4NH3+6NO --> 5N2 + 6H20


How many liters of NH3 at 32. 6 °C and 4. 25 kPa are needed to react


completely with 30. 0L of NO at STP?

A 20-liter sample of gas at STP would occupy 5.68 liters if the pressure was changed to 20 atm and the temperature was changed to 38°C.

To solve this problem, we can use combined gas law, which relates the pressure, volume, and temperature of a gas. The formula for the combined gas law is:

[tex](P_1 * V_1) / (T_1 * n_1) = (P_2 * V_2) / (T_2 * n_2)[/tex]

where P1 and P2 are the initial and final pressures of the gas [tex]V_1[/tex] and [tex]V_2[/tex] are the initial and final volumes of the gas.

At STP, the conditions are 1 atmosphere of pressure and 0°C (273 K) of temperature.

Therefore, we can use these values as our initial conditions [tex](P_1 = 1\ atm, T_1 = 273 K)[/tex] and solve for [tex]V_2[/tex], the final volume of the gas:

[tex](P_1 * V_1) / T_1 = (P_2 * V_2) / T_2\\V_2 = (P_1 * V_1 * T_2) / (P_2 * T_1)[/tex]

Substituting the given values, we get:

[tex]V_2 = (1 atm * 20 L * 311 K) / (20 atm * 273 K) \\V_2 = 5.68 L[/tex]

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The original volume of CI₂ was 19.33 L.

According to Charles' Law, the volume of a gas is directly proportional to its temperature at constant pressure. This can be expressed as V₁/T₁ = V₂/T₂, where V₁ and T₁ are the initial volume and temperature, and V₂ and T₂ are the final volume and temperature.

In this problem, we are given the initial temperature (T₁ = 836.06 K), final temperature (T₂ = 625.29 K), and final volume (V₂ = 14.509 L). We are asked to find the initial volume (V₁). To do this, we can rearrange the Charles' Law equation to solve for V₁:

V₁ = (V₂/T₂) x T₁

Plugging in the values, we get:

V₁ = (14.509 L/625.29 K) x 836.06 K

V₁ = 19.35 L

As a result, the initial volume of CI₂ was 19.33 L.

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The percent by mass of carbon in the hydrocarbon is 85.7%.

To find the percent by mass of carbon in the hydrocarbon, follow these steps:

1. Determine the mass of hydrogen in the hydrocarbon: 2.86g.
2. Calculate the mass of carbon in the hydrocarbon by subtracting the mass of hydrogen from the total mass: 20.0g (total mass) - 2.86g (mass of hydrogen) = 17.14g (mass of carbon).
3. Calculate the percent by mass of carbon by dividing the mass of carbon by the total mass of the hydrocarbon, and then multiply by 100: (17.14g carbon / 20.0g total mass) * 100 = 85.7%.

So, the percent by mass of carbon in the 20.0g hydrocarbon sample is 85.7%.

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The experimental step necessary to determine the molar mass using the freezing point depression method is to measure the freezing point of the pure solvent and then measure the freezing point of the solution. The statement 2 is correct.

The freezing point depression method is a common technique used to determine the molar mass of a solute dissolved in a solvent. The method is based on the principle that the presence of a solute lowers the freezing point of the solvent. By measuring the change in the freezing point of the solvent caused by the solute, it is possible to calculate the molar mass of the solute. Correct answer is option 2.

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--The complete Question is, which statement describes an experimental step(s) that is necessary to determine the molar mass using the freezing point depression method?

1. measure the heat of fusion of the pure solvent and then measure the heat of fusion of the pure solute.

2. measure the freezing point of the pure solvent and then measure the freezing point of the solution.

3. determine the molar mass of the solute by looking up the elements in the periodic table. 4. calculate the number of moles in a kilogram of solvent to determine its molality. --

When Owen has 28.5 grams of liquid benzene at a temperature of 287.6 K, a total of 3.809 kJ of energy is released during the freezing process.

To find the energy released when benzene freezes, we need to know its heat of fusion and the amount of benzene that freezes. The heat of fusion of benzene is 10.4 kJ/mol.

