If 5 mol of oxygen gas effuses through an opening in 10 seconds, how long will it take for the same amount of hydrogen gas to effuse under the same conditions?


( A ) 1. 6 s


( B ) 2. 5 s


( C ) 40 s


( D ) 160 s

The gas would occupy 90 L at a final pressure of 200 mmHg if the amount of gas does not change.

To determine how many liters this gas would occupy at a final pressure of 200 mmHg, we can use Boyle's Law. Boyle's Law states that the product of the initial pressure and volume of a gas is equal to the product of the final pressure and volume if the temperature and amount of gas remain constant. The formula for Boyle's Law is:

P₁ × V₁ = P₂ × V₂

Where P₁ is the initial pressure (1800 mmHg), V₁ is the initial volume (10 L), P₂ is the final pressure (200 mmHg), and V₂ is the final volume we need to find.

Rearranging the formula to find V₂:
V₂ = (P₁ × V₁) / P₂

Substituting the values:

V₂ = (1800 mmHg × 10 L) / 200 mmHg
V₂ = 18000 L·mmHg / 200 mmHg
V₂ = 90 L

So, this gas would occupy 90 L at a final pressure of 200 mmHg if the amount of gas does not change.

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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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The juice sample contains 28,920 mg of vitamin C.

The amount of iodine used in the reaction can be calculated as:

I2 used = (final buret reading - initial buret reading) * 0.005 M

I2 used = (33.08 ml - 0.24 ml) * 0.005 M = 0.16392 moles

Since 1 mole of vitamin C reacts with 1 mole of iodine, the amount of vitamin C in the juice can be calculated as:

Vitamin C = I2 used * (1 mol of vitamin C / 1 mol of I2) * (176.12 g/mol)

Vitamin C = 0.16392 * (1 / 1) * (176.12 g/mol) = 28.92 g

Converting to milligrams:

Vitamin C = 28.92 g * 1000 mg/g = 28,920 mg

Therefore, the juice sample contains 28,920 mg of vitamin C.

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The solubility of CO₂ in water at 263.4 kPa is 1.064 g/100 ml water.

We can use Henry's law to calculate the solubility of CO₂ in water at a different pressure. Henry's law states that the amount of gas dissolved in a liquid is directly proportional to the partial pressure of the gas above the liquid. Thus, we can set up the following equation:

We are given the solubility of CO₂ at a reference pressure of 101.3 kPa, which is 0.348 g/100 ml water. We want to find the solubility at a new pressure of 263.4 kPa. We can rearrange the equation above to solve for the solubility at the new pressure:

We know that the temperature is constant at 0°C, so we can assume that the solubility is directly proportional to the partial pressure. Thus, we can set up a ratio:

Plugging in the given values, we get:

Solving for the solubility at the new pressure, we get:

Therefore, the solubility of CO₂ in water at a pressure of 263.4 kPa is 1.064 g/100 ml water.

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The new molarity of the solution after dilution is 0.5 M.

To solve this problem, we can use the formula:

[tex]M_2 = M_1V_1 / V_2[/tex]

where [tex]M_1[/tex] and [tex]V_1[/tex] are the initial molarity and volume of the solution, and [tex]M_2[/tex] and [tex]V_2[/tex] are the final molarity and volume of the diluted solution.

In this case, we have:

[tex]M_1[/tex] = 1.5 M

[tex]V_1[/tex] = 2.0 L

[tex]V_2[/tex] = 6.0 L

We want to find the final molarity, [tex]M_2[/tex].

Using the formula, we can solve for [tex]M_2[/tex]:

[tex]M_2 = M_1V_1 / V_2[/tex]

Substituting the given values, we get:

[tex]M_2[/tex] = (1.5 M) × (2.0 L) / (6.0 L) = 0.5 M

Therefore, the new molarity of the solution is 0.5 M.

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Yes, substances can be the same if one has more of the same type of repeating group of atoms.

For example, polymers are made up of repeating units of the same monomer, and the number of monomers can vary, resulting in different sizes of polymers but still the same substance. Another example is isotopes, which are elements with the same number of protons but varying numbers of neutrons.

They have the same chemical properties and can form the same compounds despite having different atomic masses. Thus, substances can be identical in terms of their chemical properties even if they have different physical properties due to variations in the number of repeating groups of atoms or isotopes.

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The percent dissociation of HNO₂ comes out to be 5.2% which is shown in the below secion.

