A 2.50-l volume of hydrogen measured at â196 °c is warmed to 100 °c. calculate the volume of the gas at the higher temperature, assuming no change in pressure

The correct answer is c) The reaction is not at equilibrium and will proceed to the right.

To determine the direction in which the reaction will proceed, we need to compare the reaction quotient (Q) with the equilibrium constant (K):

If Q > K, the reaction will proceed to the left (towards reactants).

If Q = K, the reaction is at equilibrium.

If Q < K, the reaction will proceed to the right (towards products).

Given:

Q = 3.56×10⁻⁴

K = 6.02×10⁻²

Since Q (3.56×10⁴) is less than K (6.02×10⁻²), the reaction will proceed to the right towards the products to reach equilibrium.

Therefore, the correct answer is: c. The reaction is not at equilibrium and will proceed to the right.

Complete question:

Ammonia can be produced via the chemical reaction n2(g)+3h2(g)?2nh3(g) during the production process, the production engineer determines the reaction quotient to be q = 3.56×10?4. if k = 6.02×10?2, what can be said about the reaction?

a. The reaction has reached equilibrium.

b. The reaction is not at equilibrium and will proceed to the left.

c. The reaction is not at equilibrium and will proceed to the right.

d. The reaction is not at equilibrium, but it is not possible to determine whether the reaction needs to proceed right or left to reach equilibrium.

The number of hydrogen atoms in an acyclic alkane with 8 carbon atoms is 18.

Alkanes can be described as organic compounds that contain single-bonded carbon and hydrogen atoms. The general formula for Alkanes is CₙH₂ₙ₊₂. Alkanes are further subdivided into three groups: chain alkanes, cycloalkanes, and branched alkanes.

Alkanes contain carbon and hydrogen atoms with single covalent bonds, which are known as saturated hydrocarbons. All the covalent bonds between carbon and hydrogen atoms are single and have a molecular formula of CₙH₂ₙ₊₂.

The simplest and smaller alkane is methane with one carbon atom and its molecular formula is CH₄.

Given, the number of carbons in the given alkane is equal to eight. Then the value of n is equal to 8.

The number of hydrogen atoms in alkane= 2 (8) + 2 = 18

Therefore, hydrogen atoms in an acyclic alkane are 18.

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Answer:

[tex]\bold{0.580 \;\rm{moles} \times 40.0\;\rm{ g/mol}} = 23.2\; g\; of\; NaOH[/tex] will be left.

Explanation:

Given:

Aqueous sulfuric acid [tex]H_2SO_4[/tex] will react with solid sodium hydroxide NaOH to produce aqueous sodium sulfate [tex]Na_2SO_4[/tex] and liquid water [tex]H_2O[/tex].

Now, balance the molecules of the left part of the equation with the right part of the equation.  

[tex]H_2SO_4 + 2NaOH \rightarrow Na_2 SO_4 + 2H_2O[/tex]

Now, from the above-balanced equation, it is clear that  1-mole sulphuric acid equated with two moles of sodium hydroxide to balance the above equation.

Now,

The weight of sulphuric acid is 89.3 g and the weight of sodium hydroxide is 96 g.

Therefore, convert 89.3 g of [tex]H_2SO_4[/tex] and 96.0 g of [tex]NaOH[/tex] into moles by using the formula,

[tex]\rm{Moles=\dfrac{Given\;weight}{Molar Mass}}[/tex]

Known Quantity:

[tex]\rm{Molar\; mass\; of\;} H_2SO_4 = 98.1\; g/mol[/tex]

[tex]\rm{Molar\; mass\; of\; NaOH = 40.0\; g/mol}[/tex]

Hence,

[tex]\begin{aligned}\rm{Moles\;of\;H_2SO_4}&=\dfrac{89.3}{98.1}\\&=0.910\end{aligned}[/tex]

[tex]\begin{aligned}\rm{Moles\;of\;NaOH}&=\dfrac{96}{40}\\&=2.40\end{aligned}[/tex]

[tex]\rm{Theoretical \;molar\; ratio} = \dfrac{2\; moles\; NaOH}{1\; mole\; H_2SO_4}[/tex]

Therefore,

If 0.91 moles react of [tex]H_2SO_4[/tex] then the number of moles required of [tex]NaOH[/tex] for the reaction will be twice as 0.91 moles.  

Moles required of [tex]NaOH[/tex] is 1.82.

Thus,

The remaining moles of NaOH will be (2.40 - 1.82) that is 0.58.

