Chemistry balance equation
At equilibrium, the value of K = 0.00659 for the reaction: N₂ (g) + 3 H₂ (g) ← → 2 NH₃ (g). Calculate [N₂] when [NH₃] = 0.000123 M and [H₂] = 0.0275 M

If heat is going into a system, it means that energy must have come out from the surroundings.

Heat is a form of energy transfer from a hotter object to a cooler one, and the direction of heat flow is always from the hotter object to the cooler one.

Therefore, if heat is entering a system, it must be gaining energy from its surroundings, which are at a lower temperature and therefore have less thermal energy.

Conversely, if heat is leaving a system, it means that energy is being transferred from the system to its surroundings, which are at a higher temperature and therefore have more thermal energy.

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When the second cation first starts to precipitate, 80.8% of Ca²⁺ will still be in solution.

A cation is an ion with a positive charge. It is formed when an atom loses one or more of its electrons, resulting in a net positive charge. Cations are attracted to anions (ions with a negative charge) due to their opposite charges. Cations are found in many different substances, including acids, bases, and salts.

Ca₃(PO₄)₂ will be the first species that separates out of solution when solid Na₃PO₄ is introduced to the mixture. This is due to Ca3(PO4)2 having a substantially lower solubility than Ag₃PO₄ and Na₃PO₄.

The proportion of the first cation (Ca²⁺ ) still in solution when the second cation (Ag⁺) is just beginning to precipitate will depend on the starting concentrations of the two cations. In this instance, the starting concentrations of Ca²⁺  and Ag⁺ are 0.0400 M and 0.0990 M, respectively. Therefore, 80.8% of Ca²⁺  will still be in solution when its second cation first begins to precipitate.

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The mathematical variable is not a standard term in experimental design and is not typically used to describe variables in scientific experiments.

The variable that is unknown until the experiment is performed is typically the responding variable.

The responding variable is the variable that is observed and measured during the experiment, and its value changes in response to the manipulated variable. In contrast, the manipulated variable is the variable that is intentionally changed by the researcher to observe its effect on the responding variable.

The controlled variable is the variable that is kept constant throughout the experiment to ensure that any changes in the responding variable are due to the manipulated variable and not due to other factors. The mathematical variable is not a standard term in experimental design and is not typically used to describe variables in scientific experiments.

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We need to carry at least 187 CCCs on board the space shuttle to absorb all the CO2 produced by the crew during the 18-day mission.

To solve this problem, we need to use the following information:

First, we need to calculate the total amount of CO2 that will be expelled during the mission:

Total CO2 = 6 crew members x 42.0 g CO2/hour x 24 hours/day x 18 days = 136,080 g CO2

Next, we need to calculate the amount of LIOH needed to absorb this CO2. The balanced chemical equation for the reaction between CO2 and LIOH is:

CO2 + 2 LIOH → Li2CO3 + H2O

The molar mass of CO2 is 44.01 g/mol, and the molar mass of LIOH is 23.95 g/mol.

This means that 2 moles of LIOH are needed to absorb 1 mole of CO2.

So, to absorb 136,080 g of CO2, we need:

136,080 g CO2 x (1 mol CO2/44.01 g) x (2 mol LIOH/1 mol CO2) x (23.95 g LIOH/1 mol) = 139,648 g LIOH

Since each CCC contains 750 g of LIOH, we need:

139,648 g LIOH / 750 g CCC = 186.2 CCCs

Therefore, we need to carry at least 187 CCCs on board the space shuttle to absorb all the CO2 produced by the crew during the 18-day mission.

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Answer:

Unsaturated

Explanation:

A solution is unsaturated when it contains less than the maximum amount of solute that is capable of being dissolved.

To calculate the mass of KOH needed to make a 5 M solution in 185.5 mL, we need to use the formula:

mass = moles × molar mass

where moles is the amount of KOH in moles and molar mass is the mass of one mole of KOH.