The calculations of pKa is done as follows-

pKa = - log Ka

      = - log (4.0 x 10⁻⁴)

     = 3.398

Mole of NaNO₂ = mass / molar mass

                          = 0.058 g / 68.9953 g/mole

                          = 8.406 x 10⁻⁴ mole

Mole of HNO₂ = 0.110 L * 0.060 mole / L = 6.6 x10⁻³ mole.

Resulting solution is buffer solution.

pH = pKa + log [salt] / [acid]

Substituting the known values in the above formula.

pH = 3.398 + log ( 8.406 x 10⁻⁴ / 6.6 x 10⁻³ )

pH = 2.503

The pH can also be evaluated using the below expression.

pH = -log[H⁺]

-log[H] = 2.503

[H⁺]= 3.14 x 10⁻³ M

Thus

Percent of ionization = 3.14 x 10⁻³ M x 100 / 0.060 = 5.2 %

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2 moles of solute dissolved in 1 liter of solution has the greatest concentration.

Concentration refers to the quantity of solute that is dissolved in a specific amount of solution, and it is commonly measured in units such as moles per liter or grams per liter.

To determine which solution has the greatest concentration, we need to calculate the number of moles of solute present in each solution and then compare the values.

a. Concentration = 2 moles / 1 liter = 2 M

b. Concentration = 0.3 moles / 0.6 liters = 0.5 M

c. Concentration = 2 moles / 10 liters = 0.2 M

d. Concentration = 0.1 moles / 0.5 liters = 0.2 M

Comparing the concentrations, we see that solution (a) has the greatest concentration of 2 M, while the other solutions have concentrations of 0.5 M or lower.

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Answer:

you equate the question 800×7.2 divide the answer by 735.And you'll get 7.84litre then covert to 0.0m³ if the question says so to get 0.00784

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 of 4.30 mol of magnesium with an excess of oxygen produces 6.40 mol of magnesium oxide (MgO).

Magnesium oxide is a white, odorless inorganic compound composed of magnesium and oxygen atoms. It is a strong basic oxide and an important mineral component of many rocks and soils. It has a wide range of industrial uses, such as in the production of cement, ceramics, and glass. It is also used as an antacid and laxative, and as a supplement to increase dietary magnesium intake.

The energy released in this reaction can be determined using the following equation:E = ΔHf (MgO) x (6.40 mol MgO)

In this equation, ΔHf (MgO) is the molar enthalpy of formation of magnesium oxide. The molar enthalpy of formation of magnesium oxide is -601.8 kJ/mol. Therefore, the total energy released in this reaction is:

E = -601.8 kJ/mol x (6.40 mol MgO)

E = -3,854.7 kJ.To determine the number of grams of MgO produced, we can use the following equation: Mass (MgO) = (6.40 mol MgO) x (Molar mass MgO) .

The molar mass of MgO is 40.3 g/mol. Therefore, the mass of MgO produced is: Mass (MgO) = (6.40 mol MgO.

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The apparent loss of mass from 253.0 g to 250.2 g can be attributed to the involvement of a gas in the chemical reaction, either as a product or a reactant. This gas escapes the system during the reaction, causing a decrease in the observed mass, but the law of conservation of mass still holds true as the total mass is conserved in the reaction.

According to the law of conservation of mass, the total mass of the products of a chemical reaction should equal the total mass of the reactants. In your reaction, the mass went down from 253.0 g to 250.2 g, which seems to contradict this law. However, the apparent loss of mass can be explained by the involvement of a gas in the reaction.

Here's a step-by-step explanation:
1. Identify the given chemical equation. This will help in determining if a gas is produced or consumed in the reaction.
2. Examine the reactants and products to see if any of them are gases. Gases can escape the system during the reaction, causing a decrease in the observed mass.
3. If a gas is produced, this explains the apparent loss of mass. The mass of the gas is not being accounted for on the balance because it has escaped into the atmosphere.
4. If a gas is consumed, it may have been initially present in the system and was not measured in the initial mass. Once it is consumed, the mass of the system would appear to decrease.

In summary, the apparent loss of mass from 253.0 g to 250.2 g can be attributed to the involvement of a gas in the chemical reaction, either as a product or a reactant.

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6-ethyl-8-methyl-5-propyl- non-2-ene is the name of this branched alkene. The World Union of Applied and Pure Chemistry (IUPAC).

