draw a lewis structure for ammonia, nh3 . include all hydrogen atoms and show all unshared pairs and the formal charges, if any. assume that bonding follows the octet rule.

Extraction is a commonly used technique in chemistry to separate or isolate specific compounds from a mixture based on their differential solubility in different solvents.

Extraction involves dissolving the mixture in a suitable solvent and then selectively extracting the desired compound into another solvent phase.

To separate p-dichlorobenzene from p-chloroaniline by extraction, the following procedure can be employed:

1. Prepare the mixture: Combine the mixture containing p-dichlorobenzene and p-chloroaniline in a suitable container.

Ensure that the compounds are thoroughly mixed.

2. Select an appropriate solvent: Choose a solvent that is selective for dissolving p-dichlorobenzene while minimizing the solubility of p-chloroaniline.

In this case, a nonpolar organic solvent such as ether or dichloromethane would be suitable.

3. Perform extraction: Add the chosen solvent to the mixture and agitate or stir the mixture for a sufficient amount of time.

This allows the solvent to extract the p-dichlorobenzene from the mixture while leaving the p-chloroaniline behind.

4. Allow phase separation: After agitation, allow the mixture to settle so that two distinct phases form: an organic phase (containing the solvent and extracted p-dichlorobenzene) and an aqueous phase (containing the remaining p-chloroaniline).

5. Collect the organic phase: Carefully separate the organic phase from the aqueous phase using a separating funnel or by decantation.

The organic phase will contain the dissolved p-dichlorobenzene.

6. Optional repeat extraction: If necessary, the extraction process can be repeated with fresh solvent to enhance the separation efficiency.

7. Evaporate the solvent: In order to obtain pure p-dichlorobenzene, the solvent can be evaporated from the collected organic phase using techniques such as rotary evaporation or simple distillation.

This process leaves behind the p-dichlorobenzene as a solid residue.

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To calculate the pH of a buffer solution, we use the Henderson-Hasselbalch equation, which is:

pH = pKa + log([A-]/[HA])Where pH is the pH of the buffer solution, pKa is the dissociation constant of the weak acid, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the weak acid.In this case, NH3 is the weak base and NH4+ is its conjugate acid. The dissociation constant for NH4+ is 9.24 x 10^-10 at 25°C.Therefore, the pKa for NH4+ is:pKa = -log(Ka) = -log(9.24 x 10^-10) = 9.04The concentrations of NH3 and NH4+ are given as 0.17 M and 0.47 M, respectively.[A-]/[HA] = [NH3]/[NH4+] = 0.17/0.47 = 0.361Substituting the values into the Henderson-Hasselbalch equal= 8.60Therefore, the pH of the buffer system is approximately 8.60.


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In this problem, we need to find the pH of a buffer (pH = 4.45) that is produced by the reaction of strong acids as HBr that reacts with weak bases as NaC₇H₅O₂ producing its weak acid and a salt, as follows:

HBr + NaC₇H₅O₂ → HC₇H₅O₂ + NaBr

If in the reaction, the limiting reactant is the strong acid, we will produce a buffer (The aqueous mixture of a weak acid and its strong base).

Using Henderson-Hasselbalch equation, we can find the pH of this buffer:

\(pH = pKa + log \frac{[NaC_7H_5O_2]}{[HC_7H_5O_2]}\)

Where:

pKa is -log Ka = 4.20

[] could be taken as the moles of each species

After the reaction, the moles of HBr = Moles of HC₇H₅O₂ and the remaining moles of NaC₇H₅O₂ = Initial moles of NaC₇H₅O₂ - Moles of HBr.