We can calculate the moles of KOH as follows:

moles = Molarity × Volume (in liters)

First, we need to convert the volume from milliliters to liters:

185.5 mL = 0.1855 L

Now we can calculate the moles of KOH:

moles = 5 M × 0.1855 L = 0.9275 moles

The molar mass of KOH is 56.11 g/mol. Therefore, the mass of KOH needed is:

mass = 0.9275 moles × 56.11 g/mol = 52.05 g

Therefore, 52.05 grams of KOH are needed to make a 5 M solution in 185.5 mL.

The volume of the balloon once they have cooled to the outside temperature of -34.0 c is 1.53 L.

Charles' law predicts the relationship between the volumes and the temperatures of a sample of an ideal gas at different conditions. For this equation to hold true, the number of molecules and the pressure must remain constant despite changes in the environment.

Determine the volume of the balloon outside, V2. We do this by applying Charles' law, such that we relate the volume, V, and the temperature, T, of a sample of gas as

V₁ /T₁ = V₂/ T₂

at two conditions. We are given the following values for the variables:

• V₁ = 1.90 L

T₁ = 22.0+ 273.15 = 295.15 K

T₂= 34.0+273.15= 239.15 K

We proceed with the solution.

V₁/T₁ = V₂/T₂

V₁ /T₁ × T₂ = V₂

1.90 L/295.15 K x 239.15 K = V₂

1.53 L =V₂

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The complete question is

Balloons for a New Year's Eve party in Fargo, ND, are filled to a volume of 1.90 L at a temperature of 22.0 degrees Celsius and then hung outside where the temperature is -34.0 degrees Celsius. What is the volume of the balloons after they have cooled to the outside temperature? Assume that atmospheric pressure inside and outside the house is the same.

A potential problem associated with travel to another planet is : the extended time for humans to be in space.

It is highly likely that humans will travel to another planet, and Mars is currently considered the most viable option for human exploration. However, there are many potential problems associated with this endeavor, and one of the major issues is the extended time that humans would need to spend in space.

Traveling to Mars would take several months, and once there, astronauts would need to spend significant amount of time on planet before returning to Earth. This means that they would be exposed to high levels of radiation and would need to find ways to survive in harsh and unforgiving environment.

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The order of the energy levels for the Li2 molecule is:

1s < σ2s < 2s < σ*2s

The 1s orbital is the lowest in energy because it is closest to the nucleus and has the highest electron density. The σ2s orbital is next in energy because it is a bonding orbital that is formed by the overlap of two atomic 2s orbitals. The 2s orbital is higher in energy than the σ2s orbital because it is an atomic orbital that has not participated in bonding. The σ*2s orbital is the highest in energy because it is an antibonding orbital that weakens the bond between the two Li atoms.

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The initial sample had 0.010925 mol of Copper(II) oxide, or one mole.

The kinetic-molecular theory, which describes the states of matter, is based on the presumption that matter is composed of minuscule particles that are constantly in motion. This theory explains the observable properties and behaviours of solids, liquids, and gases. The container's walls and the quickly moving particles' collisions with one another are constant.

Copper(II) oxide + Sulfuric acid → Cupric sulfate + Water

One mole of Copper(II) oxide interacts with one mole of Sulfuric acid, as shown by the equation. The amount of Sulfuric acid that reacted with the Copper(II) oxide in the sample is therefore equal to the amount of Copper(II) oxide in the sample.

We must first determine how many moles of Sulfuric acid interacted with the sample:

moles Sulfuric acid = concentration × volume

moles Sulfuric acid = 0.437 mol/L × 0.025 L

moles Sulfuric acid = 0.010925 mol

Since the acid is in excess, the moles of Sulfuric acid remaining after treatment of the sample is:

moles Sulfuric acid remaining = moles Sulfuric acid added – moles Sulfuric acid reacted

moles Sulfuric acid remaining = 0.437 mol/L × 0.0354 L – 0.010925 mol

moles Sulfuric acid remaining = 0.007571 mol

To determine the number of moles of Copper(II) oxide in the original sample, we can use the following equation:

moles Copper(II) oxide = moles Sulfuric acid reacted = 0.010925 mol

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The reaction C₆H₅COOH(aq) + C₆H₅NH₂(aq) ⇌ C₆H₅COO⁻(aq) + C₆H₅NH₃⁺(aq) has an equilibrium constant of 0.3698.