The World Union of Applied and Pure Chemistry (commonly abbreviated to IUPAC) recommends a systematic approach for the terminology of organic compounds, which is referred to as the IUPAC classification of organic compounds.

The IUPAC naming criteria are occasionally followed by chemists, nevertheless, as some compounds have names that are incredibly long and difficult to pronounce. More banal names are given to these substances. 6-ethyl-8-methyl-5-propyl- non-2-ene is the name of this branched alkene.

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1. Based on the information provided, a weak acid with a pKa close to the ideal pH of 5.5 would be the best buffer for Xylanase. Acetic acid (CH3COOH) can be utilised as the weak acid component of the buffer since it has a pKa of 4.76, which is close to the ideal pH.

Acetate ion (CH3COO-), its conjugate base, can also be utilised as a buffering agent. Since Xylanase prefers an acidic pH (below 7), we would anticipate the buffer to contain more acid than base.

2. The pH of the buffer would drop if 0.1 M HCl was introduced because the weak acid would arise when the H+ ions from the HCl react with the conjugate base in the buffer.

Instead, if 0.1 M NaOH was added, the pH would rise as the weak acid in the buffer reacts with the NaOH's OH- ions to form the conjugate base. The capacity of the buffer and the quantity of HCl or NaOH injected, however, would determine how much the pH changed.

3. It would take more HCl to raise the pH by one unit in the buffer in question 1 since it contains a weak acid and its conjugate base. This is due to the fact that the conjugate acid might be created when the weak acid component of the buffer reacts with extra H+ ions to limit significant pH shifts.

On the other hand, if NaOH were to be added to the buffer, the buffer's acid component would be consumed, causing a greater pH change.

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Gay-Lussac's Laws, also known as the Pressure-Temperature Law and Volume-Temperature Law, are a set of gas laws that explain the relationship between pressure, volume, and temperature in a given gas sample.

According to Gay-Lussac's Laws, if the pressure of a gas is increased while the volume remains constant, the temperature of the gas will also increase. Similarly, if the volume of a gas is decreased while the pressure remains constant, the temperature of the gas will increase as well.

In terms of a concept map, Gay-Lussac's Laws can be placed in the center with arrows pointing to both Boyle's Law and Charles's Law, which are two other gas laws that are also related to pressure, volume, and temperature.

Boyle's Law states that the pressure of a gas is inversely proportional to its volume, while Charles's Law states that the volume of a gas is directly proportional to its temperature.

To connect these three gas laws, the arrows can be labeled with key terms such as "pressure," "volume," and "temperature," with each gas law demonstrating how changes in one variable will affect the others.

The concept map can also include real-world examples of each gas law, such as how a tire pressure gauge can be used to demonstrate Boyle's Law, or how a hot air balloon can be used to demonstrate Charles's Law.

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The true statements regarding the greenhouse effect are:

The greenhouse effect is a natural process that occurs when certain gases in the atmosphere, known as greenhouse gases, trap heat from the sun and prevent it from escaping back into space. This helps to keep the Earth's surface warm enough to support life. However, human activities, such as burning fossil fuels, have increased the concentration of greenhouse gases in the atmosphere, which has led to an enhanced greenhouse effect and global warming.

In the greenhouse effect, not all energy from the sun is absorbed by atmospheric gases. Rather, some of it is reflected back into space by the atmosphere. Additionally, the greenhouse effect does not change sunlight into carbon dioxide; rather, it is the burning of fossil fuels and other human activities that release carbon dioxide into the atmosphere. Options B, C and E are correct.

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Baking soda is actually a compound known as sodium bicarbonate, which is water-soluble. Sugar, on the other hand, is also soluble in water and other liquids that contain water.

This means that it dissolves in water and can also dissolve in other liquids that contain water, such as soda. Therefore, baking soda is indeed soluble in soda.

Sugar, on the other hand, is also soluble in water and other liquids that contain water. This includes soda, which is a carbonated beverage that typically contains a high amount of dissolved sugar.

However, the solubility of sugar in soda can depend on various factors such as the temperature of the soda, the amount of sugar present, and the type of sugar used.

In general, both baking soda and sugar are soluble in soda and can dissolve to some extent. However, the exact degree of solubility can vary depending on various factors. It is worth noting that excessive consumption of sugary soda can have negative impacts on health, so it is important to consume such beverages in moderation.

