A voltaic cell employs the redox reaction: 2 Fe3+(aq) + 3 Mg(s) → 2 Fe(s) + 3 Mg2+(aq) Calculate the cell potential at 25°C under each set of conditions. (See the table.) (

(b) [Fe3+] = 1.0 10-3 M; [Mg2+] = 2.50 M V

(c) [Fe3+] = 2.22 M; [Mg2+] = 1.3 10-3 M

Answer:

X

Explanation:

because it is the greatest

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:

The correct answer is B] Feeling the grains of salt in between your fingers.

Explanation:

Evidence that matter is too small to be seen without magnification includes feeling the grains of salt in between your fingers, because the individual grains of salt are too small to be seen with the eye. Other examples of matter that is too small to be seen without magnification include atoms, molecules, and subatomic particles.

A] Feeling the heat come off the flames of a fire, C] Watching salt disappear when it dissolves in water, and D] Watching rain fall from the sky during a thunderstorm are all examples of phenomena that are caused by matter, but they do not necessarily provide evidence that matter is too small to be seen without magnification.

If a system at equilibrium is disturbed by changing the concentration of a reactant or product, the equilibrium will shift to restore the balance.

However, once the system reestablishes equilibrium, the ratio of product to reactant will remain constant, and the value of the equilibrium constant (K) will be unchanged.

The equilibrium constant (K) is a constant value that represents the ratio of the concentrations of products to reactants at equilibrium. It is determined by the stoichiometry of the balanced chemical equation and is independent of the initial concentrations or any changes that occur during the reaction.

When the concentration of a reactant or product is altered, the system will respond by shifting the equilibrium to counteract the disturbance. This shift aims to restore the original ratio of product to reactant, corresponding to the equilibrium constant (K). As a result, once the system reaches a new equilibrium, the value of K remains constant, reflecting the unchanged ratio of product to reactant.

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The electronegativity periodic table is a chart that arranges elements according to their electronegativity, which is a measure of an atom's ability to attract electrons to itself. Electronegativity is a chemical property that reflects the relative tendency of an atom to draw electrons towards itself when it forms a chemical bond with another atom.

The electronegativity values are usually determined using the Pauling scale, which was developed by Linus Pauling and is widely used in chemistry. In this scale, the electronegativity of an element ranges from 0.7 for cesium to 4.0 for fluorine, with increasing electronegativity moving from left to right across a period and increasing as one moves down a group.

The electronegativity values can be useful in understanding chemical bonding and the behavior of molecules. For example, elements with high electronegativity values tend to form ionic bonds, while elements with low electronegativity values tend to form covalent bonds. Additionally, the electronegativity difference between two bonded atoms determines the type of bond, with larger differences indicating polar covalent bonds and smaller differences indicating nonpolar covalent bonds.

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The above are sorted into "Characteristics of Good Claim" and "Characteristics of Bad Claim" as follows:

Characteristic of a good claim:

Not a characteristic of a good claim:

A claim in literature is a statement or proposition put forward for consideration or argument, often as a central idea in an essay or research paper.

It is important in literature because it provides a clear focus and direction for the writer's analysis or argument, and helps to ensure that the writing is logical, organized, and persuasive. Claims are essential for constructing a strong argument and presenting a convincing case to the reader.

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The molar mass of the acid, given the data is 625 g/mole

This is a point is a titration reaction where the amount of the acid added is enough to neutralize the base

This simply implies that the number of mole of the acid used is equal to the number of mole of the base.

Mole = molarity × volume

Mole of base (NaOH) = 0.1 × 0.02

Mole of base (NaOH) = 0.002 mole

Molar mass = mass / mole

Molar mass of acid = 1.25 / 0.002

Molar mass of acid = 625 g/mole

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The reaction given is a redox reaction, and based on the given equilibrium constant, the standard voltage (E°) is positive, while the standard free energy change (∆G°) is negative.

The standard voltage (E°) of a redox reaction represents the tendency of the reaction to proceed in the forward direction. A positive E° indicates that the reaction is spontaneous in the forward direction, meaning that the reduction half-reaction is favored. In the given reaction, copper (Cu) is being oxidized to Cu2+, while silver ions (Ag+) are being reduced to form solid silver (Ag). Since the reaction is spontaneous in the forward direction, E° must be positive.

The standard free energy change (∆G°) of a reaction determines the spontaneity of the reaction. A negative ∆G° indicates that the reaction is thermodynamically favorable and will proceed spontaneously in the forward direction. Based on the relationship between ∆G° and the equilibrium constant (K), which is given as 3.7 x 10^15, we can determine that ∆G° is negative. The equation relating ∆G° and K is ∆G° = -RT ln(K), where R is the gas constant and T is the temperature in Kelvin. Since ln(K) is positive, ∆G° must be negative for a large equilibrium constant like \(3.7 \times 10^{15\).

Therefore, the correct description for this reaction is: (A) E° is positive and ∆G° is negative.

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Answer:

NaCI (salt) in water

Explanation:

After the salt compounds are pulled away apart the sodium chloride atoms are surrounded by water molecules. Once this happens we know that the salt is dissolved resulting in a homogeneous solution!