First, we need to determine how many moles of benzene we have:

Molar mass of benzene (C₆H₆) = 78.11 g/mol

Number of moles of benzene = 28.5 g / 78.11 g/mol = 0.3647 mol

Since the molar ratio of benzene to energy released is 1:1, the energy released when benzene freezes can be calculated as:

Energy released = moles of benzene x heat of fusion

Energy released = 0.3647 mol x 10.4 kJ/mol = 3.809 kJ

Therefore, 3.809 kJ of energy is released when the given amount of liquid benzene freezes.

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To calculate the shear stress required to deform benzene at a velocity gradient of 4900 s-1, we can use the equation:

Shear stress = viscosity x velocity gradient

Plugging in the given values, we get:

Shear stress = 0.000651 Pa. S x 4900 s-1

Shear stress = 3.19 Pa

Therefore, a shear stress of 3.19 Pa is required to deform benzene at a velocity gradient of 4900 s-1.

What is Shear stress?

Shear stress is a type of stress that occurs when a force is applied parallel to a surface or along a plane within a material. It is the result of the force causing the material to deform or change shape, with one part of the material sliding or shearing over another part.

Shear stress is often described in terms of the shear force per unit area, or shear strength, that is required to cause the material to shear or deform. The unit of measurement for shear stress is typically in pascals (Pa) or pounds per square inch (psi).

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The loss of mass could be accounted for by the release of gases such as water vapor, carbon dioxide, and sulfur dioxide during the reaction between sulfuric acid and sugar.

This is due to the fact that both sugar and sulfuric acid are organic compounds, and when heated, they undergo dehydration and decomposition reactions respectively, producing gases that escape the system. The carbon product is also likely to be less dense than the reactants, resulting in a further loss of mass.

Additionally, some of the sugar may have not fully reacted due to incomplete mixing, resulting in residual sugar that was not accounted for in the mass measurement. Overall, the loss of mass is expected in any chemical reaction, and in this case, it can be attributed to the production of gases and incomplete reaction.

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To find the molarity of a solution, we need to know the number of moles of solute (in this case, sugar) and the volume of the solution in liters. We can use the given mass of sugar and the molar mass of sugar to find the number of moles:

Mass of sugar = 194.55 g

Molar mass of sugar (C12H22O11) = 342.3 g/mol

Number of moles of sugar = Mass of sugar / Molar mass of sugar

Number of moles of sugar = 194.55 g / 342.3 g/mol

Number of moles of sugar = 0.568 mol

Now we need to convert the given volume of the solution (250 mL) to liters:

Volume of solution = 250 mL

Volume of solution = 250 mL / 1000 mL/L

Volume of solution = 0.250 L

Finally, we can use the number of moles of sugar and the volume of the solution to calculate the molarity:

Molarity = Number of moles of sugar / Volume of solution

Molarity = 0.568 mol / 0.250 L

Molarity = 2.27 M

Therefore, the molarity of the sugar solution is 2.27 M.

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The reaction of 42.7 L of chlorine gas at STP with 50.0 g of aluminum produces 150.5 g of aluminum chloride.

The balanced chemical equation for the reaction between aluminum and chlorine gas is:

2Al + 3Cl₂ -> 2AlCl₃

To use this equation to calculate the grams of aluminum chloride produced, we need to convert the given volume of chlorine gas to moles using the ideal gas law:

n = PV/RT

At STP, the pressure (P) and temperature (T) are 1 atm and 273 K, respectively. The volume (V) is given as 42.7 L. The gas constant (R) is 0.08206 L atm K⁻¹ mol⁻¹ Plugging these values in, we get:

n = (1 atm * 42.7 L) / (0.08206 L atm K⁻¹ mol⁻¹ * 273 K) = 1.694 mol

Since the stoichiometry of the balanced equation is 2:3 (2 moles of aluminum react with 3 moles of chlorine gas to produce 2 moles of aluminum chloride), we need to calculate how many moles of aluminum are needed to react with 1.694 moles of chlorine gas:

2 mol Al / 3 mol Cl₂ * 1.694 mol Cl₂ = 1.129 mol Al

Finally, we can use the molar mass of aluminum chloride (133.34 g/mol) to calculate the grams of product:

1.129 mol AlCl₃ * 133.34 g/mol = 150.5 g AlCl₃

Therefore, 150.5 g of aluminum chloride would be produced from 42.7 L of chlorine gas at STP reacting with 50.0 g of aluminum.

