How many grams of Aluminum Sulfate are produced when 4 g of Aluminum Nitrate react with 3 g of Sodium Sulfate?

Al(NO3)3 + Na2SO4 ---------> Al2(SO4)3 + NaNO3

The order for reactant A is 2 and the order for reactant B is 1. For the first reaction, the overall order of the reaction is 3 and for the second reaction, the overall order of the reaction is 5.

The order of a reaction is the sum of the exponents in the rate law expression that relates the rate of a chemical reaction to the concentrations of the reactants.

To determine the order of each reactant, we need to compare the initial rates of reaction at different concentrations while keeping the concentration of the other reactant constant.

For reactant A:

Exp#1 (0.100 M A, 0.100 M B): initial rate = k(0.100)^2(0.100) = 0.001 k

Exp#2 (0.200 M A, 0.100 M B): initial rate = k(0.200)^2(0.100) = 0.004 k

Exp#3 (0.100 M A, 0.200 M B): initial rate = k(0.100)^2(0.200) = 0.002 k

We can see that when the concentration of A doubles (Exp#1 to Exp#2), the initial rate quadruples, which indicates that A is second order. When the concentration of B doubles (Exp#1 to Exp#3), the initial rate doubles, which indicates that B is first order.

Therefore, the order for reactant A is 2 and the order for reactant B is 1.

To determine the overall order of the reaction, we add the orders of the reactants:

Overall order = 2 (order of A) + 1 (order of B) = 3

Therefore, the overall order of the reaction is 3.

For the second reaction, we can see that the rate depends on the concentration of both reactants, and we cannot determine their individual orders without further information or experiments. However, we can determine the overall order of the reaction by adding the exponents of the concentration terms in the rate law:

Overall order = 1 + 4 = 5

Therefore, the overall order of the reaction is 5.

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Around 80.5 KJ

Multiply Heat of Fusion and Mass to get the q value.

The density of sulfur dioxide gas at a temperature of 15°C and pressure of 300 torr is 0.001022 g/cm³, or 0.001022 g/mL, or 1.022 kg/m³, or 0.01022 g/L when converted to atm.

What is density?

To calculate the density of sulfur dioxide gas at a temperature of 15°C and a pressure of 300 torr, we can use the ideal gas law:

PV = nRT

where P is the pressure in atmospheres, V is the volume in liters, n is the number of moles, R is the ideal gas constant (0.08206 L·atm/(mol·K)), and T is the temperature in Kelvin.

First, we need to convert the given temperature of 15°C to Kelvin:

T = 15°C + 273.15 = 288.15 K

Next, we can rearrange the ideal gas law to solve for the number of moles:

n = PV/RT

where we can use the given pressure of 300 torr and convert it to atm by dividing by 760 torr/atm:

P = 300 torr / 760 torr/atm = 0.3947 atm

Substituting the values into the equation, we get:

n = (0.3947 atm) V / (0.08206 L·atm/(mol·K) × 288.15 K)

Now, we can use the molar mass of sulfur dioxide, which is 64.06 g/mol, to convert the number of moles to mass:

mass = n × molar mass

Finally, we can calculate the density of sulfur dioxide gas using the mass and volume:

density = mass / V

To convert the density from g/L to g/cm³, we divide by 1000.

Putting it all together, we get:

n = (0.3947 atm) V / (0.08206 L·atm/(mol·K) × 288.15 K)

n = 0.01595 V

mass = n × molar mass = 0.01595 V * 64.06 g/mol = 1.022 gV

density = mass / V = 1.022 gV / V = 1.022 g/L = 0.001022 g/cm³

Therefore, the density of sulfur dioxide gas at a temperature of 15°C and pressure of 300 torr is 0.001022 g/cm³, or 0.001022 g/mL, or 1.022 kg/m³, or 0.01022 g/L when converted to atm.

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Complete question is: The density of Sulfur dioxide gas at a temperature of 15oC and pressure of 300 torr is 0.01022 atm.

The compound NaOH as shown is a strong base.

NaOH (sodium hydroxide) is considered a strong base. A strong base is a base that dissociates completely in water to form hydroxide ions (OH-) and cations. NaOH is highly soluble in water and, when added to water, it completely dissociates into Na+ and OH- ions, which makes it a strong base.