Now,

The weight of 0.580 moles NaOH will be calculated by the below expression:

[tex]0.580 \;\rm{moles} \times 40.0\;\rm{ g/mol} = 23.2\; g\; of\; NaOH \;will\; be\; left.[/tex]

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The age of the igneous rock with a Pb-206/U-238 mass ratio of 0.372 is approximately 1.7 billion years old, determined by the amount of U-238 that has decayed to Pb-206 and using the known half-life of U-238.

To determine the age of the igneous rock using its lead to uranium mass ratio, we apply the principles of radioactive decay, specifically the decay of Uranium-238 (U-238) into Lead-206 (Pb-206). Since each decay of U-238 produces one Pb-206, we can use this information to calculate the amount of U-238 that has decayed since the rock was formed. Knowing the half-life of U-238 is approximately 4.5 billion years, and the initial quantity of U-238 present would be equal to the sum of current U-238 and the Pb-206 produced from its decay.

The given Pb-206/U-238 mass ratio is 0.372. From this ratio, we can determine the fraction of U-238 that has decayed. Using the appropriate decay equation and the half-life, we can calculate the correct answer to be approximately 1.7 billion years, which is the approximate age of the rock.

The age of the rock is approximately 2.2 billion years.

To determine the age of the rock based on the Pb-206/U-238 mass ratio, we use the concept of radioactive decay and the uranium-lead (U-Pb) dating method.

Understanding Uranium-Lead Dating:

Uranium-238 (U-238) is a radioactive isotope that decays into lead-206 (Pb-206) through a series of radioactive decay steps.

The rate of decay of U-238 to Pb-206 is characterized by a half-life, which is the time it takes for half of the U-238 atoms to decay into Pb-206.

Using the Pb-206/U-238 Mass Ratio:

The Pb-206/U-238 mass ratio in the rock gives us information about the amount of lead-206 relative to uranium-238 at the time of rock formation.

The Pb-206/U-238 ratio changes over time due to the radioactive decay of uranium-238 into lead-206.

Calculating the Age:

The age of the rock can be calculated using the formula derived from radioactive decay principles.

We use the formula:

Age = log(Pb-206/U-238 ratio) / log(e^(λt) - 1)

Where:

λ is the decay constant of uranium-238.

t is the age of the rock in years.

Substituting Values:

Given that the Pb-206/U-238 mass ratio (R) = 0.372, we can rearrange the formula to solve for the age (t).

We use the decay constant (λ) of uranium-238.

Using Decay Constants:

The decay constant (λ) for uranium-238 is approximately 1.551 × 10⁻¹⁰ per year.

Calculating Age:

Substitute the values into the age formula to calculate the age of the rock:

Calculate log(0.372) ≈ -0.430.

Substitute the values into the age equation:

Age ≈ -0.430 / (1.551 × 10⁻¹⁰) ≈ 2.77 × 10⁹ years.

Final Answer:

The calculated age of the rock is approximately 2.77 × 10⁹ years, which is equivalent to about 2.2 billion years (since 1 billion years = 10⁹ years).

Therefore, based on the Pb-206/U-238 mass ratio of 0.372 in the igneous rock, the estimated age of the rock is approximately 2.2 billion years.

An example for how the Paleozic supercontinent ice cap melted is through plants dying off, which in turn, increased the greenhouse effect.

To add, the trapping of the sun's warmth in a planet's lower atmosphere due to the greater transparency of the atmosphere to visible radiation from the sun than to infrared radiation emitted from the planet's surface is called the greenhouse effect.

Answer:

9

Explanation:

Hello,

In this case, since the third shell of electrons has  the following 3 subshells:

[tex]3s, 3p \ and \  3d[/tex]

- The [tex]s[/tex] subsehll has one orbital for one pair of electrons: [tex]s^1 \ and \   s^2[/tex].

- The [tex]p[/tex] subsehll has three orbitals for three pairs of electrons: [tex]p^1, \ p^2,\ p^3, \ p^4,\ p^5\ and \ p^6\[/tex].

- The [tex]d[/tex] subsehll has five orbital for five pairs of electrons [tex]d^1, \ d^2,\ d^3, \ d^4,\ d^5,\ d^6,\ d^7,\ d^8,\   d^9\ and \ d^{10}\[/tex].

Therefore, the total number of orbitals when n=3 is:

[tex]1+3+5=9[/tex]

Best regards.

The third shell (n=3) of an atom contains nine orbitals, which are divided into three subshells: 3s, 3p, and 3d containing 1, 3, and 5 orbitals respectively.

