What is the ΔG (kJ/mol) for a reaction at 25 Celsius that is:
Mg3(PO4)2 (s) ⇄ 3 Mg2+ (aq) + 2 PO43− (aq) ΔG0 = 137.0 kJ/mol
If there is initially 0.65 M Mg2+(aq) and 0.43 M PO43− (aq) in solution?

Answer: A terrestrial phenomenon commonly referred to as lunar eclipse transpires during an astronomical event wherein the Earth traverses its orbit between the Sun and the Moon, resulting in an obstruction and projection of its shade onto the lunar surface. During a lunar eclipse, the moon may exhibit a crimson tint owing to the Earth's atmosphere refracting sunlight towards the moon. This phenomenon is commonly referred to as a "blood moon" in academic discourse. The acquisition of photographic images of lunar eclipses necessitates the utilization of a camera and a tripod. However, it is imperative to exercise care with respect to exposure duration and ISO settings to facilitate the capture of the subdued illumination emitted by the eclipsed moon. An upcoming celestial phenomenon, which is anticipated to be observable from Earth, is the occurrence of a lunar eclipse on May 16, 2022.

Explanation:

5.82 x 10⁶ atoms of H₂O can produce 72800 L of NaOH.

An exothermic mist from the commencement of high temperatures and heat from the breakdown of incompatible combinations can emerge from the reaction of NaOH with H₂O₂, which creates a powerful oxidant known as Na₂O₂ and poses a serious thermal danger.

5.82 x 10⁶ atoms / 3 atoms per mole = 1.94 x 10⁶ moles of H₂O

1.94 x 10⁶ moles of H₂O x (4 moles of NaOH / 2 moles of H₂O) = 3.88 x 10⁶ moles of NaOH

To convert moles of NaOH to liters, we need to use the density of NaOH and its molar mass. The molar mass of NaOH is:

23 g/mol (for Na) + 16 g/mol (for O) + 1 g/mol (for H) = 40 g/mol

The density of NaOH is 2.13 g/mL. Using these values, we can calculate the volume of NaOH produced:

3.88 x 10⁶ moles of NaOH x 40 g/mol = 1.55 x 10⁸ g of NaOH

1.55 x 10⁸ g / 2.13 g/mL = 7.28 x 10⁷ mL

7.28 x 10⁷ mL / 1000 mL/L = 72800 L

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The new pressure inside the balloon will be 1.91 atm after the gas is compressed to a volume of 4.6 L.

The scientist Robert Boyle discovered this empirical relationship in 1662, which asserts that under constant temperature, the pressure (p) of a given quantity of gas changes inversely with its volume (v). As a result, the equation pv = k, a constant, is produced.

P1 x V1 = P2 x V2

where the starting pressure and volume are P1 and V1, and the ultimate pressure and volume are P2 and V2.

Plugging in the given values, we get:

1.35 atm x 6.5 L = P2 x 4.6 L

Solving for P2, we get:

P2 = (1.35 atm x 6.5 L) / 4.6 L

P2 = 1.91 atm

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Answer:

The average would be found by:

(22.59 + 27.43 + 35.642) ÷ 3 = 28.554

Rounded to 4 significant figures:

28.55

The percent composition of iron, carbon, and nitrogen in Prussian Blue is approximately 56.27%, 15.38%, and 28.35%, respectively. The exact percentage of hydrogen depends on the value of x in the formula.

When evaluating a chemical complex, knowing the percentage composition of a certain element inside it may be helpful. The formula for percent composition is 100 times the product of the element's mass and its molecular mass.

It is free of sulphur and phosphorus and contains 48% iron. Siderite-smithsonite, siderite-magnesite, and siderite-rhodochrosite are solid solutions formed when zinc, magnesium, and manganese are substituted for iron.

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The entire spectrum of electromagnetic radiation emitted by celestial bodies, helps astronomers gain complete understanding of stars and planets.

The electromagnetic spectrum encompasses a vast array of electromagnetic radiation, spanning from the elongated wavelength of radio waves to the compact wavelength of gamma rays. This includes an assortment of radiation types, including radio waves, microwaves, infrared radiation, visible light, ultraviolet radiation, X-rays and gamma rays. With each variation possessing a unique wavelength and frequency, they all traverse at the speed of light.

