A Sample Of Methane Gas In A Piston Exerts A Pressure Of 1.26 10^3 When The Volume Is 54.3 Cm. When The (2024)

Chemistry High School

Answers

Answer 1

The new gas volume is approximately 24,488 cm³.

To solve this problem, we can use the combined gas law, which relates the initial and final states of a gas sample when pressure, volume, and temperature change while the amount of gas remains constant.

The combined gas law equation is:

(P1 * V1) / T1 = (P2 * V2) / T2

Where:

P1 = Initial pressure

V1 = Initial volume

T1 = Initial temperature (which remains constant)

P2 = Final pressure

V2 = Final volume (what we need to find)

T2 = Final temperature (which remains constant)

We are given:

P1 = 1.26 × 10^3 atm

V1 = 54.3 cm³

P2 = 2.77 atm

Since the temperature remains constant, T1 = T2, we can simplify the equation to:

(P1 * V1) = (P2 * V2)

Now we can plug in the values:

(1.26 × 10^3 atm) * (54.3 cm³) = (2.77 atm) * V2

Solving for V2, we get:

V2 = (1.26 × 10^3 atm * 54.3 cm³) / (2.77 atm)

V2 ≈ 24,488 cm³

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Related Questions

Part I: Kinetic Molecular Theory (KMT) of Gases
Our fundamental understanding of ideal" gases makes the following 5 assumptions.
Describe how each of these assumptions is (or is not!) represented in the simulation.
Assumption of KMT
1. Gas particles are small and are separated by relatively large distances.
Representation in Simulation
2. Gas particles are constantly in random motion.
3. Gas particles undergo elastic collisions (like billiard balls) with each other and the walls of the container.
4. Gas particles are not attracted or repulsed by each other.
4. The average kinetic energy of gas molecules in a sample is proportional to temperature (in K).

Answers

The underlying assumptions of the Kinetic Molecular Theory (KMT) of gases are partially reflected in the simulation.

Due to its depiction of a simulation of individual particles moving freely inside the container, it reflects the earlier idea that the particles of a gas are small and widely separated. The particles exhibit unpredictable velocities and change position with time, representing the idea that the particles of a gas are always in a state of random motion.

The simulation also demonstrates the idea of ​​elastic collisions, as the particles collide with the walls of the container and with each other without any permanent damage. However, neither the ratio of the average kinetic energy to the temperature nor the absence of attractive or repulsive forces between the particles are clearly demonstrated by the simulations.

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what is the freezing point of antifreeze solution created by adding 651 grams of ethylene glycol to 2505 grams of water? kf

Answers

Water's freezing point is 0 °C, the antifreeze solution's freezing point is -7.77 °C.

The freezing point of the antifreeze solution created by adding 651 grams of ethylene glycol to 2505 grams of water depends on the value of kf, which is the freezing point depression constant of the solvent. Without knowing the value of kf, it's impossible to calculate the freezing point. However, we can use the equation ΔT = kf * molality to determine the freezing point depression, where ΔT is the change in freezing point, and molality is the number of moles of solute per kilogram of solvent. This calculation can be used to find the freezing point of the solution. First, determine the molality by dividing the moles of ethylene glycol (651 g / 62.07 g/mol = 10.48 mol) by the mass of water in kg (2505 g = 2.505 kg). This gives a molality of 4.18 mol/kg. Next, calculate the freezing point depression: ΔTf = 1.86 °C/m * 4.18 mol/kg = 7.77 °C. Since water's freezing point is 0 °C, the antifreeze solution's freezing point is -7.77 °C.

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exactly 1 mole of na2so3 contains how many moles of na s and o

Answers

Exactly 1 mole of na2so3 contains

- 1 mole of Na2SO3 contains 2 moles of Na (Na2SO3 → 2Na+)

- 1 mole of Na2SO3 contains 1 mole of S (Na2SO3 → S2-)

- 1 mole of Na2SO3 contains 3 moles of O (Na2SO3 → 3O2-)

In Na2SO3, there are two sodium ions (Na+), one sulfur ion (S2-), and three oxygen ions (O2-). To determine the number of moles of Na, S, and O in 1 mole of Na2SO3, we look at the subscripts in the chemical formula.

For Na2SO3, the subscript 2 indicates that there are 2 moles of Na for every 1 mole of Na2SO3. Therefore, 1 mole of Na2SO3 contains 2 moles of Na.

Similarly, the subscript 1 for S indicates that there is 1 mole of S in 1 mole of Na2SO3.

The subscript 3 for O indicates that there are 3 moles of O for every 1 mole of Na2SO3. Therefore, 1 mole of Na2SO3 contains 3 moles of O.

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determine which of the following pairs of reactants will result in a spontaneous reaction at 25°c. a) sn4 (aq) mg(s) b) cr3 (aq) ni(s) c) zn(s) na (aq)

Answers

As per the given details, Zn has a negative reduction potential (-0.76 V), which indicates that it is more likely to undergo oxidation.

The standard reduction potentials of the constituent elements must be taken into account in order to identify which of the given pairs of reactants will undergo a spontaneous reaction at 25°C.

The standard reduction potential gauges a species' propensity to pick up electrons and go through reduction.

The reduction potentials of the species involved in each reaction can be compared. If the species being reduced has a higher reduction potential than the species being oxidised, which is losing electrons, the reaction will occur spontaneously.