The strength of a base depends on the extent of its dissociation in water. Strong bases dissociate completely in water, while weak bases dissociate only partially. The dissociation of a base is usually represented by its base dissociation constant (Kb), which is the equilibrium constant for the reaction of the base with water to form hydroxide ions.

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The eutectic point is the lowest temperature at which the liquid phase is constant at a particular pressure.

A melting composition known as a eutectic consists of at least two components that melt and freeze at the same rates. The components combine during the crystallisation phase, operating as a single component as a result.

The eutectic is the system's lowest melting point under its own pressure; it has a matching temperature called the eutectic temperature and produces the eutectic liquid as a result. In terms of composition, eutectic liquids are located between the system's solid phases.

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K and Br since an halogen and a metal make a salt

Answer: C)Deposition (gas to solid)

Explanation: According to the phase diagram, at room temperature (25°C) and 1 atm, the sample is in the gas phase.  As the temperature decreases to 80 K, it falls below the sublimation curve. T he sublimation curve represents the conditions at which a substance can change directly from a solid to a gas or from a gas to a solid without passing through the liquid phase.

Since the sample is in the gas phase at room temperature, cooling it to 80 K would cause it to go through the process of deposition, where the gas particles directly transform into a solid without first becoming a liquid.  This is indicated by the section of the phase diagram below the sublimation curve.

The answer is A --------

The state of matter that tends to have unique factors to consider when discussing solubility compared to the other two states (solid and gas) is the liquid state.

Solubility refers to the ability of a substance (solute) to dissolve in a particular solvent to form a homogeneous mixture (solution). Various factors can affect the solubility of a substance, including temperature, pressure, and the nature of the solute and solvent.

In the case of liquids, the unique factor to consider when discussing solubility is often temperature. The solubility of many solid solutes in liquids generally increases with increasing temperature. This is because higher temperatures provide more energy to break the intermolecular forces between solute particles, allowing them to disperse more evenly throughout the solvent.

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The decay constant for this process is

The decay constant (λ) is related to the half-life (t1/2) by the following equation:

λ = ln(2) / t1/2

where

ln(2) is the natural logarithm of 2, which is approximately 0.693.

Substituting the given half-life of 3.1 min into the equation, we get:

λ = ln(2) / (3.1 min) ≈ 0.223 min^(-1)

Therefore, the decay constant for the β decay of Tl-208 is approximately 0.223 min^(-1).

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[tex] \ddots[/tex] The heat energy can be deduced as -

[tex] \odot\sf \footnotesize{Heat \:energy = Mass\: of\: substance\times Specific \:heat\times Change\: in \:temperature}\\[/tex]

[tex] \qquad :\implies\sf \boxed{\sf Q = mS\Delta T}\\[/tex]

Where-

In this instant, we are given -

[tex] \ddots[/tex] Now that we have all the required values except ∆T,so we can plug the rest of the known values into the formula and solve for ∆T -

[tex] \qquad :\implies\sf \underline{Q = mS\Delta T}\\[/tex]

[tex] \qquad :\implies\sf 14500 = 485 \times 4.18 \times \Delta T\\[/tex]

[tex] \qquad :\implies\sf 14500 = 2027.3\times \Delta T\\[/tex]

[tex] \qquad :\implies\sf \Delta T = \dfrac{14500}{2027.3}\\[/tex]

[tex] \qquad :\implies\sf \Delta T = 7.152370........\:°C\\[/tex]

[tex] \qquad :\implies\sf \underline{\boxed{\sf \Delta T=7.15\:°C}}\\[/tex]

[tex] \ddots[/tex]Correct answer - [tex]\boxed{\sf \Delta T=7.15\:°C}.[/tex]

This is an exercise in specific heat and thermal conductivity which are two important physical properties that describe how materials interact with heat. Specific heat refers to the amount of energy required to raise the temperature of a material by a given amount, while thermal conductivity refers to a material's ability to transfer heat through itself.

The formula for specific heat is Q = mcΔT, where Q is the amount of heat transferred, m is the mass of the material, c is the specific heat, and ΔT is the change in temperature. The unit of measure for specific heat is J/(g*°C).