The equilibrium constant (Kb) for the reaction can be calculated using the Ka and Kb values of the reactants and the equation:

Kw = Ka x Kb

where Kw = ion product constant of water (1.0 x 10⁻¹⁴ at 25°C).

Calculate the Kb value for aniline:

Kb = Kw/Ka = (1.0 x 10⁻¹⁴)/(4.3 x 10⁻¹⁰) = 2.33 x 10⁻⁵

Use the Kb value for aniline and the Ka value for benzoic acid to calculate the equilibrium constant (K) for the reaction:

K = Kb/Ka = (2.33 x 10⁻⁵)/(6.3 x 10⁻⁵) = 0.3698

Therefore, the equilibrium constant for the reaction C₆H₅COOH(aq) + C₆H₅NH₂(aq) ⇌ C₆H₅COO⁻(aq) + C₆H₅NH₃⁺(aq) is 0.3698.

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Answer: As they develop theories, chemists use models to attempt to explain their findings. Chemists assess the model they are using as new evidence becomes available and, if required, continue to refine it by making modifications.

Explanation:

The amount of zinc chloride that would be formed if 77.1 grams of zinc reacts is approximately 160.77 grams

To determine how many grams of zinc chloride ([tex]ZnCl_2[/tex]) would be formed if 77.1 grams of zinc (Zn) reacts, we'll use stoichiometry.

First, we need the molar masses of the substances involved:

Zn: 65.38 g/mol
[tex]ZnCl_2[/tex] : 136.29 g/mol
Now, we'll convert grams of Zn to moles:
77.1 g Zn × (1 mol Zn / 65.38 g Zn) = 1.179 moles Zn
According to the balanced chemical equation, 1 mole of Zn reacts to form 1 mole of [tex]ZnCl_2[/tex]:
1.179 moles Zn × (1 mol ZnCl₂ / 1 mol Zn) = 1.179 moles ZnCl₂

Finally, convert moles of ZnCl₂ to grams:
1.179 moles ZnCl₂ × (136.29 g ZnCl₂ / 1 mol ZnCl₂) ≈ 160.77 g ZnCl₂
So, approximately 160.77 grams of zinc chloride would be formed if 77.1 grams of zinc reacts.

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First, we need to calculate the heat gained by the cold water and the heat lost by the hot water:

Qcold = mcΔT = (30 g)(4.184 J/g-°C)(1.93°C) = 242.06 J

Qhot = mcΔT = (70 g)(4.184 J/g-°C)(-1.93°C) = -546.53 J

Since energy is conserved, we can assume that the heat gained by the cold water and calorimeter is equal to the heat lost by the hot water:

Qcold + Qcalorimeter = Qhot

Qcalorimeter = Qhot - Qcold

Qcalorimeter = -546.53 J - 242.06 J = -788.59 J

Therefore, the heat capacity of the calorimeter can be calculated as:

Ccalorimeter = Qcalorimeter / ΔT

Ccalorimeter = (-788.59 J) / (1.93°C)

Ccalorimeter ≈ -408.4 J/°C

Note that the negative sign indicates that the calorimeter loses heat when the system gains heat, which is expected since the calorimeter is absorbing some of the heat from the hot water.

a. The mass of Cr2O3 is 0.559 g Cr2O3 is the maximum amount of  Cr2O3 produced.

b.The limiting reactant is  Cr(NO3)3  because it produces less moles of Cr2O3 than Na2O.

c. The percent yield is 84%.

The balanced equation is shown below:

2 Cr(NO3)3 + 3 Na2O → Cr2O3 + 6 NaNO3

moles of Cr(NO3)3 = 1.75 g / 238.01 g/mol = 0.00735 mol

moles of Na2O = 1.75 g / 61.98 g/mol = 0.0282 mol

moles of Cr2O3 = (0.00735 mol Cr(NO3)3) × (1 mol Cr2O3 / 2 mol Cr(NO3)3) = 0.00368 mol Cr2O3 (theoretical yield)

mass of Cr2O3 = (0.00368 mol Cr2O3) × (151.99 g/mol) = 0.559 g Cr2O3

The percent yield can be calculated by dividing the actual yield by the theoretical yield and multiplying by 100%:

percent yield = (actual yield / theoretical yield) × 100%

percent yield = (0.455 g / 0.559 g) × 100% = 81.4%

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Answer:

0.0104 moles of gas in the flask.