HOPE THIS HELPED!

Answer:

NaCI (salt) in water

Explanation:

took the quiz

To deploy configuration profiles for computers from Jamf Pro, the Apple Push Notification service (APNs) must be available.

APNs is a cloud-based service provided by Apple that enables the secure transfer of data between Apple devices and servers. It is essential for communication between Jamf Pro and Apple devices during the deployment of configuration profiles.

APNs is used to initiate the connection between the devices and Jamf Pro, allowing for the transfer of the configuration profiles. Without APNs, it would be impossible to deploy configuration profiles to Apple devices using Jamf Pro.

Therefore, it is critical to ensure that APNs is available and functioning correctly before attempting to deploy configuration profiles using Jamf Pro.

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Among the given compounds, nitrogen gas (N2) has a vibration that is infrared inactive. The other compounds, H2S, 2-butyne, ethene, and 1-butyne, have vibrations that are infrared active.

Infrared spectroscopy is a technique used to identify functional groups and molecular vibrations in a compound. When a molecule absorbs infrared radiation, it undergoes a change in vibrational energy levels.

To be infrared active, a molecule must have a change in dipole moment during a vibration. This occurs when there is a net movement of charge within the molecule. Among the given compounds, H2S, 2-butyne, ethene, and 1-butyne all have polar bonds and undergo changes in dipole moment during vibrations. Therefore, they are infrared active.

On the other hand, nitrogen gas (N2) consists of a homonuclear diatomic molecule with a nonpolar bond. Since there is no change in dipole moment during the vibrations of N2, it is infrared inactive.

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The ionization energy generally decreases as you move from top to bottom down the groups of the periodic table. This trend occurs because of the following factors:

1. Atomic radius: As you move down a group, the atomic radius increases due to the addition of electron shells. This results in the outermost electrons being farther from the nucleus, which weakens the electrostatic attraction between the nucleus and the electrons.

2. Shielding effect: With the addition of electron shells, the inner electrons "shield" or "screen" the outermost electrons from the full positive charge of the nucleus. This further reduces the electrostatic attraction between the nucleus and the outermost electrons.

Due to these factors, it requires less energy to remove an electron from an atom as you move down a group, causing the ionization energy to decrease.

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Answer:

The answer is 6608J ..

Explanation:

Answer:

a. To compare the concentration of 80 g/litre NaOH solution and 3 M NaOH solution, we need to convert one of the concentrations to the other unit.

One mole of NaOH weighs 40 grams. So, to convert 80 g/litre NaOH to Molarity, we can divide 80 g/litre by 40 g/mol to get:

80 g/litre NaOH = 2 M NaOH

Therefore, 3 M NaOH has a higher concentration than 80 g/litre NaOH solution.

b. To compare the concentration of 5.3 g/litre Na2CO3 and N/10 Na2CO3 solution, we need to first understand what N/10 solution means.

N/10 Na2CO3 means that the solution contains 1/10th of the normal concentration of Na2CO3. The normal concentration of Na2CO3 is the molar concentration of Na2CO3 that corresponds to the formula weight of Na2CO3, which is 106 g/mol.

So, the normal concentration of Na2CO3 is 1 mol/L or 1 M Na2CO3.

Therefore, N/10 Na2CO3 solution has a concentration of 1/10 M Na2CO3.

Now, let's compare the two concentrations:

5.3 g/litre Na2CO3 = (5.3/106) M Na2CO3 = 0.05 M Na2CO3

Since 0.05 M Na2CO3 is greater than 1/10 M Na2CO3, the concentration of 5.3 g/litre Na2CO3 solution is higher than that of N/10 Na2CO3 solution.

Explanation:

The values of balanced chemical reaction is w = 1, x = 6, y = 3, and z = 2

To balance the chemical equation:

1. Balancing nitrogen (N):

There are three nitrogen atoms on the left side (Ba₃N₂), so we need to place a coefficient of 3 in front of NH₃:

w Ba₃N₂ + x H₂O → y Ba(OH)₂ + 3 z NH₃

2. Balancing hydrogen (H):

There are six hydrogen atoms on the left side (2 × 3), so we need to place a coefficient of 6 in front of H₂O:

w Ba₃N₂ + 6 H₂O → y Ba(OH)₂ + 3 z NH₃

3. Balancing barium (Ba):

There are three barium atoms on the left side (3 × Ba₃N₂), so we need to place a coefficient of 3 in front of Ba(OH)₂:

w Ba₃N₂ + 6 H₂O → 3 y Ba(OH)₂ + 3 z NH₃

4. Balancing oxygen (O):

There are six oxygen atoms on the right side (6 × OH), so we need to place a coefficient of 3 in front of Ba(OH)₂:

w Ba₃N₂ + 6 H₂O → 3 Ba(OH)₂ + 3 z NH₃

Now the equation is balanced with the following coefficients:

w Ba₃N₂ + 6 H₂O → 3 Ba(OH)₂ + 3 z NH₃

Therefore, w = 1, x = 6, y = 3, and z = 2 would satisfy the balanced chemical equation.

