17. An artist took two photographs of the Moon that were several days apart. Images that look like his photographs are shown above. The light part of the Moon appeared to get smaller over time. Why did this happen?

a. [C]: [tex]HCl[/tex] or any other strong acid b. [S]: Lead acetate or any other heavy metal salt c. [I]: Lead nitrate or silver nitrate

a. To confirm the presence of [tex]CO_32[/tex]-, a solution of dilute [tex]HCl[/tex] (hydrochloric acid) is added. If [tex]CO_32[/tex]- is present, it will react with the [tex]HCl[/tex] to produce [tex]CO_2[/tex] gas, which can be identified by bubbling it through limewater [tex](Ca(OH)_2)[/tex].

b. To confirm the presence of [tex]S_2[/tex]-, a solution of lead acetate [tex](Pb(CH_3COO)_2)[/tex] is added. If [tex]S_2[/tex]- is present, it will react with the lead acetate to produce a black precipitate of lead sulfide ([tex]PbS[/tex]).

c. To confirm the presence of I-, a solution of chlorine water ([tex]Cl_2[/tex] in water) is added. If I- is present, it will react with the chlorine to produce a brown color, which is due to the formation of iodine (I2).

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When the second cation first starts to precipitate, 80.8% of Ca²⁺ will still be in solution.

A cation is an ion with a positive charge. It is formed when an atom loses one or more of its electrons, resulting in a net positive charge. Cations are attracted to anions (ions with a negative charge) due to their opposite charges. Cations are found in many different substances, including acids, bases, and salts.

Ca₃(PO₄)₂ will be the first species that separates out of solution when solid Na₃PO₄ is introduced to the mixture. This is due to Ca3(PO4)2 having a substantially lower solubility than Ag₃PO₄ and Na₃PO₄.

The proportion of the first cation (Ca²⁺ ) still in solution when the second cation (Ag⁺) is just beginning to precipitate will depend on the starting concentrations of the two cations. In this instance, the starting concentrations of Ca²⁺  and Ag⁺ are 0.0400 M and 0.0990 M, respectively. Therefore, 80.8% of Ca²⁺  will still be in solution when its second cation first begins to precipitate.

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My model is distinct from the model in the image in that it takes a more thorough and all-encompassing approach to comprehending the fundamental parts of a system.

It considers the interactions between various system elements as well as the connections between those elements and their surroundings. It also looks at how the system changes over time, and how different components interact with each other.

As a result, the system may be understood more precisely, and management choices can be made with more knowledge. In order to offer a more precise and current picture of the system, my model also integrates the most recent research and technological advancements.

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The given answer statement  "there is no aluminum left" and " limiting reactants is aluminum" are correct.

In the analysis of Trial 2, it was found that there was no aluminum left after the reaction had taken place. This indicates that all of the aluminum had reacted with the copper (II) chloride and that it was the limiting reactant in the reaction.

To confirm this, the mass of each reactant was converted to moles using their respective molar masses. It was found that the aluminum had a smaller number of moles than the copper (II) chloride, indicating that it would be used up first and thus be the limiting reactant.

Therefore, the limiting reactant in Trial 2 was aluminum.

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The standard cell potential is found as +1.95 V and is a  spontaneous  reaction.

The standard cell potential (E°cell) of an electrochemical cell is given by the difference between the standard reduction potentials of the two half-cells involved.

E°cell = E°reduction (cathode) - E°reduction (anode)

The half-reactions given are:

Pb4+ + 2e− → Pb2+ (reduction)

Co3+ + e− → Co2+ (reduction)

The standard reduction potentials for these half-reactions are:

E°reduction(Pb4+/Pb2+) = -0.13 V

E°reduction(Co3+/Co2+) = +1.82 V

We then calculate as:

E°cell = E°reduction (Co3+/Co2+) - E°reduction (Pb4+/Pb2+)

E°cell = (+1.82 V) - (-0.13 V)

E°cell = +1.95 V

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Answer:

Ans: 8.9 NaCl

Explanation:

Ocean water contains 3.5 nacl by mass how much salt can be obtained from 254 g of seawater

Question: Ocean water contains 3.5% NaCl by mass. How much salt can be obtained from 254g of seawater?

The precipitation of an ionic substance from solution occurs when the ionic product exceeds the value of its solubility product at that temperature. Here the moles of Na₂S₂O₃ needed is

The solubility product of a sparingly soluble salt is defined as the product of the molar concentrations of its ions in a saturated solution of it at a given temperature.

Here the concentration of Ag⁺ ions = √Ksp = √3.3 × 10⁻¹³ = 1.81 × 10⁻¹³.