We must contrast the reduction potentials of [tex]Sn^{4+[/tex] and Mg. [tex]Sn^{4+[/tex] (aq) + Mg(s). This has a positive (+0.15 V) reduction potential, indicating a propensity to undergo reduction.

Mg has a positive reduction potential (-2.37 V), which denotes a propensity to be decreased.

Ni(s) + [tex]Cr^{3+[/tex] (aq): [tex]Cr^{3+[/tex] has a positive (+0.74 V) reduction potential, indicating a propensity to be reduced.

Zn(s) + Na+ (aq): Zn has a negative reduction potential (-0.76 V), which indicates that it is more likely to undergo oxidation.

Thus, this can be concluded regarding the given scenario.

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a sample of o2 gas occupies a volume of 344 ml at 25 degrees celsius. if pressure remains constant, what would be the new volume if the temperature changed to:

Answers

The new volume of the O2 gas would be approximately 355 ml if the temperature changed from 25 degrees Celsius to 35 degrees Celsius, assuming the pressure remains constant

Assuming the pressure remains constant, we can use the formula V1/T1 = V2/T2 to find the new volume. Converting 25 degrees Celsius to Kelvin (25 + 273 = 298K), we have:
V1 = 344 ml
T1 = 298K
If the temperature changed to 35 degrees Celsius (35 + 273 = 308K), we can solve for V2:
V1/T1 = V2/T2
344 ml / 298K = V2 / 308K
Solving for V2, we get:
V2 = (344 ml / 298K) * 308K = 355 ml (approximately)
Therefore, the new volume of the O2 gas would be approximately 355 ml if the temperature changed from 25 degrees Celsius to 35 degrees Celsius, assuming the pressure remains constant.
A sample of O2 gas occupies a volume of 344 mL at 25°C. If the pressure remains constant, we can apply Charles's Law to determine the new volume when the temperature changes. Charles's Law states that V1/T1 = V2/T2, where V1 and T1 are the initial volume and temperature, and V2 and T2 are the final volume and temperature. To use this formula, temperatures must be in Kelvin. 25°C is equivalent to 298 K. When the temperature changes to T2, substitute the known values into the equation:
(344 mL / 298 K) = (V2 / T2)
Solve for V2 by multiplying both sides by T2:
V2 = (344 mL / 298 K) × T2
To find the new volume, simply replace T2 with the desired final temperature (in Kelvin) and solve for V2.

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the ph of four different substances is shown below. substance ph shampoo 6 lemon juice 2 tomato juice 4 liquid drain cleaner 14 which substance is closest to being neutral on the ph scale? shampoo lemon juice tomato juice liquid drain cleaner

Answers

The substance closest to being neutral on the pH scale is shampoo, with a pH of 6.

A neutral pH is 7, so substances with a pH below 7 are considered acidic and those above 7 are considered basic. Lemon juice has a pH of 2, which is highly acidic, while tomato juice has a pH of 4, making it slightly acidic. Liquid drain cleaner, on the other hand, has a pH of 14, making it highly basic. Therefore, of the four substances listed, shampoo has the pH closest to neutral. The pH scale ranges from 0 to 14, with 7 being neutral. The four substances mentioned have the following pH levels: shampoo (6), lemon juice (2), tomato juice (4), and liquid drain cleaner (14). Among these substances, shampoo has a pH of 6, which is closest to the neutral pH level of 7. Therefore, shampoo is the substance that is closest to being neutral on the pH scale.

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how many half lives have passed if 87.5% of a substance has decomposed? how mamy if 99.999% has decomposed?

Answers

3 half-lives have passed for 87.5% decomposition, and 17 half-lives for 99.999% decomposition.

To determine the number of half-lives that have passed, you can use the formula N = (log N0 - log N)/log 2, where N0 is the initial amount, N is the remaining amount, and log is the logarithmic function. For 87.5% decomposition, the remaining amount is 12.5% or 0.125N0, which means that N/N0 = 0.125. Plugging this into the formula, you get N = 3. For 99.999% decomposition, the remaining amount is 0.00001N0, which means that N/N0 = 0.00001. Plugging this into the formula, you get N = 5. For 87.5% decomposition, 12.5% remains. Let x be the number of half-lives: 0.125 = (1/2)^x. Solving for x, we get x ≈ 3 half-lives. For 99.999% decomposition, 0.001% remains. Using the same formula: 0.00001 = (1/2)^y. Solving for y, we get y ≈ 17 half-lives. So, 3 half-lives have passed for 87.5% decomposition, and 17 half-lives for 99.999% decomposition.

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a 50.0 ml sample of 0.155 m hno2(aq) is titrated with 0.100 m naoh(aq). what is the ph of a solution after the addition of 25.0 ml of naoh? [ ka of hno2 = 4.5 × 10–4 ]

Answers

After the addition of 25.0 ml of 0.100 M NaOH to a 50.0 ml sample of 0.155 M [tex]HNO_{2}[/tex], the resulting solution's pH can be calculated by considering the neutralization reaction between HNO_{2} and NaOH. Using the given Ka value of HNO_{2} (4.5 × [tex]10^{-4}[/tex]), the concentration of the resulting [tex]H_{3}O^{+}[/tex] ions can be determined, and the pH can be calculated.

