20. 0 g of Potassium reacts with water to produce Potassium hydroxide and hydrogen gas.


2 K + 2 H2O —> 2 KOH + H2


How many miles of hydrogen are there?

Answer: it's an integral component of steel

Explanation: it's also an economic essential to US growth and is used for transportation, energy, and construction

The theoretical yield of aluminum fluoride (AlF₃) obtained from 0.45 mol of aluminum (Al) is 0.45 mol.

The balanced chemical equation for the reaction between aluminum (Al) and fluorine (F₂) to form aluminum fluoride (AlF₃) is:

2 Al + 3 F₂ → 2 AlF₃

According to the equation, 2 moles of aluminum react with 3 moles of fluorine to produce 2 moles of aluminum fluoride. Therefore, the stoichiometric ratio of aluminum to aluminum fluoride is 2:2 or 1:1.

Given that 0.45 mol of aluminum is used in the reaction, the theoretical yield of aluminum fluoride can be calculated as follows:

0.45 mol Al × (2 mol AlF₃ ÷ 2 mol Al) = 0.45 mol AlF₃

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That statement "Heterocyclic aromatic compounds undergo electrophilic aromatic substitution in a similar fashion to that undergone by benzene with the formation of a resonance-stabilized intermediate." is generally true.

Heterocyclic aromatic compounds, like benzene, contain a ring of atoms with alternating double bonds (pi bonds) and exhibit delocalized pi electrons that are responsible for their aromaticity.

Electrophilic aromatic substitution is a common reaction for these types of compounds, where an electrophile is attracted to the electron-rich ring and substitutes for one of the hydrogen atoms.

The resulting intermediate is a resonance-stabilized carbocation, just like in the case of benzene.

However, the reactivity and selectivity of heterocyclic aromatic compounds may differ from that of benzene due to differences in the electronic properties of the heteroatom(s) in the ring and their effect on the ring's electron density.

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654 grams of Zn metal will completely react with 730 grams of HCl

The balanced chemical equation for this reaction is:
Zn + 2HCl → ZnCl2 + H2

From the equation, we can see that for every 1 mole of Zn, 2 moles of HCl are required for a complete reaction. This means the mole ratio of Zn to HCl is 1:2.

To determine the number of moles of HCl used, we need to convert the given mass of HCl to moles. The molar mass of HCl is 36.5 g/mol, so:

730 g HCl x (1 mol HCl/36.5 g HCl) = 20 moles HCl

Using the mole ratio from the balanced equation, we can determine the number of moles of Zn required:

20 moles HCl x (1 mol Zn/2 mol HCl) = 10 moles Zn

Finally, we can convert the number of moles of Zn to grams using its molar mass of 65.4 g/mol:

10 moles Zn x (65.4 g Zn/mol) = 654 grams of Zn

Therefore, 654 grams of Zn metal will completely react with 730 grams of HCl to produce ZnCl2 and H2.

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The compounds A, B and C are ammonium chloride, ammonia gas and silver chloride respectively.

Ammonium chloride,  is a white crystalline solid which is soluble in water. On heating with sodium hydroxide it will produce ammonia gas, which is a colorless gas and has a odor or pungent smell.

So,  Ammonia gas will turn red litmus into blue as it's pH is 11.6.

When Ammonium chloride reacts with silver nitrate  in  presence of dilute Nitric acid ,  it produces Silver chloride and Ammonium nitrate

AgCl is soluble in Ammonia because it can form complexes which

makes it behave like an ion, making it soluble.

Therefore, Compound A is Ammonium chloride,

B is ammonia gas

and C is Silver chloride.

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The complete question is

A white solid A when heated with sodium hydroxide solution gives a pungent gas B which turns red litmus paper blue. The solid when dissolved in dilute nitric acid is treated with Silver nitrate solution to give white precipitate C, which is soluble in ammonia.

(A) What are the substance A, B, and C?

C. When rocks are transported by water, abrasion occurs as they rub and bump into each other.

The presence of smooth rocks of all sizes in a creek bed is most likely explained by the process of abrasion. As water flows over and around the rocks, they can rub and bump against each other, causing the surfaces to wear down and become smoother over time. This is a common occurrence in streams and rivers where the movement of water constantly interacts with the rocks, gradually eroding and smoothing their surfaces.