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We can use titration to figure out the concentration of potassium hydroxide in a water sample and the laboratory setup/investigation will dry Erlenmeyer flask and other equipment.

To determine the concentration of potassium hydroxide (KOH) in a water sample, we can use an acid-base titration with a known concentration of a strong acid, such as hydrochloric acid (HCl).

The laboratory setup for this titration would involve:

Measuring a precise volume of the water sample containing the KOH and transferring it to a clean and dry Erlenmeyer flask. Adding a few drops of a suitable indicator, such as phenolphthalein, to the Erlenmeyer flask.

Filling a burette with the HCl solution of known concentration. Titrating the HCl solution into the Erlenmeyer flask containing the water sample, slowly and carefully swirling the flask until the indicator changes color. Recording the volume of HCl solution added at the point of color change. The concepts behind this titration involve the neutralization of KOH by HCl:

KOH + HCl → KCl + H2O

The endpoint of the titration occurs when all of the KOH has been neutralized by the HCl, leaving only HCl and KCl in the solution. At this point, the indicator changes color, signaling that the titration is complete.

From the volume and concentration of the HCl solution used in the titration, we can calculate the moles of HCl added. Since the stoichiometry of the reaction is 1:1, the moles of HCl added is equal to the moles of KOH in the water sample.

Finally, we can use the volume and moles of KOH to calculate the concentration of KOH in the water sample, expressed in units of molarity (M).

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The answer to the part A, B, C, D and E are as follows-

Part A: [tex][H3O+] [OH−] pOH[/tex] pH Acidic or Basic

1.0×10−8 1.0×10−6 6.00 8.00 basic

1.0×10−5 1.0×10−9 9.00 5.00 acidic

9.7×10−9 1.0×10−5 5.00 8.99 basic

[tex]1.0×10−14 1.0×10^14 14.00 0.00 neutral\\1.0×10^−3 1.04×10^−11 11.98 2.00 acidic[/tex]

Part B:

[tex][H3O+] [OH−] pOH[/tex] pH Acidic or Basic

1.0×10−8 1.0×10−6 6.00 8.00 basic

1.0×10−5 1.0×10−9 9.00 5.00 acidic

9.7×10−9 1.0×10−5 5.00 8.99 basic

[tex]1.0×10−14 1.0×10^14 14.00 0.00 neutral\\1.0×10^−3 1.04×10^−11 11.98 2.00 acidic[/tex]

Part C:

[tex][H3O+] [OH−] pOH[/tex]pH Acidic or Basic

1.0×10−8 1.0×10−6 6.00 8.00 basic

1.0×10−5 1.0×10−9 9.00 5.00 acidic

9.7×10−9 1.0×10−5 5.00 8.99 basic

[tex]1.0×10−14 1.0×10^14 14.00 0.00 neutral\\1.0×10^−3 1.04×10^−11 11.98 2.00 acidic[/tex]

Part D:

[tex][H3O+] [OH−] pOH[/tex]pH Acidic or Basic

1.0×10−8 1.0×10−6 6.00 8.00 basic

1.0×10−5 1.0×10−9 9.00 5.00 acidic

9.7×10−9 1.0×10−5 5.00 8.99 basic

[tex]1.0×10−14 1.0×10^14 14.00 0.00 neutral\\1.0×10^−3 1.04×10^−11 11.98 2.00 acidic[/tex]

Part E:

[tex][H3O+] [OH−] pOH[/tex]pH Acidic or Basic

1.0×10−8 1.0×10−6 6.00 8.00 basic

1.0×10−5 1.0×10−9 9.00 5.00 acidic

9.7×10−9 1.0×10−5 5.00 8.99 basic

[tex]1.0×10−14 1.0×10^14 14.00 0.00 neutral\\1.0×10^−3 1.04×10^−11 11.98 2.00 acidic[/tex]

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Beer's law establishes a connection between a substance's concentration in a solution and its absorbance at a certain wavelength.

By charting the absorbance vs concentration of each solution, a Beer's law plot is created in order to calculate concentrations of a series of copper(II) sulfate solutions with known absorbances at a set wavelength. The resulting graph should be a straight line that the least-squares approach can fit with a linear equation. The molar absorptivity is represented by slope of the best-fit line, and the absorbance at zero concentration is represented by the y-intercept. By measuring the absorbance of the unknown copper(II) sulfate solutions and solving for concentration using the equation, the concentrations of unknown solutions can be found.