Explanation:

To calculate the number of moles of gas in the flask, you can use the ideal gas law equation: PV = nRT. Where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant and T is temperature.

First, you need to convert the pressure from mmHg to atm and the temperature from Celsius to Kelvin. The pressure in atm is 760.0 mmHg / 760 mmHg/atm = 1 atm. The temperature in Kelvin is 17.00°C + 273.15 = 290.15 K.

Next, you need to convert the volume from mL to L. The volume in L is 250.0 mL / 1000 mL/L = 0.2500 L.

Now you can plug all the values into the ideal gas law equation and solve for n: (1 atm)(0.2500 L) = n(0.08206 L·atm/mol·K)(290.15 K). Solving for n gives n = 0.0104 mol.

So there are approximately 0.0104 moles of gas in the flask.

According to the information, we can infer that the difference between photographs 1 and 2 originate from the translation of the Moon around the earth (option C).

To explain the difference between both images we must take into account the movement patterns of the earth and the moon. In the case of the earth, it has 2 main movements, which are rotation on its own axis and translation around the sun.

On the other hand, the moon has a translational movement around the earth, which is what causes the different lunar phases. This motion causes the moon to appear partially shadowed from the earth because the earth blocks the sunlight.

Based on the above, we can infer that the correct answer is option C because this phenomenon is caused by the translation of the moon.

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The mass of silver oxide needed to decompose in order to produce 120.6 grams of silver is 494.5 grams.

The balanced chemical equation for silver oxide breakdown is:

[tex]Ag_2O[/tex] → [tex]2 Ag[/tex] + [tex]1/2 O_2[/tex]

The equation shows that for every mole of silver oxide that decomposes, two moles of silver are created, and the molar mass of [tex]Ag_2O[/tex] is 231.74 g/mol.

Hence, using stoichiometry, we can calculate the quantity of silver oxide necessary to generate 120.6 grams of silver:

120.6 g Ag × (1 mol Ag / 107.87 g Ag) × (1 mol [tex]Ag_2O[/tex]/ 2 mol Ag) × (231.74 g [tex]Ag_2O[/tex] / 1 mol [tex]Ag_2O[/tex] )

= 494.5 g [tex]Ag_2O[/tex]

As a result, 494.5 grams of silver oxide is needed to decompose in order to produce 120.6 grams of silver.

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The energy difference of an electron's transition from n = 6 to n = 1 in a hydrogen atom is approximately -2.17 × 10⁻¹⁸ Joules.

To calculate the energy difference (deltaE) of an electron's transition from n = 6 to n = 1 in a hydrogen atom, we can use the following equation:

deltaE = -13.6 * (1/n_final^2 - 1/n_initial^2) eV
where n_initial is the initial energy level (6 in this case), n_final is the final energy level (1 in this case), and -13.6 eV is the ionization energy of hydrogen.

Converting eV to Joules, we get:
1 eV = 1.602 x 10^-19 J
Therefore, deltaE in Joules can be calculated as follows:
deltaE = -13.6 * (1/1^2 - 1/6^2) * 1.602 x 10^-19 J/eV
deltaE = -2.179 x 10^-18 J

Therefore, the energy difference (deltaE) of an electron's transition from n = 6 to n = 1 in a hydrogen atom is -2.179 x 10^-18 J.

To calculate the energy difference (ΔE) for an electron's transition from n = 6 to n = 1 in a hydrogen atom, you can use the following formula:
ΔE = -13.6 eV * (1/nf² - 1/ni²)

Where ΔE is the energy difference in electron volts (eV), nf is the final energy level (1 in this case), and ni is the initial energy level (6 in this case).
ΔE = -13.6 eV * (1/1² - 1/6²)
ΔE ≈ -13.56 eV

Now convert electron volts to Joules:
1 eV = 1.6 × 10⁻¹⁹ J
ΔE ≈ -13.56 eV * 1.6 × 10⁻¹⁹ J/eV
ΔE ≈ -2.17 × 10⁻¹⁸ J

So,  approximately -2.17 × 10⁻¹⁸ Joules is  the energy difference of an electron's transition from n = 6 to n = 1 in a hydrogen atom.

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The volume of O2 that can be released at STP from the given sample of silver oxide is 23.45 mL.

Creating the balanced chemical equation for the breakdown of silver oxide is the first step in tackling this issue:

2Ag2O(s) → 4Ag(s) + O2(g)

We can deduce from the equation that 2 moles of AgO will result in 1 mole of O2. Since Ag2O has a molar mass of 231.735 g/mol, 0.4856 g of Ag2O is equivalent to:

0.4856 g Ag2O x (1 mol Ag2O/231.735 g Ag2O) = 0.002095 mol Ag2O

Therefore, the number of moles of O2 that can be produced from 0.4856 g of Ag2O is:

0.002095 mol Ag2O x (1 mol O2/2 mol Ag2O) = 0.0010475 mol O2

1 mole of any gas takes up 22.4 L of space at STP As a result, 0.4856 g of Ag2O can generate the following amount of O2 at STP:

0.0010475 mol O2 x 22.4 L/mol = 0.02345 L or 23.45 mL

Therefore, the volume of O2 that can be released at STP from the given sample of silver oxide is 23.45 mL.

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A mass of 37.0 g of sulfur must be used to produce 25.7 L of gaseous sulfur dioxide at STP.

The molar ratio of S₈ to SO₂ is 1:8.

At STP, one mole of gas occupies 22.4 L. Therefore, 25.7 L of SO₂ gas will contain;

25.7 L / 22.4 L/mol = 1.15 mol of SO₂.

The number of moles of S₈ needed is calculated as;

= 1.15 mol SO₂ / 8 mol S₈ per 1 mol SO₂

= 0.144 mol S₈.

The mass of S₈ needed is calculated as;

0.144 mol S₈ × 256.6 g/mol = 37.0 g of S₈.

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Explanation:

The temperature change of a substance when it absorbs or loses energy can be calculated using the specific heat capacity of the substance. The specific heat capacity of water is approximately 4.18 J/(g°C), which means that it takes 4.18 joules of energy to raise the temperature of 1 gram of water by 1 degree Celsius.

To calculate the temperature change of the water sample when 50 joules of energy is added, we need to use the following equation:

q = m * c * ΔT

where q is the amount of energy absorbed by the water, m is the mass of the water sample, c is the specific heat capacity of water, and ΔT is the resulting temperature change.

Rearranging the equation to solve for ΔT, we get:

ΔT = q / (m * c)

Plugging in the values, we get:

ΔT = 50 J / (m * 4.18 J/(g°C))

We need to know the mass of the water sample to calculate the temperature change. Let's assume a mass of 10 grams:

ΔT = 50 J / (10 g * 4.18 J/(g°C))

ΔT = 1.2°C

Therefore, if 50 joules of energy is added to a 10-gram sample of water, the resulting temperature change will be approximately 1.2 degrees Celsius.

B. To plot the data on a bar chart, draw a horizontal axis for metals and a vertical axis for time to complete the reaction. Then, draw bars for each metal that represent the amount of time required to complete the reaction. The height of the bars must match the time values ​​in the table.

C. No, Emilia was not correct in her forecast. According to the data, aluminum reacted 100 seconds faster than magnesium, which reacted in 50 seconds. Thus aluminum reacts more rapidly with hydrochloric acid than magnesium.

From most reactive to least reactive, the metals are as follows:

This order is consistent with the reactivity series, which is:

We are unable to estimate the reactivity of potassium, sodium, calcium, copper, silver, or gold from this experiment because those variables are not present in the data.