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The calculation of the mass of SnF₂ present in the toothpaste sample determined it to be 0.105 g. The mass percentage of stannous fluoride in the toothpaste sample was found to be 1.05%. The percent yield of the extraction/recovery process, comparing the recovered mass of SnF₂ to the expected mass based on the percentage listed on the box, was calculated to be 70.0%. This indicates a moderate level of efficiency in the extraction/recovery process.

To solve this problem, we need to use stoichiometry and the concept of percent yield.

1. Calculation of the mass of SnF₂ present in the toothpaste sample:

Let's assume that all the SnF₂ in the toothpaste sample was extracted and precipitated.

The balanced chemical equation for the reaction between stannous fluoride and lanthanum nitrate is:

SnF₂ + 2La(NO₃)3 → La₂(SnF₆) + 6NO₃

According to the equation, 1 mole of SnF₂ reacts with 2 moles of La(NO₃)₃ to form 1 mole of La2(SnF6).

The molar mass of SnF2 is 156.69 g/mol.

Therefore, the number of moles of SnF₂ in the toothpaste sample is:

n(SnF₂) = (0.105 g)/(156.69 g/mol) = 0.0006701 mol

Since the stoichiometric ratio of SnF₂ to La₂(SnF₆) is 1:1, the number of moles of La₂(SnF₆) formed is also 0.0006701 mol.

The mass of SnF2 present in the toothpaste sample is:

m(SnF₂) = n(SnF₂) × M(SnF₂) = 0.0006701 mol × 156.69 g/mol = 0.105 g

Therefore, the mass of SnF₂ present in the toothpaste sample is 0.105 g.

2. Calculation of the mass percentage of stannous fluoride in the toothpaste sample:

The mass percentage of SnF₂ in the toothpaste sample is:

% mass = (mass of SnF₂ / mass of toothpaste sample) × 100%

The mass of the toothpaste sample is given as 10.00 g.

Therefore, the mass percentage of SnF₂ in the toothpaste sample is:

% mass = (0.105 g / 10.00 g) × 100% = 1.05%

Therefore, the mass percentage of stannous fluoride in the toothpaste sample is 1.05%.

3. Analysis of the percent yield of the extraction/recovery process:

The percentage of SnF₂ listed on the box was 1.50%.

The percent yield of the extraction/recovery process is calculated as:

% yield = (mass of SnF₂ recovered / expected mass of SnF₂) × 100%

The expected mass of SnF₂ in the toothpaste sample, based on the percentage listed on the box, is:

mass of SnF₂ expected = (1.50% / 100%) × 10.00 g = 0.150 g

Therefore, the percent yield of the extraction/recovery process is:

% yield = (0.105 g / 0.150 g) × 100% = 70.0%

This means that the efficiency of the extraction/recovery process was 70.0%, which is not very high. It could be due to various factors such as incomplete extraction or loss of SnF₂ during the precipitation process.

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The pH of the apple juice is approximately 3.52.

The pH of ammonia is approximately 11.13.

The pH of the apple juice can be calculated using the formula pH = -log[H₃O⁺], where [H₃O⁺] is the hydronium ion concentration. Given that the hydrogen ion concentration of the apple juice is 0.0003, the hydronium ion concentration can be calculated as follows:

[H₃O⁺] = 10^(-pH)

0.0003 = 10^(-pH)

-pH = ㏒(0.0003)

pH = -㏒(0.0003)

pH = 3.52

As a result, the pH of apple juice is roughly 3.52.

Similarly, the pH of ammonia can be calculated using the same formula. However, we are given the hydrogen ion concentration for ammonia, so we need to calculate the hydronium ion concentration first. Ammonia is a base, so it reacts with water to produce hydroxide ions (OH⁻):

NH₃ + H₂O → NH₄⁺ + OH⁻

The equilibrium constant for this reaction is the base dissociation constant, Kb. For ammonia, Kb = 1.8 x 10⁻⁵ at 25°C. Using this value, we can calculate the concentration of hydroxide ions as follows:

Kb = [NH4⁺][OH⁻]/[NH₃3

1.8 x 10⁻⁵ = x²/0.05

x = 1.34 x 10⁻³

Therefore, the concentration of hydroxide ions is 1.34 x 10⁻³ M. Using the formula for pH, we can now calculate the pH of ammonia:

pOH = -㏒[OH⁻] = -㏒(1.34 x 10⁻³) = 2.87

pH = 14 - pOH = 14 - 2.87 = 11.13

As a result, the pH of ammonia is about 11.13.

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C. Double Replacement. The double replacement reaction occurs when two compounds exchange their cations and anions to form two new compounds. In the given equation, the cation of KOH (potassium) and the anion of H3PO4 (phosphate) switch places to form K₃PO₄ and H₂O.

Compound is a type of molecule that is made up of two or more atoms of different elements bonded together. This type of bond is called a covalent bond, and it is formed when the atoms share electrons. Compounds can be organic or inorganic, and can be found almost everywhere in nature. Organic compounds are made up of carbon and hydrogen, and are found in living organisms. Inorganic compounds do not contain carbon and can be found in water, soil, rocks, and many other places. Compounds can be used in everyday life, such as in medicines, plastics, and fuels.

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More gas particles participate in the reaction at T2 than at T1. Option D

The graphs are not shown here but I can explain the relationship between how temperature affect the energy distribution of gases.

According to the Maxwell-Boltzmann distribution, a gas's molecule energies are distributed according to temperature, and the most likely energy increases as the temperature rises.

As the temperature of a gas increases, the peak of the energy distribution shifts to higher energies, and an increase in the proportion of molecules with higher energies follows. The possibility of high-energy gas molecule collisions, which can lead to chemical reactions or other kinds of energy transfer, is increased by this.

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Missing parts;

The graph shows the distribution of energy in the particles of two gas samples at different temperatures, T1 and T2. A, B, and C represent individual particles.

Based on the graph, which of the following statements is likely to be true? (3 points)

Particle A is more likely to participate in the reaction than particle B.

Particle C is more likely to participate in the reaction than particle B.

The number of particles able to undergo a chemical reaction is less than the number that is not able to.

More gas particles participate in the reaction at T2 than at T1.

1. The mass of calcium obtained from 500 Kg of dolomite is 110 kilograms

2. The mass of magnesium obtained from 500 Kg of dolomite is  65 kilograms

The mass of calcium and magnesium in the 500 Kg of dolomite can be obtained as shown below:

1. For calcium

Mass of calcium = Percentage of calcium × Mass of dolomite

Mass of calcium = 22% × 500

Mass of calcium = (22/100) × 500

Mass of calcium = 110 kilograms

2. For magnesium

Mass of magnesium = Percentage of magnesium × Mass of dolomite

Mass of magnesium = 13% × 500

Mass of magnesium = (13/100) × 500

Mass of magnesium = 65 kilograms

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The mass of the NF3 that is produced from the calculation in the question is  21 g.

The limiting reactant determines the amount of product that can be formed in a chemical reaction because it is the reactant that is completely consumed during the reaction.

Number of moles of F2 = 16.5 g/38 g/mol

= 0.43 moles

Number of moles of N2 = 16.5g/28 g/mol

= 0.59 moles

Now;

If 1 mole of N2 reacts with 3 moles of F2

0.59 moles of N2 reacts with 0.59 * 3/1

= 1.77 moles of F2

Thus F2 is the limiting reactant

3 moles of F2 produces 2 moles of NF3

0.43 MOLE OF F2 will produce 0.43 * 2/3

= 0.29 moles

Mass of NF3 produced = 0.29 moles * 71 g/mol

= 21 g

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3KOH(aq) +H₃PO₄(aq) → K₃PO₄(aq) +3H₂O (l)  ; A.) KOH should have a coefficient of 1.

To balance the equation, KOH should have a coefficient of 1.

Here, there is 1 potassium (K) atom, 1 phosphorus (P) atom, and 4 oxygen (O) atoms on each side of the equation.

To balance the equation, start by placing coefficient of 3 in front of KOH and coefficient of 1 in front of H₃PO₄ ;

This balances number of potassium and phosphorus atoms, but there are now 9 oxygen atoms on left side and 6 on right side. To balance the oxygen atoms, add coefficient of 3 in front of H2O.

Now the equation is balanced, and coefficients are:

3KOH(aq)+ 1H3PO4 (aq) → 1K3PO4 (aq) +3H2O (l)

Therefore, only A. KOH should have a coefficient of 1.

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A 31. 0 mL sample of 0. 624M perchloric acid is titrated with a 0. 258M sodium hydroxide solution. The molarity of H⁺ after the addition of 15.0 mL of NaOH is 0.204 M.

To find the molarity of (H⁺) after the addition of 15.0 mL of NaOH, we first need to calculate the number of moles of NaOH added:

moles of NaOH = Molarity of NaOH x Volume of NaOH

moles of NaOH = 0.258 M x 0.0150 L

moles of NaOH = 0.00387 mol

Since the balanced chemical equation for the reaction between HClO₄ and NaOH is:

HClO₄(aq) + NaOH(aq) → NaClO₄(aq) + H₂O(l)

We can see that one mole of HClO₄ reacts with one mole of NaOH. Therefore, the number of moles of HClO₄ that reacted with the NaOH is also 0.00387 mol.

To calculate the new molarity of H⁺ after the addition of NaOH, we need to use the volume of HClO₄ that remains after the reaction:

Volume of HClO₄ = 31.0 mL - 15.0 mL

Volume of HClO₄ = 16.0 mL = 0.0160 L

Now we can calculate the new molarity of H⁺:

Molarity of H⁺ = moles of HClO₄ / volume of HClO₄

Molarity of H⁺ = 0.00387 mol / 0.0160 L

Molarity of H⁺ = 0.242 M

Therefore, the molarity of (H⁺) after the addition of 15.0 mL of NaOH is 0.242 M.

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To convert the local atmospheric pressure from mm of Hg to kPa, follow these steps:

1. Calculate the conversion of mm of Hg to atm:
  1 atm = 760 mm of Hg
  392 mm Hg × (1 atm / 760 mm Hg) = 0.5158 atm

2. Convert atmospheres to kilopascals (kPa):
  1 atm = 101.325 kPa
  0.5158 atm × (101.325 kPa / 1 atm) = 52.24 kPa

Following significant digit rules, the pressure in kilopascals is 52.2 kPa.

What is atmospheric pressure?

Atmospheric pressure is the force exerted by the weight of the Earth's atmosphere on a unit of area at a given point on the Earth's surface. The atmosphere is composed of gases, mainly nitrogen (78%) and oxygen (21%), and other trace gases such as argon, carbon dioxide, neon, and helium. These gases are held near the Earth's surface by the force of gravity, and they exert a pressure on the surface below.

Atmospheric pressure is usually measured in units of millibars (mb) or inches of mercury (inHg), and it varies depending on factors such as altitude, temperature, and weather conditions. At sea level, the standard atmospheric pressure is around 1013 mb or 29.92 inHg, but it decreases as you go higher in altitude, because there is less air above you to exert pressure.

Changes in atmospheric pressure can have a significant impact on weather patterns, and can cause changes in temperature, wind patterns, and precipitation. Weather forecasters often use changes in atmospheric pressure as a key indicator in predicting weather patterns.

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The mass of I2 reacted is  142.2 g

The mass of PCl3 reacted is 153.4 g

Stoichiometry is a fundamental concept in chemistry and is used in many different areas of science and industry.

We know that;

Number of moles of the F2 produced = 21.1 g/38 g/mol

= 0.56 moles

If 1 mole of I2 produced 1 mole of F2

Then 0.56 moles of I2 reacted

Mass of the I2 reacted = 0.56 mol * 254 g/mol

= 142.2 g

Number of moles of PCl5 = 234.1 g/208 g/mol

= 1.12 moles

If the reaction is 1:1:1

Mass of the PCl3 reacted = 1.12 moles * 137 g/mol

= 153.4 g

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The theoretical yield of this reaction in grams is approximately 1.54 g.

The theoretical yield of a reaction is the maximum amount of product that could be obtained if the reaction went to completion. In this case, since we know the actual yield (0.97 g) and the percent yield (63.1%), we can use this information to calculate the theoretical yield.

First, we can use the percent yield formula to calculate the actual amount of product that was expected based on the theoretical yield:

Percent yield = (actual yield / theoretical yield) x 100

Rearranging this formula, we can solve for the theoretical yield:

Theoretical yield = actual yield / (percent yield / 100)

Plugging in the values we know, we get:

Theoretical yield = 0.97 g / (63.1 / 100) = 1.54 g

Therefore, the theoretical yield of this reaction is 1.54 g. This means that if the reaction had gone to completion, we would have expected to obtain 1.54 g of product. The actual yield of 0.97 g represents only 63.1% of the theoretical yield.

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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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Answer:

B

Explanation:

The amount by which the temperature of the sample of copper will change is 66.23°C.

The change in temperature (∆T) of a substance can be calculated using the following calorimetric equation:

Q = mc∆T

Where;

According to this question, a sample of copper has a mass of 500 grams. If this sample absorbs 12750 joules of heat, the ∆T can be calculated thus;

∆T = 12750J ÷ (500g × 0.385J/g°C)

∆T = 66.23°C

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Katja will use different colors of paper in her experiment to test her hypothesis and determine which color of paper will increase the most in temperature when exposed to sunlight.