Moles of Ag⁺ ions: (1.82 x 10⁻¹³ M) x 1.0 L = 1.82 x 10⁻¹³  mol Ag⁺

Use the stoichiometry of the reaction to find the moles of Na₂S₂O₃ needed: 1 mol Na₂S₂O₃ / 2 mol Ag⁺ = 0.5 mol Na₂S₂O₃/mol Ag⁺

Moles of Na₂S₂O₃ required: 0.5 mol Na₂S₂O₃/mol Ag⁺ x 1.82 x 10⁻¹³ mol Ag⁺ = 9.1 x 10⁻¹⁴ mol Na₂S₂O₃

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The difference in Pa between the pressure in the bag and the atmospheric pressure is 1.505 kPa.

To obtain the difference in pressure, we first need to know the atmospheric pressure near sea level. This is 760 mm Hg. When we convert this to pascals, we will have,  101.32472 kPa.

Now, the difference in pressure will be obtained by subtracting the atmospheric pressure in the bag from the atmospheric pressure near sea level and this is:

101.32472 kPa -  99.82 kPa

= 1.505 kPa.

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To calculate the amount of heat energy needed to cause the rise in temperature, you can use the formula:
Q = mcΔT

Where Q represents the heat energy, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.
Given:
m = 100 grams
c = 4.18 J/g°C
Initial temperature (T1) = 4°C
Final temperature (T2) = 37°C
First, find the change in temperature (ΔT):
ΔT = T2 - T1 = 37°C - 4°C = 33°C
Now, plug the values into the formula:
Q = (100 g) × (4.18 J/g°C) × (33°C)
Q = 13794 J

So, the amount of heat energy needed to cause this rise in temperature is 13,794 Joules.

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Answer:

A. setting standards and governing the cleanliness of water used by Americans

Explanation:

The responsibilities of the Environmental Protection Agency (EPA) is to make sure that:

The mass of the platinum ring is 76.0 grams.

We need to know the molar mass of platinum and the number of platinum atoms in the ring.

The molar mass of platinum (Pt) is 195.08 g/mol.

The number of platinum atoms in the ring is 2.35×10^23.

Now we can use the following formula to calculate the mass of the ring:

mass = (number of atoms) x (atomic mass) / Avogadro's number

where Avogadro's number is 6.022 x 10^23 mol^-1.

Substituting the values:

mass = (2.35×10^23 atoms) x (195.08 g/mol) / (6.022 x 10^23 mol^-1)

mass = 76.0 g

Therefore, the mass of the platinum ring is 76.0 grams.

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The partial pressure of each gas are:

First, we shall determine the mole of 8.00 g of methane, CH₄ and 18.0 g of ethane, C₂H₆. Details below:

For methane, CH₄

Mole = mass / molar mass

Mole of CH₄ = 8 / 16

Mole of CH₄ = 0.5 mole

For ethane, C₂H₆

Mole = mass / molar mass

Mole of C₂H₆ = 18 / 30

Mole of C₂H₆ = 0.6 mole

Next, we shall determine the total mole. Details below:

PV = nRT

3.70 × 10 = n × 0.0821 × 293

Divide both sides by (0.0821 × 293)

n = (3.70 × 10) / (0.0821 × 293)

n = 1.52 mole

Finally, we shall determine the partial pressure of each gas. Details below:

For methane, CH₄

Partial pressure = (Mole / total mole) × total pressure

Partial pressure of CH₄ = (0.5 / 1.52) × 3.70

Partial pressure of CH₄ = 1.22 atm

For ethane, C₂H₆

Partial pressure = (Mole / total mole) × total pressure

Partial pressure of C₂H₆ = (0.6 / 1.52) × 3.70

Partial pressure of C₂H₆ = 1.46 atm

For propane, C₃H₈

Partial pressure of C₃H₈ = Total pressure - (Partial pressure of CH₄ + Partial pressure of C₂H₆)

Partial pressure of C₃H₈ = 3.7 - (1.22 + 1.46)

Partial pressure of C₃H₈ = 1.02 atm

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The ideal gas law states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the temperature in Kelvin. Rearranging the equation, we get:

n = PV/RT

For the container of helium:

n = (201.6 kPa) x (3.9 L) / [(8.31 J/mol*K) x (298 K)] = 0.0688 mol

Now, using the same equation for the container of xenon:

n = (201.6 kPa) x (3.9 L) / [(8.31 J/mol*K) x (298 K)] = 0.0688 mol

Therefore, there are also 0.0688 moles of xenon gas present in the container.

i) The equation for the reaction between hydrochloric acid and zinc is:

[tex]Zn + 2HCl → ZnCl2 + H2[/tex]

ii) n(HCl) = C × V = 1.0M × 0.05 L = 0.05 moles

iii) The volume of gas evolved at STP is 0.544 L or 544 mL.