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Ionic compounds have ionic bonds and are well defined as substances that are able to conduct electricity and exist in a solid state. Thus, options b and d are correct.

Ionic compounds are defined as a substance that shows ionic bondings and is characterized by the presence of the electrostatic force that helds the ions of the compound.

The charged ions carry the electrical charge in the solution. The ionic compounds are generally present as solids that are due to the structure of the crystal lattice that contains an alternative negative and positive charge.

Therefore, options b and d. ionic compounds can conduct electricity is correct.

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Your question is incomplete, but most probably your full question was, From the following statements write those applicable to ionic compounds?

a) Usually do not dissolve in water.

b) Conduct electricity in the molten state and aqueous solution.

c) Generally not a conductor of electricity.

d) Exists in the solid state​

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:

5

Explanation:

Density = mass divided by volume

15 divided by 3 is 5

Carbonation is a crucial process in rock breakdown and formation of karst topography, involving the formation of calcium bicarbonate through precipitation of carbon dioxide and dissolved substances. This process affects rocks' physical properties and regulates atmospheric carbon dioxide concentration.

Calcium bicarbonate produced in the chemical weathering process of carbonation causes the rock to become weak and break down. This occurs when rainwater reacts with carbon dioxide and turns into a weak carbonic acid solution that can dissolve rocks. As a result, carbonation is an essential process in the breakdown of rocks and formation of karst topography.The chemical formula of calcium bicarbonate is Ca(HCO3)2. It is formed when rainwater, which contains carbon dioxide, reacts with rocks that contain calcium carbonate (CaCO3) like limestone and marble. The reaction is as follows:

CaCO3 + H2CO3 → Ca(HCO3)2

The carbonic acid solution reacts with the rock and breaks it down into calcium bicarbonate and other dissolved substances. Calcium bicarbonate is carried away by groundwater and eventually deposits to form stalactites, stalagmites, and other types of cave formations.

This chemical weathering process of carbonation not only affects the physical properties of rocks but also plays a significant role in the carbon cycle of the Earth. Carbonation helps to regulate the concentration of carbon dioxide in the atmosphere by removing it and storing it underground in the form of calcium carbonate deposits.

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Answer:

5 moles

Explanation:

By convert grams to moles u need to divided the number of grams to the molar mass

220 ÷ 44.01 = 4.9 ≈ 5.0

Metal solids tend to be good conductors because of the property of electron mobility.

Metal cations are the building blocks of metallic solids, and a delocalized "sea" of valence electrons holds them all together. Metallic solids are good conductors of heat and electricity because of the mobility of their electrons. Metal atoms bound together by metallic bonds form metallic solids, which are solids. These linkages act as expansive molecular orbitals that cover the entire solid.

The electrons in metallic solids are hence delocalized. They are not only linked together by two atoms in a sigma bond. Metals are effective heat and electricity conductors because they contain at least one free electron per atom, which means that it is not bound to any one atom and is therefore free to travel about the metal.

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Anhydrous magnesium sulfate removes water from organic liquids.

Anhydrous magnesium sulfate is used in steam distillation to remove any water present in the organic liquid being distilled. Water can interfere with the distillation process and cause inaccurate results. Anhydrous magnesium sulfate is a desiccant that can absorb water from the organic liquid, thereby ensuring that the steam distilled product is dry and free from any water. It is added to the organic liquid before distillation and is commonly used in the separation and purification of essential oils from plants. By removing any water, anhydrous magnesium sulfate helps to ensure that the resulting steam distilled product is of high quality and purity.

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The conductivity of metalloids can be compared to the conductivity of metals and nonmetals because metalloids conduct electricity better than nonmetals, but not as well as metals (Option C).

The expression metalloids is a term used to denote chemical elements that have features resembling metals such as an acceptable electrical conductivity, but they are not metals (e.g. boron, germanium, antimony, arsenic, polonium, etc).

Therefore, with this data, we can see that metalloids are similar to metals in electrical conductivity but they are not metals because they do not fit all properties of metals.

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I actually do not know the answer can I help tommorow if you are fine with it tooo sleeeepy have 14 exams tommorow have to prepare.

Sorry:(

The heat energy needed to change the temperature of  15 g of the sample by 8°C is given 303 J. Then, the specific heat of the substance is 2.52 J/°C g

Calorimetry is an analytical  technique used to determine the heat energy absorbed or released by  a system. The calorimetric equation relating the heat energy q with the mass of the substance m, specific heat c and temperature difference ΔT is given as:

q = m c ΔT.

Given that the mass of the sample = 15 g

temperature difference   ΔT = 8 °C

specific heat c = ?

The heat energy absorbed by the sample is calculated as follows:

303 J= 15 g ×  C × 8 °C

Then, C = 303 J /( 15 g× 8 °C) = 2.52 J/°C g

Therefore, the specific heat capacity of the substance is 2.52 J/°C g.

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Answer:

2.5 J/gC degrees

Explanation: