What happened to the concentration of the ions as the water evaporates

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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A teaspoon of salt has more particles (approximately 6.20 x 10^22) than a teaspoon of sugar (approximately 7.41 x 10^21).

To compare the number of particles in a teaspoon of salt and a teaspoon of sugar, we need to understand the concept of moles.

A mole is a unit of measurement used to express the amount of a substance, and it corresponds to approximately 6.022 x 10^23 particles.

The number of moles in a given mass of a substance can be calculated using the formula:

moles = mass / molar mass.

The molar mass of common table salt (NaCl) is 58.44 g/mol, while the molar mass of table sugar (C12H22O11) is 342.3 g/mol.

Considering that a teaspoon of salt typically weighs about 6 grams and a teaspoon of sugar weighs about 4.2 grams, we can calculate the moles of each substance:

Moles of salt = 6 g / 58.44 g/mol ≈ 0.103 moles
Moles of sugar = 4.2 g / 342.3 g/mol ≈ 0.0123 moles

Now, to find the number of particles in each substance, we multiply the moles by Avogadro's number (6.022 x 10^23 particles/mol):

Particles of salt = 0.103 moles x 6.022 x 10^23 particles/mol ≈ 6.20 x 10^22 particles
Particles of sugar = 0.0123 moles x 6.022 x 10^23 particles/mol ≈ 7.41 x 10^21 particles

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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. --

39.4 mL of 0.586 M Ba(OH)2 solution is required to neutralize 10 mL of vinegar containing 2.78 g of acetic acid.

The balanced chemical equation for the reaction between barium hydroxide and acetic acid is:

Ba(OH)2(aq) + 2CH3COOH(aq) ⟶ Ba(CH3COO)2(aq) + 2H2O(l)

The molar mass of acetic acid (CH3COOH) is 60.05 g/mol.

First, we need to determine the moles of acetic acid present in 10 mL of vinegar:

molar mass of acetic acid = 60.05 g/mol

mass of acetic acid in 10 mL of vinegar = 2.78 g

moles of acetic acid = mass/molar mass = 2.78 g / 60.05 g/mol = 0.0463 mol

From the balanced chemical equation, we see that 1 mole of Ba(OH)2 reacts with 2 moles of CH3COOH. Therefore, the moles of Ba(OH)2 required to react with 0.0463 mol of CH3COOH is:

moles of Ba(OH)2 = (0.0463 mol CH3COOH) / 2 = 0.0231 mol Ba(OH)2

Now, we can use the definition of molarity to find the volume of 0.586 M Ba(OH)2 solution needed to provide 0.0231 mol Ba(OH)2:

Molarity = moles of solute / volume of solution in liters

volume of solution in liters = moles of solute / Molarity

volume of Ba(OH)2 solution needed = 0.0231 mol / 0.586 mol/L = 0.0394 L

Finally, we convert the volume of Ba(OH)2 solution from liters to milliliters:

volume of Ba(OH)2 solution needed = 0.0394 L * (1000 mL/L) = 39.4 mL

Therefore, 39.4 mL of 0.586 M Ba(OH)2 solution is required to neutralize 10 mL of vinegar containing 2.78 g of acetic acid.

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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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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 molarity of the diluted ammonium chloride solution is 0.09475 M.

The molarity of the diluted ammonium chloride solution, we can use the equation:

[tex]M_1V_1 = M_2V_2[/tex]

here [tex]M_1[/tex] is the initial molarity

[tex]V_1[/tex] is the initial volume,

[tex]M_2[/tex] is the final molarity, and

[tex]V_2[/tex] is the final volume.

[tex]M_1[/tex] = 3.79 M (from the initial solution)

,[tex]V_1[/tex]  = 50.0 mL = 0.050 L (from the initial solution)

[tex]V_2[/tex] = 2.0 L (the final volume after dilution)

For  [tex]M_2[/tex] , we get:

[tex]M_2[/tex]  = ( [tex]M_1[/tex] × ,[tex]V_1[/tex] ) /  [tex]V_2[/tex]

[tex]M_2[/tex]  = (3.79 M × 0.050 L) / 2.0 L

[tex]M_2[/tex]  = 0.09475 M

Therefore, the molarity of the diluted ammonium chloride solution is 0.09475 M.

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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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1. Neopentyl bromide will react more slowly than 1-bromobutane with sodium iodide in acetone. The difference in reactivity is due to steric hindrance. Neopentyl bromide is a primary alkyl bromide with three bulky methyl groups attached to the primary carbon, which creates significant steric hindrance.

2. Allyl bromide will react faster than allyl chloride, and 1-bromobutane will react faster than 1-chlorobutane with sodium iodide in acetone. The nature of the leaving group affects the rate of the SN2 reaction.