The concentration of hydrochloric acid is 1.0M, which means that there is 1 mole of hydrochloric acid in 1 liter (1000 cm³) of solution. The volume of the hydrochloric acid used is 50 cm³, which is 0.05 liters.

According to the stoichiometry of the reaction, each mole of hydrochloric acid produces one mole of hydrogen ions, so the number of moles of hydrogen ions in the solution is also 0.05 moles.

The volume of gas evolved can be calculated from the ideal gas law:

PV = nRT

where P is the pressure of the gas, V is the volume of the gas, n is the number of moles of the gas, R is the gas constant (0.0821 L·atm/K·mol), and T is the temperature of the gas in Kelvin. At standard temperature and pressure (STP), the pressure is 1 atm and the temperature is 273 K. The molar volume of a gas at STP is 22.4 L/mol.

From the equation for the reaction, we know that one mole of hydrogen gas is produced for every two moles of hydrochloric acid used. Therefore, the number of moles of hydrogen gas produced is:

n(H2) = 0.5 × n(HCl) = 0.5 × 0.05 moles = 0.025 moles

Using the ideal gas law, we can calculate the volume of hydrogen gas produced at STP:

V(H2) = n(H2) × RT/P = 0.025 mol × 0.0821 L·atm/K·mol × 273 K/1 atm = 0.544 L

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The pressure of hydrogen gas formed is 709.5 mmHg.

Partial pressure is the pressure exerted by a single gas component in a mixture of gases, assuming all other gases are held constant.

In this case, the hydrogen gas is formed by a chemical reaction.

To calculate the partial pressure of hydrogen gas, we need to subtract the vapor pressure of water from the total pressure of the gas collected.

The vapor pressure of water at 18.0 °C is 15.5 mmHg.

Therefore, the partial pressure of hydrogen gas can be calculated as:

Partial pressure of hydrogen gas = Total pressure - Vapor pressure of water

Partial pressure of hydrogen gas = 725.0 mmHg - 15.5 mmHg = 709.5 mmHg

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We would need about 16 hydrogen atoms so that we can convert the compound to a saturated fat.

In animal products like meat and dairy, saturated fat is a form of dietary fat that is normally solid at room temperature. It is known as being "saturated" because each molecule of fat has the most hydrogen atoms possible, giving it a stable structure.

We can see this by counting the number of double bonds in the fat and there are eight of them so sixteen hydrogen atoms are needed for saturation.

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When an equilibrium system is put under stress, Le Chatelier's principle can be used to forecast changes in equilibrium concentrations.

Thus, The adjustments required to reach equilibrium might not be as obvious if we have a mixture of reactants and products that have not yet reached equilibrium.

In this situation, we can compare the Q and K values for the system to forecast changes.

By adding or withdrawing one or more of the reactants or products, an equilibrium chemical system can be momentarily moved out of equilibrium. After that, additional adjustments are made to the reactant and product concentrations in order to bring the system back to equilibrium.

Thus, When an equilibrium system is put under stress, Le Chatelier's principle can be used to forecast changes in equilibrium concentrations.

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Answer:

Explanation:

According to the Law of Conservation of Mass, matter cannot be created or destroyed in a chemical reaction. Therefore, the mass of the reactants must be equal to the mass of the products.

If 15 grams of reactant went into the reaction, then the mass of the products formed must also be 15 grams. This assumes that the reaction is complete and no reactants are left unreacted.

It is important to note that this applies to closed systems where there is no loss or gain of mass. In real-world situations, some mass may be lost due to factors such as evaporation or incomplete reactions, which can affect the accuracy of the calculations.

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The surface area of the rectangular prism is 0.034 square meters.

For a rectangular prism with length l, width w, and height h, the surface area is:

Surface area = 2lw + 2lh + 2wh

Substituting the given values, we get:

Surface area = 2(10 cm x 5 cm) + 2(10 cm x 8 cm) + 2(5 cm x 8 cm)

Surface area = 100 cm² + 160 cm² + 80 cm² = 340 cm²

We can use dimensional analysis. So the conversion factor is:

1 m² / 10,000 cm²

Multiplying the surface area by this conversion factor, we get:

Surface area = 340 cm² x (1 m² / 10,000 cm²)

Surface area = 0.034 m²

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--The complete Question is, What is the surface area of a rectangular prism that has a length of 10 cm, a width of 5 cm, and a height of 8 cm? Use dimensional analysis to convert the answer to square meters--

a) The amount of N₂O₅ is lowered to 2.5 g during the course of around 4.41 × 10⁴  seconds or 12.25 hours.

b) 9.71 L of O₂ are generated at 745 mmHg and 45 °C.

a) To solve for the time required for the quantity of N₂O₅ to be reduced to 2.5 g, use the first-order integrated rate law:

ln[N₂O₅]t/[N₂O₅]0 = -kt

where [N₂O₅]t = concentration of N₂O₅ at time t, [N₂O₅]0 = initial concentration of N₂O₅, k = rate constant, and t = time.