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--The complete Question is, Use Beer's law plot and best-fit line to determine the concentrations for a series of copper(II) sulfate solutions with known absorbances at a fixed wavelength. --

The temperature of this 282.8 g copper sample changed by approximately 6.78 degrees Celsius.

To find the temperature change of a 282.8 g sample of copper that releases 175.1 calories of heat with a specific heat capacity of 0.092 cal/(g·°C), we can use the following formula:
q = mcΔT

where:
q = heat released (calories)
m = mass of the sample (grams)
c = specific heat capacity (cal/(g·°C))
ΔT = temperature change (°C)

Step 1: Plug in the given values into the formula.
175.1 = (282.8)(0.092)(ΔT)

Step 2: Solve for ΔT.
ΔT = 175.1 / (282.8× 0.092)

Step 3: Calculate the value of ΔT.
ΔT ≈ 6.78 °C

So, the temperature of this 282.8 g copper sample changed by approximately 6.78 degrees Celsius.

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The temperature of the gas at standard pressure is 219.02 °C.

Gay-Lussac's law states that the pressure exerted by a given quantity of gas varies directly with the absolute temperature of the gas.

It is expressed as;

P₁/T₁ = P₂/T₂

We know that the pressure (P1) is 499.0 mmHg at a temperature (T1) of 50.0°C. We want to find the temperature (T2) at standard pressure (P2 = 1 atm = 760 mmHg). We also know that the volume (V1) is constant, so we can write:

P₁/T₁ = P₂/T₂

Solving for T2, we get:

T2 = (P2 × T1)/P1

T2 = (760 mmHg × 323.15 K)/499.0 mmHg

T2 = 492.172 K

Converting this temperature to °C, we get:

T2 = 492.172 K - 273.15

T2 = 219.02 °C

Therefore, the temperature is 219.02 °C.

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Answer:

492.17 K (2 d.p.) = 219.02 °C (2 d.p.)

Explanation:

To find the final pressure inside the steel tank, we can use Gay-Lussac's law since the volume is constant.

[tex]\boxed{\sf \dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}}[/tex]

where:

As we are solving for the final temperature, rearrange the equation to isolate T₂:

[tex]\sf T_2=\dfrac{P_2T_1}{P_1}[/tex]

Convert the initial temperature from Celsius to Kelvin by adding 273.15:

[tex]\implies \sf T_1=50+273.15=323.15\;K[/tex]

The standard pressure is 1 atm = 760 mmHg.

Therefore, the values to substitute into the equation are:

Substitute the values into the equation and solve for T₂:

[tex]\implies \sf T_2=\dfrac{760 \cdot 323.15}{499}[/tex]

[tex]\implies \sf T_2=\dfrac{245594}{499}[/tex]

[tex]\implies \sf T_2=492.172344689...[/tex]

[tex]\implies \sf T_2=492.17\;K\;(2\;d.p.)[/tex]

Therefore, the temperature at standard pressure for a gas with a pressure of 499.0 mmHg at 50.0 °C is 492.17 K (or 219.02 °C).

If the decomposition of hydrogen peroxide into water and oxygen gas is an exothermic reaction, we would expect the final temperature to be lower than the initial temperature of 28˚C.

This is due to the fact that energy is released from the system during an exothermic reaction in the form of heat into the surroundings. In other words, the energy of the reactants is more than that of the products, and the excess energy is released into the environment.

As a result, the environment's temperature will rise, while the system's temperature will fall. This indicates that the reaction's final temperature will be lower than its 28° C starting point.

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The boiling points of the hydrocarbons can be ranked as follows;

1. 4

2. 2

3. 3

4. 1

The size of the molecules and the nature of the intermolecular interactions between the molecules essentially determine the boiling points of hydrocarbons.

Because they have more electrons and a larger surface area available for intermolecular interactions like Van der Waals forces, larger hydrocarbon molecules typically have higher boiling points.

Additionally, polar hydrocarbons and those that can form hydrogen bonds have higher boiling points than non-polar hydrocarbons because of stronger intermolecular forces.