The third shell (n=3) of an atom contains nine orbitals. Electron shells are divided into subshells, which are made up of orbitals. For n=3, the subshells are 3s, 3p, and 3d. The 3s subshell contains only one orbital, the 3p subshell contains three orbitals, and the 3d subshell contains five orbitals. So, if you add up the number of orbitals in each subshell (1 for 3s, 3 for 3p, and 5 for 3d), you will find that the third shell has a total of nine orbitals.

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Answer: Option (C) is the correct answer.

Explanation:

Entropy means the degree of randomness in a substance or object. This means more is the kinetic energy of particles of an object more will be its entropy.

For example, powdered sugar will have more number of particles and when it is added in a hot cup of coffee then its molecules will gain kinetic energy.

As a result, more number of collisions will take place due to which rate of reaction will also increase. Hence, powdered sugar will readily dissolve in the coffee.

Therefore, we can conclude that powdered sugar in a hot cup of coffee systems possesses the highest entropy.

Answer:

K⁺ and NO₃⁻ are the spectator ions.

Explanation:

First, we will consider the molecular equation because is the easiest to balance.

K₂CrO₄(aq) + Ba(NO₃)₂(aq) → BaCrO₄(s) + 2 KNO₃(aq)

Then, we will write the full ionic equation, which includes all the ions and the molecular species.

2 K⁺(aq) + CrO₄²⁻(aq) + Ba²⁺(aq) + 2 NO₃⁻(aq) → BaCrO₄(s) + 2 K⁺(aq) + 2 NO₃⁻(aq)

Finally, we will write the net ionic equation, which includes only the ions that participate in the reaction and the molecular species. The missing ions are the spectator ones.

CrO₄²⁻(aq) + Ba²⁺(aq) → BaCrO₄(s)

Final answer:

In the reaction between potassium chromate and barium nitrate, the spectator ions are NO3- and K+, as they appear unchanged on both sides of the chemical equation.

Explanation:

The question involves identifying the spectator ion(s) in the reaction between potassium chromate (K2CrO4) and barium nitrate (Ba(NO3)2). When these two aqueous solutions are mixed, a precipitation reaction occurs, forming barium chromate (BaCrO4) as the precipitate, and potassium nitrate (KNO3) remains in solution. Given that spectator ions are those ions that do not participate in the actual chemical reaction but are present in the same form on both sides of the equation, the NO3-(aq) and K+(aq) ions are the spectator ions in this reaction. They are present on both sides of the equation and remain unchanged.

Redox and neutralization reactions both involve the transfer of particles. Redox reactions involve the transfer of electrons, with one component gaining and another losing electrons. Neutralization reactions involve the transfer of protons or hydrogen ions, where an acid donates a proton that a base accepts.

A redox reaction and a neutralization reaction both involve the transfer of particles, but they differ in the type of particle that is transferred. In a redox reaction (which stands for reduction-oxidation), the key particles involved are electrons. During this type of reaction, one atom loses electrons (oxidation) and another atom gains electrons (reduction). For example, when copper reacts with silver nitrate in solution, silver is reduced (gains electrons) and copper is oxidized (loses electrons).

On the other hand, a neutralization reaction is a type of reaction between an acid and a base. Here, the primary particles involved are protons (or hydrogen ions, H+). An acid donates a proton (H+) and a base receives it. For example, when hydrochloric acid (HCl) reacts with sodium hydroxide (NaOH), HCl donates a proton to OH-, neutralizing both the acid and base to form water and a salt.

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Explanation:

Rate of dissolving is the rate at which a solute is able to dissolve in a solvent.

Some factors which affect the rate of dissolving are as follows.

More surface area : When there are more number of particles then it means there is more surface area of solute present in the solution. Thus, there will be more number of collisions between the solute and solvent molecules. As a result, rate of dissolving increases.

High temperature : More is the increase in temperature more will be the kinetic energy gained by molecules. Thus, this will lead to greater number of collisions and as a result, rate of dissolving increases.

Lot of agitation : When we stir a solution vigorously or create a disturbance then there will be increase in number of collisions which will also lead to increase in rate of dissolving.

Thus, we can conclude that the rate of dissolving would increase when there is:

Answer: Mantle

Explanation:

The formula for freezing point depression is:

ΔT = Kf * m         --->1

Where,

ΔT = change in temperature = 2.88 degrees Celcius

Kf = freezing point molar constant of solvent

m = molality (moles solute/mass solvent)

First we calculate for molality since we are given the mass of solute and solvent.

