A sample of helium gas occupies 2.65 l at 1.20 atm. what pressure would
this sample of gas exert in a 1.50-l container at the same temperature?
(use boyle's law: v1p1=v2p2)

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

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

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

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

What is atmospheric pressure?

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

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

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

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

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

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

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

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

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Answer: Double Replacement

Explanation:

Two elements are being switched around in this reaction, H and K, so it is a double replacement. The K from potassium hydroxide replaces the H in hydrobromic acid, becoming potassium bromide, and the H from hydrobromic acid replaces the K in potassium hydroxide, becoming water.

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

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

moles of NaOH = Molarity of NaOH x Volume of NaOH

moles of NaOH = 0.258 M x 0.0150 L

moles of NaOH = 0.00387 mol

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

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

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

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

Volume of HClO₄ = 31.0 mL - 15.0 mL

Volume of HClO₄ = 16.0 mL = 0.0160 L

Now we can calculate the new molarity of H⁺:

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

Molarity of H⁺ = 0.00387 mol / 0.0160 L

Molarity of H⁺ = 0.242 M

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

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This problem can be solved using Boyle's Law, which posits that the pressure of any given gas will be inversely proportional to its volume when temperature is kept as a constant.

Mathematically Boyle's Law can be expressed as:

P₁V₁ = P₂V₂

where P₁ and V₁ are the initial pressure and volume, respectively, and P₂ and V₂ are the final pressure and volume, respectively.

We are given that:

P₁ = 1.15 atm

V₁ = 640 mL

T = 23.9 °C (which is 297.05 K, using the Kelvin temperature scale)

We need to find V₂ when P₂ = 1.43 atm.

Using Boyle's Law, we can set up the following equation:

P₁V₁ = P₂V₂

(1.15 atm)(640 mL) = (1.43 atm)(V₂)

Solving for V₂:

V₂ = (1.15 atm)(640 mL) / (1.43 atm)

V₂ = 514.69 mL

Therefore, the final volume of the methane gas is 514.69 mL when compressed at constant temperature until its pressure is 1.43 atm.

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

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

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

1. For calcium

Mass of calcium = Percentage of calcium × Mass of dolomite

Mass of calcium = 22% × 500

Mass of calcium = (22/100) × 500

Mass of calcium = 110 kilograms

2. For magnesium

Mass of magnesium = Percentage of magnesium × Mass of dolomite

Mass of magnesium = 13% × 500

Mass of magnesium = (13/100) × 500

Mass of magnesium = 65 kilograms

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

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

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

= 0.43 moles

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

= 0.59 moles

Now;

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

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

= 1.77 moles of F2

Thus F2 is the limiting reactant

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

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

= 0.29 moles

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

= 21 g

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186.48 g of [tex]Na_2CS_3[/tex] were generated throughout the reaction.

The balanced chemical equation is:

[tex]3CS_2[/tex] + 6NaOH → [tex]2Na_2CS_3[/tex] + NaOH + [tex]3H_2O[/tex]

The molar mass of [tex]CS_2[/tex] is 76.14 g/mol, and the molar mass of [tex]2Na_2CS_3[/tex] is 192.23 g/mol.

To find the limiting reactant, we need to calculate the number of moles of each reactant. Using the given mass of [tex]CS_2[/tex]:

110.88 g [tex]CS_2[/tex] / 76.14 g/mol = 1.456 mol [tex]CS_2[/tex]

Using the given number of moles of [tex]NaOH[/tex]:

3.12 mol [tex]NaOH[/tex]

We can see that [tex]CS_2[/tex] is the limiting reactant, since it has fewer moles than [tex]NaOH[/tex]. Therefore, we will use the amount of [tex]CS_2[/tex] to calculate the amount of [tex]Na_2CS_3[/tex] produced.

From the balanced equation, we can see that 3 mol of [tex]CS_2[/tex] produces 2 mol of [tex]Na_2CS_3[/tex]. So, 1.456 mol of [tex]CS_2[/tex] will produce:

(2 mol [tex]Na_2CS_3[/tex] / 3 mol [tex]CS_2[/tex]) * 1.456 mol [tex]CS_2[/tex] = 0.971 mol [tex]Na_2CS_3[/tex]

Now, we can use the molar mass of [tex]Na_2CS_3[/tex] to calculate the mass produced:

0.971 mol [tex]Na_2CS_3[/tex] * 192.23 g/mol = 186.48 g [tex]Na_2CS_3[/tex]

Therefore, the mass of [tex]Na_2CS_3[/tex] produced in the reaction is 186.48 g.

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Answer:

The volume of 9.0 M copper (II) sulfate stock solution needed to prepare 3.0 L of a 5.0 M solution is 1.667 L

Explanation:

Dilution is a process by which the concentration of a solute in solution is reduced by adding more solvent.

In other words, dilution is the procedure followed to prepare a less concentrated solution from a more concentrated one, and it simply consists of adding more solvent.

In a dilution the amount of solute does not vary. What varies in a dilution is the volume of the solvent: as more solvent is added, the concentration of the solute decreases, as the volume (and weight) of the solution increases.

The equation used in this case is:

Ci * Vi = Cf * Vf

where

Ci: initial concentration

Vi: initial volume

Cf: final concentration

Vf: final volume

In this case:

Ci: 9 M

Vi: ?

Cf: 5 M

Vf: 3 L

Answer:

B

Explanation:

An atom becomes negatively charged when it gains electrons. The number of electrons an atom needs to gain to become negatively charged depends on the number of protons in its nucleus, which determines its atomic number and the number of electrons it normally has in its neutral state.