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The student incorrectly calculated the mass of the sample of pyrite by assuming the density of pyrite to be 2 g/cm³, which is actually the density of water. The actual density of pyrite is about 5 g/cm³, so the actual mass of the 10 cm³ sample would be 50 g.

The student likely confused the concept of density, which is the mass per unit volume of a substance, with the specific gravity, which is the ratio of the density of a substance to the density of water.

Pyrite has a specific gravity of about 5, meaning that its density is about 5 times greater than that of water. Therefore, the mass of a 10 cm³ sample of pyrite would be 5 times greater than the mass of a 10 cm³ sample of water, or 50 g.

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1.21 grams of hydrogen are required to produce 0.6 moles of methane (CH₄) in the given reaction.

The given reaction is:

C + 2H₂ → CH₄

We can see that 2 moles of hydrogen (H₂) are required to produce 1 mole of methane (CH₄) according to the balanced chemical equation. Therefore, to produce 0.6 moles of methane, we will need 2 times as many moles of hydrogen, which is:

number of moles of hydrogen = 2 × number of moles of methane

number of moles of hydrogen = 2 × 0.6 moles

number of moles of hydrogen = 1.2 moles

To convert the number of moles of hydrogen to grams, we need to use the molar mass of hydrogen, which is approximately 1.008 g/mol. Thus, the mass of hydrogen required can be calculated as:

mass of hydrogen = number of moles of hydrogen × molar mass of hydrogen

mass of hydrogen = 1.2 moles × 1.008 g/mol

mass of hydrogen = 1.21 g

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The complete question is:

For the reaction C+2H₂ → CH₄, how many grams of hydrogen are required to produce 0.6 moles of methane, CH₄?

When given 20.76 moles of H2 and plenty of O2, you can make 20.76 moles of H2O.

To determine how many moles of H2O can be produced from 20.76 moles of H2 and plenty of O2, we'll use the balanced chemical equation provided: 2H2 + 1O2 --> 2H2O.

Step 1: Identify the limiting reactant. In this case, we have plenty of O2, so H2 is the limiting reactant.

Step 2: Determine the mole ratio between the limiting reactant (H2) and the product (H2O). According to the balanced equation, the mole ratio is 2H2 to 2H2O, or 1:1.

Step 3: Calculate the moles of H2O produced. Since the mole ratio is 1:1, the number of moles of H2O produced will be the same as the number of moles of H2 available. Thus, you can produce 20.76 moles of H2O.

In summary, when given 20.76 moles of H2 and plenty of O2, you can make 20.76 moles of H2O.

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The answer to the multistep calculation, expressed using the appropriate number of significant figures, is 24.3 feet.

In order to determine the appropriate number of significant figures in the answer, we need to follow the rules of significant figures for addition and subtraction.

When adding or subtracting numbers, the answer should be rounded to the same number of decimal places as the measurement with the least number of decimal places.

Here, the measurement with the least number of decimal places is 10.0 feet, which has one decimal place. Therefore, we should round the final answer to one decimal place as well.

Now, let's perform the calculation:

87.95 feet x 0.277 feet + 5.02 feet - 1.348 feet + 10.0 feet = 24.3108725 feet

Rounding to one decimal place, the final answer is:

24.3 feet

Therefore, the answer to the multistep calculation, expressed using the appropriate number of significant figures, is 24.3 feet.

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The complete combustion of one gallon of gasoline containing 2370.0 grams of octane produces 6888.2 grams of carbon dioxide.

In 2017, people in the US generated approximately 9.85 x 10¹⁴ grams of carbon dioxide by burning 143 billion gallons of gasoline.



1. Write the balanced chemical equation for the combustion of octane:
  2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O

2. Determine the molecular weight of octane (C₈H₁₈) and carbon dioxide (CO₂):
  C₈H₁₈: (8 x 12.01) + (18 x 1.01) = 114.23 g/mol
 CO₂: (1 x 12.01) + (2 x 16.00) = 44.01 g/mol

3. Use stoichiometry to find the grams of CO₂ produced from the combustion of 2370.0 grams of octane:
  (2370.0 g octane) x (16 mol CO₂/ 2 mol octane) x (44.01 g CO₂ / mol CO₂) = 6888.2 g CO₂

4. Calculate the total grams of CO₂ generated by burning 143 billion gallons of gasoline in the US in 2017:
  (143 billion gallons) x (6888.2 g CO₂ / gallon) = 9.85 x 10¹⁴ grams of CO₂

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The order of the reaction in ligand is zeroth order, as changing the ligand concentration from 1.0 mM to 200 mM does not affect the reaction rate. The rate equation is: rate = k[substrate], where k is the rate constant.

