what happens during an exothermic reaction

The most stable contributor to the resonance hybrid formed when anisole undergoes nitration is structure C.

This is because the negative charge is delocalised to both the oxygen atom and the ring structure of the anisole, which makes the molecule more stable. Anisole directs nitration through resonance, where the delocalised electrons present in the aromatic ring contribute to the electron density around the electrophile at centre and facilitate the electrophilic attack.

Thus, structure C, which has the most delocalised electrons and the most resonance stabilisation, will be the most stable contributor to the resonance hybrid formed when anisole undergoes nitration.

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The most stable contributor to the resonance hybrid formed when anisole undergoes

nitration

is Structure A, which is

ortho

-substituted.

This is because anisole is a meta-directing compound, meaning that it prefers the intermediate formed when a nitronium ion attacks the

meta

-position of an aromatic ring.

Structure A is

ortho-substituted

, meaning that it is the most likely intermediate formed when anisole undergoes nitration.

Structure A is the most stable intermediate of the three structures, since it is the only one with two equivalent aromatic carbocations in the

resonance

hybrid.

The reason why anisole is meta-directing is because of the electron-withdrawing oxygen atom, which increases the electron density of the ring opposite to it (the

meta

-position).

This increased electron density of the meta-position makes it more likely for a nitronium ion to attack the meta-position, as opposed to the ortho- and para-positions.

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Answer:

309.1 mL

What is molarity?

The most common way to express solution concentration is molarity (M), which is defined as the amount of solute in moles divided by the volume of solution in litres: M = moles of solute/litres of solution.

To calculate volume required for the solution, we first must find number of moles (n) required. To do this, we can divide mass in grams, by molar mass. Molar mass can be found on a standard IUPAC Periodic Table (International Union of Pure and Applied Science).

n[(NH₄)₂SO₄] = m/MM = 449.4/[(14.01+1.008×4)×2+32.07+16.00×4]

= 3.4006 mol

Now we have moles of solute, as well as molarity, (from the question - 11M) Thus, we can calculate litres of solution.

Volume = moles ÷ molarity = 3.4006 / 11 = 0.309 L = 309.1 mL

The rate of reaction, time taken to displace all of the water in a measuring cylinder will be greater for lumps of calcium carbonate or powder so that Carbon dioxide can escape.

Gases that do not dissolve in water can be trapped in water using a technique known as displacement of water.

This method necessitates the use of a delivery tube, a measurement cylinder, and a water basin.

We begin by filling a measuring cylinder with water and inverting it in a water basin. The gas is then bubbled into the inverted measurement cylinder through a delivery line. The water level falls as gas bubbles up and fills the inverted measuring cylinder.

The faster the response, the higher the quantity of mass lost. The quantity of mass loss, as demonstrated by the issue, is decreasing over time. It may be deduced from this that the pace at which the product particles are created is slowing and finally stopping.

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We would require 13.5 moles of AgNO₃ to make 6.75 moles of  Cu(NO₃)₂1

The balanced chemical equation for the reaction between silver nitrate (AgNO₃) and copper (Cu) is:

2AgNO₃ + Cu -> Cu(NO₃)₂ + 2Ag

According to the stoichiometry of the reaction, 2 moles of AgNO₃ react with 1 mole of Cu to produce 1 mole of Cu(NO₃)₂ and 2 moles of Ag. Therefore, if 6.75 moles of Cu(NO₃)₂ are produced, the number of moles of Cu that reacted is also 6.75 moles.

To determine the number of moles of AgNO₃ required, we can use the ratio of moles of Cu to moles of AgNO₃ in the balanced equation. This ratio is 1:2, meaning that for every mole of Cu, 2 moles of AgNO₃ are needed.

Therefore, to produce 6.75 moles of Cu(NO₃)₂, we would need 2 × 6.75 = 13.5 moles of AgNO3.

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Considering the ideal gas law, if you have 0.857 moles of gas with a volume of 1.97 L at a pressure of 0.942 atm, the temperature is 26.407 K.

Ideal gas refers to a hypothetical gas composed of molecules that do not attract or repel each other and are approximated by point particles that themselves have no volume.