To calculate the pH of the solution after the addition of NaOH, we need to determine the number of moles of HNO_{2} and NaOH reacted in the titration. The initial moles of HNO_{2} can be calculated by multiplying the initial concentration (0.155 M) by the initial volume (50.0 ml). Similarly, the moles of NaOH added can be obtained by multiplying the concentration (0.100 M) by the volume added (25.0 ml). Since HNO_{2} and NaOH react in a 1:1 ratio, the moles of HNO_{2} remaining after the reaction will be the difference between the initial moles and the moles of NaOH added.

Next, we can calculate the concentration of HNO_{2} after the reaction by dividing the moles of HNO_{2} remaining by the final volume (75.0 ml). Using the given Ka value of HNO_{2} (4.5 × [tex]10^{-4}[/tex]), we can set up an expression for the equilibrium constant and solve for the concentration of H_{3}O^{+} ions, which is equal to the concentration of HNO_{2} after the reaction. Finally, the pH can be calculated by taking the negative logarithm (base 10) of the concentration. By following these steps, the pH of the solution after the addition of NaOH can be determined based on the given information.

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Both H2O and H2PO4? are amphoteric.
Part A
Write an equation to show how H2PO4? can act as an acid with H2O acting as a base.
Part B
Write an equation to show how H2PO4? can act as a base with H2O acting as an acid.

Answers

Both equations demonstrate the amphoteric nature of [tex]H_2PO_4^-[/tex], as it can act as both an acid and a base depending on the nature of the other species involved in the reaction.

Part A:

[tex]H_2PO_4^- (aq) + H_2O (l) -- > H_3O^+ (aq) + HPO_4^{2-} (aq)[/tex]

In this equation, [tex]H_2PO_4^-[/tex] acts as an acid by donating a proton (H⁺) to water ([tex]H_2O[/tex]), which acts as a base. The result is the formation of hydronium ion ([tex]H_3O^+[/tex]) and the conjugate base, [tex]H_2PO_4^-[/tex].

Part B:

[tex]H_2PO_4^- (aq) + H_2O (l) < -- > OH^- (aq) + H_3PO_4 (aq)[/tex]

In this equation, [tex]H_2PO_4^-[/tex]⁻ acts as a base by accepting a proton (H⁺) from water ([tex]H_2O[/tex]), which acts as an acid. The result is the formation of hydroxide ion (OH⁻) and the conjugate acid, [tex]H_3PO_4[/tex].

Water, being a neutral molecule, can act as both an acid and a base, depending on the reaction conditions.

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Considering the limiting reactant concept, how many moles of copper(I) sulfide are produced from the reaction of 1.00 mole of copper and 1.00 mole of sulfur?
2 Cu(s) + S(s) Cu2S(s)
a. 2.00 mol
b. 1.00 mol
c. 0.500 mol
d. 1.50 mol
e. none of the above

Answers

To determine the moles of copper(I) sulfide produced from the reaction of 1.00 mole of copper and 1.00 mole of sulfur, we need to identify the limiting reactant. Thus, the correct answer is b. 1.00 mol.

First, we calculate the moles of copper and sulfur:

Moles of copper (Cu) = 1.00 mole

Moles of sulfur (S) = 1.00 mole

Next, we compare the stoichiometric coefficients of copper and sulfur in the balanced equation: 2 Cu + S -> Cu2S. The ratio of moles of copper to sulfur is 2:1. Therefore, for every 2 moles of copper, we need 1 mole of sulfur. Since we have equal moles of copper and sulfur, the reactants are present in the stoichiometric ratio. Therefore, neither reactant is in excess or limiting. As a result, the balanced reaction will consume all 1.00 mole of copper and 1.00 mole of sulfur, producing 1.00 mole of copper(I) sulfide.

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If ksp=1. 05×10−2, what is the molar solubility of kclo4?

Answers

The molar solubility of KClO₄ if Ksp 1.05 × 10⁻² is 0.102 M.

Ksp or solubility product constant is a thermodynamic equilibrium constant. It's the product of the ion concentrations in the solution that are in equilibrium with a solid, which has a certain solubility.

For the substance KClO₄, its Ksp value is 1.05 × 10⁻², and the molar solubility of KClO₄ is required to be calculated.

The molar solubility of a substance in water is given by the concentration of ions that are dissolved in water at equilibrium with undissolved solute (solid) in the solution.

To determine the molar solubility of the substance KClO₄ from Ksp, the equation is given as below:

Ksp = [K⁺][ClO₄⁻]

Let x be the molar solubility of KClO₄.

Therefore,

Ksp = x²x

= √(Ksp)

= √(1.05 × 10⁻²)

= 0.102 M

So, the KClO₄ solubility of KClO₄ is 0.102 M.

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Select the atom in each compound that does not follow the octet rule. Highlight the appropriate atoms by clicking on them. Part G
Select the atom in each compound that does not follow the octet rule.
Highlight the appropriate atoms by clicking on them.
NO
XeF4
OPBr3
BF3
ICl2

Answers

The octet rule states that atoms tend to gain, lose, or share electrons in order to have a full outer shell of eight electrons.

In the compound NO, the nitrogen atom does not follow the octet rule because it only has seven valence electrons. In XeF4, the xenon atom does not follow the octet rule because it has twelve valence electrons. In OPBr3, the phosphorus atom does not follow the octet rule because it has ten valence electrons. In BF3, the boron atom does not follow the octet rule because it only has six valence electrons. In ICl2, the iodine atom does not follow the octet rule because it only has seven valence electrons. It's important to note that some elements, such as hydrogen and helium, only need two valence electrons to have a full outer shell.

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Draw one Lewis structure for each of the following molecules, Determine the molecular shape (for example: trigonal pyramid). Indicate bond angles:
A) NCl3 B) COCI2 C) SF6 D) Tecl4

Answers

A) [tex]NCl_3[/tex]: N with three Cl atoms attached to it in a trigonal pyramid shape and Approximately 107 degrees.

B) [tex]COCl_2[/tex]:C double bonded to O and single bonded to two Cl atoms in a trigonal planar shape and Approximately 120 degrees.

C) [tex]SF_6[/tex]:S with six F atoms attached to it in an octahedral shape and 90 degrees.

D) [tex]TeCl_4[/tex]: Te with four Cl atoms attached to it in a tetrahedral shape and Approximately 109.5 degrees.

What is Lewis structure?

Lewis structure, also known as Lewis dot structure or electron dot structure, is a representation of a molecule or ion that shows the arrangement of atoms and their valence electrons.

A) [tex]NCl_3:[/tex]

Lewis Structure:

Cl

|

N - Cl

|

Cl

Molecular Shape: Trigonal Pyramidal Bond Angles: The bond angle between each Cl-N-Cl bond is approximately 107 degrees.

B) [tex]COCl_2:[/tex]

Lewis Structure:

Cl

|

O = C - Cl

|

Cl

Molecular Shape: Trigonal Planar Bond Angles: The bond angle between each Cl-C-Cl bond is approximately 120 degrees.

C) [tex]SF_6:[/tex]

Lewis Structure:

F F

| |

F - S - F

| |

F F

Molecular Shape: Octahedral Bond Angles: The bond angle between each F-S-F bond is approximately 90 degrees.

D)[tex]TeCl_4:[/tex]

Lewis Structure:

Cl

|

Cl - Te - Cl

|

Cl

Molecular Shape: Tetrahedral Bond Angles: The bond angle between each Cl-Te-Cl bond is approximately 109.5 degrees.

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Which of the following is recommended in moving something​ heavy?
A.
Pushing
B.
Reaching
C.
Leaning
D.
Pulling

Answers

When moving something heavy, the recommended method is to either push or pull the object. When moving something heavy, the most effective methods are pushing or pulling the object.

Pushing involves exerting force on the object in a forward direction, using your body weight and leg muscles for leverage. This method is suitable when you have enough space in front of the object and can maintain a stable posture while pushing.

On the other hand, pulling involves applying force in a backward direction, typically using a handle or a rope attached to the object. This method is useful when you need to move the object over a longer distance or when there are obstacles in the way. It allows you to utilize your upper body strength to generate force and overcome the resistance of the heavy object.

Reaching and leaning are not recommended techniques for moving something heavy as they may result in strain or injury. Reaching out to move a heavy object can put excessive stress on your back and arms, increasing the risk of muscle strain. Leaning against a heavy object without proper support or stability can lead to imbalance or loss of control, posing a safety hazard.

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match the following definitions with the terms. - oxidizing agent - oxidative reactions in which a carboxylate group is removed to form carbon dioxide - free energies of reactants is greater than the free energies of products in a reaction. - chemical process that occur within a living organism in order to maintain life a. the reaction is exergonic b. anabolism c. metabolism d. the reaction is endergonic e. species that is oxidized f. it is the species that is reduced g. oxidative decarboxylation

Answers

Oxidizing agent is matched with e. species that is οxidized

What is an Oxidizing Agent?

An οxidizing agent (οften referred tο as an οxidizer οr an οxidant) is a chemical species that tends tο οxidize οther substances, i.e. cause an increase in the οxidatiοn state οf the substance by making it lοse electrοns.

οxidative reactiοns in which a carbοxylate grοup is remοved tο fοrm carbοn diοxide: g. οxidative decarbοxylatiοnfree energies οf reactants is greater than the free energies οf prοducts in a reactiοn: d. the reactiοn is endergοnicchemical prοcess that οccur within a living οrganism in οrder tο maintain life: c. metabοlismthe reactiοn is exergοnic: a. the reactiοn is exergοnicanabοlism: the term "anabοlism" is nοt included in the prοvided definitiοns.it is the species that is reduced: f. it is the species that is reduced

Matching with available οptiοns:

a. the reactiοn is exergοnic

b. nοt prοvided in the definitiοns

c. metabοlism

d. the reactiοn is endergοnic

e. species that is οxidized

f. it is the species that is reduced

g. οxidative decarbοxylatiοn

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cylinder with a moveable piston contains 0.569 mol of gas and has a volume of 215 ml . part a what will its volume be if an additional 0.352 mol of gas is added to the cylinder? (assume constant temperature and pressure.)

Answers

The volume of the cylinder will be 0.580 L when an additional 0.352 mol of gas is added to the cylinder.


