A 7.32 l tire contains 0.448 mol of gas at a temperature of 28°c. what is the pressure (in atm) of the gas in the tire?

The mass of the ancient gold medallion is 360 grams.

To calculate the mass of the gold medallion, we need to use the specific heat capacity of gold, which is 0.129 J/g°C. We also need to know the initial temperature of the medallion.

Let's assume the initial temperature of the gold medallion is 20.0°C (room temperature). The heat absorbed by the gold medallion can be calculated using the following formula:

Q = m * c * ΔT

Substituting the given values, we get:

576 J = m * 0.129 J/g°C * 25.0°C

Solving for m, we get:

m = 360 g

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Answer is B) 1:2

The electron configuration of magnesium is 1s2 2s2 2p6 3s2, which means it has two valence electrons that it can lose to form a cation with a +2 charge.

The electron configuration of sulfur is 1s2 2s2 2p6 3s2 3p4, which means it has six valence electrons that it can gain to form an anion with a -2 charge.

Since magnesium can form a cation with a +2 charge and sulfur can form an anion with a -2 charge, the ratio of metal cationic (+) atom to nonmetal anionic (-) atom in the compound will be 1:2. Therefore, the answer is b. 1:2.

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A total of 28,000 grams of calcium oxide will be produced.

To find out how many grams of calcium oxide will be produced in a closed vessel containing 20.0 kg of calcium and 20.0 kg of oxygen gas, follow these steps:

1. Convert the given masses into moles using the molar mass of each element:
  - For calcium (Ca): 20,000 g / 40.08 g/mol ≈ 499 moles
  - For oxygen (O2): 20,000 g / 32 g/mol ≈ 625 moles

2. Determine the limiting reactant using the stoichiometry of the balanced equation:
  - The stoichiometric ratio of Ca to O2 is 2:1, so 625 moles of O2 would require 1,250 moles of Ca, but there are only 499 moles of Ca available. Therefore, calcium is the limiting reactant.

3. Calculate the moles of calcium oxide (CaO) produced using the stoichiometry of the balanced equation:
  - The ratio of Ca to CaO is 1:1, so 499 moles of Ca will produce 499 moles of CaO.

4. Convert the moles of calcium oxide back to grams using the molar mass:
  - The molar mass of CaO is 56.08 g/mol (40.08 g/mol for Ca + 16 g/mol for O). Therefore, 499 moles of CaO * 56.08 g/mol ≈ 28,000 grams of calcium oxide will be produced.

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The effect of increased temperature on the rate of dissolution of a solid solute is; Increased temperature increases the rate of dissolution of a solid solute. Option A is correct.

This is because at higher temperatures, the kinetic energy of the solvent molecules increases, leading to more frequent and more energetic collisions with the solute particles. This increased kinetic energy can overcome the intermolecular forces holding the solute together, leading to more rapid dissolution.

The rate of dissolution refers to how quickly a solute dissolves in a solvent to form a homogeneous solution. It is usually expressed as the amount of solute that dissolves per unit time, typically in grams per second or moles per minute.

Hence, A. is the correct option.

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The cost of electricity for the device left on for 8.0 hours is 3.52 cents.

To find the cost of electricity for the device, first, we need to calculate the power consumption, then the total energy consumed, and finally the cost.

1. Calculate the power consumption:

Power (P) = Voltage (V) x Current (I)
P = 110 volts x 0.50 amperes = 55 watts

2. Calculate the total energy consumed:

Energy (E) = Power (P) x Time (t)
E = 55 watts x 8.0 hours = 440 watt-hours = 0.44 kilowatt-hours (kWh)

3. Calculate the cost:

Cost = Energy (E) x Rate
Cost = 0.44 kWh x 8 cents/kWh = 3.52 cents

The cost of electricity for the device left on for 8.0 hours is 3.52 cents.

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To make a 1.5 L solution of 0.35 M HCl using 16.0 M HCl, you will need 32.81 mL of concentrated acid.

1. Use the dilution formula: M1V1 = M2V2

2. M1 is the initial concentration (16.0 M), V1 is the volume of concentrated acid needed, M2 is the final concentration (0.35 M), and V2 is the final volume (1.5 L).