Moles HBr:

22.5mL = 0.0225L * (0.200mol/L) = 4.5x10⁻³ moles HBr

Moles NaC₇H₅O₂:

50.0mL = 0.0500L * (0.250mol/L) = 0.0125 moles NaC₇H₅O₂

That means the moles after the reaction of the species of the buffer are:

Moles HC₇H₅O₂ = Moles HBr = 4.5x10⁻³ moles

Moles NaC₇H₅O₂ = 0.0125 moles - 4.5x10⁻³ moles = 8.0x10⁻³ moles

Replacing in H-H equation:

\(pH = 4.20+ log \frac{[8.0x10^{-3}]}{[4.5x10^{-3}]}\)

pH = 4.45

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1. The differential equation satisfied by y(t) is:  dy/dt = 0.6 kg/min

The amount of salt in the tank after t minutes can be represented by the function y(t). We need to find the differential equation that y satisfies.
Initially, the tank contains 100 L of pure water, which means there is no salt in the tank. As time passes, a solution with a salt concentration of 0.3 kg/L is added at a rate of 7 L/min. The salt concentration in the tank will increase with the addition of this solution.
At the same time, the solution is drained from the tank at a rate of 5 L/min. This will result in a decrease in the salt concentration in the tank.
To find the differential equation satisfied by y(t), we need to consider the rate of change of salt in the tank.
Rate of change of salt in the tank = Rate of salt added - Rate of salt drained
The rate of salt added is given by the product of the concentration of the solution (0.3 kg/L) and the rate at which the solution is added (7 L/min). So, the rate of salt added = 0.3 kg/L * 7 L/min.

The rate of salt drained is given by the product of the concentration of the solution (0.3 kg/L) and the rate at which the solution is drained (5 L/min). So, the rate of salt drained = 0.3 kg/L * 5 L/min.

Therefore, the differential equation satisfied by y(t) is:
dy/dt = (0.3 kg/L * 7 L/min) - (0.3 kg/L * 5 L/min)

Simplifying the equation:
dy/dt = 2.1 kg/min - 1.5 kg/min
dy/dt = 0.6 kg/min

So, the differential equation satisfied by y(t) is:
dy/dt = 0.6 kg/min

2. The amount of salt in the tank after 40 minutes is 24 kilograms.

To find the amount of salt in the tank after 40 minutes, we can solve the differential equation.

dy/dt = 0.6 kg/min

Integrating both sides with respect to t:
∫dy = ∫0.6 dt

Integrating, we get:
y = 0.6t + C

To find the value of C, we can use the initial condition that the tank initially contains 100 L of pure water, which means there is no salt. So, at t = 0, y = 0.

Substituting these values into the equation:
0 = 0.6(0) + C
C = 0

Therefore, the equation becomes:
y = 0.6t

Now, we can find the amount of salt in the tank after 40 minutes by substituting t = 40 into the equation:
y = 0.6(40)
y = 24 kg


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The decay constant of the radioactive isotope, which is roughly -0.044 years⁻¹, causes a 10 kilogram mass to decompose to 5 kg in 17 years.

The decay of a radioactive isotope can be described by the exponential decay equation:

\(\[N(t) = N_0 \cdot e^{-kt}\]\)

where:

N(t) is the quantity of the isotope at time t,

N₀ is the initial quantity of the isotope,

k is the decay constant,

t is the time in years.

In this case, we have:

N₀ = 10 kg (initial quantity),

N(t) = 7 kg (quantity after 17 years),

t = 17 years.

Substituting these values into the equation, we get:

\(\[7 = 10 \cdot e^{-17k}\]\)

To find the decay constant (k), we can rearrange the equation as follows:

\(-17k = \ln\left(\frac{7}{10}\right)\)

Taking the natural logarithm (ln) of both sides:

\(-17k = \ln\left(\frac{7}{10}\right)\)

Dividing both sides by -17:

\(k = \frac{\ln\left(\frac{7}{10}\right)}{-17}\)

Using a calculator and rounding to three decimal places, we find:

k ≈ -0.044.

Therefore, the decay constant of the isotope is approximately -0.044 years⁻¹.

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Complete question :

A 10-kg quantity of a radioactive isotope decays to 7 kg after 17 years. Find the decay constant of the isotope. (Use decimal notation. Give your answer to three decimal places.) k= years-1

Answer:

it takes about an hour

Explanation:

On average, it takes about an hour for a person to metabolize around 14 grams of pure ethanol—the amount of alcohol contained in one standard drink—which amounts to roughly 12 ounces of 5% ABV beer, 5 ounces of wine, or 1.5 ounces of 80 proof liquor.