The ideal gas law equation, PV = nRT, relates the pressure, volume, amount of gas (in moles), and temperature of an ideal gas. Assuming constant temperature and pressure, we can use this equation to solve for the final volume of the cylinder when an additional 0.352 mol of gas is added.
First, we need to find the initial pressure of the gas in the cylinder. We can use the ideal gas law and the given values of n, V, and T to solve for P:
P = nRT/V
P = (0.569 mol)(0.0821 L•atm/mol•K)(T)/(0.215 L)
P = 13.2 atm
Next, we can use the combined gas law equation, P1V1 = P2V2, to solve for the final volume of the cylinder when the additional 0.352 mol of gas is added:
P1V1 = P2V2
(13.2 atm)(0.215 L) = (0.569 mol + 0.352 mol)(0.0821 L•atm/mol•K)(T)/V2
Solving for V2:
V2 = (0.921 mol)(0.0821 L•atm/mol•K)(T)/(13.2 atm)
V2 = 0.580 L
Therefore, the volume of the cylinder will be 0.580 L when an additional 0.352 mol of gas is added to the cylinder.

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what charge in coulombs passes through a cell if 2.3×10^-7 moles of electrons are transferred in this cell? select the correct answer below: a)0.022C b)0.41C c)1.5C d)7.2 C

Answers

The charge in coulombs is a) 0.022 C

What is electric charge?

Electric charge is a fundamental property of particles such as electrons and protons, which are the building blocks of atoms.

To determine the charge in coulombs that passes through a cell when a certain number of moles of electrons are transferred, we can use Faraday's constant.

Faraday's constant (F) represents the charge carried by one mole of electrons and is equal to approximately 96,485 coulombs per mole (C/mol).

In this case, we have[tex]2.3*10^{-7 }[/tex]moles of electrons transferred. To calculate the charge in coulombs, we can multiply the number of moles by Faraday's constant:

Charge (C) = ([tex]2.3*10^{-7 }[/tex] mol) * (96,485 C/mol)

Calculating this expression:

Charge (C) = 22.222 C

Therefore, the correct answer is: a) 0.022 C

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Chemical compound that leads to formation of photochemical smog in the troposphere when it reacts with other compounds in the presence of sunlight. A) Carbon dioxide B) Methane C) Nitrogen oxides D) Ozone

Answers

The chemical cοmpοund that leads tο the fοrmatiοn οf phοtοchemical smοg in the trοpοsphere when it reacts with οther cοmpοunds in the presence οf sunlight is:

C) Nitrοgen οxides

What is Nitrοgen οxides ?

Nitrοgen οxides (NOx), which include nitrοgen mοnοxide (NO) and nitrοgen diοxide (NO₂), play a significant rοle in the fοrmatiοn οf phοtοchemical smοg.

In the presence οf sunlight, nitrοgen οxides react with vοlatile οrganic cοmpοunds (VOCs) and οther pοllutants tο fοrm grοund-level οzοne (O3) and οther harmful pοllutants, cοntributing tο the fοrmatiοn οf smοg. Nitrοgen οxides are οften emitted by vehicles, pοwer plants, and industrial prοcesses.

What are the uses οf NO₂?

NO₂ is used as an intermediate in the manufacturing οf nitric acid, as a nitrating agent in the manufacturing οf chemical explοsives, as a pοlymerizatiοn inhibitοr fοr acrylates, as a flοur bleaching agent, and as a rοοm temperature sterilizatiοn agent.

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what was the maximum temperature displayed on the thermometer after the addition of the naoh solution to the hcl solution in the flask?

Answers

To determine the maximum temperature, carefully record the initial temperature and monitor the thermometer during the reaction until the temperature peaks and begins to decrease.

The maximum temperature displayed on the thermometer after the addition of the NaOH solution to the HCl solution in the flask cannot be determined without specific data from the experiment. The temperature change depends on factors like the concentration and volume of the solutions, as well as the initial temperature. However, when an acid (HCl) reacts with a base (NaOH), an exothermic neutralization reaction occurs, producing heat and causing the temperature to increase. To determine the maximum temperature, carefully record the initial temperature and monitor the thermometer during the reaction until the temperature peaks and begins to decrease. The temperature change depends on factors like the concentration and volume of the solutions, as well as the initial temperature.

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draw lewis structures of cnno2, showing all resonance forms, based on the following two possible skeletal structures for it. be sure to add all lone pairs and non-zero formal charges. do not add arrows between the structures. do not delete the boxes around the structures.

Answers

The atom's valence electrons are represented by Lewis Dot structures. An atom has the same number of electrons as its atomic number.

Resonance form :

Reverberation is the delocalisation of π electrons (present either in type of unsaturation or in type of solitary sets of electrons) and the subsequent designs are known as Resounding designs.

In other words, resonance is the process of moving electrons freely from one atom to another in a given structure under the condition that

the molecule's bonding framework must not change.The general charge of the framework should stay same.

Lewis structure =

: O :

.. ║

:O: ------- N ----- C ≡ N :

Lewis structure :

A very simplified representation of a molecule's valence shell electrons is known as a Lewis Structure. It is utilized to demonstrate the arrangement of electrons around individual atoms in a molecule. Electrons are displayed as "specks" or for holding electrons as a line between the two iotas.

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1. what is the molarity of a solution made by dissolving 3.00 moles of nacl in enough water to make 6.00 liters of the solution?

Answers

To find the molarity of a solution, you need to divide the number of moles of the solute by the volume of the solution in liters. In this case, you have 3.00 moles of NaCl dissolved in 6.00 liters of water, so:


Molarity = 3.00 moles NaCl / 6.00 L solution
Molarity = 0.50 M
Therefore, the molarity of the solution is 0.50 M.

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What structural features do cyanide and thiamine have in common that makes them both capable of catalyzing the benzoin condensation?