Astronomers rely heavily on the electromagnetic spectrum to study celestial bodies like stars and planets. This spectrum enables them to detect and measure various forms of electromagnetic radiation emitted by these objects. The range of frequencies in this radiation spans from radio waves with longer wavelengths to gamma rays with shorter ones.

Different types of electromagnetic radiation provide distinct information about celestial objects, including their temperature, composition, and movement. For instance, visible light allows astronomers to observe the color and brightness of stars and planets, while radio waves can identify magnetic fields or map out gas and dust distribution in galaxies. This knowledge helps answer fundamental questions about the universe's nature such as how galaxies and stars form, evolve over time, and what causes the universe's expansion.

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20.0 moles of oxygen react with Antimony to form 13.3 moles of Antimony (III) Oxide. We want to calculate how many moles of Antimony (III) Oxide will be formed from 20.0 moles of oxygen. This is a stoichiometry problem.

The link between the proportional amounts of components participating in a reaction or generating a compound is known as stoichiometry, and it is often expressed as a ratio of whole integers.

Assuming a balanced chemical equation, the stoichiometric ratio between Antimony (III) Oxide  and oxygen can be used to determine the number of moles of Antimony (III) Oxide  formed.

For example, the balanced equation for the reaction of Antimony with O2 to form Antimony (III) Oxide  is:

4 Antimony + 3 O2 → 2 Antimony (III) Oxide

From this equation, it can be seen that 3 moles of oxygen react with 2 moles of Antimony (III) Oxide . Therefore, if there are 20.0 moles of O2, then the number of moles of Antimony (III) Oxide  formed would be:

20.0 moles oxygen × (2 moles Antimony (III) Oxide  / 3 moles oxygen) = 13.3 moles Antimony (III) Oxide.

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The same components always present the same ratios in a compound. Water always has two hydrogen atoms for oxygen, and carbon dioxide always contains two oxygen atoms for every one of carbon.

Considering that each of them is composed of two similar atoms, elements include hydrogen, oxygen, and nitrogen. Water is not an element since it only contains two hydrogen atoms and one oxygen atom.

Pure substances include both elements and compounds. Substances made up of many atom types are known as compounds. The simplest compounds are called elements because they only contain one kind of atom.

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The volume of O₂ produced when 8.52 g of Na₂O₂ is added to excess water is approximately 1.17 liters at STP.

When it comes to the ideal gas law, the underlying assumption is that the gas is in a state of thermodynamic equilibrium. This means that its molecules are not interacting with each other except through perfectly elastic collisions.

Equation:

We can use stoichiometry to determine the volume of oxygen gas produced from the reaction of 8.52 g of Na₂O₂.

First, we need to convert the mass of Na₂O₂ to moles using its molar mass:

8.52 g Na₂O₂ × (1 mol Na₂O₂ / 77.98 g Na₂O₂) = 0.1093 mol Na₂O₂

According to the balanced chemical equation, 1 mole of Na₂O₂ produces 1/2 mole of O₂:

2 Na₂O₂ + 2 H₂O → 4 NaOH + O₂

1 mol Na₂O₂ produces 1/2 mol O₂

So, the number of moles of O₂ produced is:

0.1093 mol Na₂O₂ × (1/2 mol O₂ / 1 mol Na₂O₂) = 0.05465 mol O₂

Finally, we can use the ideal gas law to determine the volume of O₂ produced at standard temperature and pressure (STP), which is defined as 0°C (273 K) and 1 atmosphere (1 atm) of pressure:

PV = nRT

At STP, the pressure and temperature are known, so we can rearrange the equation to solve for V:

V = nRT / P

Plugging the values, we get:

V = (0.05465 mol) × (0.08206 L atm mol⁻¹ K⁻¹) × (273 K) / (1 atm)

V ≈ 1.17 L

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The new volume of the helium gas is 0.39 L.

The new volume of the helium gas is calculated by applying the following formula as shown below:

n₁V₁ = n₂V₂

where;

the final number of moles is calculated as;

n₂ = 0.51 g/4 g/mol

n₂ = 0.1275 mol

The final volume of the gas;

V₂ =(n₁V₁)/n₂

V₂ = (0.1 x 0.5 L )/ (0.1275)

V₂ = 0.39 L

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Sodium bicarbonate is the limiting reactant and just 0.65 moles of CO₂ can be produced.