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T), related by the ideal gas law. This law is an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar gas constant:

P×V = n×R×T

In this case, you know:

Replacing in the ideal gas law:

0.942 atm× 1.97 L = 0.857 moles× 0.082 (atmL)÷(molK)× T

Solving:

(0.942 atm× 1.97 L)÷ (0.857 moles× 0.082 (atmL)÷(molK))= T

26.407 K= T

Finally, the temperature is 26.407 K.

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The amount of solid barium sulfate formed is 1.42g and 7.02 moles of sodium chloride are formed.

When 38.0 mL of 0.160 M barium chloride reacts with 54.0 mL of 0.065 M sodium sulfate, aqueous sodium chloride, and solid barium sulfate form. To determine the number of grams of solid barium sulfate formed, we must first calculate the amount of barium chloride and sodium sulfate used.

The amount of barium chloride used is equal to the product of the molarity and the volume, or 38.0 mL x 0.160 M = 6.08 mmol.

Similarly, the amount of sodium sulfate used is 54.0 mL x 0.065 M = 3.51 mmol.

The reaction between barium chloride and sodium sulfate is an exchange reaction:

BaCl₂(aq) + Na₂SO₄(aq) -> 2NaCl(aq) + BaSO₄(s)

We know that for every mole of barium chloride that is used, one mole of barium sulfate is formed, and for every mole of sodium sulfate used, two moles of sodium chloride are formed.

Therefore, 6.08 mmol of barium sulfate is formed and 7.02 mmol of sodium chloride is formed.

The number of grams of solid barium sulfate formed is equal to the product of the molar mass of barium sulfate and the number of moles of barium sulfate formed.

233.43 g/mol x 6.08 mmol = 1419.25 mg = 1.42g

In conclusion, when 38.0 mL of 0.160 M barium chloride reacts with 54.0 mL of 0.065 M sodium sulfate, 1.42g of solid barium sulfate, and 7.02 mmol of aqueous sodium chloride form.

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0.0279 mole of the excess reactant is left over if 1.98 mole of chlorine react with 1.98 moles of nitrogen.

The mole is defined as the amount of substance of a system which contains as many elementary entities as there are atoms in 0.012 kilogram of carbon 12. It is expressed as "mol". It contains 6.022 X1023 entities like particles, atoms, ions, molecules, etc. of the given substance. It measures the number of atoms, ions, or molecules. It is a standard scientific unit for measuring large quantities of very small entities such as atoms, molecules, or other specified particles. It explains the number of atoms or other particles in a mole is the same for all substances.

If 1. 98 mole of chlorine reacts with 1. 98 moles of nitrogen. then, the we can calculate the amount of reactant left over with the help of the concept for mole.

Moles = 1.98 / 70.9

          = 0.0279

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The correct question is,

3Cl2 + 2N2 → 2N2Cl3

Molar masses:

chlorine gas = 70. 9 g/mole

nitrogen gas = 28. 02 g/mole

dinitrogen trichloride = 134. 27 g/mole

1. 98 mole of chlorine reacts with 1. 98 moles of nitrogen. How many moles of the excess reactant is left over?

the concentration of ClCH2CH2Cl in the vessel 5.80 seconds later would be 1.49 M.

The second-order rate law for the given reaction is:

rate = k[ClCH2CH2Cl]

where k is the rate constant.

We can use the integrated rate law for a second-order reaction to solve for the concentration of ClCH2CH2Cl at a later time:

1/[ClCH2CH2Cl]t = 1/[ClCH2CH2Cl]0 + kt

where [ClCH2CH2Cl]t is the concentration of ClCH2CH2Cl at time t, [ClCH2CH2Cl]0 is the initial concentration, and k is the rate constant.

Plugging in the values given in the problem, we get:

1/[ClCH2CH2Cl]t = 1/1.22 + (0.743 M-1 s-1)(5.80 s)

1/[ClCH2CH2Cl]t = 0.8205

[ClCH2CH2Cl]t = 1.220 M / 0.8205

[ClCH2CH2Cl]t = 1.487 M

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Answer:

D

Explanation:

The difference between a reflected and diffracted sound wave is:

A. A reflected sound wave changes direction as it passes through an opening, whereas a diffracted sound wave bounces back to the place of origin.

Reflection occurs when a sound wave hits a surface and bounces back in the opposite direction. When a sound wave is reflected, it changes direction but does not necessarily change its wavelength or frequency. This is why we can hear echoes in a room with reflective surfaces.