Molar mass of pentane = 72.15 g / mol

molality m= (0.123 g / 72.15 g / mol) / (2.493 x 10^-3 kg)

m = 0.684 molal

Going back to equation 1:

ΔT = Kf * m

2.88 = Kf * 0.684

Kf = 4.21 degC / molal

Value for Kf in question 1 given that m = 0. 829 mol/kg:

2.88 = Kf * 0.829

Kf = 3.47 degC / molal

The cell reaction that occurs is as follows:

 [tex]\boxed{{\text{Mg + C}}{{\text{u}}^{{\text{2 + }}}} \rightleftarrows {\text{M}}{{\text{g}}^{2 + }}{\text{ + Cu}}}[/tex]

The line notation of cell is as follows:

[tex]\left. {{\text{Mg}}\left( {\text{s}} \right)} \right|{\text{M}}{{\text{g}}^{2 + }}\left( {{a_{{\text{M}}{{\text{g}}^{2 + }}}}} \right)\left\| {{\text{C}}{{\text{u}}^{2 + }}\left( {{a_{{\text{C}}{{\text{u}}^{2 + }}}}} \right)\left| {{\text{Cu}}\left( {\text{s}} \right)} \right.} \right.[/tex]

The standard emf value of the cell is [tex]\boxed{2.7{\text{ V}}}[/tex].

Further Explanation:

Redox reaction:

It is a type of chemical reaction in which the oxidation states of atoms are changed. In this reaction, both reduction and oxidation are carried out at the same time. Such reactions are characterized by the transfer of electrons between the species involved in the reaction.

The general representation of a redox reaction is,

 [tex]{\text{X}} + {\text{Y}} \to {{\text{X}}^ + }+{{\text{Y}}^ - }[/tex]

The oxidation half-reaction can be written as:

[tex]{\text{X}} \to {{\text{X}}^ + } + {e^ - }[/tex]

The reduction half-reaction can be written as:

[tex]{\text{Y}} + {e^ - } \to {{\text{Y}}^ - }[/tex]  

Here, X is getting oxidized and its oxidation state changes from  to +1 whereas B is getting reduced and its oxidation state changes from 0 to -1.

The element which has higher oxidation potential is oxidized at anode and the element with the less oxidation potential is reduced at cathode in the cell.

The standard oxidation potential for [tex]{\text{Mg/M}}{{\text{g}}^{2 + }}[/tex] is  [tex]+ 2.363{\text{ V}}[/tex].

The standard oxidation potential for [tex]{\text{Cu/C}}{{\text{u}}^{2 + }}[/tex] is [tex]- 0.337{\text{ V}}[/tex].

Since [tex]{\text{Mg}}[/tex] has higher oxidation potential thus the oxidation of [tex]{\text{Mg}}[/tex] takes place at anode and reduction of [tex]{\text{C}}{{\text{u}}^{2 + }}[/tex] takes place at cathode.

The oxidation half-reaction of [tex]{\text{Mg/M}}{{\text{g}}^{2 + }}[/tex] can be written as:

[tex]{\text{Mg}} \to {\text{M}}{{\text{g}}^{2 + }} + 2{e^ - }[/tex]     ......(1)

The reduction half-reaction [tex]{\text{Cu/C}}{{\text{u}}^{2 + }}[/tex] can be written as:

[tex]{\text{C}}{{\text{u}}^{2 + }} + 2{e^ - } \to {\text{Cu}}[/tex]     ......(2)

Add reaction (1) and (2) and eliminate common terms to determine the net reaction for the given cell.

[tex]{\text{Mg + C}}{{\text{u}}^{{\text{2 + }}}} \rightleftarrows {\text{M}}{{\text{g}}^{2 + }}{\text{ + Cu}}[/tex]           ......(3)

The expression of the cell is written as follows:

[tex]\left. {{\text{Mg}}\left( {\text{s}} \right)} \right|{\text{M}}{{\text{g}}^{2 + }}\left( {{a_{{\text{M}}{{\text{g}}^{2 + }}}}} \right)\left\| {{\text{C}}{{\text{u}}^{2 + }}\left( {{a_{{\text{C}}{{\text{u}}^{2 + }}}}} \right)\left| {{\text{Cu}}\left( {\text{s}} \right)} \right.} \right.[/tex]                             ......(4)

The expression to calculate the standard emf of the cell is as follows:

[tex]E_{{\text{cell}}}^0 = E_{{\text{anode}}}^0 - E_{{\text{cathode}}}^0[/tex]                  ......(5)

Substitute [tex]+ 2.363{\text{ V}}[/tex] for [tex]E_{{\text{anode}}}^0[/tex] and [tex]- 0.337{\text{ V}}[/tex] for [tex]E_{{\text{cathode}}}^0[/tex] in equation (5).

