The temperature of a sample of nitrogen dioxide
at a fixed volume is changed, causing a change in
pressure from 732.7 kPa to 238.5 kPa. If its new
temperature is 934 K, what was its original
temperature in kelvins?

Answers

Answer 1

The original temperature (K) of the sample of nitrogen dioxide at a fixed volume is 2,873.33K.

How to calculate temperature?

The temperature of a substance at a fixed volume can be calculated using the following formula;

Pa/Ta = Pb/Tb

Where;

Pa and Ta = initial pressure and temperature respectivelyPb and Tb = final pressure and temperature respectively

According to this question, the temperature of a sample of nitrogen dioxide at a fixed volume is changed, causing a change in pressure from 732.7 kPa to 238.5 kPa. The temperature can be calculated as follows:

732.7/Ta = 238.5/934

0.255Ta = 732.7

Ta = 2,873.33K

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Related Questions

Which tool likely made these marks?
hammer
file
saw
O screwdriver

Answers

Answer: Saw

Explanation:

Pretty obvious

Answer:

Which tool likely made these marks?

O hammer

O file

O saw

O screwdriver

Explanation:

You're welcome.

2 term Symbol: What information do Explain the you get from the formula Caca3​

Answers

The chemical formula CaCO₃ tells us the type and number of atoms present in the chemical compound.

What is chemical formula?

Chemical or molecular formula is a notation indicating the number of atoms of each element present in one molecule of a substance.

Chemical formula is a way of presenting information about the chemical proportions of atoms that constitute a particular chemical compound or molecule, using chemical element symbols, numbers.

According to this question, the chemical formula of calcium carbonate is given: CaCO₃. This chemical formula tells us that it contains calcium, carbon and oxygen atoms in the proportion 1:1:3.

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What mass of HI should be present in 0.200L of solution to obtain a solution with each of the following pH's?
pH=1.20
pH=1.75
pH=2.85

Answers

The mass of HI should be present in 0.200l of solution to obtain a solution with pH value's,

(a) pH value is 1.20 the mass is 1.08g

(b) pH value is 1.75 the mass is 0.0066g

(c) pH value is 2.85 the mass is 0.00012g

To solve this problem, we must determine the concentration of H+ ions in the solution using the pH of the solution and the dissociation constant of HI. The concentration of HI and the mass of HI required to make the solution may then be calculated.

The dissociation reaction for HI is:

HI(aq) ↔ H+(aq) + I-(aq)

The dissociation constant, Ka, for this reaction, is:

Ka = [H+][I-]/[HI]

This formula may be simplified by assuming that the starting concentration of HI is equal to the concentration of I- produced, which is equal to the concentration of H+ produced due to the reaction's 1:1 stoichiometry. This results in:

Ka = [H+]^2/[HI]

Solving for [H+], we get:

[H+] = sqrt(Ka*[HI])

Taking the negative log of both sides gives us the pH of the solution:

pH = -log[H+] = -log(sqrt(Ka*[HI]))

pH= -0.5*log(Ka) - 0.5*log([HI])

Rearranging this equation, we get:

[HI] = 10^(-(pH + 0.5*log(Ka)))/V

where V is the volume of the solution.

Now we can calculate the mass of HI required for each pH:

(a) For pH = 1.20:

Ka for HI is 1.3 x 10^-10. Substituting this value into the equation above, we get:

[HI] = 10^(-(1.20 + 0.5*log(1.3 x 10^-10)))/0.200L ≈ 0.0042 M

The mass of HI required is:

mass = concentration x volume x molar mass

     = 0.0042 mol/L x 0.200 L x 127.91 g/mol

     ≈ 1.08 g

Therefore, approximately 1.08 grams of HI is required to prepare a solution with a pH of 1.20.

(b) For pH = 1.75:

[HI] = 10^(-(1.75 + 0.5*log(1.3 x 10^-10)))/0.200L ≈ 0.00026 M

mass = 0.00026 mol/L x 0.200 L x 127.91 g/mol ≈ 0.0066 g

Therefore, approximately 0.0066 grams of HI is required to prepare a solution with a pH of 1.75.

(c) For pH = 2.85:

[HI] = 10^(-(2.85 + 0.5*log(1.3 x 10^-10)))/0.200L ≈ 0.0000047 M

mass = 0.0000047 mol/L x 0.200 L x 127.91 g/mol ≈ 0.00012 g

Therefore, approximately 0.00012 grams of HI is required to prepare a solution with a pH of 2.85.

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the co2 produced during cellular respiration can react with water to form the acid carbonic acid. thus, one can measure the rate of cellular respiration by using the ph indicator phenolphthalein. in procedure 12.3, what color is the solution expected to be after the ph indicator is first added? according to the experimental protocol, how should the naoh be added and how much should be added to the solution?

Answers

In procedure 12.3, when the pH indicator phenolphthalein is first added, the solution is expected to be colorless. This is because phenolphthalein is a colorless compound in acidic solutions and only turns pink or red in basic solutions.

To measure the rate of cellular respiration using phenolphthalein, we need to add a small amount of NaOH to the solution after adding the pH indicator. The NaOH will react with the carbonic acid produced by the cellular respiration, increasing the pH of the solution and causing the phenolphthalein to turn pink or red.
According to the experimental protocol, we should add 1-2 drops of NaOH at a time while monitoring the color change of the solution. We should continue adding NaOH until the solution turns pink or red, indicating that the pH has become basic. However, we should be careful not to add too much NaOH, as this could cause the pH to become too basic and interfere with the accuracy of our measurements.
Overall, by using phenolphthalein as a pH indicator and carefully adding NaOH, we can accurately measure the rate of cellular respiration and better understand the metabolic processes occurring within living organisms.

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choose the compound that should have the highest melting point according to the ionic bonding model. group of answer choices cao srcl2 ki cas

Answers

The compound with the highest melting point according to the ionic bonding model would be CaO (calcium oxide).

Ionic bonding occurs between atoms with a large difference in electronegativity, resulting in the transfer of electrons from the metal to the non-metal. In CaO, calcium (a metal) loses two electrons to oxygen (a non-metal), resulting in the formation of Ca2+ and O2- ions. These ions are held together by strong electrostatic forces, forming an ionic lattice structure.The strength of the electrostatic forces between the ions is directly related to the size of the charges on the ions and the distance between them. Ca2+ has a larger charge than the other cations listed (Sr2+, K+) and O2- has a smaller radius than the other anions listed (Cl-, S2-), meaning the electrostatic forces between Ca2+ and O2- are stronger.This results in a higher melting point for CaO as more energy is required to break the strong electrostatic forces holding the ions together. In addition, CaO has a higher lattice energy (the energy required to separate a mole of a solid ionic compound into its gaseous ions) than the other compounds listed, further contributing to its higher melting point.

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For ungrouped binary data, explain why when # is near 1 , residuals are necessarily 1< either small and positive or large and negative. What happens when %; is near O?

Answers

For ungrouped binary data, when the proportion (#) is near 1, residuals are necessarily either small and positive or large and negative. This is because binary data can only take on two values, such as 0 and 1. When the proportion is near 1, it means that most of the data points are positive (1), and only a few are negative (0).

In this case, the residuals will be small and positive for the data points close to 1, as their predicted values are close to the actual values. However, the residuals for the data points close to 0 will be large and negative, as their predicted values are far from the actual values.

On the other hand, when the proportion (%) is near 0, it means that most of the data points are negative (0), and only a few are positive (1). In this case, the residuals will be small and negative for the data points close to 0, as their predicted values are close to the actual values. However, the residuals for the data points close to 1 will be large and positive, as their predicted values are far from the actual values.

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How many bonds can a hydrogen atom form?

Answers

A Hydrogen atom can form 1 bond

Calculate the molar solubility and the solubility in g/L of each salt at 25 degreeC: (a) PbF2 Ksp = 4. 0 x 10^-8 ______ x 10^___ M ______ g/L (b) Ag2C03 Ksp = 8. 1 x 10^-12 ____ x 10^____ M ______ x 10^_____ g/L (c) Bi2S3 Ksp = 1. 6 x 10-72 ______ x 10^____ M _____ x 10^_____ g/L Enter all of your answers in scientific notation except the solubility of (a)

Answers

The Molar solubility and the solubility of each salt at 25°C.

(a) PbF₂ Ksp = 4.0 x 10⁻⁸ ,  1.8 x 10⁻⁷ M, 4.41 x 10⁻⁵  g/L

(b) Ag₂CO₃ Ksp = 8.1 x 10⁻¹²,  1.2 x 10⁻⁴ M,  0.0398 g/L

(c) Bi₂S₃ Ksp = 1.6 x 10⁻⁷² , 3.2 x 10⁻¹⁶M,  1.65 x 10⁻¹³ g/L

(a) PbF₂:

Ksp = [Pb₂+][F-]²

Let x be the molar solubility of PbF₂. Then, [Pb2+] = x and [F-] = 2x. Substituting into the Ksp expression and solving for x:

4.0 x 10⁻⁸ = x*(2x)²

x = 1.8 x 10⁻⁷ M

To convert to g/L, we need to multiply by the molar mass of PbF₂ (245.2 g/mol):

solubility = 1.8 x 10^-7 * 245.2 = 4.41 x 10⁻⁵ g/L

(b) Ag₂CO₃:

Ksp = [Ag+]²[CO₃²⁻]

Let x be the molar solubility of Ag₂CO₃. Then, [Ag+] = 2x and [CO₃²⁻] = x. Substituting into the Ksp expression and solving for x:

8.1 x 10⁻¹² = (2x)² * x

x = 1.2 x 10⁻⁴ M

To convert to g/L, we need to multiply by the molar mass of Ag2CO3 (331.8 g/mol):

solubility = 1.2 x 10⁻⁴ * 331.8 = 0.0398 g/L

(c) Bi₂S₃:

Ksp = [Bi³⁺]²[S²⁻]³

Let x be the molar solubility of Bi₂S₃. Then, [Bi3+] = 2x and [S2-] = 3x. Substituting into the Ksp expression and solving for x:

1.6 x 10⁻⁷² = (2x)²*(3x)³

x = 3.2 x 10⁻¹⁶

To convert to g/L, we need to multiply by the molar mass of Bi₂S₃ (514.2 g/mol):

solubility = 3.2 x 10⁻¹⁶ * 514.2 = 1.65 x 10⁻¹³ g/L

In summary, using the suitable Ksp formula and solving for the unknown variable, we can compute the molar solubility and solubility in g/L of salt at a particular temperature. The molar solubility is represented in M units, but the solubility in g/L is calculated by multiplying the molar solubility by the salt's molar mass. The Ksp value indicates the salt's dissolving equilibrium constant and gives information on the relative solubility of various salts under the same circumstances.

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A 0. 001 in. BCC iron foil is used to separate a high hydrogen gas from a low hydrogen gas at 650 °C. 5 ×108 H atoms/cm3 are in equilibrium on one side of the foil, and 2 × 103 H atoms/cm3 are in equilibrium on the other side. Determine (a) the concentration gradient of hydrogen; and (b) the flux of hydrogen through the foil

Answers

The negative sign indicates that the concentration gradient is in the direction of high to low hydrogen concentration. The flux of hydrogen through the foil is 4.3 × [tex]10^5[/tex] atoms/([tex]cm^2.s[/tex]) from the high hydrogen gas to the low hydrogen gas.

J = -D (dC/dx)

a) The concentration gradient of hydrogen can be calculated as follows:

dC/dx = (C2 - C1)/x

dC/dx = (2 × 10³ - 5 × [tex]10^8[/tex])/(0.001 × 2.54 × [tex]10^{-4}[/tex]) = -7.8 × [tex]10^{14}[/tex] atoms/[tex]cm^4[/tex]

(b) The flux of hydrogen through the foil can be calculated using Fick's first law:

J = -D (dC/dx)

D = D0 exp(-Q/RT)

D = 1.6 ×[tex]10^{-6}[/tex]exp(-44,200/8.31/923) = 5.5 × 10^-10 [tex]cm^2/s[/tex]

Substituting the calculated concentration gradient, we get:

J = -D (dC/dx) = -5.5 × [tex]10^{-10}[/tex] × (-7.8 × [tex]10^{14}[/tex]) = 4.3 × [tex]10^5[/tex] atoms/([tex]cm^2.s[/tex])

Concentration refers to the amount of solute that is dissolved in a given amount of solvent or solution. It is an essential concept in chemistry and plays a vital role in many processes such as synthesis, reaction, and separation. The concentration of a solution can affect its properties and behavior. For example, a more concentrated solution may have a higher boiling point or freezing point than a less concentrated one.

There are several ways to express the concentration of a solution, including molarity, molality, mass percent, mole fraction, and parts per million (ppm). Molarity is the most commonly used unit and is defined as the number of moles of solute dissolved per liter of solution. Molality is another unit that measures the number of moles of solute per kilogram of solvent.

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Consider a monatomic ion that has a charge of 1+ and the electron configuration of [Kr]4d105s2.
Write the element symbol and charge for this monatomic ion:
How many unpaired electrons are there in the ground state of this ion?

Answers

The monatomic ion with a charge of 1+ and the electron configuration of [Kr]4d105s2 is the ion of silver, Ag+. There is one unpaired electron in the ground state of this ion.

The electron configuration [Kr]4d105s2 corresponds to the neutral atom of silver (Ag). When silver loses one electron to form a 1+ ion, the electron is removed from the 5s orbital, leaving the ion with the electron configuration [Kr]4d105s1. The 4d and 5s orbitals are close in energy, so there is a possibility for one of the unpaired electrons in the 4d orbital to be promoted to the 5s orbital, resulting in a fully filled 4d subshell and one unpaired electron in the 5s orbital. In this case, since only one electron is removed from the neutral atom, there will be one unpaired electron in the ground state of the Ag+ ion.

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A sample of brass with a mass of 28. 75 grams changes from an initial temperature of 19. 8°C

to a final temperature of 78. 4°C. Calculate the change in thermal energy, and state whether

heat was gained or lost

Answers

The change in thermal energy for this process is 3,097.26 J and the brass sample gained heat.

To calculate the change in thermal energy, we need to use the equation:

[tex]$\Delta Q = mC \Delta T$[/tex]

where:

[tex]$\Delta Q$[/tex]is the change in thermal energy

[tex]$m$[/tex]is the mass of the sample

[tex]$C$[/tex] is the specific heat capacity of brass

[tex]$\Delta T$[/tex] is the change in temperature

The specific heat capacity of brass is typically around 0.38 J/g°C.

Plugging in the given values, we get:

[tex]$\Delta Q = (28.75 \text{ g}) \times (0.38 \text{ J/g°C}) \times (78.4°C - 19.8°C) = 3,097.26 \text{ J}$[/tex]

Since the temperature increased, the sample gained thermal energy. Therefore, the change in thermal energy for this process is 3,097.26 J and the brass sample gained heat.

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an investigator is studying the reduction of rhodium ions in solution. this is a new process and data for the half-cell potential is not available. the standard cell potential for the following reaction is 1.71 v. what is the standard reduction potential for the rh4 /rh3 couple?

Answers

The standard reduction potential for the Rh4+/Rh3+ couple is dependent on the concentration ratio of Rh3+ to Rh4+, which can be determined experimentally. E° = 1.71 V + 0.059 V * ln([Rh3+]/[Rh4+])

To determine the standard reduction potential for the rhodium ions (Rh4+/Rh3+), we can use the Nernst equation, which relates the standard reduction potential to the half-cell potential:


Ecell = E°cell - (RT/nF)ln(Q)
Where E°cell is the standard cell potential, R is the gas constant, T is the temperature, n is the number of electrons transferred in the reaction, F is the Faraday constant, and Q is the reaction quotient.
For the reduction of Rh4+ to Rh3+ in solution, the balanced equation is:
Rh4+ + e- → Rh3+
The number of electrons transferred (n) is 1. The reaction quotient (Q) can be expressed as the concentration of Rh4+ over the concentration of Rh3+:
Q = [Rh3+]/[Rh4+]
Since the investigator is studying a new process and data for the half-cell potential is not available, we can assume that the half-cell potential for the reduction of Rh4+ to Rh3+ is equal to the standard reduction potential for the couple (E°).
Therefore, we can rearrange the Nernst equation and solve for E°:
E° = Ecell + (RT/nF)ln(Q)
Substituting the given values, we get:
E° = 1.71 V + (0.0257 V/K)(298 K)/1 * ln([Rh3+]/[Rh4+])
Simplifying, we get:
E° = 1.71 V + 0.059 V * ln([Rh3+]/[Rh4+])
Thus, the standard reduction potential for the Rh4+/Rh3+ couple is dependent on the concentration ratio of Rh3+ to Rh4+, which can be determined experimentally.

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Physical, Chemical, or Therapeutic Incompatibility?:
Synergism between propofol and alcohol.

Answers

In this case, propofol and alcohol both have depressant effects on the central nervous system, which means that when taken together, their combined effects are more potent than if they were taken separately.

The synergism between propofol and alcohol can be classified as a type of chemical incompatibility. This is because when these two substances are combined, they can have a greater effect than if they were taken separately, potentially leading to dangerous interactions and adverse effects.

However, it is important to note that this chemical incompatibility can also lead to physical and therapeutic incompatibility, as the combined effects of propofol and alcohol can cause physical symptoms and may not be suitable for certain therapeutic applications.

In this case, propofol and alcohol both have depressant effects on the central nervous system, which means that when taken together, their combined effects are more potent than if they were taken separately. This can lead to increased sedation, respiratory depression, and other potential risks.

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when two ions from an ionic bond, what type of compound is formed?

Answers

Answer:

ionic compound

Explanation:

When two ions come together to form a compound, they form an ionic compound. Ionic compounds are made up of positively charged ions (cations) and negatively charged ions (anions) that are held together by electrostatic attraction.

For example, sodium chloride (NaCl) is an ionic compound formed when a sodium ion (Na+) and a chloride ion (Cl-) come together. The sodium ion loses an electron to become a positively charged ion, while the chloride ion gains an electron to become a negatively charged ion. The oppositely charged ions attract each other, forming a crystal lattice structure.

Other examples of ionic compounds include magnesium oxide (MgO), potassium iodide (KI), and calcium carbonate (CaCO3).

select the compound(s) can undergo an aldol addition reaction in the presence of aqueous sodium hydroxide? question 18 options: 2,2-bromopentanal pentanal 2-methyl pentanal 3-chloropentanal

Answers

Option 1 is correct compound. 2-bromopentanal and pentanal do not have alpha-hydrogens and therefore cannot undergo aldol addition reaction in aqueous sodium hydroxide.

Out of the given options, only 2-methyl pentanal and 3-chloropentanal can undergo aldol addition reaction in the presence of aqueous sodium hydroxide. This is because both of these compounds have alpha-hydrogens (hydrogens attached to the carbon adjacent to the carbonyl group), which are necessary for the aldol reaction to occur.

Aldol condensations play an important role in the creation of organic compounds because they provide a dependable way to create carbon-carbon bonds. For instance, the Robinson annulation reaction sequence results in aldol condensation, and the Wieland-Miescher ketone product is an essential component in a variety of chemical synthesis processes.

In the aldol reaction between 2-bromopentanal pentanal  and a 2-methyl pentanal, the mechanism for the aldol addition product and the aldol condensation product is described .

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The complete question is

select the compound(s) can undergo an aldol addition reaction in the presence of aqueous sodium hydroxide? options: 1. ,2-bromopentanal pentanal option 2.  2-methyl pentanal option 3. 3-chloropentanal

the protein content of most plasma membranes is, on average, about 50p% by weight. myelin has a protein content of about 18% , whereas the internal membranes of mitochondria may be composed of 75u% protein.label the membrane proteins on the diagram.