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

The combined gas law is expressed as:

(P1V1)/T1 = (P2V2)/T2

where P1 is the initial pressure, V1 is the initial volume, T1 is the initial temperature, P2 is the final pressure, V2 is the final volume, and T2 is the final temperature.

In this problem, we are given the initial volume V1 as 2.32 liters and the initial temperature T1 as 24.0°C. We need to find the final volume V2 when the temperature is raised to 48.0°C.

We can set up the equation as:

(P1V1)/T1 = (P2V2)/T2

Since the pressure remains constant, we can cancel it out:

V1/T1 = V2/T2

We can rearrange this equation to solve for V2:

V2 = (V1/T1) x T2

Substituting the given values, we get:

V2 = (2.32 L/297.15 K) x 321.15 K

V2 = 2.86 L

Therefore, the new volume of the balloon is 2.86 liters when heated to 48.0°C.

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The student collected 2.63 x 10¹⁰ grams of ammonium phosphate.

To determine the mass of the sample collected, we need to know the molar mass of ammonium phosphate, which is (NH₄)₃PO₄. The molar mass of (NH₄)₃PO₄ can be calculated by adding the atomic masses of the constituent atoms:

Molar mass of (NH₄)₃PO₄ = (3 x molar mass of NH₄) + (1 x molar mass of PO₄)

= (3 x 18.04 g/mol) + (1 x 94.97 g/mol)

= 149.99 g/mol

The number of moles of (NH₄)₃PO₄ in the sample can be calculated by dividing the number of molecules by Avogadro's number (6.022 x 10²³):

Number of moles of (NH₄)₃PO₄ = 9.52 x 10²⁵ molecules / 6.022 x 10²³ molecules/mol

= 15.8 mol

Finally, we can calculate the mass of the sample using the formula:

Mass = Number of moles x Molar mass

= 15.8 mol x 149.99 g/mol

= 2.63 x 10¹⁰ g

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8.32 grams of CaCl2 are required to produce 750 ml of a 0.100 M CaCl2 solution.

The number of moles of solute that can dissolve in 1 L of a solution is known as molarity or molar concentration.

Volume of solution (in litres) / Number of solutes (in moles), or

C = n / V

According to the question,

V = 750 ml and C = 0.100 M

Let's convert millilitres to litres for this.

We know 1 L = 1000 ml

Consequently, 750 ml equals (750/1000) L, or 0.75 L.

So, V = 0.75 L

We know that C = n / V

So, n = C x V

n = 0.100 x 0.75 = 0.075

The solute contains n moles in total.

CaCl2 thus has a mole count of 0.075 moles.

In 750 ml of solution, this demonstrates that there are 0.075 moles of CaCl2.

We must know the molar mass of CaCl2 in order to calculate the mass of CaCl2 in grams.

To do this, we must use the periodic table to determine the atomic masses of each atom.

CaCl2 consists of 1 Ca and 2 Cl atoms.

Atomic mass of Ca is 40.08 g and that of Cl is 35.45, so 2 x 35.45 = 70.90 g.

We obtain the mass of CaCl2 in grams by averaging these measurements.

Hence, mass of CaCl2 = 40.08 + 70.90 = 110.98 g

Thus, 110.98 g equals 1 mole of CaCl2.

Therefore, 0.075 moles of CaCl2 will weigh 8.3235 g, which is rounded to 8.32 g, or 0.075 x 110.98 g.

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The complete question is

How many grams of calcium chloride will be needed to make 750 mL of a 0.100 M CaCl2 solution?

The solubility product constant (Ksp) for Ag2CrO4 is given by the following equation:

Ag2CrO4(s) ⇌ 2Ag+(aq) + CrO4^(2-)(aq)

The expression for Ksp is:

Ksp = [Ag+]^2[CrO4^(2-)]

where [Ag+] and [CrO4^(2-)] are the concentrations of the silver ion and chromate ion in the equilibrium mixture, respectively.