On the other hand, thermal conductivity is measured in terms of the amount of heat that is transferred through a material per unit time and area, given a temperature difference. It is expressed as the amount of heat transferred per second, per square meter, per meter of material thickness, when the temperature difference between the extremes is one Kelvin. Its formula is Q/t = -kA(∆T/∆x), where Q/t is the heat transfer rate, k is the thermal conductivity, A is the cross-sectional area, ∆T is the temperature difference, and ∆ x is the thickness of the material.

These properties are useful for understanding how materials interact with heat in a variety of situations, from building design to heating and cooling equipment manufacturing.

Now to calculate the temperature rise of 485 g of liquid water when 14.5 kJ of heat is added to it, we can use the formula:

Q = mcΔT

We must know that it has a quantity of heat of 14.5 Kj, with a mass of 485 g. The specific heat capacity of water is 4.18 J/(g °C).

First, we need to convert the heat added to joules:

Q = 14.5 KJ × (1000 J/1 KJ)

Q = 14500 J

We can then solve for ΔT. We clear the formula.

ΔT = Q / (m × c)

We substitute our data in the formula and solve the temperature change:

ΔT = Q / (m × c)

ΔT = (14500 J)/(485 g × 4.18 J/(g·°C))

ΔT ≈ 7.15 °C

The water would increase its temperature by approximately 7.15°C if 14.5 kJ of heat were added.

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I think it is 2,4-dimethylpentane

[tex]CO_{2}[/tex]The following compounds' mole fractions (X) are (a)0.986 (b)0.750 (c)0.811 for the given solutions.

[tex]H_{2}SO_{4}[/tex] mass is 19.4 g.

[tex]H_{2}SO_{4}[/tex]'s molecular weight is 98.08 g/mol.

It's molecular weight is 19.4 g/98.08 g/mol, or 0.1979 mol.

Density times volume is 1.00 g/mL times 0.251 L and 251 g for water mass.

[tex]H_{2} O[/tex] has a molecular weight of 18.02 g/mol.

Water moles are equal to 251 g / 18.02 g/mol, or 13.93 mol.

The solution's total moles are equal to 0.1979 mol plus 13.93 mol, or 14.13 mol.

Sulphuric Acid's mole fraction is equal to 0.1979 mol/14.13 mol, or 0.014.

Water mole fraction is equal to 13.93 mol / 14.13 mol, or 0.986 mol.

KBr's mass is 35.7 g.

KBr has a molecular weight of 119 g/mol.

The formula for KBr is 35.7 g/119 g/mol, which equals 0.300 mol.

16.2 g of water in mass

Water has a molecular weight of 18.02 g/mol.

Water moles are equal to 16.2 g / 18.02 g/mol, or 0.899 mol.

The solution has a total of 1.199 moles (0.300 mol + 0.899 mol).

The mole fraction of KBr is equal to 0.300 mol/1.199 mol, or 0.250

Water mole fraction is equal to 0.899 mol / 1.199 mol, or 0.750 moles.

[tex]CO_{2}[/tex] mass = 233 g

It has a molecular weight of 44.01 g/mol.

Its moles are equal to 233 g / 44.01 g/mol, or 5.291 mol.

Water volume equals 0.409 L.

Water has a molecular weight of 18.02 g/mol.

(density × volume) / molecular weight (1.00 g/mL 409 mL) / 18.02 g/mol = 22.71 mol = number of moles of water

The solution's total moles are equal to 5.291 mol plus 22.71 mol, or 28.00 mol.

[tex]CO_{2}[/tex] mole fraction = 5.291 moles / 28.00 moles = 0.189

[tex]H_{2} O[/tex] mole fraction is 22.71 mol/28.00 mol, or 0.811 moles.

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To draw the two circles, we would need to draw a smaller circle with a diameter of 2,532.5 miles (representing the moon) and a larger circle with a diameter of 75,974.4 miles (representing the Earth) that is 30 times larger than the smaller circle.

If we assume that the larger circle represents the Earth, then the diameter of the Earth would be 30 times the diameter of the smaller circle representing the moon. Let's say that the diameter of the smaller circle is x. Then the diameter of the larger circle (Earth) would be 30 times x or 30x.