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Cyanide and thiamine do not have any structural features in common that enable them to catalyze the benzoin condensation.

In fact, cyanide is a potent poison that inhibits cellular respiration by binding to cytochrome c oxidase in the mitochondria, while thiamine is a vitamin that plays an essential role in energy metabolism as a cofactor for several enzymes. The benzoin condensation is a reaction that involves the condensation of two molecules of benzaldehyde in the presence of a base catalyst, typically NaOH or KOH, to form benzoin. While thiamine can act as a coenzyme for some enzymes that catalyze the benzoin condensation, it does not have any catalytic activity on its own and is not structurally similar to cyanide.

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calculate the pka of lactic acid (ch3ch(oh)cooh) given the following information. 40.0 ml of 0.2 m koh are added to 100. ml of a 0.500 m lactic acid solution producing a ph of 3.134.

Answers

The pKa of lactic acid [tex](CH_3CH(OH)COOH)[/tex] can be calculated by determining the concentration of its conjugate base (lactate) and the concentration of the undissociated lactic acid using the Henderson-Hasselbalch equation.

By measuring the pH of the solution after adding a known amount of KOH, the pKa can be determined to be approximately 3.86. To calculate the pKa of lactic acid, we can use the Henderson-Hasselbalch equation:

[tex]\[ \text{pH} = \text{pKa} + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \][/tex]

where pH is the measured pH, pKa is the desired value, [tex][A^-][/tex] is the concentration of the conjugate base (lactate), and [HA] is the concentration of the undissociated acid (lactic acid).

Initially, we have 100 ml of a 0.500 M lactic acid solution, which corresponds to 0.500 moles of lactic acid. When 40.0 ml of 0.2 M KOH is added, it reacts with the lactic acid in a 1:1 ratio to form lactate. Thus, 0.020 moles of lactic acid are neutralized, leaving 0.480 moles of lactic acid remaining.

The total volume of the solution after mixing is 140 ml (100 ml + 40 ml). By dividing the moles of lactate by the total volume, we can calculate the concentration of lactate, which is 0.020 moles / 0.140 L = 0.143 M.

Using the Henderson-Hasselbalch equation and the measured pH of 3.134, we can rearrange the equation to solve for pKa:

[tex]\[ \text{pKa} = \text{pH} - \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) = 3.134 - \log\left(\frac{0.143}{0.480}\right) \approx 3.86 \][/tex]

Therefore, the pKa of lactic acid is approximately 3.86.

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The reactants zinc and hydrochloric acid are involved in a redox reaction. Which of the following is a product of the reaction?
Select the correct answer below:
Cl2(g)
H2(g)
ZnCl(aq)
none of the abov

Answers

The correct answer is ZnCl(aq). When zinc reacts with hydrochloric acid, a redox reaction takes place. In this reaction, zinc acts as a reducing agent and donates electrons to hydrogen ions in hydrochloric acid, which act as an oxidizing agent.

As a result, hydrogen ions are reduced to hydrogen gas (H_{2}), while zinc is oxidized to form zinc ions (Zn2+) that react with chloride ions in hydrochloric acid to form zinc chloride (ZnCl_{2)}. The chemical equation for this reaction is:
Zn(s) + 2 HCl(aq) → ZnCl_{2}(aq) + H_[2}(g)
Therefore, the product of the reaction is ZnCl_{2}, which is an aqueous solution of zinc chloride. It is important to note that Cl_{2}(g) is not a product of this reaction because there is no evidence of the formation of chlorine gas during the reaction. Hence, the correct answer is ZnCl(aq).

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Suppose we tune the temperature and pressure of a container of gallium to its triple point at a temperature T=302 K, and pressure p=101 kPa. The densities of the phases of gallium are (i) solid: 5.91 g/cm^3 (ii) liquid: 6.05 g/cm (ii) gas: 0.116 g/cm^3.
If we slightly increase the pressure, which phase is stabilized in equilibrium? Que (a) Solid (b) Gas (c) Liquid

Answers

At the triple point, all three phases of gallium can exist in equilibrium. However, if we slightly increase the pressure, one phase will become more stable than the others. In this case, we can use the densities of the phases to determine which phase will be stabilized.

Since the density of the solid phase is greater than that of the liquid and gas phases, increasing pressure will stabilize the solid phase. Therefore, the answer to the question is (a) Solid. It is important to note that this is assuming the temperature remains constant. If the temperature were to increase or decrease, the answer may change depending on the phase diagram of gallium at that temperature and pressure.
At the triple point (T=302 K, p=101 kPa), all three phases of gallium (solid, liquid, and gas) coexist in equilibrium. If we slightly increase the pressure, the phase with the highest density will be stabilized, as it can withstand the increased pressure better.
Comparing the densities of the phases:
(i) Solid: 5.91 g/cm^3
(ii) Liquid: 6.05 g/cm^3
(iii) Gas: 0.116 g/cm^3
The liquid phase has the highest density (6.05 g/cm^3). Therefore, upon a slight increase in pressure, the liquid phase of gallium will be stabilized in equilibrium. So, the answer is (c) Liquid.

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most acidic and least acidic of the following acids: a) ch3ccl2co2h b) ch3ch co2h c) ch3chchco2h d) ch3ch2co2h

Answers

The order of acidity from most acidic to least acidic is: a) CH3CCl2CO2H, b) CH3CHCO2H, c) CH3CHCHCO2H, d) CH3CH2CO2H.

