 1 liter of a 10% NaCl solution, the following procedure should be followed:Measure out 100g of NaCl using a balance.

Measure the weight of an empty 250-mL volumetric flask.Add the NaCl to a 250-mL beaker and add a small amount of distilled water to it to dissolve the NaCl.Carefully pour the dissolved NaCl solution into the 250-mL volumetric flask. Add distilled water to the mark on the flask to make up the volume. Stopper the flask and invert it several times to mix the solution.Measure the weight of the 1-Liter volumetric flask.Add the 250-mL volumetric flask solution to a 1-Liter volumetric flask.Add distilled water to the mark on the flask to make up the volume.

Stopper the flask and invert it several times to mix the solution.The final volume of the solution will be 1 liter of a 10% NaCl solution.PrecautionsEnsure the NaCl has completely dissolved before adding more water to avoid making a less concentrated solution.Measure the weight of the volumetric flask before and after adding the solution to calculate the volume of solution that was added.Use distilled water to prepare the solution.

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Answer:

20.0/1 x 1mole/58.44=0.34

                 (Atomic mass of salt)

Been a minute since I've done this, but this is how I remember doing it, soo..hope this helps!

The  speed of the fallen tree from the hill in km/hr would be 46.4 km/hr.

simple unit conversion is applied here so, The problem here is about converting the speed of an object from one unit to another. Specifically, we are to convert from m/s to km/hr.

Speed is the function of distance and time. Thus, the distance needs to be converted to km from m and the time to hr from s.

13 m/s is equivalent to 13 m distance and 1 second time.

remember that:

1000 m = 1 km

13 m = 13/1000

 = 0.013 km

3600 seconds = 1 hr

1 second = 1/3600

  = 0.00028 hour

The speed of the fallen tree in km/hr will be then:

0.013/ 0.00028  = 46.4 km/hr

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Glass is not a mineral because it lacks a crystalline structure and is formed through a different process. It can be many different colors, is not limited to occurring naturally, and does not occur in rocks.

Minerals are naturally occurring, inorganic substances that have a specific crystal structure. Glass, on the other hand, is an amorphous material, meaning it does not have a regular, repeating crystal structure like minerals do. Instead, glass has a disordered arrangement of atoms, which gives it its unique properties.

While some minerals can be transparent, glass is not limited to being transparent and can be many different colors. Additionally, glass is not formed through geological processes like minerals are. Instead, it is produced by melting a mixture of silica, soda, and lime at high temperatures and then cooling it rapidly. This manufacturing process allows for the creation of glass in various shapes, sizes, and colors, making it versatile for different applications.

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Answer:

Option A.

Explanation:

Balanced equation is:

2SO₂ (g) + O₂(g)  →  2SO₃(g)

First of all, we need to determine the limiting reagent. We convert mass to moles:

544.5 g . 1 mol / 64.06g = 8.5 moles of sulfur dioxide

160 g . 1mol / 32g = 5 moles of oxygen

Ratio is 1:2. 1 mol of oxygen needs 2 moles of sulfur dioxide

5 moles of oxygen may react to (5 . 2) /1 = 10 moles of SO₂

We only have 8.5 moles of SO₂ but we need 10 moles. In conclussion limiting reagent is SO₂.

Ratio is 2:2. 2 moles of SO₂ can prdouce 2 moles of SO₃

Then 8.5 moles of SO₂  must produce 3 moles of SO₃

We convert mass to moles, to determine the theoretical yield (100 % yield reaction) → 8.5 mol . 80.06 g /mol = 680.51 g

Formula for percent yield is: (Produced yield / Theoretical yield) . 100

(382 g / 680.51g) . 100 = 56.1 %

Answer:

Boil so the water can evaporate

Explanation:

Answer: Crystallization/ evaporation.

This process collects the solute- in this case the salt crystals.

The salt is dissolved so we can't filter it, it won't be caught on filter paper. Instead we heat the water to evaporate it and then leave it, making the salt crystals form.

hope it helps.

Given:

Crystal shield (CuSO4 . 5H2O) dissolves in water in the large tank.CuSO4′5H2O(s) → CuSO4 (aq) + 5H2O(l)0.01 mm = thickness of film layer24.3 g crystal/100 g water at 275 K, density 1140 kg/m3Diffusion of CuSO4 in dilute aqueous solution 3.6x10-10 m2/s at 275 K

Concentration can be taken at an average of 55.93 kmol/m3, water density 1000 kg/m3The mass of CuSO4 that dissolves per unit area per unit time is called flux.The transfer rate per unit area from the crystal surface to the bulk solution is given by;\(\frac{dn}{dt} = D.

A\frac{dc}{dx}\)Where, D is diffusion coefficient A is the area of the interfacec is the concentration of the diffusing species in the filmx is the direction of diffusion

From the given data, crystal density can be calculated as;\(\frac{24.3}{100+24.3}=0.1955\)

Now, we can calculate the mass transfer flux from the given data;\(\frac{55.93}{159.61}=\frac{C_{CuSO_4}}{1140}\)\(C_{CuSO_4}=39.76 kmol/m^3\)\(J = -D\frac{dc}{dx}\)

Since the inner film is saturated, \(c_0 = 39.76 kmol/m^3\)

At the outer surface, there is no CuSO4 present and the concentration gradient is;\(\frac{39.76-0}{0.01*10^{-3}}= 3.976*10^6 kmol/m^4\)\(J = -3.6*10^{-10}*3.976*10^6\)\(= -1.43456*10^{-3} kmol/m^2s\)

Thus, the mass transfer flux is -1.43456 * 10^-3 kmol/m^2s.

The transfer rate per unit area from the crystal surface to the bulk solution is -1.43456 * 10^-3 kmol/m^2s.

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The concentration of carbon dioxide in the atmosphere is a small fraction of 1%.Explanation:The Earth's atmosphere comprises a thin layer of gases that serve as a blanket over the Earth's surface, enabling life on the planet. Nitrogen (78%) and oxygen (21%) make up the majority of the Earth's atmosphere, while other gases such as argon, carbon dioxide, and neon make up the remainder.

Carbon dioxide (CO2) is one of the most well-known greenhouse gases, accounting for around 0.04% of the Earth's atmosphere. Because of its role in climate change, it is regarded as the most significant greenhouse gas. CO2 concentrations are at an all-time high in Earth's atmosphere as a result of human activity, particularly the burning of fossil fuels like coal, oil, and gas. The concentration of carbon dioxide in the atmosphere is a small fraction of 1%. Despite its low concentration, it plays an important role in climate change by trapping heat and warming the planet. This effect, known as the greenhouse effect, has been amplified by human activity, resulting in a rapid rise in global temperatures and significant climate impacts.

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