The order of the reaction in substrate is first order, as doubling the substrate concentration (from 5 mM to 10 mM) leads to a doubling of the reaction rate so the or.

To find the MSDS for decahydronaphthalene, one can search for it on the website of the manufacturer or supplier. Alternatively, one can search for it on the website of the National Institute for Occupational Safety and Health (NIOSH), which provides a database of MSDSs for various chemicals.

It is important to consult the MSDS before handling or using the chemical, as it contains information on its physical and chemical properties, hazards, and precautions for safe use and disposal.

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All of these elements are in the same period, we can focus on the groups. As we move from left to right, the atomic radius decreases. Therefore, lithium (Li) would have the largest atomic radius among the elements you listed.

To know which of these elements has the largest atomic radius: lithium (Li), beryllium (Be), boron (B), carbon (C), nitrogen (N), oxygen (O), fluorine (F), and neon (Ne).

The largest atomic radius can be determined by examining the elements in the periodic table. Atomic radius generally decreases across a period (from left to right) and increases down a group (top to bottom).

Considering the elements you provided:
- Li (lithium) is in Group 1 and Period 2
- Be (beryllium) is in Group 2 and Period 2
- B (boron) is in Group 13 and Period 2
- C (carbon) is in Group 14 and Period 2
- N (nitrogen) is in Group 15 and Period 2
- O (oxygen) is in Group 16 and Period 2
- F (fluorine) is in Group 17 and Period 2
- Ne (neon) is in Group 18 and Period 2

Since all of these elements are in the same period, we can focus on the groups. As we move from left to right, the atomic radius decreases. Therefore, lithium (Li) would have the largest atomic radius among the elements you listed.

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1. Xylene will react with bromine solution to undergo electrophilic aromatic substitution, where bromine will replace one of the hydrogen atoms on the aromatic ring.

2. Xylene will not react with KMnO₄ under normal conditions as it is a relatively stable aromatic compound.

3. Xylene can react with AlCl₃ and CHCl₃ under Friedel-Crafts conditions to form a substituted product. AlCl₃ acts as a Lewis acid, facilitating the reaction by generating a carbocation intermediate, which is then attacked by the chloride ion from CHCl3 to form a substituted product.

In summary, xylene will undergo electrophilic aromatic substitution with bromine solution, will not react with KMnO₄, and can undergo Friedel-Crafts reaction with AlCl₃ and CHCl₃ to form a substituted product.

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The freezing point of the solution is approximately 1.06306 °C

The freezing point of a solution of 0.300 mol of lithium bromide in 525 mL of water would be lower than the freezing point of pure water. The exact freezing point depression can be calculated using the formula ΔTf = Kf·m, where ΔTf is the freezing point depression, Kf is the freezing point depression constant of water (1.86 °C/m), and m is the molality of the solution. To find the molality of the solution, we need to convert the volume of water to mass using its density (1 g/mL), which gives us 525 g of water. Then, we can calculate the molality as:

molality = moles of solute/mass of solvent in kg
             = 0.300 mol / 0.525 kg
             = 0.571 mol/kg

Substituting this value into the freezing point depression formula, we get:

ΔTf = 1.86 °C/m x 0.571 mol/kg
      = 1.06306 °C

Therefore, the freezing point of the solution would be lowered by 1.06 °C compared to pure water.

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Two balloons can repel each other without touching by becoming charged through friction, resulting in a net repulsive force between them due to the interaction of their charges.

This phenomenon is governed by Coulomb's law & can be explained by the behavior of atoms and molecules at a microscopic level.

Now, when the two balloons are brought near each other, the negatively charged balloon repels the electrons in the other balloon, causing the atoms in the balloon to shift slightly.

This results in a slight imbalance of charge, with one side of the balloon becoming positively charged & the other becoming negatively charged.

The positively charged side of the balloon is attracted to the negatively charged balloon, while the negatively charged side is repelled by it. This creates a net repulsive force between the two balloons, causing them to move away from each other without touching.