Answers

 The plasma membrane is the outer layer of a cell, composed of about 50% protein by weight. The proteins within the plasma membrane are crucial for various functions, such as transportation of molecules, signal transduction, and cell adhesion. These membrane proteins can be integral spanning the entire membrane or peripheral associated with the membrane surface.

The regarding the protein content in various membranes, including plasma membranes, myelin, and the internal membranes of mitochondria. Unfortunately, I cannot label a diagram, but I can provide an explanation for each membrane type. Myelin is an insulating layer that covers the axons of nerve cells, composed of about 18% protein by weight. The main proteins in myelin are myelin basic protein (MBP) and myelin proteolipid protein (PLP). Myelin helps increase the speed of nerve impulse transmission by preventing signal loss through insulation. Mitochondrial Internal Membrane The internal membranes of mitochondria, known as the cristae, are composed of about 75% protein by weight. These proteins are vital for the production of energy in the cell through a process called oxidative phosphorylation. Key proteins include electron transport chain complexes and ATP synthase, which facilitate the generation of ATP, the cell's main energy currency. I hope this information helps you understand the protein content and significance of these different membrane types.

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what can be said about the favorability of the overall reaction? view available hint(s)for part b what can be said about the favorability of the overall reaction? this reaction is thermodynamically favorable. this reaction is thermodynamically neutral. this reaction is thermodynamically unfavorable. there is not enough information to determine thermodynamic favorableness.

Answers

The thermodynamic favorableness of a reaction can be determined by looking at the sign of the ΔG value. If the ΔG value is negative, the reaction is thermodynamically favorable, meaning that the products are more stable than the reactants.

If the ΔG value is positive, the reaction is thermodynamically unfavorable, meaning that the products are less stable than the reactants. If the ΔG value is zero, the reaction is thermodynamically neutral, meaning that the reactants and products are equally as stable.

Without knowing the ΔG value of a reaction, it is impossible to determine whether the reaction is thermodynamically favorable, unfavorable, or neutral. Knowing the ΔG value is important because it allows us to determine whether a reaction will occur spontaneously or not.

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If 3.28g of a gas occupies a volume of 6.22 liters at a pressure of 845mmHg and a temperature of 378k

A) how many moles of gas exist in the container?
B) what is the molar mass of the gas?

SHOW YOUR WORK!!!!

Answers

0.22 moles of gas exist in the container and the molar mass of the gas is 15g/mol.

The Ideal gas law is the equation of state of a hypothetical ideal gas. It is a good approximation to the behaviour of many gases under many conditions, although it has several limitations. The ideal gas equation can be written as

                                PV = nRT

where,

P = Pressure

V = Volume

T = Temperature

n = number of moles

Given,

Mass = 3.28g

Volume = 6.22 L

Temperature = 378K

Pressure = 845 mm Hg

PV = nRT

845 × 6.22 = n × 62.36 × 378

number of moles = 0.22 moles

Moles = mass / molar mass

Molar mass = mass / moles

= 3.28 / 0.22

= 15 g/mol

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5. Commercial airplanes have a cruising altitude between 9000 m and 12,000 m. At this altitude, air pressure is less than 0.3 atm. How has technology made flying at this altitude safe?

Answers

Commercial airplanes have a cruising altitude between 9000 m and 12,000 m. At this altitude, air pressure is less than 0.3 atm. Technology has made flying at this altitude safe by air pressurization systems.

Pressurization systems constantly pump fresh, outside air into the fuselage. To control the interior pressure, and allow old, stinky air to exit, there is a motorized door called an outflow valve located near the tail of the aircraft.  Larger aircraft often have two outflow valves.

The valves are automatically controlled by the aircraft’s pressurization system. If higher pressure is needed inside the cabin, the door closes. To reduce cabin pressure, the door slowly opens, allowing more air to escape. It’s one of the simplest systems on an aircraft.

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Please help on question 9

Answers

Ethanol (structure b) would have a higher boiling point because it able to form more hydrogen bonds than diethyl ether (structure a)

How does polarity affect boiling point?

Through its effect on intermolecular forces, polarity influences boiling point. Intermolecular forces, such as those that govern a substance's boiling point, are the attracting or repulsive interactions that exist between molecules.

Oxygen molecule has London dispersion forceswater has hydrogen bondsHBr has dipole interaction

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For each of the following pairs, predict which substance possesses the larger entropy per mole.

PART A:

Compare 1 mol of NO(g) at 300 ∘C, 0. 01 atm and 1 mol of NO2(g) at 300 ∘C, 0. 01 atm.

When comparing NO(g)and NO2(g), one mole of________ at 300 ∘Cand 0. 01 atm possesses the larger entropy per mole.

at 300 ∘C and 0. 01 atm possesses the larger entropy per mole.