3. Plug in the values: (16.0 M)(V1) = (0.35 M)(1.5 L)

4. Solve for V1: V1 = (0.35 M)(1.5 L) / 16.0 M

5. V1 = 0.0328125 L, which is equal to 32.81 mL.

6. So, 32.81 mL of concentrated 16.0 M HCl is needed to make the 1.5 L solution of 0.35 M HCl.

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An object in motion will continue to move at a constant velocity unless acted upon by an external force. This principle is known as Newton's First Law of Motion, also referred to as the law of inertia.

Inertia is the tendency of an object to resist changes in its state of motion.

Similarly, an object at rest will remain at rest unless acted upon by an external force. This means that if an object is not moving, it will continue to stay still until a force is applied to it.

Newton's First Law of Motion is a fundamental concept in physics that explains how objects behave when in motion or at rest. It is important to understand this law because it helps us to predict how objects will move and interact with each other.

Additionally, it is also essential in the design and engineering of machines and structures that require a thorough understanding of motion and force.

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The standard entropy change for the reaction [tex]2Mg(s)+O_2(g)\rightarrow 2MgO(s)[/tex] is -405.6 J/(K⋅mol).

Entropy is a measure of the randomness or disorder in a system. It is a thermodynamic property that can be used to measure the amount of energy that is unavailable for work in a thermodynamic process. Entropy is closely related to the second law of thermodynamics and can be used to assess the direction of a thermodynamic process. Entropy is also a measure of the amount of information contained in a system. High entropy systems have more randomness and disorder, while low entropy systems have less.

The entropy change for the reaction [tex]2Mg(s)+O_2(g) \rightarrow 2MgO(s)[/tex] is calculated using the following equation: [tex]\Delta S^\circ = \Sigma S^\circ products -\Sigma S^\circ reactants[/tex]

Substituting the values from the table:

[tex]\Delta S^\circ = (2 \times 27.00 J/(Kmol)) - (32.70 J/(Kmol) + 205.0 J/(Kmol))\\\Delta S^\circ = -405.6 J/(Kmol) .[/tex]

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The mass of NaCl required to raise the boiling point of 4.73 L of water by 1.00°C is 25.3 g.

The boiling point elevation (ΔTb) is given by the equation ΔTb = Kb × molality, where Kb is the boiling point elevation constant for water (0.51°C/m) and molality is the concentration of solute in mol/kg of solvent. To calculate the molality, we need to convert the volume of water to mass (assuming a density of 1 g/mL) and calculate the number of moles of water. We have:

To raise the boiling point by 1.00°C, we need to find the molality that gives a ΔTb of 1.00°C. Rearranging the equation above, we get:

Now we can calculate the mass of NaCl required to achieve this molality:

Therefore, the mass of NaCl required to raise the boiling point of 4.73 L of water by 1.00°C is 25.3 g (since 550 g is more than the mass of water).

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The molar enthalpy for the reaction of calcium chloride is -22,982.5 J/mol.

Calcium chloride is a chemical compound that is commonly used as a drying agent due to its hygroscopic properties. In this question, we are given the amount of calcium chloride and asked to determine the molar enthalpy for its reaction with excess potassium.

The given temperature change of the solution in the calorimeter can be used to calculate the heat released or absorbed during the reaction.

To begin, we need to determine the number of moles of calcium chloride in the given amount of 4 g. Using the molar mass of calcium chloride (110.98 g/mol), we can calculate that 4 g of calcium chloride is equal to 0.036 moles. Since the reaction is with excess potassium, we can assume that all the calcium chloride will react.

Next, we can use the heat capacity of the solution and the temperature change to calculate the heat released or absorbed during the reaction. Assuming that the solution is mainly water, we can use the specific heat capacity of water (4.18 J/g°C) to calculate the heat absorbed by the solution.

The mass of the solution is the sum of the mass of calcium chloride and the mass of water, which is 4 g + 7.5 g = 11.5 g. The temperature change is 32 °C - 15 °C = 17 °C. Therefore, the heat absorbed by the solution is:

Q = m x c x ΔT = 11.5 g x 4.18 J/g°C x 17 °C = 827.37 J

Since the reaction is exothermic (heat is released), the molar enthalpy can be calculated using the following equation:

ΔH = -Q/n

where n is the number of moles of calcium chloride. Plugging in the values, we get:

ΔH = -827.37 J/0.036 mol = -22,982.5 J/mol

Therefore, the molar enthalpy for the reaction of calcium chloride is -22,982.5 J/mol.