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The number of molecules of BF₃ present in 2 grams of BF₃ is 1.776×10²² molecules (1st option)

We'll begin by calculating the number of mole of 2 grams of BF₃. Details below:

Mole = mass / molar mass

Mole of BF₃ = 2 / 67.81

Mole of BF₃ = 0.02949 mole

Finally, we shall determine the number of molecules of BF₃. This is shown below:

Avogadro's hypothesis suggest that:

1 mole of BF₃ = 6.022×10²³ molecules

Therefore,

0.02949 mole of BF₃ = 0.02949 × 6.02×10²³

0.02949 mole of BF₃ = 1.776×10²² molecules

Thus, the number of molecules of BF₃ is 1.776×10²² molecules (1st option)

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The intital common temperature of copper and water is 9.5°C, under the condition that 10.0 g of ice at -10.0C is placed into 300.0 g of water in a 200.0-g copper calorimeter.

Now to evaluate the initial common temperature of the water and copper calorimeter, we have to apply the formula
m1c1(Tk - Ti) + m2c2(Tk - Ti)
= mcopperccopper(Tk - Ti)

Here,
m1 = mass of water,
c1 =specific heat capacity of water,
m2 = mass of copper calorimeter,
c2 = specific heat capacity of copper calorimeter, mcopper = mass of copper block
ccopper =specific heat capacity of copper.

Here, this equation to evaluate Ti
Ti = (m1c1Tk + m2c2Tk - mcopperccopperTk - m1c1Ti - m2c2Ti) / (m1c1 + m2c2 - mcopperccopper)

Staging the given values into this equation
Ti = (-300.0 g)(4.18 J/g°C)(18.0°C) + (200.0 g)(0.385 J/g°C)(18.0°C) + (10.0 g)(0.385 J/g°C)(18.0°C) / [(300.0 g)(4.18 J/g°C) + (200.0 g)(0.385 J/g°C) - (10.0 g)(0.385 J/g°C)]
Ti = 9.5°C

Hence, the initial common temperature of the water and copper calorimeter was 9.5°C.
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Answer:

9. B

10. Loses

Explanation:

9. Conduction is The procedure by which thermal energy or electricity is directly transported through a substance without the material moving when there is a variance in temperature between adjacent parts. Only choice B shows this process.

10. In exothermic reactions, energy/heat is lost. Exothermic reactions are characterized by a negative delta H, such as the delta H for the reaction show.

a. We will be able to release 37,827 kJ. b. You can cook a maximum of 19 hamburgers without exceeding the limit of 0.750 kg of carbon dioxide.

a. We need to use the balanced chemical equation and the enthalpy change of the reaction.

Therefore, moles  [tex]CO_2[/tex] produced are[tex]0.750 kg / 44.01 g/mol = 17.03 mol.[/tex]

The enthalpy change of the reaction is -2220.1 kJ/mol. Thus, the maximum number of kilojoules that can be released is:

[tex]\Delta Hrxn * moles of[/tex] [tex]CO_2[/tex] = [tex]-2220.1 kJ/mol * 17.03 mol = -37,827 kJ[/tex]

We need to reverse the sign of the answer, giving us 37,827 kJ.

b. If it requires 1900 kJ to cook one hamburger, we can divide the maximum number of kilojoules that can be released by the energy required to cook one hamburger:

37,827 kJ / 1900 kJ/hamburger = 19.91 hamburgers

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The pressure of a gas that occupies 50.5L at 69.0°C is 6.95 atm.

The pressure of an ideal gas can be calculated using Avogadro's equation as follows;

PV = nRT

Where;

According to this question, 12.5 mol of an ideal gas occupies 50.5 L at 69.00°C. The pressure can be calculated as follows:

P × 50.5 = 12.5 × 0.0821 × 342

50.5P = 350.9775

P = 6.95 atm

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Johannes Brsted and Thomas M. Lowry, two chemists, identified the Bromsted-Lowry acids and bases as a particular kind of acid-base reaction in 1923.

Acids are substances that give a base a proton (H+), whereas bases are substances that take a proton from an acid. In a Bromsted-Lowry acid-base reaction, the acid gives the base a proton in order to create the conjugate base and the conjugate acid, two new compounds.

Nitric acid (HNO3), sulfuric acid (H2SO4), and hydrochloric acid (HCl) are a few examples of acids. Sodium hydroxide (NaOH), ammonium hydroxide (NH4OH), and potassium hydroxide (KOH) are a few examples of bases.

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