Several factors influence the rate at which alcohol is absorbed into the bloodstream. For example, food in the stomach slows gastric emptying and alcohol absorption.

Answer:

76 grams is the correct answer to this question

The most accurate statement considering the dipole moment is: c. The O-Cl bonds are all polar, but due to the linear shape of the molecules, the individual dipoles cancel to yield no net dipole moment for either molecule.

The most accurate statement considering the dipole moment is option c. In this case, the molecules in question have linear shapes, and all the O-Cl bonds are polar.

A polar bond occurs when there is an unequal distribution of electron density between two atoms, resulting in a separation of positive and negative charges. However, even though the O-Cl bonds are polar, the linear molecular structure leads to the cancellation of the individual dipole moments.

The dipole moment of a molecule is determined by both the magnitude and direction of its constituent bond dipoles. In this scenario, the linear shape causes the dipole moments to point in opposite directions, effectively canceling each other out.

As a result, there is no net dipole moment for either molecule. This cancellation of dipole moments due to molecular geometry is known as "vector sum" or "vector cancellation."

Thus, option c accurately describes the absence of a net dipole moment in the given molecules despite having polar O-Cl bonds.

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The actual settings for a group policy are downloaded from the SYSVOL share on a domain controller.

In a Windows Active Directory domain, the SYSVOL share on a domain controller is responsible for storing and replicating the domain's public files and data, including group policy settings. The SYSVOL share contains a folder structure that includes Group Policy Objects (GPOs), which are used to define and enforce various settings and configurations for users and computers in the domain.

When a client computer or user logs on to the domain, it contacts a domain controller to retrieve the group policy settings applicable to its organizational unit (OU) or site. The client requests the GPO information from the SYSVOL share, and the domain controller responds by sending the necessary policy files to the client. These files contain the specific group policy settings that dictate the behavior and configurations of the client computer or user within the domain.

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The balanced chemical equation for the reaction between sodium and oxygen to form sodium oxide is: 4Na + O2 → 2Na2O According to the balanced equation, 4 moles of sodium react with 1 mole of oxygen to produce 2 moles of sodium oxide.

Therefore, the stoichiometric ratio is 4:1:2 (sodium:oxygen:sodium oxide). If you have 8 moles of sodium, you can use the stoichiometry to determine the amount of sodium oxide produced. Since the ratio of sodium to sodium oxide is 4:2, you would expect to produce half the number of moles of sodium oxide compared to the moles of sodium. Therefore, with 8 moles of sodium, you would expect to produce 4 moles of sodium oxide.

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Glucokinase is inhibited by high levels of its product, glucose-6-phosphate. When glycolysis is inhibited through glucokinase, glucose-6-phosphate builds up, shutting down glucokinase. This prevents glucose from being metabolized and instead activates glucose-6-phosphate metabolism in the liver, where it is needed in the blood and other tissues. This is an example of feedback inhibition.

Glucokinase is a key enzyme involved in the first step of glycolysis, catalyzing the phosphorylation of glucose to form glucose-6-phosphate. However, when glucose levels are high, such as during a high-carbohydrate meal, the concentration of glucose-6-phosphate increases. This accumulation of glucose-6-phosphate acts as an allosteric inhibitor, binding to the active site of glucokinase and inhibiting its activity. As a result, glycolysis is inhibited at this step.

The buildup of glucose-6-phosphate due to glucokinase inhibition serves an important regulatory function in the liver. Glucose-6-phosphate is a precursor for glycogen synthesis, which helps store glucose for later use. Additionally, glucose-6-phosphate can be further metabolized through the pentose phosphate pathway to generate reducing equivalents in the form of NADPH, which is required for various biosynthetic reactions and cellular processes.

By inhibiting glycolysis and promoting glucose-6-phosphate metabolism, the liver ensures that glucose is directed toward glycogen synthesis and other essential metabolic pathways rather than being metabolized for immediate energy production. This regulation helps maintain appropriate blood glucose levels and ensures a steady supply of glucose for other tissues that depend on it.