Diffraction, on the other hand, occurs when a sound wave passes through an opening or around an obstacle and changes direction. When a sound wave is diffracted, it spreads out and changes its wavelength and frequency. This is why we can hear sound around corners or through a partially open door.

Therefore, option D is the correct answer: A reflected sound wave bounces back to the place of origin, a diffracted sound wave changes direction as it passes through an opening.

The steps in scientific learning are observations, hypotheses, predictions, testing, analysis, and conclusion.

The steps of scientific learning are as follows:

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Every substance has some physical properties and chemical properties. The new substance is solid and the new substance forms into thin flat sheets can be considered as the inference based on the observations.

A physical property is defined as an any property that is measurable, whose value describes a state of a physical system. The changes in the physical properties of a system can be used to describe the changes between momentary states of the substance. These properties are  referred to as observables. Physical properties are not modal properties.

A chemical property is defined as any of a material's properties that becomes evident during or after a chemical reaction that is, any quality that can be established only by changing a substance's chemical identity.

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The calculated value of ∆H will be lower than the actual value due to heat loss. This is because the heat lost to the surroundings is not accounted for in the calculation, resulting in a lower measured temperature change than what would have occurred if no heat was lost.

If there is significant heat loss to the surroundings during a neutralization reaction in an open polystyrene coffee-cup calorimeter, the calculated value of ∆H (enthalpy change) would be lower than the actual value.

The enthalpy change for the reaction is calculated using the following equation: ∆H = q / n

where q is the heat released or absorbed by the reaction, n is the number of moles of the limiting reactant, and ∆H is the enthalpy change per mole of the limiting reactant.

In an open calorimeter, heat can be lost to the surroundings through conduction, convection, and radiation. The extent of heat loss depends on various factors such as the ambient temperature, the specific heat capacity of the surroundings, and the rate of heat transfer.

To minimize heat loss, the experiment should be conducted as quickly as possible, with good stirring to ensure uniform mixing of the reactants, and the calorimeter should be well-insulated to reduce heat loss to the surroundings. If these precautions are not taken, the calculated value of ∆H will be lower than the actual value due to heat loss.

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A. Thiele modulus for the three pellets is 1.78, 2.74, and 6.04 for conversion 3. 4 mm 36.9, 5.1 mm 31.7, and 9.5 mm 25.6 respectively

B. The yield of butadiene using catalyst particles with a diameter of 1. 0 and 7. 0 mm in a reactor of the same length is 0.118 and 0.073 respectively.

C. The Thiele modulus (φ) will rise for all the pellets if both rate constants are increased by a factor of 4, as the reactions speed up.

a) To calculate the Thiele modulus (φ) for each pellet, we can use the following equation:

φ² / (6 x Th) = -(1/2)ln(1-X)

where,

φ is the particle diameter

Th is the Thiele modulus

X is the conversion.

The conversion for the first pellet, which has a 3.4 mm diameter, is 36.9%. Thus,

Th = φ² / [tex](6\times(-1/2)ln(1-X))[/tex]

Th = [tex]3.4^2 / (6\times(-1/2)ln(1-0.369))[/tex]

Th = 1.78

The conversion for the first pellet, which has a 5.1 mm diameter, is 31.7%. Thus,

Th = φ² / [tex](6\times(-1/2)ln(1-X))[/tex]

Th = [tex]5.1^2 / (6\times(-1/2)ln(1-0.317))[/tex]

Th = 2.74

The conversion for the first pellet, which has a 9.5 mm diameter, is 25.6%. Thus,

Th =φ²/ [tex](6\times(-1/2)ln(1-X))[/tex]

Th = [tex]9.5^2 / (6\times(-1/2)ln(1-0.256))[/tex]

Th = 6.04

B) In order to compute the butadiene yield utilizing catalyst particles with diameters of 1.0 and 7.0 mm, we must first know the conversion at the reactor's end.

We may infer from the information provided that the conversion is a diminishing function of pellet size.

Hence, we may interpolate the conversion values for the two particle sizes:

Conversion at 1.0 mm

= 36.9 x (1 - ((1.0 - 3.4) / (9.5 - 3.4))

= 26.6%

Conversion at 7.0 mm

= 25.6 x (1 - ((7.0 - 3.4) / (9.5 - 3.4))

= 16.5%

Thus, we must account for both the conversion and the selectivity when calculating the butadiene yield.