[tex]\begin{aligned}E_{{\text{cell}}}^0&=\left({ + 2.363{\text{ V}}}\right)-\left( { - 0.337{\text{ V}}} \right)\\&= 2.7{\text{ V}}\\\end{aligned}[/tex]        

Learn more:

1. Which occurs during the redox reaction? https://brainly.com/question/1616320

2. Oxidation and reduction reaction: https://brainly.com/question/2973661

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Electrochemistry

Keywords: Mg/M2+, Cu/Cu2+, half-cell reaction, standard emf of the cell, oxidation state, reduction, oxidation, redox reaction, transfer of electrons, reducing agents, oxidizing agents.

To calculate the standard emf of a cell with Mg/Mg2+ and Cu/Cu2+ at 25 °C, subtract the anode potential (-2.37 V) from the cathode potential (+0.34 V) to get +2.71 V, indicating a spontaneous reaction. The line notation for the cell is Mg(s) | Mg2+(aq) || Cu2+(aq) | Cu(s).

To calculate the standard emf of a cell using the Mg/Mg2+ and Cu/Cu2+ half-cell reactions at 25 °C, we use the standard reduction potentials from the electrochemical series. The standard reduction potential for the Mg2+ to Mg half-cell is -2.37 V and for the Cu2+ to Cu half-cell is +0.34 V. Because oxidation occurs at the anode and reduction at the cathode, the Mg/Mg2+ cell will act as the anode and the Cu/Cu2+ cell as the cathode.

The overall cell reaction, under standard conditions, is derived by combining the half-reactions:

Mg(s) -> Mg2+(aq) + 2e- (Oxidation)

Cu2+(aq) + 2e- -> Cu(s) (Reduction)

When combined, the overall reaction is:

Mg(s) + Cu2+(aq) -> Mg2+(aq) + Cu(s)

The standard cell potential (Ecell) is calculated by taking the difference between the standard reduction potentials of the cathode and anode:

Ecell = Ecathode(red) - Eanode(ox) = 0.34 V - (-2.37 V) = +2.71 V

The positive value of Ecell indicates that the reaction is spontaneous under standard conditions. The line notation for the cell is:

Mg(s) | Mg2+(aq) || Cu2+(aq) | Cu(s)

Final answer:

To identify the beaker showing HA behaving as a weak acid, we need to look for the beaker in which the concentration of H3O+ and A- is relatively low, indicating partial dissociation.

Explanation:

In the given question, we are asked to identify the beaker that shows the acid HA behaving as a weak acid in water.

A weak acid is one that only partially dissociates in water, creating a small amount of hydronium ions (H3O+) and the conjugate base (A-). Strong acids, on the other hand, completely dissociate into hydronium ions and the conjugate base.

To identify the beaker showing HA behaving as a weak acid, we need to look for the beaker in which the concentration of H3O+ and A- is relatively low, indicating partial dissociation.

Answer:

-contain hydroxide ion or produce it in a solution

-taste sour

Explanation:

Answer: A saturated solution is a solution with as much dissolved solute as it can hold at a given temperature.

Explanation:

Saturated solution: It is a solution in which solute is dissolved in its maximum amount in a solvent at a given temperature. Further addition of solute will not get dissolve in the solution. For : Soda is a saturated solution of carbon-dioxide in water.

Unsaturated solution: It is a solution in which addition of more solute will get easily dissolve in a solution at a given temperature. In unsaturated solution the amount of solute present in less than the amount of solute present in saturated solution. For example: Pinch of salt in a glass of water.

Answer:

saturated

Explanation:

Final answer:

To find the number of excess electrons on a conducting sphere with a net charge of -4.8 × 10⁻¹⁷ C, the total charge is divided by the charge of a single electron, resulting in approximately 300 excess electrons.

Explanation:

The question asks about the number of excess electrons on a conducting sphere with a net charge of –4.8 × 10⁻¹⁷ C. To find the number of excess electrons, we need to divide the total charge by the charge of a single electron, which is approximately 1.6 × 10⁻¹⁹ C. The calculation is as follows

Number of electrons = Total charge / Charge of one electron

= (-4.8 × 10⁻¹⁷ C) / (1.6 × 10⁻¹⁹ C)

= 3 × 10² electrons.