Limiting reactant is the reactant that gets consumed completely in a chemical reaction, thereby limiting the amount of product that can be formed.

Acetic acid and sodium bicarbonate react, and the balanced chemical equation for this reaction is:

CH₃COOH + NaHCO₃ -> CH₃COONa + CO₂ + H₂O

Calculating the quantity of product that can be created from each reactant and comparing them to discover which one generates the least will help us identify the limiting reactant. The reactant that produces less product is the limiting reactant.

The provided amounts of each reactant must first be converted to moles:

0.80 moles of acetic acid

0.65 moles of sodium bicarbonate

Then, we may determine how much CO₂ can be created from each reactant using stoichiometry. The balanced equation reveals that 1 mole of acetic acid and 1 mole of sodium bicarbonate combine to create 1 mole of carbon dioxide.

Using acetic acid:

0.80 moles CH₃COOH x (1 mole CO₂ / 1 mole CH₃COOH) = 0.80 moles CO₂

About sodium bicarbonate:

0.65 moles NaHCO₃ x (1 mole CO₂ / 1 mole NaHCO₃) = 0.65 moles CO₂

As 0.65 moles of CO₂ are fewer than 0.80 moles of CO₂, sodium bicarbonate is the limiting reactant. Just 0.65 moles of CO₂ can be produced as a result.

Dimensional analysis:

0.80 moles CH₃COOH x (1 mole CO₂ / 1 mole CH₃COOH) x (44.01 g CO₂ / 1 mole CO₂) = 35.21 g CO₂

0.65 moles NaHCO₃ x (1 mole CO₂ / 1 mole NaHCO₃) x (44.01 g CO₂ / 1 mole CO₂) = 28.61 g CO₂

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The general trend in boiling points across the period is that they increase from left to right. This is because as you move from left to right, the number of protons in the nucleus increases, causing the attraction between the electrons and the nucleus to increase.

In an open system, this is referred to as atmospheric pressure. Pressure rises as boiling points and the amount of energy needed to boil liquids rise.

The standard boiling point should rise as intermolecular tensions grow. This is because it takes more energy to separate molecules during the transition from liquid to gas when intermolecular interactions are stronger.

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A solution with a pH of 9.7 contains the hydronium ion. [tex]1x10^-9.7[/tex]M, which is equivalent to approximately [tex]2.01x10^-10 M.[/tex]

You must be aware of the hydronium ion concentration expressed in moles per litre (molarity) in order to calculate the pH of an aqueous solution. The pH is then calculated using the equation pH = - log [H3O+].

At the far end, there are no more than 1M hydrogen ions, which leads to a pH of no more than 0. The pH value is limited to 14, whereas the other end contains no more than 1M of hydroxide ions.

Question:

[tex]The [H3O+] of a solution with pH = 9.7 is 9.7 M O 1x 109.7M O2 x 10-10 M O 5 x 10-5M 09.7 x 10-1 M The [H3O+] of a solution with pH = 9.7 is 09.7 M O 1x 109.7 M 2 x 10-10 M. 0 5x 10-5m 09.7 10-1 M.[/tex]

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The value of (approximate) z-score for an area of 0.1500 to the left of the z-score in a standard normal distribution is approximately -1.04.

If the area to the left of a z-score is 0.15 this means that 15% of the area under the standard normal distribution curve is to the left of the z-score. To find the approximate z-score, we can use a table of standard normal probabilities or a calculator that has a normal distribution function.

Using a standard normal distribution table, we can look up the closest probability to 0.15 in the table, which is 0.1492. The corresponding z-score is approximately -1.04.

This is how you arrive at the approximate z-score for an area of 0.1500 to the left of the z-score in a standard normal distribution which is approximately -1.04.

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Filtration is the process of passing a liquid containing contaminants through a filter to get a cleaner liquid.

After filtration impurities are left on the filter when the liquid goes through it, resulting in a cleaner combination. The procedure of separating one ingredient from another in a compound is used in the purification of chemical compounds or any materials. A mixed sample is turned into fresh samples, each of which contains a single ingredient.Separation procedures are then utilised to separate these compounds. These techniques alter the relative proportions of components in a combination.The type of the compounds and the impurities in them influence the purification of organic chemicals and compounds.

Purification of compounds can be done in other methods as well such as distillation, chromatography, sublimation and crystallisation.