Overall, the inhibition of glycolysis at high levels of glucose-6-phosphate through feedback inhibition of glucokinase represents an adaptive mechanism of the liver to coordinate glucose metabolism and homeostasis in response to fluctuating glucose levels in the body.

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To generate 50.0 g of sodium chloride according to the given chemical equation, we need 91.12 g of sodium chlorate.

To calculate the grams of sodium chlorate required to generate 50.0 g of sodium chloride, we first need to use the balanced chemical equation to determine the molar ratio of sodium chlorate to sodium chloride. From the equation 2NaClO3 → 2NaCl + 3O2, we can see that for every 2 moles of sodium chlorate, 2 moles of sodium chloride are produced.
The molar mass of sodium chloride is 58.44 g/mol, and so 50.0 g of sodium chloride corresponds to 50.0 g / 58.44 g/mol = 0.8557 moles.
Since the molar ratio of sodium chlorate to sodium chloride is 2:2, or simply 1:1, we know that we need 0.8557 moles of sodium chlorate to generate 50.0 g of sodium chloride.
The molar mass of sodium chlorate is 106.44 g/mol, and so to convert moles to grams, we can simply multiply the number of moles by the molar mass. Therefore, we need:
0.8557 moles x 106.44 g/mol = 91.12 g of sodium chlorate.
Therefore, to generate 50.0 g of sodium chloride according to the given chemical equation, we need 91.12 g of sodium chlorate.

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If you observe that during the distillation some of the boric acid solution has begun to travel toward the cobalt complex solution, and has made it halfway through the Tygon tubing :

Step 1: Stop the distillation process immediately. This will stop the further mixing of the two solutions and prevent further damage.

Step 2: Discard the distillate already collected in the collection flask.

Step 3: Disassemble the distillation setup.

Step 4: Thoroughly rinse the distillation flask, distillation column, condenser, and Tygon tubing with a sufficient amount of distilled water. Rinse everything multiple times until there is no trace of boric acid or cobalt complex remaining. This will prevent the contamination of the next experiment.

Step 5: Set up the distillation process again using new, clean Tygon tubing.

Step 6: Refill the distillation flask with fresh boric acid solution and repeat the distillation process from the beginning.

During the distillation, one should ensure that no leakages occur in the distillation setup. Leaks can cause the two solutions to mix with each other, leading to contamination of the final distillate.

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Answer:

Negative

Explanation:

Any atom gaining electrons will get a negative change since the charge of the electron is negative.

Answer:

A.) They would have to prey on other organisms

Explanation:

The answers are.....

Animals that prey on rabbits would begin to prey on other organisms.

And...

Animals that prey on rabbits would starve.

And...

Animals that prey on rabbits would leave the area in search of more rabbits.

My son attends K12 and had this question on his unit 5 test. These were his answer choices and they were correct.

At 457.3 K temperature, the vapour pressure will be higher, for the given substance and its heat of vapourization.

The pressure that a vapour exerts on its condensed phases in a closed system when they are in thermodynamic equilibrium with one another at a specific temperature is known as vapour pressure. A liquid's evaporation rate can be determined by looking at the equilibrium vapour pressure.

Calculation:

The Clausius-Clapeyron equation is as follows:

ln (P2/P1) = ∆Hvap/R * (1/T1 - 1/T2)

ln (8.008) =  (26.64 * 10³J/mol) / (8.314J/mol/K) * {(1/287K) - (1/T2) }

2 = (3.2 * 10³/K) * {(T2 - 287K)/ (287T2 * K)

0.000625/K * 287T2 * K = T2 - 287K

179.375K = T2 -287K

T2 = 457.3K.

Hence, at 457.3 K temperature, the vapour pressure will be higher.

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The ΔH for the given reactions are:

To find the ΔH for the given reaction, using Hess's Law, which states that the ΔH of an overall reaction is equal to the sum of the ΔH values for each individual reaction involved in the process:

2 Al + (3/2) O₂ → Al₂O₃ ΔH=-1670 kJ (multiplied by 2)

Fe₂O₃ → 2 Fe + (3/2) O₂ ΔH=+824 kJ (reversed)

2 Fe + (3/2) O₂ → Fe₂O₃ ΔH=-824 kJ (multiplied by 2)

2 Al2O₃ → 4 Al + (3/2) O₂ ΔH=+3340 kJ (reversed)

Adding the two equations obtained above, then the overall reaction:

2 Al + Fe₂O₃ → 2 Fe + Al₂O₃ ΔH=+1670-824=+846 kJ

Therefore, the ΔH for the given reaction is +846 kJ.