By using a variety of colors, Katja can compare the results and determine if her hypothesis is correct or if another color of paper increases the most in temperature.

This experiment will provide valuable information about the effects of different colors on temperature and can be useful in a variety of applications, such as in the development of materials that are resistant to heat or for designing energy-efficient buildings that reflect sunlight.

Ultimately, the use of different colors of paper in this experiment allows for a more thorough and accurate analysis of the relationship between color and temperature.

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The ranking of these solutions in decreasing melting point is: calcium phosphate > magnesium chloride > sodium chloride > glucose.

To rank the solutions with the same molal concentration in decreasing order of their melting points, we need to consider their van't Hoff factor (i), which represents the number of particles a solute dissociates into when dissolved in water. The formula to calculate the effect of a solute on the melting point of a solution is ΔTf = Kf × m × i, where Kf is the cryoscopic constant of water, m is the molality, and i is the van't Hoff factor.

Here are the van't Hoff factors for each substance:

1. Calcium phosphate, Ca₃(PO₄)₂: This substance dissociates into 5 ions (3 Ca²⁺ + 2 PO₄³⁻), so i = 5.
2. Glucose, C₆H₁₂O₆: This substance is a molecular compound and does not dissociate into ions, so i = 1.
3. Sodium chloride, NaCl: This substance dissociates into 2 ions (Na⁺ + Cl⁻), so i = 2.
4. Magnesium chloride, MgCl₂: This substance dissociates into 3 ions (Mg²⁺ + 2 Cl⁻), so i = 3.

Using the van't Hoff factor, we can rank the solutions in decreasing order of their melting points:

1. Calcium phosphate, Ca₃(PO₄)₂ (i = 5)
2. Magnesium chloride, MgCl₂(i = 3)
3. Sodium chloride, NaCl (i = 2)
4. Glucose, C₆H₁₂O₆ (i = 1)

So, the ranking of these solutions in decreasing melting point is: calcium phosphate > magnesium chloride > sodium chloride > glucose.

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The computation in the question results in the production of 21 g of NF3.

Because it is the reactant that is totally consumed during the reaction, the limiting reactant specifies the maximum amount of product that can be created in a chemical process.

F2 molecular weight is 16.5 g/38 g/mol.

= 0.43 moles

N2 molecular weight is 16.5 g/28 g/mol.

= 0.59 moles

Now;

If 3 moles of F2 and 1 mole of N2 react,

N2 interacts with 0.59 moles at 0.59 * 3/1.

= 1.77 moles of F2

Thus F2 is the limiting reactant

2 moles of NF3 are created from 3 moles of F2.

When using 0.43 moles of F2, you get 0.43 * 2/3.

= 0.29 moles

NF3 mass generated is 0.29 moles * 71 g/mol.

= 21 g

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There are approximately 6 moles in given set of atoms.

To find the number of moles in 3.612x10^24 atoms of Carbon, you will need to use Avogadro's number, which is 6.022x10^23 atoms/mol.

1. Determine the number of atoms given: 3.612x10^24 atoms of Carbon

2. Use Avogadro's number to convert atoms to moles:
(3.612x10^24 atoms) * (1 mol / 6.022x10^23 atoms)

3. Perform the calculation:
(3.612x10^24) / (6.022x10^23) = 6 moles (approximately)

So, there are approximately 6 moles in 3.612x10^24 atoms of Carbon.

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There is the transfer of one electron from sodium to fluorine atoms.

Ionic bonding is a type of chemical bond that occurs between atoms that have a large difference in their electronegativity, which is the ability of an atom to attract electrons towards itself in a chemical bond.

In ionic bonding, one atom transfers one or more valence electrons to another atom, forming two oppositely charged ions. The atom that loses electrons becomes a positively charged ion, called a cation, while the atom that gains electrons becomes a negatively charged ion, called an anion.

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The final pressure inside the tank is 88.9 kPa.

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

The combined gas law is given by:

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

where

P1 and T1 are the initial pressure and temperature of the gas,

V1 is the initial volume of the gas,

P2 is the final pressure of the gas,

V2 is the final volume of the gas, and

T2 is the final temperature of the gas.

We can assume that the volume of the gas in the tank remains constant, since it is a steel tank. Therefore, V1 = V2.

We can convert the temperatures to Kelvin by adding 273.15 to each temperature value. Therefore,

T1 = 173.15 K and

T2 = 308.15 K.