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Full question:

What can you conclude after learning about filtration of solutions of organic compounds? Name other methods of purification of compounds.

The molecular equation for the reaction of aqueous iron(III) nitrate and aqueous lithium hydroxide is :

Fe(NO3)3(aq) + 3LiOH(aq) ⟶ Fe(OH)3(s) + 3LiNO3(aq)

By removing the spectator ions from the molecular equation—ions that appear on both sides of the equation but do not participate in the reaction—we can derive the balanced net ionic equation for the reaction. The spectator ions in this instance are Li+ and NO3-.

The net ionic equation is :

Fe3+(aq) + 3OH-(aq) ⟶ Fe(OH)3(s)

Be aware that solid Fe(OH)3 is created when the ions Fe3+ and OH- mix. Equations with (aq) for aqueous species and (s) for solid species denote the physical states.

Answer:

1. To calculate the number of moles of Cl- ion given, we need to first calculate the total number of Cl- ions in 1250 mL of Ringer's solution:

1250 mL x (109 mEq/L / 1000 mL/L) = 136.25 mEq of Cl-

Next, we can convert mEq to moles using the molecular weight of Cl-:

136.25 mEq x (1 mol / 1000 mEq) = 0.13625 mol of Cl-

Therefore, 0.13625 moles of Cl- ion were given.

2. To calculate the volume of 10% NaOH solution containing 2.2 g of NaOH, we need to use the definition of percent concentration:

10 g NaOH / 100 mL solution = 2.2 g NaOH / x mL solution

Solving for x, we get:

x = (2.2 g NaOH) / (10 g NaOH / 100 mL) = 22 mL

Therefore, 22 mL of 10% NaOH solution would contain 2.2 g of NaOH.

3. To calculate the number of moles of BaCl2 in 600 mL of a 2.5 M solution, we can use the definition of molarity:

2.5 mol/L = x mol / 0.6 L

Solving for x, we get:

x = 1.5 mol of BaCl2

Therefore, 1.5 moles of BaCl2 are in 600 mL of a 2.5 M solution.

4. To calculate the molarity of a solution of 29.2 g of NaCl in 134 mL of solution, we need to first convert the mass of NaCl to moles:

29.2 g NaCl x (1 mol / 58.44 g) = 0.499 mol NaCl

Next, we can calculate the volume of the solution in liters:

134 mL x (1 L / 1000 mL) = 0.134 L

Finally, we can calculate the molarity:

0.499 mol / 0.134 L = 3.72 M

Therefore, the molarity of the solution is 3.72 M.

5. To calculate the mass of Na2CO3 needed to prepare a 400 mL of 3.00 M solution of Na2CO3, we can use the molarity formula:

Molarity = moles of solute / liters of solution

Rearranging the formula to solve for moles of solute:

moles of solute = Molarity x liters of solution

moles of Na2CO3 = 3.00 M x 0.400 L = 1.20 moles of Na2CO3

Next, we can convert moles of Na2CO3 to grams using the molar mass of Na2CO3:

1.20 moles Na2CO3 x (105.99 g / 1 mole) = 127.2 g of Na2CO3

Therefore, 127.2 grams of Na2CO3 are needed to prepare a 400 mL of 3.00 M solution of Na2CO3.

6. To calculate the molarity of NaNO3 solution made by diluting 33.0 ml of a 5.50 M solution to a final volume of 250. mL, we can use the dilution formula:

M1V1 = M2V2

where M1 is the initial molarity, V1 is the initial volume, M2 is the final molarity, and V2 is the final volume.

Rearranging the formula to solve for M2:

M2 = M1V1 / V2

M2 = (5.50 M) x (33.0 mL) / (250. mL)

M2 = 0.726 M

Therefore, the molarity of NaNO3 solution made by diluting 33.0 ml of a 5.50 M solution to a final volume of 250. mL is 0.726 M.

7. To calculate the grams of mannitol given in one hour, we can use the definition of percent concentration:

20 g mannitol / 100 mL solution = x g mannitol / 100 mL solution

Simplifying the equation, we get:

x = 20 g mannitol

Therefore, 20 g of mannitol are given in 100 mL of solution. Since the patient receives 100 mL of solution every hour, the amount of mannitol given in one hour is also 20 g.

8. To calculate the microliters of a 4000.0 ppm Fe stock solution

Na sodium has one valence electron and has the electron configuration [Ne] 3s1.