The second reaction's selectivity ([tex]S_2[/tex]) may be represented as

[tex]k_2 / (k_1 + k_2)[/tex] = 0.8 / 1.8

= 0.444 since the reactions occur one after the other.

For the 1.0 mm particle, the yield of butadiene can be calculated as:

Yield

= Conversion x Selectivity

= 0.266 x 0.444

≈ 0.118

For the 7.0 mm particle, the yield of butadiene can be calculated as:

Yield

= Conversion x Selectivity

= 0.165 x 0.444

≈ 0.073

c) The Thiele modulus (φ) will rise for all the pellets if both rate constants are increased by a factor of 4, as the reactions speed up. For the same reactor duration and circumstances, this will lead to a drop in conversion and yield.

Using the Thiele modulus equation from section (a), we can recalculate the conversion and yield using the new rate constants to determine the magnitude of this effect.

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Answer:

Explanation:

need the equation

Explanation:

SPD (s-block elements) are the elements located in the leftmost two columns of the periodic table. These elements include the alkali metals (Group 1A) and the alkaline earth metals (Group 2A). These elements are the most reactive of all elements and are characterized by their low ionization energies, high reactivities and relatively large atomic radii. The s-block elements are the only elements to form covalent bonds with other atoms. They also tend to form ionic bonds with other elements and have a tendency to form multiple bonds with each other. Because of their low ionization energies, they are very reactive, which is why they are often used in industrial applications, such as in the production of fertilizers, explosives, and drugs. The s-block elements are also important in the biochemical processes of living organisms. For example, sodium, potassium and calcium ions are essential in the regulation of the nervous system and muscular contraction. Magnesium is also important for energy production and muscle relaxation. In summary, the s-block elements are the most reactive of all the elements and have many important uses in industry and biochemistry. They are characterized by their low ionization energies, high reactivities, and relatively large atomic radii.

The Cationic molecular ions are more abundant than Anionic molecular ions because capacity of cationic to stabilise unpaired electrons make synthesis of cationic molecular ions less energetically demanding.

The capacity of cationic molecular ions to stabilise unpaired electrons makes the synthesis of cationic molecular ions less energetically demanding. The most advantageous reason for the abundance of cationic molecular ions is that the cosmic ray has so much energy that it is largely unaffected by the comparatively little energy necessary to ionise molecules. There is ample of radiation from a broadband source like the Sun to drive the molecules to high energy states from which they can be triggered and spontaneously emit. As a result, cationic molecular ions outnumber anionic ones.

Because of the low temperature and density of interstellar space, the most prevalent cationic molecular ion is H3+. Because its two electrons are sole valence electrons in the system, this H3+ is also thought to be a major source of cationic molecular ions.

Another reason for their prevalence is their capacity to stabilise unpaired electrons, which makes the production of cationic molecule ions less energetically demanding.

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We may find the size of a typically Regular shaped object by adding those following phases (area of both the bottom of the item and its height).

The volume of the thing is equivalent to the circumference of the bottom (longest times width times height of the object), or L x W x H. In this illustration, the length units we utilised were centimetres. Constant cross - sectional object volume A solid's volume can be calculated by multiplying its length, breadth, and height all at once.

V is equal to l times w. Regular forms feature interior (inner) slopes which are all equal as well as equal sides. Shapes that are irregular include curves and sides of any size and length. Below are a variety of shapes, both regular and infrequent: Pentagon regularly, regularly.

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A portion of the supernatant liquid was filtered and analysed after the precipitate settled. the ksp is 5.0 ×

[tex] {10}^{ - 4} [/tex]

The balanced chemical equation for the dissolution of AB2 in water is:

AB2 (s) ⇌ A2+ (aq) + 2B- (aq)

The solubility product expression for AB2 is:

Ksp = [A2+][B-]

[A2+] = 0.100 M × 0.5000 L / (0.5000 L + 0.5000 L) = 0.0500 M

[B-] = 0.205 M × 0.5000 L / (0.5000 L + 0.5000 L) = 0.1025 M

[A2+]_ppt = [A2+]_initial - [A2+]_filtered

[A2+]_ppt = 0.0500 M - 2.3 × 10M

[A2+]_ppt = 0.0500 M

[B-]_ppt = 2 × [A2+]_ppt

[B-]_ppt = 2 × 0.0500 M

[B-]_ppt = 0.100 M

Now, we can use these values to calculate the Ksp:

Ksp = [A2+][B-]^2

Ksp = (0.0500 M)(0.100 M)

Ksp = 5.0 ×

[tex] {10}^{ - 4} [/tex]

Therefore, the Ksp of AB2 is 5.0 ×

[tex] {10}^{ - 4} [/tex]

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1.3 Liters of a 0. 352M solution of CaSO4 would contain 62. 1g CaSO₄. So, option C is correct answer.