This is best explained because it _______________

choices: NO(g), occupies a larger volume, NO2(g), has more freedom of movement in aq solution & is more complex molecule with more vibrational degrees of freedom

PART B:

Compare 1 mol of H2O(g) at 100 ∘C, 1 atm and 1 mol of H2O(l) at 100 ∘C, 1 atm.

PART C:

Compare 0. 5 mol of O2(g) at 298 K, 20-L volume and 0. 5 CH4(g) at 298 K, 20-L volume.

PART D

Compare 100 g Na2CO3(s) at 30 ∘C and 100 g Na2CO3(aq) at 30 ∘C

Answers

When comparing NO(g) and NO₂(g) at 300 ∘C and 0.01 atm, one mole of NO₂(g) possesses the larger entropy per mole. When comparing H₂O(g) and H₂O(l) at 100 ∘C and 1 atm, one mole of H₂O(g) possesses the larger entropy per mole. When comparing O₂(g) and CH₄(g) at 298 K and 20-L volume, one mole of O₂(g) possesses the larger entropy per mole. When comparing Na₂CO₃(s) and Na₂CO₃(aq) at 30 ∘C, they have the same entropy per mole .

PART A: When comparing NO(g) and NO₂(g) at 300 ∘C and 0.01 atm, one mole of NO₂(g) possesses the larger entropy per mole.

This is best explained because NO₂(g) is a more complex molecule than NO(g) with more vibrational degrees of freedom. This means that NO₂(g) has more ways in which it can store energy, leading to a higher entropy.

PART B: When comparing H₂O(g) and H₂O(l) at 100 ∘C and 1 atm, one mole of H₂O(g) possesses the larger entropy per mole.

This is because in the gaseous state, H₂O molecules have more freedom of movement and can occupy a larger volume, leading to a higher entropy.

PART C: When comparing O₂(g) and CH₄(g) at 298 K and 20-L volume, one mole of O₂(g) possesses the larger entropy per mole. This is because O₂(g) is a diatomic molecule with more degrees of freedom than CH₄(g), which has a more complex structure with fewer degrees of freedom.

PART D: When comparing Na₂CO₃(s) and Na₂CO₃(aq) at 30 ∘C, they have the same entropy per mole since the state of matter (solid or aqueous) does not affect the entropy of a substance.

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a 50.0 ml sample of 0.200 m sodium hydroxide is titrated with 0.200 m nitric acid. calculate the ph in the titration after the addition of 60.0 ml of 0.200 mhno3 . express your answer to two decimal places.

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The pH value in the titration after the addition of 60.0 ml of 0.200 m HNO₃ is 1.74. At the end point, all the base has reacted with the acid and the solution is neutral.

To solve this problem, we need to use the concept of titration and the equation for the reaction between sodium hydroxide (NaOH) and nitric acid (HNO₃):
NaOH + HNO₃ → NaNO₃ + H₂O
In this reaction, NaOH is a base and HNO₃ is an acid. During titration, we add the acid slowly to the base until the reaction is complete.
We can use the equation:
moles of NaOH = moles of HNO₃
to calculate the amount of HNO₃ required to react with the NaOH in the sample. We can then use the remaining amount of HNO₃ added to the solution after the end point to calculate the pH.
First, let's calculate the number of moles of NaOH in the sample:
moles of NaOH = concentration x volume
moles of NaOH = 0.200 M x 0.0500 L
moles of NaOH = 0.0100 mol
Since the molar ratio of NaOH to HNO₃ is 1:1, we know that we need 0.0100 mol of HNO₃ to react completely with the NaOH. Let's see how much HNO₃ we added to the solution after 60.0 ml:
moles of HNO₃ = concentration x volume
moles of HNO₃ = 0.200 M x 0.0600 L
moles of HNO₃ = 0.0120 mol
Since we only needed 0.0100 mol of HNO₃ to react with the NaOH, we have 0.0020 mol of HNO₃ left in the solution. To calculate the pH, we need to find the concentration of H⁺ ions in the solution. This can be done using the equation:
[H⁺] = moles of HNO₃ left / total volume of solution
Total volume of solution = volume of NaOH + volume of HNO₃ added
Total volume of solution = 0.0500 L + 0.0600 L
Total volume of solution = 0.1100 L
[H⁺] = 0.0020 mol / 0.1100 L
[H⁺] = 0.0182 M
To find the pH, we can use the equation:
pH = -log[H⁺]
pH = -log(0.0182)
pH = 1.74
Therefore, the pH Value in the titration after the addition of 60.0 ml of 0.200 M HNO3 is 1.74.

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sometimes patients are only allowed to have ice chips. ice melts down to 1/2 the volume of water. you gave your patient 220 ml of ice at 9:00 a.m. and 150 ml at 11:00 a.m. he ate all of the ice. how many ml of water did he drink?

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When patients are only allowed to have ice chips, it means that they are not allowed to have any fluids except for the ice chips. Ice melts down to 1/2 the volume of water, which means that if you have 220 ml of ice, it will melt down to 110 ml of water. Similarly, if you have 150 ml of ice, it will melt down to 75 ml of water.