Atoms frequently form bonds with other atoms that enable them to add electrons to their valence shells. For instance, the chemical sodium chloride (NaCl) can be created by the bonding of an atom with a single valence electron, like sodium, and an atom with seven valence electrons, like as chlorine. Chlorine absorbs the one valence electron from sodium and then adds eight electrons to its valence shell.

A subatomic particle with an electrical charge that is negative is called an electron. complemented by protons and neutrons, it is one of the primary building blocks of atoms.

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The four types of physical weathering are:

1.) Frost action or freeze-thaw weathering

2.) Exfoliation or unloading

3.) Thermal expansion and contraction

4.) Abrasion

These types of weathering are similar in that they all involve physical processes that break down rocks and other geological materials without the involvement of chemical reactions.

Frost action occurs when water seeps into cracks in rocks and freezes, causing the water to expand and exert pressure on the surrounding rock. This process repeats over time and can cause the rock to fracture and break apart.

Exfoliation occurs when rocks that were once buried deep beneath the earth's surface are exposed to lower pressures as overlying rocks erode away. As the pressure decreases, the rocks expand and crack, causing outer layers of rock to peel away like layers of an onion.

Thermal expansion and contraction occurs when rocks are repeatedly subjected to heating and cooling cycles. This can cause the rock to expand and contract, leading to the development of cracks and fractures.

Abrasion occurs when rocks and other geological materials are subjected to repeated rubbing or grinding by other materials, such as wind-blown sand, water, or other rocks. This process can cause rocks to wear down and break apart over time.

In summary, these four types of physical weathering are similar in that they all involve physical processes that break down rocks and other geological materials, but each type operates under different conditions and results in different forms of weathering.

Answer:

below

Explanation:

The four types of physical weathering are:

1. Freeze-thaw weathering

2. Exfoliation

3. Abrasion

4. Thermal expansion and contraction

These types of physical weathering are similar in that they all involve the breakdown of rocks and other geological features due to physical forces rather than chemical reactions. In each case, the physical stress causes the rock to crack or break apart. Additionally, they all can occur in various types of climates and environmental conditions.

The vapor pressure of methanol at 40°C is 0.234 atm.

Only two types of alcohol are methanol and ethanol. Ethanol, sometimes referred to as ethyl alcohol, has a chemical composition of two carbon atoms. Methanol, sometimes referred to as methyl alcohol, is made up of just one carbon atom.

ln(P2/P1) = (ΔHvap/R) x (1/T1 - 1/T2)

ΔGvap = -RTln(Pvap/P) = ΔHvap - TΔSvap

ΔGvap = -RTln(Pvap/P) = -166.6 kJ/mol

ΔSvap = S(g) - S(l) = 239.9 J/mol-K - 126.8 J/mol-K = 113.1 J/mol-K

ΔHvap = ΔGvap + TΔSvap = -166.6 kJ/mol + (298.15 K)(113.1 J/mol-K) = -134.6 kJ/mol

Now we can use the Clausius-Clapeyron equation to find the vapor pressure of methanol at -30°C and 40°C.

At -30°C, we have:

T1 = 25°C + 273.15 = 298.15 K

T2 = -30°C + 273.15 = 243.15 K

ΔHvap = -134.6 kJ/mol

R = 8.314 J/mol-K

ln(P2/5 atm) = (-134.6 kJ/mol / 8.314 J/mol-K) x (1/298.15 K - 1/243.15 K)

P2 = 0.0038 atm

Therefore, the vapor pressure of methanol at -30°C is 0.0038 atm.

At 40°C, we have:

T1 = 25°C + 273.15 = 298.15 K

T2 = 40°C + 273.15 = 313.15 K

ΔHvap = -134.6 kJ/mol

R = 8.314 J/mol-K

ln(P2/5 atm) = (-134.6 kJ/mol / 8.314 J/mol-K) x (1/298.15 K - 1/313.15 K)

P2 = 0.234 atm

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The birth weight of cats at the 90th percentile would be approximately 3.7 ounces. b. The birth weight of cats at the 10th percentile would be approximately 2.3 ounces.

There are three main methods of calculating weight:

1. Balance Beam Scale: A balance beam scale is an example of a mechanical weighing system. It uses a set of calibrated weights to measure the weight of an object.