The solution has a concentration of 0.35M, it means that there ar 0.35M moles of CaSO₄ are there is the solution.

So, we can write is as,

Concentration = Moles/volume in Liters, this can be written as,

Concentration = Mass/Molar mass/volume in Liters

The molar mass of CaSO₄ is given to be 135.14 g/mol and the given mass of the compound is 62.1 grams. Now, putting all values,

0.35 = 62.1/(135.14V)

Volume = 1.35 Liters.

So, the volume of the solution required is found to be 1.35L hence option C is correct.

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Answer:

so the mass of sulphur dioxide = 0.208334 × 22.4 L = 4.6666816 L = 4666.6816 ml Therefore the volume of sulphur dioxide is 4666.6816 ml .

Explanation:

Reaction A is the equilibrium between Fe₃+ and SCN- ions to form FeSCN₂+ ions, which results in an orange color solution.

If SCN- is added to the mixture, according to Le Chatelier's principle, the equilibrium will shift to the right to consume the added SCN- ions. This means that more FeSCN₂+ ions will form, resulting in a darker orange color solution, possibly even turning red if enough SCN- is added.

On the other hand, if SCN- is removed from the mixture, the equilibrium will shift to the left to replace the removed SCN- ions. This means that the FeSCN₂+ ions will dissociate to form Fe₃+ and SCN- ions, resulting in a lighter orange or even yellow color solution.

If FeSCN₂+ is added to the mixture, it would not have any significant effect on the color of the solution, as it is already in equilibrium and adding more product (FeSCN₂+ ions) will not shift the equilibrium in any particular direction. Therefore, the color of the solution would remain orange.

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A limiting reagent is a chemical reactant that controls the amount of product produced. The limiting reagent provides the lowest product yield estimated from the available reagents (reactants).

No additional product will develop once the limiting reactant has been completely consumed. The theoretical yield of the reaction is the maximum quantity of product that might be created depending on the limiting reactant. The 'percent yield' is calculated by comparing the actual yield to the theoretical yield..

To calculate the "expected yield" of the product, multiply the reaction equivalents for the limiting reagent by the product's stoichiometric factor. This is the estimated number of millimoles. To get the predicted mass of the product, multiply this value by the MW of the product.

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Answer: 355

Explanation: Given the molecular weights:

M

r

N

a

O

H

=

40

g

m

o

l

M

r

N

a

2

S

O

4

=

142

g

m

o

l

The analogy of the moles will be held constant:

n

N

a

O

H

n

N

a

2

S

O

4

=

2

1

n

N

a

O

H

n

N

a

2

S

O

4

=

2

For each one, substitute:

n

=

m

M

r

Therefore:

n

N

a

O

H

n

N

a

2

S

O

4

=

2

m

N

a

O

H

M

r

N

a

O

H

m

N

a

2

S

O

4

M

r

N

a

2

S

O

4

=

2

200

40

x

142

=

2

200

⋅

142

40

x

=

2

200

⋅

142

=

2

⋅

40

x

x

=

200

⋅

142

2

⋅

40

=

100

⋅

142

40

=

10

⋅

142

4

=

1420

4

=

=

710

2

Given the molecular weights:

M

r

N

a

O

H

=

40

g

m

o

l

M

r

N

a

2

S

O

4

=

142

g

m

o

l

The analogy of the moles will be held constant:

n

N

a

O

H

n

N

a

2

S

O

4

=

2

1

n

N

a

O

H

n

N

a

2

S

O

4

=

2

For each one, substitute:

n

=

m

M

r

Therefore:

n

N

a

O

H

n

N

a

2

S

O

4

=

2

m

N

a

O

H

M

r

N

a

O

H

m

N

a

2

S

O

4

M

r

N

a

2

S

O

4

=

2

200

40

x

142

=

2

200

⋅

142

40

x

=

2

200

⋅

142

=

2

⋅

40

x

x

=

200

⋅

142

2

⋅

40

=

100

⋅

142

40

=

10

⋅

142

4

=

1420

4

=

=

710

2

= 355 GMS

To solve this issue, we must use stoichiometry to calculate the relationship between the production of CO2 and the amount of C3H8.