To determine the relative acidity of the given acids, we need to consider the stability of the corresponding conjugate bases. The more stable the conjugate base, the stronger the acid.

a) CH3CCl2CO2H: This acid has two electron-withdrawing chlorine atoms attached to the carboxylic acid group, which stabilizes the resulting carboxylate anion. Therefore, it is more acidic than the other options.

b) CH3CHCO2H: This acid has one electron-withdrawing methyl group attached to the carboxylic acid group. It is less acidic than option (a) but more acidic than options (c) and (d).

c) CH3CHCHCO2H: This acid has an additional alkyl group attached to the carboxylic acid group. The presence of the alkyl group further destabilizes the conjugate base, making it less acidic than the previous options.

d) CH3CH2CO2H: This acid has no additional substituents attached to the carboxylic acid group, making it the least acidic among the given options.

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Superglue fuming
This chemical treatment produces a white-appearing permanent fingerprint

Answers

Superglue fuming is a chemical treatment that results in a white-appearing permanent fingerprint. It involves exposing a fingerprint to cyanoacrylate vapors, which react with the moisture present in the print, creating a visible white residue.

Superglue fuming is a commonly used method in forensic investigations to enhance and preserve latent fingerprints. The process involves placing an item containing the fingerprint in a sealed chamber along with a small amount of liquid superglue. The superglue releases cyanoacrylate vapors that adhere to the moisture and fatty acids present in the print, forming a durable and visible white deposit.

The white residue left by the superglue fuming process provides a contrast against the surface of the object, making the fingerprint more visible and easier to photograph or lift using various techniques. The resulting fingerprint is considered permanent because the superglue bonds with the moisture and forms a hard, solid material that can withstand handling and further processing.

Overall, superglue fuming is an effective method for developing latent fingerprints, providing investigators with valuable evidence in forensic analysis.

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calculate the heat released when 0.300 mol of steam at 158 degrees c is cooled to the ice at -83 degrees c.

Answers

The heat released when 0.300 mol of steam at 158°C is cooled to ice at -83°C is approximately -9,183.3 kJ.

How to calculate the heat released?

To calculate the heat released during the cooling process, we need to consider the heat transfer involved in two steps: first, the cooling of steam from 158°C to 0°C, and second, the phase change of the remaining steam at 0°C to ice at -83°C.

Step 1: Cooling of steam from 158°C to 0°C

The heat released during this step can be calculated using the formula:

q₁ = n × C₁ × ΔT

where

n = number of moles of steam

C₁ = molar specific heat capacity of steam

ΔT = change in temperature

Using the molar specific heat capacity of steam (C₁ = 36.9 J/(mol·°C)) and the temperature change (ΔT = 158°C - 0°C = 158°C), we can calculate q₁:

q₁ = 0.300 mol × 36.9 J/(mol·°C) × 158°C = 1,748.94 J

Step 2: Phase change from steam at 0°C to ice at -83°C

The heat released during this step can be calculated using the formula:

q₂ = n × ΔH_fusion

where

ΔH_fusion = molar enthalpy of fusion

The molar enthalpy of fusion for water is 6.01 kJ/mol. Therefore, q₂ can be calculated as:

q₂ = 0.300 mol × 6.01 kJ/mol = 1.803 kJ

The total heat released is the sum of q₁ and q₂:

Total heat released = q₁ + q₂ = 1,748.94 J + 1.803 kJ = 1,748.94 J + 1,803 J = -9,183.3 J ≈ -9,183.3 kJ

Therefore, the heat released when 0.300 mol of steam at 158°C is cooled to ice at -83°C is approximately -9,183.3 kJ.

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if the element with atomic number 63 and atomic mass 212 decays by alpha emission. what is the atomic number of the decay product

Answers

if the element with atomic number 63 and atomic mass 212 decays by alpha emission. The new element formed after alpha decay will have an atomic number of 61

Alpha emission occurs when an atomic nucleus emits an alpha particle, which consists of two protons and two neutrons. During alpha decay, the atomic number and atomic mass of the parent nucleus decrease by 2 and 4, respectively. In this case, the parent nucleus has an atomic number of 63 and an atomic mass of 212. When the parent nucleus undergoes alpha decay, it emits an alpha particle (2 protons and 2 neutrons). As a result, the atomic number decreases by 2, and the atomic mass decreases by 4. Therefore, the atomic number of the decay product is 63 - 2 = 61. The new element formed after alpha decay will have an atomic number of 61. It's important to note that the specific element with atomic number 61 cannot be determined solely from the given information. The identity of the element can be determined by considering its atomic number, which is 61 in this case, and consulting the periodic table to find the corresponding element with that atomic number.

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Which one of the following pairs of 0.100 mol L -1 solutions, when mixed, will produce a buffer solution?
• A. 50. mL of aqueous CH3COOH and 25. mL of aqueous HCI
• B. 50. mL of aqueous CH3COOH and 100. mL of aqueous NaOH
• C. 50. mL of aqueous NaOH and 25. mL of aqueous HCI
• D. 50. mL of aqueous CH3COONa and 25. mL of aqueous NaOH
© E. 50. mL of aqueous CH3COOH and 25. mL of aqueous CH3COONa

Answers

The pair of solutions that will produce a buffer solution is E, 50 mL of aqueous CH3COOH and 25 mL of aqueous CH3COONa. A buffer solution is a solution that can resist changes in pH when small amounts of acid or base are added to it. A buffer solution contains a weak acid and its conjugate base or a weak base and its conjugate acid.