1. Neopentyl bromide will react more slowly than 1-bromobutane with sodium iodide in acetone. The difference in reactivity is due to steric hindrance. Neopentyl bromide is a primary alkyl bromide with three bulky methyl groups attached to the primary carbon, which creates significant steric hindrance. InIn contrast, 1-bromobutane only has one methyl group attached to the primary carbon. In the SN2 reaction, the nucleophile approaches the primary carbon from the backside and displaces the leaving group. The bulky methyl groups in neopentyl bromide create a greater steric hindrance, making it more difficult for the nucleophile to approach the primary carbon from the backside and displace the leaving group. This results in a slower reaction rate compared to 1-bromobutane.

2. Allyl bromide will react faster than allyl chloride, and 1-bromobutane will react faster than 1-chlorobutane with sodium iodide in acetone. The nature of the leaving group affects the rate of the SN2 reaction. In general, a good leaving group is one that can stabilize the negative charge that is formed when it departs. Halogens are good leaving groups because they can stabilize the negative charge through resonance. However, chlorine is a weaker leaving group than bromine because it is larger and has a weaker bond to the carbon. Therefore, it is more difficult to displace the leaving group in allyl chloride and 1-chlorobutane than in allyl bromide and 1-bromobutane, leading to slower reaction rates. Overall, the order of reactivity in SN2 reactions is typically: primary > secondary > tertiary, and iodide > bromide > chloride as nucleophiles, and chloride < bromide < iodide as leaving groups.
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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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The heat energy transferred to 2.3g of copper is 133.01 J.

To calculate the heat energy transferred to the copper, we can use the formula:

q = mcΔT

where q is the heat energy transferred, m is the mass of the substance (2.3 g), c is the specific heat capacity (0.385 J/g·°C), and ΔT is the change in temperature (174.0°C - 23.0°C).

And;

ΔT = 174.0°C - 23.0°C = 151.0°C

Now, plug the values into the formula:

q = (2.3 g) × (0.385 J/g·°C) × (151.0°C)

q = 133.0085 J

Round the answer to two decimal places:

q = 133.01 J

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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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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 reaction between powdered zinc and hydrochloric acid would be faster than between a chunk of zinc and hydrochloric acid.

When a chunk of zinc is turned into powdered zinc, the surface area of the zinc increases. This allows for more contact points between the zinc and hydrochloric acid, resulting in a faster reaction rate.

The increased surface area provides more opportunities for the acid to interact with the zinc particles, accelerating the formation of zinc chloride and hydrogen gas, which are the products of this reaction.

In summary, converting the zinc chunk into a powdered form will speed up the reaction between zinc and hydrochloric acid due to the increase in surface area, leading to a more efficient and faster chemical process.

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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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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 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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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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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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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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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 mass in micrograms of 5. 00 scruples approximately 149,166.67 µg.

The known values are: 1 scruple = 20 grains, 1 ounce = 480 grains, and 1 oz = 28.34 g.

To find the mass of 5.00 scruples, first convert scruples to grains by multiplying by 20, then convert grains to ounces by dividing by 480, and finally convert ounces to grams by multiplying by 28.34.

The calculation is as follows:

5.00 scruples x 20 grains/scruple x 1 ounce/480 grains x 28.34 g/1 oz x 1,000,000 µg/1 g = 149,166.67 µg

Therefore, the mass of 5.00 scruples is 149,166.67 µg.

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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).

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 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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Soils can come in many different colors, but reddish-brown is a common hue that can indicate the presence of iron oxides. These oxides can give the soil a rusty appearance, and are often found in soils that have been weathered over time.

Sandy soils that are reddish-brown in color are often found in arid regions, where the soil has been weathered by wind and water. These soils may be low in nutrients and organic matter, but can be ideal for certain types of plants that are adapted to dry conditions.

Clay soils that are reddish-brown in color are often found in areas with high rainfall, where the clay has been weathered by water and minerals have leached out. These soils can be rich in nutrients, but may be difficult to work with due to their tendency to become compacted and heavy.

Loamy soils that are reddish-brown in color are a combination of sand, clay, and silt particles, and are often considered the ideal type of soil for gardening and farming. These soils are typically rich in nutrients, but also drain well and are easy to work with.

Overall, the reddish-brown color of soil can provide valuable information about the characteristics and composition of the soil, which can help gardeners, farmers, and other professionals make informed decisions about how to manage and use the land.

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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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A net ionic equation for the reaction that occurs when sodium carbonate (aq) and excess hydroiodic acid are combined.

CO₃²⁻(aq) + 2H⁺(aq) -> H2O(l) + CO2(g)

The balanced equation for the reaction between sodium carbonate (Na2CO3) and excess hydroiodic acid (HI) is:

Na2CO3(aq) + 2HI(aq) → 2NaI(aq) + CO2(g) + H2O(l)

The ionic equation is:

2Na⁺(aq) + CO₃²⁻(aq) + 2H⁺(aq) + 2I⁻(aq) -> 2Na⁺(aq) + 2I⁻(aq) + H2O(l) + CO2(g)

The spectator ions are Na+ and CO32-.

Next, we can cancel out the spectator ions (Na⁺ and I⁻) to get the net ionic equation:

CO₃²⁻(aq) + 2H⁺(aq) -> H2O(l) + CO2(g)

That's the net ionic equation for the reaction between sodium carbonate and excess hydroiodic acid.

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