To find the ΔH for the given reaction, to use the same approach as above. Write the required reactions and their corresponding ΔH values as follows:

H₂ + O₂ → H₂O₂ ΔH=-188 kJ (multiplied by 2)

H₂O₂ → 2 H₂O + O₂ ΔH=+496 kJ (reversed)

Adding the two equations obtained above, then the overall reaction:

2 H₂O₂ → 2 H₂O + 2 O₂ ΔH=+308 kJ

Therefore, the ΔH for the given reaction is +308 kJ.

To find the ΔH for the given reaction, use the same approach as above:

1/2 N₂ + 1/2 O₂ → NO ΔH=+90.0 kJ (multiplied by 2)

2 NO → N₂ + 2 O₂ ΔH=-180.0 kJ (reversed)

1/2 N₂ + O₂ → NO₂ ΔH=+34.0 kJ

Adding the two equations obtained above, then the overall reaction:

NO + 1/2 O₂ → NO₂ ΔH=-146.0 kJ

Therefore, the ΔH for the given reaction is -146.0 kJ.

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Answer:

Explanation:

The mole ratios of Mg to O2 to MgO in this question is 2:1:2, as seen from the balanced equation. However, since different elements have different masses, we cannot use the mole ratios for the grams of Mg and O2. Instead, we need to convert the grams of Mg and O2 to moles.

To convert the grams of Mg and O2 to moles, we first need to find their molar masses. The molar mass is essentially the atomic masses of all the elements within the molecule.

Magnesium has an atomic mass of approximately 24.305 u.  

Mg molar mass = 24.305 g/mol

Oxygen has an atomic mass of approximately 15.999 u.

O2 molar mass = 2*15.999 = 31.998 g/mol

We multiply by 2 here, because there are two atoms of oxygen per molecule of O2.

Now, to convert from grams to moles, we simply need to divide the substance's mass by the molar mass.

Mg: 2.04 g / (24.305 g/mol) = 49.5822 mol

O2: 12.3 g / (31.998 g/mol) = 0.384 mol

Since our values have been expressed in moles now, we can utilize the mole ratios. Looking at the mole ratio, for every 2 moles of Mg, there are 2 moles of MgO. Therefore, moles of Mg is equal to moles of MgO. That means there will be 49.5822 moles of MgO.

Now that we have the number of moles of MgO, we need to convert it back to grams. Once again, we need the molar mass of MgO.

MgO molar mass = 24.305 + 15.999 = 40.304 g/mol

To find the grams of MgO, we multiply the number of moles by its molar mass.

49.5822 * 40.304 = 1998.360989 grams --> 1998 grams

Approximately 189.52 grams of helium must be added to the cylinder to reach its maximum pressure allowed.

To solve this problem, we need to consider the ideal gas law, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

First, we need to convert the temperature from Celsius to Kelvin:

T = 147°C + 273.15 = 420.15 K

We can calculate the number of moles of nitrogen gas using the ideal gas law:

n = PV / RT

Given:

P = 1310 mmHg (we'll convert this to atm)

V = 550 L

R = 0.0821 L·atm/(K·mol)

T = 420.15 K

Converting pressure from mmHg to atm:

P = 1310 mmHg / 760 mmHg/atm = 1.7237 atm

Calculating the number of moles of nitrogen gas:

n = (1.7237 atm) * (550 L) / (0.0821 L·atm/(K·mol) * 420.15 K)

n ≈ 47.38 moles

Now, we need to determine the number of grams of helium gas needed to reach the maximum pressure allowed. Since helium is an ideal gas, we can use the same equation and solve for the mass (m) using the molar mass of helium (4 g/mol):

m = n * M

Given:

n = 47.38 moles

M (molar mass of helium) = 4 g/mol

Calculating the mass of helium:

m = 47.38 moles * 4 g/mol

m ≈ 189.52 grams

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Answer:

alpha particles

Explanation:

alpha particles the least penetrating but potentially most damaging and gamma rays the most penetrating. A beta particle, also called beta ray or beta radiation, is a high-energy, high-speed electron or positron emitted by the radioactive decay of an atomic nucleus during the process of beta decay.