Now, the patient ate all of the ice given to him at both 9:00 a.m. and 11:00 a.m., which means that he consumed all of the water that was in the ice. Therefore, the patient consumed 110 ml + 75 ml = 185 ml of water.
It is important to note that when patients are only allowed to have ice chips, it is because they may have medical conditions that restrict their fluid intake. Therefore, it is crucial to monitor their fluid intake carefully and ensure that they are getting the appropriate amount of fluids they need to maintain their health.
In conclusion, if a patient is only allowed to have ice chips, and they consume 220 ml of ice at 9:00 a.m. and 150 ml of ice at 11:00 a.m., then they will have consumed 185 ml of water. It is important to monitor their fluid intake carefully to ensure they receive the proper amount of fluids to maintain their health.

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Consider the reactionHCl(g)+ NH3(g)NH4Cl(s)Using the standard thermodynamic data in the tables linked above, calculate the equilibrium constant for this reaction at 298.15K.

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To calculate the equilibrium constant for the reaction HCl(g) + [tex]NH_{3}[/tex](g) → [tex]NH_{4}[/tex]Cl(s) at 298.15K using standard thermodynamic data, we need to use the equation:

ΔG° = -RTlnK

where ΔG° is the standard Gibbs free energy change for the reaction, R is the gas constant, T is the temperature in Kelvin, and K is the equilibrium constant.

Using the standard thermodynamic data for the formation of NH4Cl(s), HCl(g), and NH3(g) from their elements, we can calculate the standard Gibbs free energy change for the reaction as follows:

ΔG° = ΔG°f(NH4Cl) - [ΔG°f(HCl) + ΔG°f(NH3)]
ΔG° = (-314.42 kJ/mol) - [(-92.31 kJ/mol) + (-16.45 kJ/mol)]
ΔG° = -205.66 kJ/mol

Substituting the values into the equation above, we get:

-205.66 kJ/mol = -8.314 J/mol-K x 298.15 K x lnK

Solving for K, we get:

lnK = (-205.66 kJ/mol) / (-8.314 J/mol-K x 298.15 K)
lnK = 29.46
K = e^(29.46)
K = 1.29 x 10^12

Therefore, the equilibrium constant for the reaction HCl(g) + NH3(g) → NH4Cl(s) at 298.15K is 1.29 x 10^12.

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under standard conditions (298 k and 1 atm), which statement is true? diamond converts to graphite spontaneously graphite converts to diamond spontaneously none of the above how can the spontaneity of the reaction be reversed? increase the temperature decrease the temperature none of the above

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Under standard conditions (298 K and 1 atm), neither diamond nor graphite spontaneously converts to the other form. The conversion between diamond and graphite is a slow process that requires high temperature and pressure, and cannot occur spontaneously under standard conditions.

To reverse the spontaneity of the reaction, the temperature and/or pressure conditions can be changed. For example, if the temperature is increased to a sufficiently high value and the pressure is also increased, diamond can convert to graphite spontaneously. On the other hand, if the temperature is decreased to a low value and the pressure is also decreased, graphite can convert to diamond spontaneously.
The conversion between diamond and graphite is a type of phase transition, which involves a change in the arrangement of atoms in a material. In general, phase transitions occur when the energy of the system is lowered by changing the arrangement of its constituents. For diamond and graphite, the energy difference between the two forms is relatively small, which makes the conversion between them possible at high temperatures and pressures.
In summary, under standard conditions, neither diamond nor graphite spontaneously converts to the other form. To reverse the spontaneity of the reaction, the temperature and/or pressure conditions can be changed. The conversion between diamond and graphite is a type of phase transition that occurs when the energy of the system is lowered by changing the arrangement of its constituents.

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Consider the Bohr model of the atom. Which transition would correspond to the largest wavelength of light absorbed? Select one: O n=2 to n=6 n=6 to n=10 O n=1 to n=5 O n=6 to n=3 O n=4 to n=1

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The transition that would correspond to the largest wavelength of light absorbed is from n=1 to n=5.

According to the Bohr model, when an electron moves from a lower energy level (n=1) to a higher energy level (n=5), it absorbs light with a specific wavelength.

The larger the difference between the energy levels, the longer the wavelength of light absorbed. In this case, the transition from n=1 to n=5 has the largest difference in energy levels, resulting in the largest wavelength of light absorbed.

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for the following exothermic reversible reaction at equilibrium, how will adding ch4 affect it? c(s) 2h2(g) rightwards harpoon over leftwards harpoon with blank on top ch4(g)

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Adding CH4 to the exothermic reversible reaction at equilibrium will cause the reaction to shift to the left, favoring the formation of C(s) and 2H2(g).

To understand how adding CH4 will affect the equilibrium, we can use Le Chatelier's Principle, which states that if a change is made to a system at equilibrium, the system will adjust itself to counteract that change and restore the equilibrium.