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31.45 g of dilute trioxonitrate (V) acid containing 10% W/W of pure acid will be required to dissolve 2.5 g of chalk.

We need to use balanced chemical equation of the reaction between calcium carbonate and trioxonitrate (V) acid to determine the number of moles of acid required to dissolve 2.5 g of chalk.

[tex]CaCO_3 + 2HNO_3 → Ca(NO_3)_2 + CO_2 + H_2O[/tex]

From the equation, one mole of [tex]CaCO_3[/tex] reacts with two moles of [tex]HNO_3[/tex]. The molar mass of CaCO3 is 100.09 g/mol.

[tex]Number\ of\ moles\ of\ CaCO_3 = 2.5 g / 100.09 g/mol = 0.02498 mol[/tex]

[tex]Number\ of\ moles\ of HNO_3 = 2 * 0.02498 = 0.04996 mol[/tex]

Now, we can calculate the mass of dilute trioxonitrate (V) acid containing 10% W/W of pure acid required to provide 0.04996 mol of [tex]HNO_3[/tex].

Assuming the density of the dilute trioxonitrate (V) acid is 1.1 g/cm3, the mass of the acid required will be:

[tex]Mass\ of\ acid = (0.04996 mol * 63.01 g/mol) / 0.1 = 31.45 g[/tex]

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2.4453  ×  10⁹ KJ energy was evolved  when the total volume of hydrogen gas needed to fill the hindenburg was 2.09 × 10⁸ l at 1.00 atm and 25.1°

According to the given data,  

Volume of the hydrogen gas = 2.09 × 10⁸ L

Pressure of the gas = P = 1 atm

Temperature of the gas =T = 25.1 °C =298.1 K

We know that, for an ideal gas equation

PV=nRT

1 atm ×2.09 × 10⁸ L = n × 0.0820 atmL/molK × 298.1 K

⇒n = 1 atm ×2.09 × 10⁸ L/  0.0820 atmL/molK × 298.1 K

⇒n = 0.0855 ×10⁸ mol

ΔH for hydrogen gas is =-286 kJ/mol

For  0.0855 ×10⁸ mol energy evolved when hydrogen gas is burned =

0.0855 ×10⁸ mol × (-286 KJ/mol) = -2.4453  ×  10⁹ KJ

Therefore, 2.4453  ×  10⁹ KJ energy was evolved when it was burned.

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The complete question is-

The total volume of hydrogen gas needed to fill the hindenburg was 2.09 × 108 l at 1.00 atm and 25.1°. how much energy was evolved when it burned?

Answer:

388.88 mmHg (2 d.p.)

Explanation:

To find the final pressure when the volume is kept constant, we can use Gay-Lussac's law.

[tex]\boxed{\sf \dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}}[/tex]

where:

The values to substitute into the equation are:

Substitute the values into the equation and solve for P₂:

[tex]\implies \sf \dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}[/tex]

[tex]\implies \sf \dfrac{745}{523 }=\dfrac{P_2}{273}[/tex]

[tex]\implies \sf P_2=\dfrac{745 \cdot 273}{523 }[/tex]

[tex]\implies \sf P_2=\dfrac{203385}{523 }[/tex]

[tex]\implies \sf P_2=388.88145315...[/tex]

[tex]\implies \sf P_2=388.88\;mmHg\;(2\;d.p.)[/tex]

Therefore, the final pressure would be 388.88 mmHg if a gas is cooled from 523 K to 273 K and the volume is kept constant, starting with an initial pressure of 745 mmHg.

You would need to walk at a brisk pace for about 1 hour and 40 minutes to burn off the calories obtained from eating the cheeseburger.

25 g of protein and 31 g of carbohydrates release 4 cal/g, which equals 240 calories. 25 g of fat release 9 cal/g, which equals 225 calories. So, the total calories in the cheeseburger are 465.

Now, to burn off 465 calories at a rate of 280 cal/h, we need to divide 465 by 280, which equals 1.66 hours or approximately 1 hour and 40 minutes.

In summary, to burn off the calories obtained from a cheeseburger containing 25 g of protein, 25 g of fat, and 31 g of carbohydrates, you would need to walk at a brisk pace for about 1 hour and 40 minutes.

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