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a. There are 1.85 moles in 2.00 × 10² g of silver (Ag).

b. There are 0.618 moles in 37.1 g of silicon dioxide (SiO₂)

The molar mass is the mass in grams of 1 mole of particles, that is, the mass in grams of 6.02 × 10²³ particles. The units are g/mol.

We want to calculate the number of moles represented by different masses of different substances. In each case, the conversion factor between mass and moles is the molar mass.

The molar mass of silver is 107.87 g/mol.

2.00 × 10² g × (1 mol/107.87 g) = 1.85 mol

The molar mass of silicon dioxide is 60.08 g/mol.

37.1 g × (1 mol/60.08 g) = 0.618 mol

a. There are 1.85 moles in 2.00 × 10² g of silver (Ag).

b. There are 0.618 moles in 37.1 g of silicon dioxide (SiO₂)

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Up to 428.6 ml volume, you should dilute 75 ml of a 10.0 M H₂SO₄ solution to obtain a 1.75 m H₂SO₄ solution

Initial volume of H₂SO₄ solution (V1) = 75 ml

Initial molarity of H₂SO₄ solution (M1) = 10.0

Final molarity of H₂SO₄ solution (M2) = 1.75

Final volume of H₂SO₄ solution (V2) = ?

To find out the final volume (V2) we will use the following equation

   M1V1 = M2V2

Rearrange the equation for final volume (V2)

  V2 = M1V1 / M2

Put the values in the above equation

   V2 = 10.0 M × 75 ml / 1.75 M

   V2 = 750 / 1.75 ml

   V2 = 428.6 ml

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Answer:

bbecause it make the answer

Cycloalkanes are capable of forming geometric (cis-trans) isomers when they contain substituents on their carbon atoms. Geometric isomers are compounds with the same molecular formula but different spatial arrangements.

In cis isomers, the substituents are located on the same side of the cycloalkane, while in trans isomers, they are on opposite sides. For a cycloalkane to form cis-trans isomers, it must have at least two carbon atoms with substituents. The smallest cycloalkane capable of forming geometric isomers is cyclohexane due to its flexibility in adopting different conformations.

The cis isomer has both substituents on the same side of the ring, making them spatially closer to each other. In contrast, the trans isomer has substituents on opposite sides of the ring, causing them to be further apart.

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I think it's C

Atomic number

Answer:

wwwwwwwwww

Explanation:

The growth of the balloon after being taken out of liquid nitrogen is a result of the increase in temperature, which leads to an increase in the kinetic energy and speed of the gas molecules, causing them to spread out and expand, resulting in the expansion of the balloon.

The phenomenon you are describing, where a small, partially inflated balloon grows in size after being taken out of liquid nitrogen, can be explained by the principles of gas expansion due to temperature change.

When the balloon is submerged in liquid nitrogen, it is exposed to an extremely low temperature. As a result, the air molecules inside the balloon lose thermal energy and their average kinetic energy decreases. This decrease in kinetic energy causes the molecules to slow down and move closer together, leading to a decrease in pressure and volume of the gas inside the balloon.

When the balloon is removed from the liquid nitrogen and placed on the table, it starts to warm up. As the temperature increases, the air molecules regain thermal energy and their average kinetic energy rises. This increase in kinetic energy causes the molecules to move faster and spread out, leading to an increase in pressure and volume of the gas inside the balloon.

Furthermore, gases typically exhibit a property known as Thermal expansion, which means they expand when heated and contract when cooled. As the temperature of the air inside the balloon rises, the gas expands, causing the balloon to grow in size.

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Answer:

the answer is A

Explanation:

hope it hlp

Answer:

12

Explanation:

6 moles multiplied 2 atoms of O2

That gives 12 atoms