To determine the value of Q, the reaction quotient, we need to determine the concentrations of Ag+ and CrO4^(2-) in the mixture of 5.00 mL of 0.0040 M AgNO3 and 4.00 mL of 0.0024 M K2CrO4. To do this, we need to make some assumptions:

1. The volumes of the two solutions are additive, so the total volume is 9.00 mL.

2. The AgNO3 and K2CrO4 solutions react completely to form Ag2CrO4.

First, we need to determine the moles of Ag+ and CrO4^(2-) in each solution:

For the AgNO3 solution:

moles of Ag+ = (0.0040 M) x (0.00500 L) = 2.0 x 10^-5 mol

For the K2CrO4 solution:

moles of CrO4^(2-) = (0.0024 M) x (0.00400 L) = 9.6 x 10^-6 mol

Since the AgNO3 and K2CrO4 react in a 1:1 ratio to form Ag2CrO4, the limiting reactant is K2CrO4. Therefore, all of the CrO4^(2-) is used up in the reaction, and the concentration of CrO4^(2-) in the equilibrium mixture is zero.

The concentration of Ag+ in the equilibrium mixture is:

[Ag+] = moles of Ag+ / total volume of mixture

[Ag+] = (2.0 x 10^-5 mol) / (9.00 x 10^-6 L)

[Ag+] = 2.22 M

Now, we can calculate the value of Q:

Q = [Ag+]^2[CrO4^(2-)] = (2.22 M)^2(0 M) = 0

Since Q is equal to zero and Ksp is greater than zero (1.8 x 10^-10), the reaction is not at equilibrium and Ag2CrO4 will precipitate from the solution.

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The first two carboxylic acids described contain a benzene ring, which is not susceptible to the malonic ester synthesis.

The malonic ester synthesis requires a compound with a methyl group adjacent to both carboxylate groups, and a benzene ring does not fulfill this requirement. The last two carboxylic acids described cannot be prepared by the malonic ester synthesis because an SN₂ reaction cannot be performed on compounds with bulky substituents or with two or more halogen atoms attached to the same carbon atom.

The synthesis requires the use of an alkyl halide that can undergo an SN₂ reaction with sodium ethoxide, but benzyl bromide, bromobenzene, and dibromobenzene are not suitable for this type of reaction. Additionally, the bromide required for the synthesis is unstable, which further complicates the reaction.

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0.18 g of a divalent metal was completely dissolved in 250 cc of acid solution containing 4. 9 g H₂SO₄ per liter. 50 cc of the residual acid solution required 20 cc of N/10 alkali for complete neutralization. The atomic weight of metal is 45 g/mol.  

First, we need to determine the moles of H₂SO₄ present in 250 cc of the acid solution:

4.9 g/L = 0.0049 g/cc

0.0049 g/cc x 250 cc = 1.225 g of H₂SO₄

Next, we can calculate the number of moles of H₂SO₄ that were neutralized by the alkali solution:

20 cc of N/10 NaOH = 0.002 mol NaOH

Since the reaction is:

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

then 1 mol of H₂SO₄ reacts with 2 mol of NaOH, therefore 0.004 mol of H₂SO₄ reacted with 0.002 mol of NaOH.

So, the remaining number of moles of H₂SO₄ is:

0.004 mol - 0.002 mol = 0.002 mol

Now we can calculate the moles of metal present in the solution:

0.18 g / atomic weight = moles of metal

We can use the remaining H₂SO₄ to find the number of moles of metal:

1 mol of H₂SO₄ reacts with 1 mol of metal, so the number of moles of metal is equal to the number of moles of H₂SO₄ remaining:

0.002 mol H₂SO₄ = 0.002 mol metal

Now we can solve for the atomic weight:

0.18 g / 0.002 mol = 90 g/mol

Since the metal is divalent, we need to divide by 2 to get the atomic weight:

90 g/mol / 2 = 45 g/mol

Therefore, the atomic weight of the metal is 45 g/mol.

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According to the question the q(gas) in J for this compression process is 0J.

Gas is a state of matter in which particles are spread out and have enough energy to move around freely. Gas is composed of molecules in constant motion and takes the shape and volume of its container. Gas can be either naturally occurring or man-made and is found in the atmosphere. Examples of naturally occurring gases include oxygen, nitrogen, and carbon dioxide. Man-made gases include helium, chlorine, and hydrogen. Gas is often used as a source of energy and is burned to produce heat, which can be used to power machines and vehicles. Gas is also used in many industries, such as in the production of chemicals and plastics.

In this case, n = 0.100 moles,
[tex]C_v[/tex] = (3/2)R = (3/2)(8.314 J/mol K) = 12.471 J/mol K, and
T₁ = 298.0 K,
T2 = 298.0 K.
Therefore, q(gas)
= nCv (T₂- T₁)
= 0.100 mol × 12.471 J/mol K × (298.0 K - 298.0 K)
= 0 J.