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The difference in temperature from Saturday to Sunday is 9 degrees Celsius. This means that the temperature increased by 9 degrees from Saturday to Sunday.

To show the difference in temperature from Saturday to Sunday, we can use the equation:

Difference = Sunday temperature - Saturday temperature

Given that the temperature on Saturday is -13° and on Sunday it is -4°, we can calculate the difference in temperature using the above equation as follows:

Difference = -4° - (-13°)

Difference = -4° + 13°

Difference = 9°

The number line is a graphical representation of numbers where we can visualize their position relative to each other. Starting from -13° on the number line and moving 9 units to the right, we reach -4°, which represents the temperature on Sunday. This visualization confirms that the difference between the two temperatures is 9 degrees.

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The pressure of the gas increases.

When 10 liters of a gas at 1 atm is compressed to 3 liters at constant temperature, the property of the gas that changes is the pressure of the gas increases. This is due to the fact that the volume of the gas has decreased while the number of gas particles remains constant. As the particles are now confined to a smaller space, they collide more frequently with the walls of the container, resulting in an increase in pressure.

The number of moles of gas and the mass of the gas remain constant because the compression occurs at a constant temperature, indicating that there is no change in the amount of gas particles. The size of the gas particles does not change either, as this is a property of the gas molecules themselves and is not influenced by external factors like pressure or temperature.

In summary, when a gas is compressed at a constant temperature, the pressure of the gas increases due to the decrease in volume. This relationship is described by Boyle's Law, which states that the pressure and volume of a gas are inversely proportional to each other at a constant temperature.

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Answer:

Hello! This lab is all about heat, which is a measure of energy in a system. In the first part, we'll be examining what happens to a gas when the temperature changes. For this part, you will need a small mouth bottle, a coin, a large container, cold water, and measuring cups. In the second part, we'll be using the idea of energy transfer to move water. For this part, you will need food coloring, water, a large bowl, a small glass, cling film, a small object, and sunny days. Follow the procedures carefully and answer the questions provided to understand the concepts of heat and energy transfer. Don't hesitate to reach out if you have any questions!

C water = 1 cal/g ℃

One of the biggest problems of human impact on the environment is the excessive use of non-renewable resources, such as fossil fuels, which release harmful gases and contribute to climate change.

This problem can be solved by designing a device or process that can harness renewable energy sources, such as solar or wind power, and provide a sustainable alternative to traditional energy sources.

By solving this problem, not only will the environment benefit from reduced carbon emissions, but also the people who rely on these resources. For instance, communities that are vulnerable to the effects of climate change, such as extreme weather conditions, will be better equipped to adapt and withstand these impacts.

The criteria and constraints of designing such a device or process would include factors such as cost, efficiency, scalability, and environmental impact. The solution would need to be cost-effective and efficient, while also being able to provide a significant amount of energy to meet the needs of communities.

Additionally, it would need to be environmentally friendly and have minimal negative impact on ecosystems.

One possible solution could be the development of solar-powered devices that can be used in homes, schools, and businesses to generate electricity. Another solution could be the installation of wind turbines in areas with high wind speeds to generate energy on a larger scale.

Overall, by designing devices or processes that harness renewable energy sources, we can mitigate the negative impacts of non-renewable energy sources on the environment and provide sustainable alternatives for the benefit of both the environment and society.

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The theoretical yield of oxygen from the oxides present in a 1.00 kg sample of lunar soil will depend on the composition of the soil. However, we can make some assumptions based on the known composition of lunar soil.

Lunar soil is known to contain various oxides, including silicon dioxide (SiO2), aluminum oxide (Al2O3), iron oxide (FeO and Fe2O3), titanium dioxide (TiO2), and others. These oxides can be chemically processed to release oxygen gas.

The stoichiometry of the chemical reactions involved will depend on the specific oxides present in the soil. However, for the purposes of estimation, we can assume that all the oxides present in the soil are converted to their respective metals and oxygen gas.