To find the correct distance from Earth to the moon at this scale, we need to know the actual distance from Earth to the moon, which is approximately 238,855 miles or 384,400 kilometers. If we divide this distance by the scale factor of 30, we get:

238,855 miles / 30 = 7,961.8 miles

Therefore, the diameter of the smaller circle (moon) would be approximately 7,961.8 miles / π = 2,532.5 miles (rounded to one decimal place). And the diameter of the larger circle (Earth) would be 30 times that or 75,974.4 miles

So, to draw the two circles, we would need to draw a smaller circle with a diameter of 2,532.5 miles (representing the moon) and a larger circle with a diameter of 75,974.4 miles (representing the Earth) that is 30 times larger than the smaller circle.

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The final temperature is -160°C

We can use the combined gas law, which relates the pressure, volume, and temperature of a gas:

(P₁V₁)/T₁ = (P₂V₂)/T₂

Where

In this case, we can assume that the pressure of the gas is constant, since it is not given in the problem statement. So we can simplify the equation to:

(V₁/T₁) = (V₂/T₂)

Where

We are given that the initial volume (V₁) is 80 L and the final volume (V₂) is twice that, or 160 L. We are also given that the initial temperature (T₁) is -40°C. To find the final temperature (T₂), we can plug these values into the equation:

(V₁/T₁) = (V₂/T₂)

(80 L)/(-40°C) = (160 L)/T₂

Simplifying:

-2 L/°C = (160 L)/T₂

Multiplying both sides by -1°C/2 L (the reciprocal of -2 L/°C):

1/2 = (T₂)/(160 L) x (-1°C/2 L)

1/2 = -T₂/320

Multiplying both sides by -1 to isolate T₂:

-1/2 = T₂/320

T₂ = -160°C

Therefore, the final temperature is -160°C.

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The above is about the movement of lithospheric plates. See explanation and attached image for details.

The movement of lithospheric plates is driven by convection currents in the Earth's mantle, which are caused by heat generated from the Earth's core. These currents cause the lithospheric plates to move, and the motion can result in a variety of geological phenomena, including earthquakes, volcanic activity, and the formation of mountain ranges and oceanic trenches.

There are three main types of plate boundaries, and each one results in a different type of movement of lithospheric plates:

Divergent Boundaries: At divergent boundaries, lithospheric plates move away from each other. This movement is caused by the upwelling of hot material from the mantle, which pushes the plates apart. As the plates move away from each other, magma rises up to fill the gap between them, creating new crust. Divergent boundaries are where new oceanic crust is formed.

Convergent Boundaries: At convergent boundaries, lithospheric plates move towards each other. There are three types of convergent boundaries: oceanic-oceanic, oceanic-continental, and continental-continental. At oceanic-oceanic and oceanic-continental convergent boundaries, one plate is forced beneath the other, creating a subduction zone. This movement is caused by the sinking of a denser plate beneath a less dense plate. As the denser plate sinks, it melts and can trigger volcanic activity. At continental-continental convergent boundaries, the plates are too buoyant to subduct, so they instead buckle and push up, forming mountain ranges.

Transform Boundaries: At transform boundaries, lithospheric plates move past each other. This movement is caused by the lateral movement of the convection currents in the mantle. Transform boundaries can create large faults, which can lead to earthquakes.

Overall, the movement of lithospheric plates is a complex and dynamic process, driven by the movement of material within the Earth's mantle.

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Component form of the vector v is as follows: 4 3 1.5 1 Using the standard basis vectors I and j), express the vector w as follows: 3 two 1 4 pp . 1 3 w 3.5 C. V plus w= d. Determine the vector v's magnitude

What does "vector" mean?

Latin word for "carrier" is "vector." Point A is transported to point B by vectors. The orientation of the vectors AB is the direction in which point A is moved in relation to point B, and the amplitude of the vector is the width of the line connecting the two locations A and B. The terms Euclidean vectors and spatial vectors are also used to refer to vectors.

A vector space is what?

A vector space, also known as a linear space, is a collection of things called vectors that can be added to and multiplied ("scaled") by figures called scalars in the fields of mathematics, physics, and engineering.

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Determine the amount of KOH present in the resulting solution. KOH was initially 0.00875 mol, then 0.00525 mol of it interacted with HCl. As a result, 0.00875 mole - 0.00525 mol (= 0.00350 mol of KOH is left. The resulting solution has a volume of 70.0 mL (35.0 mL plus 35.0 mL).

The titrant (NaOH), which is added gradually throughout the course of a titration, is added to the unknown substance. The equivalency point is the moment at which precisely the right quantity of titrant (NaOH) has indeed been added that react to the entire analyte (HCl).