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The molar solubility of [tex]AgBr[/tex] in [tex]2.8 x 10^-2 M AgNO3[/tex] solution is [tex]7.1 x 10^-7 M[/tex].

To calculate the molar solubility of [tex]AgBr[/tex] in [tex]2.8 x 10^-2 M AgNO3[/tex] solution, we need to use the common ion effect. The [tex]Ag+[/tex] ion is a common ion in both [tex]AgBr and AgNO3[/tex]. When we add [tex]AgNO3[/tex] to a solution containing AgBr, it adds more [tex]Ag+[/tex] ions to the solution and causes a shift in the equilibrium to the left. The solubility of [tex]AgBr[/tex]decreases due to this effect.

The balanced equation for the dissolution of [tex]AgBr[/tex] is:

[tex]AgBr(s) ⇌ Ag+(aq) + Br-(aq)[/tex]

The Ksp expression for AgBr is:

Ksp = [Ag+][Br-] = 5.0 x 10^-13

Let x be the molar solubility of [tex]AgBr[/tex]in [tex]2.8 x 10^-2 M AgNO3[/tex]solution. Then the concentration of [tex]Ag+[/tex] ion is[tex][Ag+] = 2.8 x 10^-2 + x[/tex], and the concentration of [tex]Br-[/tex] ion is[tex][Br-] = x[/tex].

Substituting these values into the Ksp expression, we get:

[tex]Ksp = (2.8 x 10^-2 + x)(x) = 5.0 x 10^-13[/tex]

Simplifying the equation and neglecting x in comparison to [tex]2.8 x 10^-2[/tex], we get:

[tex]x^2 = 5.0 x 10^-13x = sqrt(5.0 x 10^-13) = 7.1 x 10^-7 M[/tex]

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The enthalpy of the reaction can be obtained as 118 kJ/mol.

Reaction enthalpy, also known as heat of reaction or ΔHrxn, is the change in enthalpy that occurs during a chemical reaction. It is defined as the difference between the enthalpy of the products and the enthalpy of the reactants.

We have;

Enthalpy of reaction = Bonds broken - Bonds formed

Enthalpy of reaction = [4(413) + 2(495) - [2(799) + 2(463)

= [1652 + 990] - [1598 + 926]

=2642 - 2524

= 118 kJ/mol

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To find another element with similar properties, the student should look on the periodic table of the elements for an element In the same group.

The student should look for an element in the same group as krypton on the periodic table of elements. Elements in the same group have similar chemical properties because they have the same number of valence electrons, which determine how the element interacts with other elements.

Krypton is a noble gas located in group 18 (also known as group 8A) of the periodic table. Other elements in this group, such as neon, helium, and xenon, also have very low chemical reactivity due to their full valence electron shells.

Elements in the same period as krypton may have similar atomic properties, but they are not guaranteed to have similar chemical properties since their valence electron configurations differ.

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Globular clusters are usually found in the halo of a galaxy. The answer is d.

Globular clusters are dense, spherical collections of stars that orbit a galactic center. They are typically composed of tens of thousands to hundreds of thousands of stars and are some of the oldest known objects in the universe.

Globular clusters are usually found in the halo of a galaxy, which is the outermost region of a galaxy that surrounds the disk.

This is because they are thought to have formed early in the history of the galaxy, when the halo was still being formed.

In contrast, stars in the disk of a galaxy are typically younger and more spread out, with less dense collections of stars. The nucleus of a galaxy is the central region, which usually contains a supermassive black hole and dense concentrations of stars.

Therefore, the correct answer is d. halo.

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The final volume of the gas is 3.75 L. This result can be explained by the fact that as the temperature of the gas increased, the kinetic energy of its particles also increased, causing them to move faster and occupy a larger volume.

According to the ideal gas law, PV = nRT, the volume of a gas is directly proportional to its temperature.

Therefore, if the temperature of a gas is increased while its pressure and amount remain constant, its volume will also increase.

In this case, the initial volume of the gas is 2.5 L and its temperature is increased from 200 K to 300 K. To find the final volume of the gas, we can use the following equation:

V2 = (T2/T1) x V1

where V1 is the initial volume of the gas, T1 is the initial temperature of the gas, T2 is the final temperature of the gas, and V2 is the final volume of the gas. Plugging in the values, we get:

V2 = (300 K/200 K) x 2.5 L

V2 = 3.75 L

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