Using its molar mass, we must first determine the moles of C3H8: C3H8 has a molar mass of 44.11 g/mol, which is calculated by multiplying 3 (12.01 g/mol) by 8 (1.01 g/mol). C3H8 moles are equal to 11.89 g / 44.11 g/mol, or 0.27 mol. Next, we apply the balanced chemical equation's C3H8 to CO2 mole ratio: Three molecules of CO2 are produced from one mol of C3H8. the following will be the moles of CO2 produced: The formula for moles of CO2 is: 0.27 mol C3H8 x (3 mol CO2/1 mol C3H8) = 0.81 mol CO2. Finally, we use its molar mass to translate the moles of CO2 into grammes: Molar mass of CO2 equals 12.01 g/mol plus 2 (16.00 g/mol), or 44.01 g/mol. mass of CO2 = 44.01 g/mol x 0.81 mol CO2 to get 35.64 g CO2. As a result, 11.89 g of C3H8 will result in 35.64 g of CO2.

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Answer:

0.965 grams.

Explanation:

To calculate the mass of Na₂SO4, we can use the formula:

mass = moles x molar mass

First, let's calculate the number of moles of Na₂SO4 in 435.2 mL of 0.0156 M solution:

moles = volume (in L) x molarity

moles = 0.4352 L x 0.0156 mol/L

moles = 0.0067872 mol

The molar mass of Na₂SO4 is 142.04 g/mol, so we can calculate the mass of Na₂SO4 as:

mass = moles x molar mass

mass = 0.0067872 mol x 142.04 g/mol

mass = 0.965 g

Therefore, the mass of Na₂SO4 used to prepare 435.2 mL of 0.0156 M Na₂SO4 solution is 0.965 grams.

Answer: 18.5%

Explanation:

The chemical formula is C₂₀H₂₄N₂.

To determine the empirical formula, we first need to calculate the number of moles of each element in the given compounds:

moles of carbon: 1.321 g CO₂ x (1 mol CO2/44.01 g CO₂) x (1 mol C/1 mol CO₂) = 0.030 mol C

moes of hydrogen: 0.325 g H₂O x (1 mol H₂O/18.015 g H₂O) x (2 mol H/1 mol H₂O) = 0.036 mol H

moles of nitrogen: 0.0421 g N x (1 mol N/14.01 g N) = 0.003 mol N

Next, we can calculate the mole ratio of each element by dividing the number of moles of each element by the smallest number of moles:

C: 0.030 mol C / 0.003 mol N = 10

H: 0.036 mol H / 0.003 mol N = 12

N: 0.003 mol N / 0.003 mol N = 1

Therefore, the empirical formula is C₁₀H₁₂N.

To find the molecular formula, we need to know the molar mass of the compound. We can calculate the molar mass of the empirical formula C₁₀H₁₂N as:

molar mass = (10 x 12.01 g/mol) + (12 x 1.01 g/mol) + (1 x 14.01 g/mol) = 162.22 g/mol

We can then calculate the molecular formula by dividing the molar mass of the compound by the molar mass of the empirical formula:

molecular formula = molar mass of compound / molar mass of empirical formula

= (324 g/mol) / (162.22 g/mol) = 1.998

Since the molecular formula should be a whole number, we can round it to the nearest whole number to get 2.

Therefore, the molecular formula is C20H24N2.

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Answer:

One mole of calcium carbonate reacts with 2 moles of HCl forms one mole of calcium chloride, water and carbon dioxide each. The theoretical yield of calcium chloride from 16.5 g of calcium carbonate is 18.3 g

Explanation:

As a result, when the gas is cooled to a temperature below 45 °C, it will condense into a liquid condition. The pressure affects the precise temperature at which the gas will condense.

The slope of the solid-liquid equilibrium line for carbon dioxide is positive; when pressure is increased, the melting point rises.

Three triple points exist (a). (b) In an atmosphere, monoclinic is more stable than rhombic. (c) Sulfur changes state from liquid to vapour and back to liquid when heated from 80°C to 200°PC.

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