What took place during titration: One mole of NaOH interacts with one mole of HCl inside the reaction between the two substances. NaOH with HCl equals NaCl plus H2O. (NaOH and HCl have a mole ratio of 1:1.) • The NaOH concentration is 0.1 M, or 0.1 molecules per litre.

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The elements that have an outer electron configuration of ms2nd2 are located in the d-block of the periodic table and include some of the transition metals and lanthanides.

To determine which elements in the periodic table have this outer electron configuration, you can look at the position of the d-block elements in the table. The d-block elements are located in the middle of the table and include the transition metals. These elements have partially filled d orbitals, which can accommodate up to 10 electrons.

Elements in the d-block with an atomic number of 21 through 30 (scandium through zinc) have an outer electron configuration of d10s2 and do not fit the ms2nd2 configuration. However, elements in the d-block with an atomic number of 39 through 48 (yttrium through cadmium) have an outer electron configuration of d10s2p1 and can have the ms2nd2 configuration by removing the single electron in the p orbital. Elements in the d-block with an atomic number of 57 through 80 (lanthanum through mercury) also have the possibility of having an outer electron configuration of ms2nd2.

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The tennis ball has a kinetic energy of around 56.8 J. The aircraft has a kinetic energy of around 4.43 x 10⁹ J. The hot air balloon travels at a speed of around 9.0 m/s.

How much effort must be put into slowing down a 750 kg automobile from 100 km/h to 50 km/h. We know that the of this automobile at 50.0 km/h is 72,300 Joules from the last example problem.

Ek = 0.5 x 0.058 kg x (46 m/s)²

Ek = 0.5 x 0.058 kg x 2116 m²/s²

Ek = 56.8468 J

Ek = 0.5mv²

Ek = 0.5 x 2.15 x 10⁵ kg x (255 m/s)²

Ek = 0.5 x 2.15 x 10⁵ kg x 65025 m²/s²

Ek = 4.433 x 10⁹ J

Ek = 0.5mv²

v = √(2Ek/m)

v = √(2 x 76550 J / 1890 kg)

v = √(81.011 J/kg)

v = 9.0 m/s (approx.)

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Answer:

-191 kJ

Explanation:

The given reaction is:

N₂(g) + 3H₂(g) → 2NH₃(g) ΔH = -76.4 kJ/mol

From the balanced equation, we can see that the stoichiometric ratio between hydrogen (H₂) and ammonia (NH₃) is 3:2. This means that 3 moles of hydrogen react to produce 2 moles of ammonia.

To determine the heat energy when 5.0 g of hydrogen (H₂) burns, we need to follow these steps:

Step 1: Calculate the moles of hydrogen (H₂)

Using the molar mass of hydrogen (H₂), which is 2 g/mol, we can calculate the moles of hydrogen (H₂) in 5.0 g of hydrogen:

Moles of H₂ = Mass of H₂ / Molar mass of H₂

Moles of H₂ = 5.0 g / 2 g/mol

Moles of H₂ = 2.5 mol

Step 2: Use the stoichiometry of the reaction

Based on the stoichiometry of the reaction, we know that 3 moles of hydrogen (H₂) react to produce 2 moles of ammonia (NH₃), and the enthalpy change (ΔH) is -76.4 kJ/mol.

Step 3: Calculate the heat energy

The heat energy for 2.5 moles of hydrogen (H₂) can be calculated using the given enthalpy change (ΔH) and the stoichiometry of the reaction:

Heat energy = Moles of H₂ x ΔH

Heat energy = 2.5 mol x -76.4 kJ/mol

Heat energy = -191 kJ (rounded to three significant figures)

So, the heat energy when 5.0 g of hydrogen (H₂) burns is -191 kJ (rounded to three significant figures), and the negative sign indicates that the reaction is exothermic, releasing heat.

The percent yield of copper(I) sulfide in this experiment is 31.83%.

What is percent yield?

To calculate the percent yield, we need to compare the actual yield (the amount of product that was obtained in the experiment) with the theoretical yield (the amount of product that should have been obtained if the reaction had gone to completion).