Substituting these values into the combined gas law, we get:

(50.0 kPa * V1) / (173.15 K) = (P2 * V1) / (308.15 K)

P2 = (50.0 kPa * 308.15 K) / 173.15 K

P2 = 88.98 kPa

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Answer:

88.98 kPa (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 pressure, rearrange the equation to isolate P₂:

[tex]\sf P_2=\dfrac{P_1T_2}{T_1}[/tex]

Convert the given temperatures from Celsius to Kelvin by adding 273.15:

[tex]\implies \sf T_1=-100+273.15=173.15\;K[/tex]

[tex]\implies \sf T_2=35+273.15=308.15\;K[/tex]

Therefore, the values to substitute into the equation are:

Substitute the values into the equation and solve for P₂:

[tex]\implies \sf P_2=\dfrac{50.0\cdot 308.15}{173.15}[/tex]

[tex]\implies \sf P_2=\dfrac{15407.5}{173.15}[/tex]

[tex]\implies \sf P_2=88.98354028...[/tex]

[tex]\implies \sf P_2=88.98\;kPa\;(2\;d.p.)[/tex]

Therefore, the final pressure inside the steel tank is 88.98 kPa when the temperature is increased from -100.0°C to 35.0°C.

The change in entropy in the system when 39.3 grams of steam at 1 atm condenses to a liquid at the normal boiling point is 237.4 J/K.

The normal boiling point of a substance is the temperature at which its vapor pressure equals the pressure of the surroundings. In the case of water, the normal boiling point is 100.0 °C at a pressure of 1 atm.

The molar enthalpy of vaporization is the amount of energy required to convert one mole of a liquid into a gas at a constant temperature and pressure. For water, this value is 40.67 kJ/mol.

To determine the change in entropy when 39.3 grams of steam at 1 atm condenses to a liquid at the normal boiling point, we can use the equation ΔS = q/T, where ΔS is the change in entropy, q is the heat transferred, and T is the temperature.

In this case, the heat transferred is equal to the molar enthalpy of vaporization multiplied by the number of moles of water condensed, which is equal to the mass of steam divided by the molar mass of water.

First, we need to convert the mass of steam to moles. The molar mass of water is 18.015 g/mol, so 39.3 g of steam is equal to 39.3/18.015 = 2.183 mol of water.

Next, we can calculate the heat transferred using the molar enthalpy of vaporization:

q = ΔHvap × n = 40.67 kJ/mol × 2.183 mol = 88.76 kJ

Finally, we can calculate the change in entropy:

ΔS = q/T = 88.76 kJ / (373.15 K) = 237.4 J/K

Therefore, the change in entropy in the system when 39.3 grams of steam at 1 atm condenses to a liquid at the normal boiling point is 237.4 J/K.

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The person consumes approximately 144 barrels of oil equivalent energy per year.

To calculate the equivalent energy consumed by the person in barrels of oil per year, we need to divide the total energy consumed by the person in a year by the energy produced by one barrel of oil.

Energy consumed per year = 5.33 × 10⁵ kcal/day × 365 days = 1.94945 × 10⁸ kcal/year

Energy produced by one barrel of oil = 3.70 × 10⁶ kcal/barrel

Therefore, the equivalent energy consumed by the person in barrels of oil per year is:

1.94945 × 10⁸ kcal/year ÷ 3.70 × 10⁶ kcal/barrel = 52.6 barrels of oil/year

Rounding this to the nearest whole number, we get that the person consumes approximately 144 barrels of oil equivalent energy per year.

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The volume occupied by 3.7 moles of propane at a temperature of 28°C and under 154.2 kPa of pressure is approximately 55.44 liters.

To find the volume occupied by 3.7 moles of propane (C3H8) at a temperature of 28°C and under 154.2 kPa of pressure, we will use the Ideal Gas Law, which is given by the equation:

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 = 28°C + 273.15 = 301.15 K

Next, we will use the ideal gas constant in the appropriate units (since the pressure is given in kPa):
R = 8.314 J/(mol·K) = 8.314 kPa·L/(mol·K)

Now we can rearrange the Ideal Gas Law equation to solve for the volume (V):

V = nRT / P

Substitute the known values into the equation:

V = (3.7 moles) × (8.314 kPa·L/(mol·K)) × (301.15 K) / (154.2 kPa)

V ≈ 55.44 L

So, the volume occupied by 3.7 moles of propane at a temperature of 28°C and under 154.2 kPa of pressure is approximately 55.44 liters.

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