Find the initial concentration of N₂O₅:

n(N₂O₅) = m/M = 80.0 g / 108.01 g/mol = 0.7413 mol

[N₂O₅]0 = n/V = 0.7413 mol / 0.153 L = 4.846 M

where M = molar mass of N₂O₅ and V = volume of the solution.

Substituting the given values into the equation:

ln([N₂O₅]t / 4.846 M) = -6.2×10⁻⁴ s⁻¹ × t

When the quantity of N₂O₅ is reduced to 2.5 g, the concentration is:

n(N₂O₅) = m/M = 2.5 g / 108.01 g/mol = 0.02314 mol

[N₂O₅]t = n/V = 0.02314 mol / 0.153 L = 0.151 M

Substituting this concentration into the equation and solving for t:

ln(0.151 M / 4.846 M) = -6.2×10⁻⁴ s⁻¹ × t

t = 4.41 × 10⁴ s

Therefore, it takes approximately 4.41 × 10⁴ seconds or 12.25 hours for the quantity of N₂O₅ to be reduced to 2.5 g.

b) The balanced equation for the reaction shows that 1 mole of N₂O₅ produces 1/2 mole of O₂:

N₂O₅ → N₂O₄ + 1/2 O2

Therefore, the number of moles of O₂ produced can be calculated using the stoichiometry:

n(O₂) = 1/2 × n(N₂O₅) = 1/2 × 0.7413 mol = 0.3707 mol

The ideal gas law can be used to calculate the volume of O₂ produced at 745 mmHg and 45°C:

PV = nRT

where P = pressure, V = volume, n = number of moles, R = gas constant, and T = temperature in Kelvin.

Convert the pressure to atm and the temperature to Kelvin:

P = 745 mmHg / 760 mmHg/atm = 0.980 atm

T = 45°C + 273.15 = 318.15 K

Substituting the values and solving for V:

V = nRT/P = (0.3707 mol) × (0.08206 L·atm/mol·K) × (318.15 K) / (0.980 atm) = 9.71 L

Therefore, the volume of O₂ produced at 745 mmHg and 45°C is 9.71 L.

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The results of the lab activity showed that the larger the mass of the sun, the more likely at least one planet will fall into the habitable zone.

The mass of the sun affects the orbits of planets in a solar system. When the mass of the sun is larger, the gravitational force between the sun and the planets is stronger, causing the planets to move at a slower pace around the sun.

Conversely, when the mass of the sun is smaller, the gravitational force is weaker, causing the planets to move at a faster pace.

Additionally, when Earth is closer to the sun, the gravitational force is stronger, causing its orbit to become faster, while a farther distance from the sun results in a slower orbit.

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1.71 L of [tex]H^2[/tex] gas can be produced at STP from the given reaction.

According to the equation, 1 mole of hydrogen gas [tex]H^2[/tex]  is created for every 2 moles of sodium (Na) that react with extra water.

Using the molar mass of Na, we can get the number of moles from the given amount of sodium (3.60 g):

3.60 g Na × (1 mol Na / 22.99 g Na) = 0.157 mol Na

Since the reaction requires 2 moles of Na to produce 1 mole of [tex]H^2[/tex]  the number of moles of [tex]H^2[/tex] produced is

0.157 mol Na × (1 mol [tex]H^2[/tex] / 2 mol Na) = 0.079 mol [tex]H^2[/tex]

Now, to calculate the volume of hydrogen gas produced at STP (standard temperature and pressure), we can use the ideal gas law

PV = nRT

Where

Solving for V, we get:

V = nRT/P = (0.079 mol)(0.08206 L atm/(mol K))(273 K)/(1 atm) = 1.71 L

Therefore, 1.71 L of % [tex]H^2[/tex] gas can be produced at STP from the given reaction.