For example, the reaction for the conversion of silicon dioxide to silicon metal and oxygen gas is:

SiO2(s) + 2 C(s) → Si(s) + 2 CO(g)

From this reaction, we can see that for every 1 mole of SiO2, 1 mole of oxygen gas is produced. The molar mass of SiO2 is 60.08 g/mol, so in a 1.00 kg sample of lunar soil, there are:

1000 g / 60.08 g/mol = 16.65 moles of SiO2

Therefore, the theoretical yield of oxygen gas from the SiO2 present in the soil is:

16.65 moles of O2 (since 1 mole of SiO2 produces 1 mole of O2)

Similarly, we can calculate the theoretical yield of oxygen gas from the other oxides present in the soil using their respective stoichiometric equations. Adding up the oxygen yields from each oxide will give us the total theoretical yield of oxygen from the soil.

Note that the actual yield of oxygen will likely be less than the theoretical yield due to inefficiencies and losses during the processing of the soil.

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Answer:

NaHCO3 + HCl → NaCl + H2O + CO2

Explanation:

In constructive interference, the waves add together.

Interference is the phenomenon where two or more waves interfere with each other to form a resultant wave of greater, lower or same amplitude.

In constructive interference, the waves combine in such a way that their amplitudes are reinforced, resulting in a wave with a larger amplitude than the individual waves.

So, the correct answer is: a. they add

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The pH of the buffered solution is approximately 4.48.

The Henderson-Hasselbalch equation, which connects the pH of a buffered solution to the acid's pKa and the full concentrations of both the acid and its conjugate base as given by the situation, can then be used.

pH = pKa + log([conjugate base]/[acid])

In order to determine the pH of a buffered solution made by combining 21.5 g of benzoic acid ([tex]C_7H_6O_2[/tex]) and 37.7 g of sodium benzoate ([tex]NaC_7H_5O_2[/tex]) in water, we first need to figure out the buffer system's equilibrium constant (Ka). The benzoic acid's Ka value is  [tex]6.3 * 10^{-5}[/tex].

Substituting the values into the Henderson-Hasselbalch equation:

[tex]pH = pKa + log([NaC_7H_5O_2]/[C_7H_6O_2]) \\pH = 4.2 + log(37.7/21.5)[/tex]

pH = 4.2 + 0.28

pH = 4.48

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The pH of the solution where [tex][OH⁻][/tex]=0.00030 M is 11.48. This indicates that the solution is basic, or alkaline, since the pH is greater than 7.

To determine the pH of a solution where[tex][OH⁻][/tex]=0.00030 M, we can use the relationship between the concentrations of hydrogen ions and hydroxide ions in water, which is defined by the equation[tex]Kw = [H⁺][OH⁻].[/tex]At 25°C, the value of Kw is [tex]1.0 x 10^-14[/tex].

If we substitute the concentration of hydroxide ions given in the question ([tex][OH⁻][/tex]=0.00030 M) into this equation, we can solve for the concentration of hydrogen ions:

[tex]Kw = [H⁺][OH⁻]\\1.0 x 10^-14 = H⁺\\[H⁺] = 3.3 x 10^-12 M[/tex]

Now that we know the concentration of hydrogen ions, we can use the formula for pH, which is defined as [tex]pH = -log[H⁺][/tex], to find the pH of the solution:

[tex]pH = -log(3.3 x 10^-12)[/tex]

pH = 11.48

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To convert moles of sodium chloride to grams, we multiply by its molar mass of 58.44 g/mol. Therefore, the amount of sodium chloride produced is 0.363 mol x 58.44 g/mol = 21.2 grams.

To determine the limiting reagent in this reaction, we need to calculate the moles of both reactants. From the given information, we know that the mass of iron (II) chloride is 23 grams, and its molar mass is 126.75 g/mol.

Therefore, the number of moles of iron (II) chloride is 23 g/126.75 g/mol = 0.1815 mol.

Next, we calculate the number of moles of sodium phosphate. Since there are two molecules of sodium phosphate for every three molecules of iron (II) chloride, we need to multiply the moles of iron (II) chloride by the ratio of the coefficients. Therefore, the number of moles of sodium phosphate is (0.1815 mol x 2/3) = 0.121 mol.

Since there are fewer moles of sodium phosphate than iron (II) chloride, sodium phosphate is the limiting reagent. This means that all of the sodium phosphate will be used up in the reaction, and any remaining iron (II) chloride will be left over.