The reaction is represented as:

C(s) + 2H2(g) ⇌ CH4(g)

In this case, the change is an increase in the concentration of CH4. According to Le Chatelier's Principle, the equilibrium will shift in the direction that decreases the concentration of CH4. In this reaction, that means the equilibrium will shift to the left, favoring the formation of the reactants C(s) and 2H2(g). This will continue until a new equilibrium is established.

So, the exothermic reversible reaction that is at equilibrium will move to the left when CH4 is added, favoring the synthesis of C(s) and 2H2(g).

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write the half-reaction and the reaction quotient for the following: a) hydrogen gas electrode: pt(s)|h2(g)|h (aq) b) ag(s)|agcl(s)|cl-(aq) c) pt(s)|fe2 (aq), fe3 (aq) d) cu(s)|cu2 (aq)

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The half-reaction describes the oxidation or reduction process that occurs at an electrode during an electrochemical reaction. It shows the transfer of electrons between the species involved in the reaction. a) Half-reaction: H2(g) → 2H+(aq) + 2e-; Reaction quotient: Q = [H+]^2 / p(H2)

b) Half-reaction: AgCl(s) + e- → Ag(s) + Cl-(aq); Reaction quotient: Q = [Ag+][Cl-] / [AgCl]

c) Half-reaction: Fe2+(aq) → Fe3+(aq) + e-; Reaction quotient: Q = [Fe3+]/[Fe2+]

d) Half-reaction: Cu(s) → Cu2+(aq) + 2e-; Reaction quotient: Q = [Cu2+]/[Cu]

The reaction quotient (Q) is a measure of the relative concentrations of the species involved in the reaction at a particular point in time, and can be used to predict the direction of the reaction (whether it will proceed forward or backward). When Q is equal to the equilibrium constant (K), the reaction is at equilibrium.

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The pOH of an aqueous solution of 0. 522 M acetylsalicylic acid (aspirin), HC9H7O4, is _______The hydroxide ion concentration of an aqueous solution of 0. 522 M hypochlorous acid is ______[OH^-]= _____M

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Acetylsalicylic acid is a weak acid, so we need to use the Ka value to find the concentration of hydroxide ions. Ka of acetylsalicylic acid (HC₉H₇O₄) is 3.27 × 10⁻⁴ The hydroxide ion concentration of an aqueous solution of 0. 522 M hypochlorous acid is M [OH⁻-]= 2.28 ×  10⁻⁴ M

HC₉H₇O₄ + H₂O ⇌ H₃O+ + C₉H₇O₄⁻

Initial: 0.522 M 0 0

Change: -x +x +x

Equilibrium: 0.522 - x x x

Ka = [H₃O+][C₉H₇O₄⁻] / [HC₉H₇O₄]

3.27 × 10⁻⁴ = [tex]x^2[/tex] / (0.522 - x)

As x is very small compared to 0.522, we can approximate the denominator to be 0.522:

3.27 × 10⁻⁴ =[tex]x^2[/tex] / 0.522

x = 1.13 × 10⁻³ M

pOH = -log [OH⁻]

pOH = -log (1.13 × 10⁻³)

pOH = 2.95

The hydroxide ion concentration of an aqueous solution of hypochlorous acid (HOCl) can be found using the Kb value. We can use the fact that Kw = Ka × Kb to find Kb for HOCl.

Kw = Ka × Kb

1.0 × 10⁻¹⁴ = 3.0 × 10⁻⁸ × Kb

Kb = 3.33 × 10⁻⁷

HOCl + H₂O ⇌ H₃O+ + OCl⁻

Initial: 0.522 M 0 0

Change: -x +x +x

Equilibrium: 0.522 - x x x

Kb = [OH⁻][OCl⁻] / [HOCl]

3.33 × 10⁻⁷ = x² / (0.522 - x)

As x is very small compared to 0.522, we can approximate the denominator to be 0.522:

3.33 × 10⁻⁷ =[tex]x^2[/tex]/ 0.522

x = 2.28 ×  10⁻⁴ M

[OH⁻] = 2.28 ×  10⁻⁴ M

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Calculate the molarity of an aqueous solution of NaOH if its pH is measured and found to be 10.00

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The molarity of the aqueous solution of NaOH is 1.0 x [tex]10^-^1^0[/tex] mol/L if its pH is measured and found to be 10.00, as the pH of a solution is a measure of the concentration of hydrogen ions ([H+]) present in the solution.

pH = -log[H+]

10.00 = -log[H+]

[H+] = [tex]10^-^p^H[/tex]

[H+] =     [tex]10^-^1^0[/tex]

[H+] = 1.0 x[tex]10^-^1^0[/tex] mol/L

Since NaOH is a strong base, it dissociates completely in water to give Na+ and OH- ions. The concentration of hydroxide ions ([OH-]) in the solution is equal to the concentration of NaOH:

[OH-] = [NaOH]

The concentration of hydroxide ions to find the molarity of the NaOH solution:

Molarity = moles of solute / volume of solution (in liters)

Molarity = [OH-] = [NaOH]

Molarity = 1.0 x[tex]10^(^-^1^0^)[/tex]mol/L

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