The balanced chemical equation for the reaction between copper and sulfur to form copper(I) sulfide is:

Cu + S →  [tex]Cu_{2}S[/tex]

The molar mass of Cu is 63.55 g/mol, and the molar mass of S is 32.06 g/mol. The molar mass of  [tex]Cu_{2}S[/tex]  is 159.17 g/mol.

First, we need to calculate the theoretical yield of copper(I) sulfide using the amount of copper used in the experiment:

5 g Cu × (1 mol Cu / 63.55 g Cu) × (1 mol [tex]Cu_{2}S[/tex] / 1 mol Cu) × (159.17 g  [tex]Cu_{2}S[/tex] / 1 mol [tex]Cu_{2}S[/tex] ) = 12.57 g  [tex]Cu_{2}S[/tex]

So the theoretical yield of copper(I) sulfide is 12.57 g.

The actual yield obtained in the experiment is 4 g.

The percent yield is then:

percent yield = (actual yield / theoretical yield) × 100%

percent yield = (4 g / 12.57 g) × 100%

percent yield = 31.83%

Therefore, the percent yield of copper(I) sulfide in this experiment is 31.83%.

What is theoretical yield ?

The theoretical yield is the amount of product that would be obtained in a chemical reaction if it went to completion, meaning that all the limiting reactant was used up and no product was lost. It is calculated using stoichiometry, which involves balancing the chemical equation for the reaction and using the coefficients to determine the mole ratio between the reactants and products.

Theoretical yield is often used as a reference value to compare with the actual yield obtained in an experiment, which is the amount of product actually obtained from the reaction. The percent yield can then be calculated by dividing the actual yield by the theoretical yield and multiplying by 100%.

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The standard change in Gibbs free energy of the reaction at 25 ∘C is -1.69 kJ/mol.

What is standard change?

To find the standard change in Gibbs free energy of the reaction, we need to use the following equation:

ΔG° = -RT ln(K)

where ΔG° is the standard change in Gibbs free energy, R is the gas constant (8.314 J/mol·K), T is the temperature in Kelvin (25 °C = 298 K), and K is the equilibrium constant.

To find K, we need to use the equilibrium partial pressures:

K = (PC × PD) / (PA × PB²)

where PA, PB, PC, and PD are the equilibrium partial pressures of A, B, C, and D, respectively.

Substituting the values, we get:

K = (5.47 atm × 5.63 atm) / (5.63 atm × (5.00 atm)²)

K = 0.6176

Now we can calculate the standard change in Gibbs free energy:

ΔG° = -RT ln(K)

ΔG° = -(8.314 J/mol·K) × (298 K) × ln(0.6176)

ΔG° = -1,690 J/mol or -1.69 kJ/mol

Therefore, the standard change in Gibbs free energy of the reaction at 25 ∘C is -1.69 kJ/mol.

What is free energy?

Free energy, also known as Gibbs free energy, is a thermodynamic quantity that represents the amount of energy in a system that is available to do work at a constant temperature and pressure. It is denoted by the symbol G and is expressed in units of joules (J) or calories (cal).

In simple terms, free energy is the energy that can be used to do work. It is defined by the equation:

ΔG = ΔH - TΔS

where ΔH is the change in enthalpy (heat content) of the system, ΔS is the change in entropy (disorder) of the system, and T is the absolute temperature in Kelvin.

If ΔG is negative, the reaction is spontaneous and can proceed without the input of external energy. If ΔG is positive, the reaction is non-spontaneous and requires energy input to proceed. If ΔG is zero, the system is at equilibrium.

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Answer:

Neutrons and Protons

Explanation:

Different elements can have subatomic particles of varying sizes. The size of an atom is defined by the size of its electron cloud, which is composed of electrons, and the size of its nucleus, which is composed of protons and neutrons. The atomic number and subsequently the identity of an element are determined by the number of protons in the nucleus. The quantity of protons and neutrons in the nucleus determines its size. The quantity of electrons in the electron cloud and the energy levels they are located at define its size. The size of atoms can differ depending on the element due to differences in the amount of protons, neutrons, and electrons.