To calculate the amount of sodium chloride produced, we need to use the stoichiometric coefficients from the balanced equation.

For every 2 moles of sodium phosphate used, 6 moles of sodium chloride are produced. Therefore, since we have 0.121 mol of sodium phosphate, we can produce (0.121 mol x 6/2) = 0.363 mol of sodium chloride.

Finally, to convert moles of sodium chloride to grams, we multiply by its molar mass of 58.44 g/mol. Therefore, the amount of sodium chloride produced is 0.363 mol x 58.44 g/mol = 21.2 grams.

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The reaction did not produce as much as expected based on the theoretical yield. However, the percentage yield can be calculated by dividing the actual yield by the theoretical yield and multiplying by 100, which gives a value of 53.75%.

According to the balanced equation, the theoretical yield of water produced from the combustion of 20 grams of methane is 80 grams. This is calculated by first finding the moles of methane used (20g / 16.04 g/mol = 1.247 mol) and then using the stoichiometric ratio to determine the moles of water produced (2 moles of H2O for every 1 mole of CH4), which gives 2.494 mol of water. Finally, converting the moles of water to grams gives a theoretical yield of 80 grams.

However, the actual yield of water obtained from the reaction was only 43 grams, which is significantly less than the theoretical yield. This could be due to a variety of reasons, such as incomplete combustion of methane, loss of product during collection or transfer, or errors in measurement or calculation.

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Rounded to the nearest whole number, y is 42. Therefore, the molecular formula of the hydrocarbon is C4H42.

To find the molecular formula of the hydrocarbon, we first need to determine the molecular weight. We know that the mass ratio between hydrogen and carbon is 1:10, which means that for every 1 gram of hydrogen, there are 10 grams of carbon in the molecule.

Let's assume that we have x number of carbon atoms and y number of hydrogen atoms in the molecule. The molecular weight can then be expressed as:

Molecular weight = (x x atomic weight of carbon) + (y x atomic weight of hydrogen)

Since the mass ratio between hydrogen and carbon is 1:10, we can write:

y = 10x

Now, we can substitute y in the equation for molecular weight:

Molecular weight = (x x atomic weight of carbon) + (10x x atomic weight of hydrogen)

Molecular weight = x(atomic weight of carbon + 10 x atomic weight of hydrogen)

We also know that one liter of hydrogen at 127°C and 1 atm pressure weighs 2.8 g. Using the ideal gas law, we can calculate the number of moles of hydrogen in one liter:

PV = nRT

n = PV/RT

n = (1 atm x 1 L) / (0.0821 L.atm/mol.K x 400 K)

n = 0.0305 mol

The molecular weight of the hydrocarbon can be calculated as follows:

Molecular weight = 2.8 g / 0.0305 mol

Molecular weight = 91.80 g/mol

Now, we can solve for x in the equation for molecular weight:

91.80 g/mol = x(12.01 g/mol + 10 x 1.01 g/mol)

91.80 g/mol = 12.01x + 10.10x

91.80 g/mol = 22.11x

x = 4.15

Since x represents the number of carbon atoms in the molecule, we can round it to the nearest whole number, which is 4. Similarly, y can be calculated as:

y = 10x = 41.5

Rounded to the nearest whole number, y is 42. Therefore, the molecular formula of the hydrocarbon is C4H42.

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The scenario described in the question is an example of the negative impact that can result from introducing exotic organisms into an ecosystem.

Exotic organisms, also known as invasive species, are non-native species that are introduced to an ecosystem and can outcompete native species, disrupt natural ecological processes, and cause harm to the environment and economy.

When the exotic organisms are released into the local environment, they have no natural predators, and their population can increase unchecked, causing a decrease in biodiversity.

This is because the invasive species may outcompete and displace native species, reduce the availability of resources, and alter the habitat. The result is a homogenization of the ecosystem, where there are fewer different types of species and less overall diversity.

In summary, introducing exotic organisms can have a negative impact on biodiversity in an ecosystem, which can have cascading effects on the health and stability of the ecosystem. It is important to carefully manage and monitor the introduction of exotic organisms to prevent these negative impacts.

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A concentration of 0.30 M [[tex]H^{+}[/tex]] in the water would indicate a strong acid for the given solution of [tex]H_{2} X[/tex].