Therefore, the sphere has approximately 300 excess electrons.

The jogger's speed is first converted to meters per second, and then the time for the 483-meter run is calculated using the speed and the formula time = distance/speed, resulting in approximately 49.38 seconds.

The student's question involves converting units and calculating time based on a given rate of speed. The jogger runs 107 yards in 10.00 seconds, and we need to determine the time it would take for a 483-meter run at the same rate.

Since 1 yard is roughly equivalent to 0.9144 meters, the jogger's speed in meters per second (m/s) can be calculated as: (107 yards × 0.9144 m/yard) / 10 s = 9.77928 m/s.

Now, to find the time for a 483-meter run at the same speed, we use the equation time = distance/speed, which gives us: 483 m / 9.77928 m/s = approximately 49.38 seconds.

At 21.5 °C and 1.00 atm, the molar volume of helium gas is expected to be slightly higher than 22.41 L, the molar volume of an ideal gas at STP due to the increase in temperature and the behavior of helium closely resembling an ideal gas.

If you experiment to generate helium gas at 21.5 °C (which is 294.65 K) and 1.00 atm, you would expect the molar volume of the helium to be close to the value for an ideal gas under standard temperature and pressure conditions. This is because helium behaves relatively closely to an ideal gas due to its small, non-polar, monatomic nature. At standard temperature (0 °C or 273.15 K) and pressure (1 atm), the molar volume of an ideal gas is 22.41 L. However, because the experiment is carried out at a slightly higher temperature, the molar volume will be slightly higher than 22.41 L due to the direct relationship between temperature and volume described by Charles's law.

To calculate the molar volume at the given conditions, you would use the ideal gas law equation: PV = nRT. You can rearrange this equation to solve for molar volume (V/n) and find that V/n = RT/P. With R as the ideal gas constant (0.0821 L·atm/(K·mol)), T as the absolute temperature in Kelvin, and P as the pressure in atm, you can calculate the molar volume of helium at 21.5 °C and 1.00 atm.

The change in enthalpy for 1 mole of SiC, heated from 25°C to 1000°C, is calculated to be 1306.5 J using the constant pressure molar heat capacity formula.

Calculating the change in enthalpy of SiC:

1. Identifying the relevant information:

Substance: 1 mole of SiC (silicon carbide)

Temperature change: 25°C to 1000°C

Constant pressure molar heat capacity (cp): 1.34 J/mol°C

2. Recalling the enthalpy equation:

ΔH = n * cp * (T2 - T1)

where:

ΔH is the change in enthalpy (J)

n is the number of moles

cp is the constant pressure molar heat capacity (J/mol°C)

T1 is the initial temperature (°C)

T2 is the final temperature (°C)

3. Applying the equation to the given information:

n = 1 mole

cp = 1.34 J/mol°C

T1 = 25°C

T2 = 1000°C

4. Substituting the values and calculating ΔH:

ΔH = 1 mole * 1.34 J/mol°C * (1000°C - 25°C)

ΔH = 1306.5 J

Therefore, the change in enthalpy for 1 mole of SiC when heated from 25°C to 1000°C is 1306.5 J.

Complete question:

Calculate the change in the enthalpy and the change in entropy when 1 mole of SiC is heated from 25°C to 1000°C·The constant pressure molar heat capacity ofSiC varies with temperature as cp = 50.79 + 1.97 x 10^-3T-4.92 x 10^6T^-2 + 8.20 x 10^9 T-3 J/mol-K

Final answer:

To calculate the density of solid argon at 40 K, we first determine the edge length of the unit cell using the atomic radius, then calculate the volume of the unit cell. We then find the mass of argon in the unit cell using its atomic weight and Avogadro's number and divide the mass by the volume to find the density.

Explanation:

To calculate the density of solid argon, we can follow these steps:

By using the appropriate equations and constants, we can find the value for the density of solid argon at 40 K.

The refrigerant exits a compressor as a high-pressure vapor due to the mechanical energy applied during compression, which raises both pressure and temperature. Hence, correct option C.

The state of refrigerant as it exits a compressor in a vapor compression system is high-pressure vapor. During the compression phase in vapor compression cooling systems, mechanical energy is applied to the refrigerant, causing both pressure and temperature to rise. This process changes the state of the refrigerant from a low-pressure vapor, which it is when it enters the compressor, to a high-pressure vapor as it exits the compressor. The high-pressure vapor is subsequently cooled and condensed in the condenser, transferring heat to the surroundings and turning into a high-pressure liquid ready to go through the cycle again.