In this case, CH3COOH is a weak acid and CH3COONa is its conjugate base. When they are mixed, they form a buffer solution. Aqueous refers to a solution in which the solvent is water. The other options do not contain a weak acid and its conjugate base or a weak base and its conjugate acid, so they will not produce a buffer solution. It's important to note that buffer solutions are commonly used in laboratory settings and in the human body to maintain a stable pH. They are important in chemical and biological reactions, and the ability to identify which solutions will produce a buffer is crucial in these fields.

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A Sample Of Methane Gas In A Piston Exerts A Pressure Of 1.26 10^3 When The Volume Is 54.3 Cm. When The (2024)

FAQs

What will be the pressure exerted by 16 g of methane in a 250 ml container at 300 K using ideal gas equation? ›

The ideal gas equation predicts that, P=nRTV=1×0.0821×3000.250=98.52atm.

What is the pressure exerted by 6 gram of methane gas in a 0.03 m cube vessel at 129 degree Celsius? ›

The temperature given is 129°C. Therefore, the pressure exerted by 6.0g of methane gas in a 0.03 m^3 vessel at 129°C is approximately 41559.33 Pa.

What is the density of methane at 2.0 atm pressure and 27? ›

d=PMRT=2×160.082×300=1.30gL−1.

What is the volume occupied by 8g of methane gas at 40c and 842 torr pressure? ›

Calculate the volume: V ≈ 0.5 mol * 0.0821 L atm/mol K * 313.15 K / 1.1079 atm ≈ 11.6 L. So, the volume occupied by 8 g of methane gas at 40°C and 842 torr pressure is approximately 11.6 liters.

What is the volume occupied by 16.0 g of methane gas at STP? ›

Molecular weight of methane = 16 which means 16 g of methane occupy 22.4L volume at STP.

What is the pressure exerted by one mole of methane in a 0.25 Litre container at 300 K using Wonderwall equation ›

⇒ P + 36.048 = 118.87 ⇒ P = 82.82 a t m . Pressure exerted by 1 mole of methane in 0.25 litre container at 300 K using Van der Waal's equations?

What is the pressure exerted by 3.2 g of methane? ›

=8.31×104Pa .

What volume will 30.0 g of methane gas CH4 occupy at STP? ›

Thus, the volume of the methane gas (CH4) is 41.91 L.

What volume does 8 grams of methane occupies at STP? ›

Assertion :8 grams of methane occupies 11.2 litres of volume at 273 K and 1 atm pressure. Reason: One mole of any gas at STP occupies 22.4 litres of volume. Calculate the weight of CO2, which occupies a volume of 11.2 dm3 at STP.

What is the weight of methane in a 9l cylinder at 16 atm and 27 degree celsius? ›

Hence, the weight of methne is 93.6 g.

How to calculate methane density? ›

  1. Step 1: Given data. T = 27 ° C = 300 K P = 2 atm. Step 2: Calculating the mass of carbon. Mass of methane ( ) ...
  2. Step 2: Calculating the mass of carbon. Mass of methane ( ) [Mass of Carbon= 12 g / mol ] [Mass of Hydrogen= 1 g / mol ] Thus, ...
  3. Step 3: Evaluating density. We also know that. R = 0 . 082 L atm / K mol.

What is the density of methane at 20 degrees? ›

The Density of Methane (natural gas) can thus be derived from first principles in the chart below, using a molar mass of 16 g/mol, and then flexing the temperature and pressure. This shows how methane at 1 bar of pressure and 20ºC has a density of 0.67 kg/m3.

What is the volume occupied by 32g methane gas at STP? ›

Let the volume of methane gas be “V”. Thus, 32 grams of methane at S.T.P will occupy 44.82 dm³.

What is the volume occupied by 40 grams of methane at STP? ›

= 2 x 8 = 16 g16 g of hydrocarbon occupy at S.T.P. = 22.4. lit∴ 40 g of hydrocarbon occupy at S.T.P = 22.4 x 40/16 = 56 lits.

What is the volume occupied by 24 g methane gas at STP? ›

∴ 1.5 moles( 24.0g) of the gas will occupy =22.4×1.51=33.6l.

What 16 grams of methane gas occupies? ›

Hence, 16 g of methane occupies 22.4 litres at STP.

What is the pressure exerted by 32g of O2 in a 22l container at 30c? ›

Expert-Verified Answer. Final answer: The pressure exerted by 32g of O₂ in a 22.0L container at 30.0 degrees Celsius is 30.575 atm.

How much water is produced by combustion of 16 gram of methane in following chemical reaction? ›

According to the above equation, one mole of CH4 (16 g) on combustion gives one mole of carbon dioxide and 2 moles of water and 2 moles of water = 2 x [2 + 16] = 36 g of water.

What will be pressure exerted by a mixture of 3.2 g of methane and 4.4 g of carbon dioxide contained in a 9 dm3 flask at 27C? ›

Hence, the total pressure exerted by the mixture is=8. 314×104Pa. Q. What will be the pressure exerted by a mixture of 3.2g of methane and 4.4g of carbon dioxide contained in a 9dm3 flask at 27oC?

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