2. Digital Scale: A digital scale uses an electronic or digital readout to display the weight of an object.

3. Calipers: Calipers are devices used to measure the distance between two points, such as thickness or diameter. They come in various sizes and shapes, depending on the type of measurement the user wants to take.

Birth weight is the weight of a baby at the time of birth. It is usually measured soon after delivery, with a special baby scale, though sometimes the baby's weight is estimated. The normal range of birth weight is anywhere between 5 lbs 8 oz and 10 lbs, though preterm or premature babies may be significantly lighter. A baby's birth weight is important because it can provide a clue to the baby's overall health. High birth weights may indicate an underlying medical condition, and low birth weights can be a sign of premature or difficult delivery and health risks associated with such a delivery.

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We can see here that matching the label to the correct description, we will have to have a complete part of the description and have a full understanding of what ball-and-stick model is all about.

A specific kind of molecular model called the ball-and-stick model uses balls to represent atoms and sticks to represent the connections connecting them to depict the three-dimensional structure of molecules.

It is a physical model that enables researchers to see the configuration of atoms within a molecule and comprehend their interrelationships.

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James Ensor's Christ's Entry into Brussels (1889) was painted as a sarcastic response to Rodin's "Burghers of Calais." Therefore, the correct answer is Rodin's "Burghers of Calais."

"Burghers of Calais" is a sculpture by Auguste Rodin that depicts six burghers of Calais, a city in France, who offered their lives to save their fellow citizens during the Hundred Years' War.

The sculpture is a representation of heroism, sacrifice, and patriotism. James Ensor, however, was critical of this type of heroic representation and sought to challenge it through his own art.

In "Christ's Entry into Brussels," Ensor depicts a grotesque and chaotic scene of Christ's triumphant entry into the city. The painting is a satirical commentary on the hypocrisy of organized religion and the political corruption in Brussels at the time. It can be seen as a direct response to the heroism and idealism of Rodin's "Burghers of Calais."

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The total pressure in the container will become 4.02 atm when 0.240 mol of a third gas is added.

In accordance with Dalton's law of partial pressures:

P_total = P_A + P_B = 0.352 atm + 0.715 atm = 1.067 atm

When 0.240 mol of a third gas is further introduced, the overall number of moles of gas in the vessel jumps to:

n_total = n_A + n_B + n_3 = (P_A * V)/(R * T) + (P_B * V)/(R * T) + n_3

for no change in volume

n_total = (P_total * V)/(R * T) + n_3

P_total = (n_total - n_3) * (R * T)/V

Substituting the given value:

P_total = (0.352 + 0.715 + 0.240) * (0.0821 * (25 + 273.15))/7.95

P_total = 1.307 x 3.079

P_total =  4.024 atm

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[tex]16H_{1}[/tex]+ + [tex]2Cr_{2O_{72} }[/tex]- + [tex]3C_{2H_{60} }[/tex] →4 [tex]Cr_{3}[/tex]+ + 3 [tex]CO_{2}[/tex] + 11 [tex]H_{2} O[/tex] is balanced equation. 3 is the coefficient in front of carbon dioxide in the balanced chemical equation

Define an equation with a balanced formula.

A balanced chemical equation, in which the masses of the reactants and products are equal, contains the same amount of atoms of each element on both sides of the equation.

Varied elements' atom counts in the reactants and products of unbalanced chemical equations are varied. Because the coefficients become the powers of the concentrations of the products and reactants, a balanced equation is crucial when utilizing a constant. The constant is wrong if the equation is unbalanced.

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The temperature of the ideal gas sample is approximately 358.4 degrees Celsius

The ideal gas law is expressed as:

PV = nRT

Where P is pressure, V is volume, n is the amount of substance, T is temperature and R is the ideal gas constant ( 0.08206 Latm/molK )

Given that:

Amount of gas n = 3.24 mol

Pressure P = 2.19 atm

Volume of the gas V = 76.67 L

Temperature T = ?

To determine the temperature, substitute the values into the equation above.

PV = nRT

T = PV / nR

T = ( 2.19 atm × 76.67 L ) / ( 3.24 mol × 0.08206 Latm/molK )

T = 167.9073 / 0.2658744

T = 631.5286 K

Convert to degrees Celsius

T = ( 631.529 − 273.15 ) °C

T = 358.4°C

Therefore, the temperature is 358.4 degrees Celsius.

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