A strong acid is one that completely dissociates in water, meaning it donates all of its hydrogen ions ([tex]H^{+}[/tex]) to the solution.

For the given acid, [tex]H_{2} X[/tex], the dissociation equation would be:
[tex]H_{2} X[/tex] → 2[tex]H^{+}[/tex] + [tex]X^{2-}[/tex]

Since it's a strong acid, we assume that all molecules will dissociate, resulting in two moles of [tex]H^{+}[/tex] for every mole of [tex]H_{2} X[/tex]. Therefore, to calculate the concentration of [[tex]H^{+}[/tex]] in the solution:

[[tex]H^{+}[/tex]] = 2 × (concentration of [tex]H_{2} X[/tex])

Given the concentration of [tex]H_{2} X[/tex] is 0.15 M:

[[tex]H^{+}[/tex]] = 2 × 0.15 M

[[tex]H^{+}[/tex]] = 0.30 M

So, a concentration of 0.30 M [[tex]H^{+}[/tex]] in the water would indicate a strong acid for the given solution of [tex]H_{2} X[/tex].

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The student's task is to determine the density of the four mystery liquids using the mass and volume measurements.

Density is a physical property that describes the amount of mass per unit volume.

The formula for density is density = mass/volume. Once the density of each liquid is determined, the student can compare it to known densities of different substances to identify the liquid.

This information can be useful in various fields such as chemistry, pharmacology, and environmental science.

The student may also use this data to calculate other properties of the liquids such as viscosity, surface tension, and boiling point. Overall, measuring mass and volume is a fundamental method in scientific research and analysis.

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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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You should use about 25.51 mL of concentrated H3PO4 to prepare 125 mL of a 3.00 M H3PO4 solution.

To prepare 125 mL of a 3.00 M H3PO4 solution using concentrated H3PO4 (14.7 M), you can use the dilution formula:

M1 × V1 = M2 × V2

Where M1 is the initial molarity (14.7 M), V1 is the volume of the concentrated solution needed, M2 is the final molarity (3.00 M), and V2 is the final volume (125 mL).

Rearrange the formula to solve for V1:

V1 = (M2 × V2) / M1

V1 = (3.00 M × 125 mL) / 14.7 M

V1 ≈ 25.51 mL

Therefore, you should use approximately 25.51 mL of concentrated H3PO4 to prepare 125 mL of a 3.00 M H3PO4 solution.

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The addition of ammonia increase the solubility of slightly soluble salt AgCl as : ammonia forms very soluble complex ion by coordinating to Ag and removing it from solution. This shifts the solubility equilibrium to right.

When ammonia (NH₃) is added to a solution containing AgCl, it can coordinate with silver ions (Ag+) to form a complex ion called [Ag(NH₃)₂]+, which is highly soluble in water. This complex ion removes the Ag+ ions from the solution, thereby decreasing the concentration of Ag+ in the solution. According to Le Chatelier's principle, this will shift the equilibrium of AgCl dissolution reaction to the right, resulting in increase in the solubility of AgCl.

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Starting with 8.00 g of Phosphorus-32 (32P) with a half-life of 14.0 days, after 98.0 days, 0.125 g of  32P will remain.

The half-life of a radioactive isotope is the time required for half of the original sample to decay. In this case, the half-life of 32P is 14.0 days, which means that after 14.0 days, half of the 32P will decay, leaving 4.00 g.

To find out how much 32P remains after 98.0 days, we need to determine the number of half-lives that have passed. Dividing 98.0 days by 14.0 days gives us 7.

Therefore, after 7 half-lives, the amount of 32P that remains can be calculated as:

Amount remaining = (1/2)⁷ x 8.00 g = 0.125 g

Therefore, after 98.0 days, 0.125 g of 32P will remain.

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The hydroxide concentration of the LiOH solution is 0.000016 M.

We know that:

pOH = -log[OH-]

We can rearrange this equation to solve for [OH-]:

[OH-] = 10^(-pOH)

Substituting the given pOH value of 4.80, we get:

[OH-] = 10^(-4.80)

[OH-] = 1.58 × 10^(-5)

Rounding to two decimal places, the hydroxide concentration of the LiOH solution is:

[OH-] = 0.000016 (rounded to two decimal places)

Therefore, the hydroxide concentration of the LiOH solution is 0.000016 M.

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