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The value of K at equilibrium, for the reaction is 0.0049

First, we shall list out the given parameters from the question. This is shown below:

Equilibrium constant is defined as:

Equilibrium constant = [Product]ᵐ / [Reactant]ⁿ

Where

With the above formula, we can obtain the equilibrium constant, K as follow:

Equilibrium constant, K = [B₂][AC] / [AB₂C]

K = (0.007 × 0.0118) / 0.0168

K = 0.0049

Thus, the equilibrium constant, K for the reaction is 0.0049

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Answer:

70.91 kJ

Explanation:

The amount of energy (in kJ) required to increase the temperature of 255 g of water from 25.2 C to 90.5 C can be calculated using the formula:

Q = m * c * ΔT

Where Q is the amount of energy, m is the mass of the substance, c is the specific heat, and ΔT is the change in temperature.

Substituting the given values:

m = 255 g

c = 4.184 J/g C

ΔT = (90.5 - 25.2) C = 65.3 C

Q = 255 g * 4.184 J/g C * 65.3 C

Q = 70905.564 J

Q = 70.91 kJ (rounded to two decimal places)

Therefore, the amount of energy required to increase the temperature of 255 g of water from 25.2 C to 90.5 C is 70.91 kJ.

The average speed of a gas molecule is proportional to the square root of its temperature and inversely proportional to the square root of its molar mass. Therefore, we can use the following equation to find the average speed of a CO2 molecule at the same temperature:

v2/v1 = sqrt(M1/M2)

where v1 and v2 are the average speeds of the oxygen and CO2 molecules, respectively, M1 and M2 are the molar masses of oxygen and CO2, respectively.

The molar mass of oxygen (O2) is 32 g/mol, and the molar mass of CO2 is 44 g/mol.

We are given that the average speed of an oxygen molecule is 4.37 × 10^4 cm/s at 25°C. We can convert the temperature to Kelvin by adding 273.15 to get:

T = 25°C + 273.15 = 298.15 K

Now we can solve for v2:

v2 = v1 * sqrt(M1/M2)

v2 = 4.37 × 10^4 cm/s * sqrt(32 g/mol / 44 g/mol)

v2 = 3.67 × 10^4 cm/s

Therefore, the average speed of a CO2 molecule at the same temperature is 3.67 × 10^4 cm/s.

The feather is the least thick of the bunch. This is due to the fact that density is defined as mass per unit volume.

A feather has a relatively low mass compared to its volume, due to its porous and lightweight nature. Marble, coin, and phone all have substantially higher densities than a feather since they are constructed of denser materials such as stone, metal, and plastic/electronics.

As a result, when the density of these things is compared, the feather is the least dense.

Because it has a significantly smaller mass than the other objects listed, the feather would be the least dense. Because density is defined as mass per unit volume, an object with a lower mass and the same or greater volume has a lower density.

The stone, coin, and phone all have greater masses and thus higher densities than the feather. However, because density varies based on the exact material used, the relative densities of these things may change if they are made of different materials.

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C. It is absorbed and emitted in discrete chunks.

The energy in light is carried by particles called photons, which behave both like waves and like particles. According to the theory of quantum mechanics, photons can only be absorbed or emitted in discrete amounts of energy, known as quanta. This means that the energy in light is not continuous, but rather comes in specific packets or chunks. This phenomenon is known as quantization, and it has important implications for many areas of physics, including atomic and molecular physics, as well as the study of electromagnetic radiation.

Answer: C it is absorbed and emitted in descrete chunks.

Explanation:

photons of light are emitted or absorbed as electrons change energy levels

Answer:

Explanation:

The change in entropy of a system can be determined by comparing the entropy of the reactants to the entropy of the products. The reaction that leads to an increase in the number of moles of gas or particles will generally have a positive change in entropy.

I. 2N2(g) + O2(g) → 2N2O(g)

The reactants have 3 moles of gas, while the product also has 3 moles of gas. Therefore, there is no change in the number of moles of gas, and the change in entropy is likely to be small.

II. CaCO3(s) → CaO(s) + CO2(g)

The reactant is a solid, while the products are a solid and a gas. The formation of a gas from a solid leads to an increase in the number of moles of particles, and therefore an increase in entropy.

III. Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)

The reactants consist of a solid and a liquid, while the products consist of an aqueous solution and a gas. The formation of a gas leads to an increase in the number of moles of particles, and therefore an increase in entropy.

Therefore, reactions II and III have a positive change in entropyentropy

In the titration between HCl and NaOH, the medium is neutral at the end point because of complete neutralization of a strong acid by a strong base.

Neutralization is a chemical reaction in which acid and base react to form salt and water. Hydrogen (H⁺) ions and hydroxide (OH⁻ ions) react with each other to form water.

The strong acid and strong base neutralization have a pH value of 7.

The beaker gets warm which indicates that the reaction between acid and base is an exothermic reaction releasing heat energy into the surroundings.

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