Here's a more detailed step-by-step calculation for the theoretical yield and percent yield of chromium (Cr) in the given reaction:

Given: Mass of chromium(III) oxide (Cr2O3) = 80.0 g Mass of carbon (C) = 8.00 g Actual yield of Cr = 21.7 g

Step 1: Calculate the molar mass of Cr2O3 and C. Molar mass of Cr2O3 = 2 x (51.996 g/mol) + 3 x (15.999 g/mol) = 151.996 g/mol Molar mass of C = 12.011 g/mol

Step 2: Convert the masses of Cr2O3 and C to moles. Moles of Cr2O3 = Mass of Cr2O3 / Molar mass of Cr2O3 = 80.0 g / 151.996 g/mol = 0.527 mol (rounded to three decimal places)

Moles of C = Mass of C / Molar mass of C = 8.00 g / 12.011 g/mol = 0.666 mol (rounded to three decimal places)

Step 3: Determine the limiting reactant. The limiting reactant is the one that is completely consumed and determines the maximum amount of product that can be formed. In this case, we compare the moles of Cr2O3 and C to see which one is limiting.

From the balanced equation: Cr2O3 + 3C -> 2Cr + 3CO

We can see that 1 mol of Cr2O3 requires 3 moles of C to react completely and produce 2 moles of Cr. Therefore, the limiting reactant is C, as we have 0.666 mol of C, which is less than the 0.527 mol of Cr2O3.

Step 4: Calculate the theoretical yield of Cr. The theoretical yield of Cr is the maximum amount of Cr that can be obtained based on the limiting reactant.

Moles of limiting reactant (C) = 0.666 mol Molar mass of Cr = 51.996 g/mol

Theoretical yield of Cr = Moles of limiting reactant (C) x Molar mass of Cr = 0.666 mol x 51.996 g/mol = 34.65 g (rounded to two decimal places)

Step 5: Calculate the percent yield of Cr. The percent yield is a measure of how much of the theoretical yield was actually obtained.

Actual yield of Cr = 21.7 g Theoretical yield of Cr = 34.65 g

Percent yield = (Actual yield / Theoretical yield) x 100% = (21.7 g / 34.65 g) x 100% = 62.7% (rounded to three significant figures)

Therefore, the percent yield for the reaction is approximately 62.7%.

The molality of a solution is determined by the amount of solute (in moles) and the mass of the solvent (in kilograms). To convert the mass of NaCl to moles, the molar mass of NaCl is 58.44 g/mol. The number of moles of NaCl is 25 g / 58.44 g/mol = 0.427 mol. The molality of the solution is 0.213 mol/kg.

The amount of a solute dissolved in a solvent is indicated by the chemical term "molality," which is commonly defined in terms of moles of solute per kilogramme of solvent. Because it takes into account variations in the volume of the solution owing to temperature and pressure, it differs from molarity, which quantifies the quantity of a solute in moles per litre of solution.

To calculate the molality of a solution, we need to know the amount of solute (in moles) and the mass of the solvent (in kilograms).

In this case, we are given:

Mass of solute (NaCl) = 25 g

Mass of solvent (water) = 2.0 kg

To calculate the amount of solute in moles, we need to convert the mass of NaCl to moles using its molar mass:

Molar mass of NaCl = 58.44 g/mol

Number of moles of NaCl = (25 g) / (58.44 g/mol) = 0.427 mol

Now we can calculate the molality of the solution:

Molality = (number of moles of solute) / (mass of solvent in kg)

Molality = (0.427 mol) / (2.0 kg) = 0.213 mol/kg

Therefore, the molality of the solution is 0.213 mol/kg.

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The Ammonium Chloride solution at 0.25 M has a pH of 2.67.

As a result, the weak basic (Chlorine) in the solution is overpowered by the conjugate acid (Ammonium cation), making the solution mildly acidic. According to the equation pH =log[Hydrogen ion], an acidic solution has a pH lower than 7. Aqueous ammonium chloride solution has a pH that is less than 7.

Ammonium cation + Water ⇌ Nitrogen trihydride + Hydronium ion

Kb = [Nitrogen trihydride][Hydronium ion] / [Ammonium cation]

[Nitrogen trihydride] = [Hydronium ion] = x

[Ammonium cation] = 0.25 - x

Kb = [Nitrogen trihydride][Hydronium ion] / [Ammonium cation]

1.8 × 10–5 = x² / (0.25 - x)

1.8 × 10–5 = x² / 0.25

x² = 4.5 × 10–6

x = 2.12 × 10–3

pH = -log[Hydronium ion] = -log(2.12 × 10–3) = 2.67

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