In general, if an atom gains n electrons, it will have a negative charge of -n. For example, if an oxygen atom (atomic number 8) gains two electrons, it will have a negative charge of -2.

Therefore, the least number of electrons an atom must have in order to have a negative charge would be one more than the number of protons in its nucleus, since adding one electron will give it a charge of -1. For example, if the atom has 6 protons, it would need 7 electrons to have a negative charge of -1.

This corresponds to the element carbon, which has atomic number 6 and normally has 6 electrons in its neutral state. Adding one electron to a carbon atom would give it a negative charge of -1.

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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The molar mass of HA is approximately 168.48 g/mol.

To calculate the molar mass of HA, we need to use the balanced chemical equation for the reaction between HA and NaOH:

[tex]HA + NaOH[/tex] → [tex]NaA + H2O[/tex]

From the equation, we can see that 1 mole of HA reacts with 1 mole of NaOH to produce 1 mole of NaA. At the equivalence point of the titration,

[tex]moles of NaOH = (0.500 mol/L) * (0.0321 L) = 0.01605 mol[/tex]

Since the initial solution was prepared by dissolving 2.70 g of HA in 100.0 ml of water, we can calculate the initial concentration of HA in units of moles per liter:

[tex]moles\ of HA = (2.70 g / molar\ mass\ of HA) = (0.0270 kg / molar\ mass\ of HA)[/tex]

[tex]initial\ concentration\ of\ HA = moles\ of\ HA / (0.100 L) = moles\ of\ HA / 1000 mL[/tex]

Setting the moles of NaOH equal to the moles of HA, we can solve for the molar mass of HA:

moles of NaOH = moles of HA

[tex]0.01605\ mol = (0.0270 kg / molar\ mass\ of HA) / 0.100 L[/tex]

[tex]molar\ mass\ of\ HA = (0.0270 kg / 0.01605 mol) / 0.100 L = 168.48 g/mol[/tex]

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The Haber process involves the following steps:

Preparation of reactants; Compression of gases; Mixing of gases; Reaction; Separation of ammonia; Separation of ammonia

The Haber process is a method used to manufacture ammonia (NH3) from nitrogen gas (N2) and hydrogen gas (H2). The process is named after its inventor, German chemist Fritz Haber, who developed the process in the early 20th century.

The Haber process involves the following steps

The Haber process is an important industrial process for the production of ammonia, which is a vital ingredient in the production of fertilizers and many other chemical compounds.

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To determine if the piston will fail, we need to calculate the volume of the gas at the higher temperature and see if it exceeds the maximum volume of the piston.

First, we need to assume that the gas behaves ideally and follows the gas laws. We can use the ideal gas law, 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.

We know the initial volume of the gas is 0.075 L and the initial temperature is 171°C, which is 444 K (since we need to convert to Kelvin). We don't know the pressure or the number of moles, but we can assume they remain constant.

Next, we need to calculate the final volume of the gas when it is heated to 5934°C, which is 6207 K. We know that the pressure and number of moles remain constant, so we can rearrange the ideal gas law to solve for V:

V = nRT/P

We can plug in the values for n, R, P, and T, and solve for V:

V = (n x R x 6207 K) / P

Now we need to check if this final volume exceeds the maximum volume of the piston, which is 0.885 L. If it does, then the piston will fail.

To convert the final volume from liters to cubic centimeters (cc), we can multiply by 1000:

V = (n x R x 6207 K x 1000) / P

V = (n x 8.31 J/mol K x 6207 K x 1000) / P

V = (n x 51476870 J/mol) / P

Assuming the pressure remains constant, we can set the initial and final volumes equal to each other and solve for n:

n x 8.31 J/mol K x 444 K = n x 51476870 J/mol x 6207 K

n = (8.31 J/mol K x 444 K) / (51476870 J/mol x 6207 K)

n = 2.34 x 10^-7 mol

Now we can plug in the value for n and solve for the final volume:

V = (2.34 x 10^-7 mol x 8.31 J/mol K x 6207 K x 1000) / 1 atm

V = 1.42 cc

Since the final volume of the gas is only 1.42 cc, which is much smaller than the maximum volume of the piston (0.885 L or 885 cc), the piston will not fail.

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The mass of I2 reacted is  142.2 g

The mass of PCl3 reacted is 153.4 g

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

We know that;

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

= 0.56 moles

If 1 mole of I2 produced 1 mole of F2

Then 0.56 moles of I2 reacted

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

= 142.2 g

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

= 1.12 moles

If the reaction is 1:1:1

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

= 153.4 g

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Answer:

B. Combustion.

Explanation:

Looking at the given equation, we can see that methane (CH4) reacts with oxygen (O2) and releases water (H2O) and carbon dioxide (CO2). This matches the definition of a combustion reaction. Therefore, the answer is B. Combustion.

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

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

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

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

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

SnF₂ + 2La(NO₃)3 → La₂(SnF₆) + 6NO₃

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

The molar mass of SnF2 is 156.69 g/mol.

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

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

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

The mass of SnF2 present in the toothpaste sample is:

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

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

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

The mass percentage of SnF₂ in the toothpaste sample is:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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A total of 0.1102 moles of Ag will be produced from 3.50 g of Cu.

To determine the number of moles of Ag produced from 3.50 g of Cu, we need to use stoichiometry.

From the balanced chemical equation, we see that 1 mole of Cu reacts with 2 moles of Ag to produce 1 mole of Cu(NO₃)₂ and 2 moles of Ag.

First, we need to convert 3.50 g of Cu to moles by dividing by its molar mass, which is 63.55 g/mol.

3.50 g Cu / 63.55 g/mol = 0.0551 mol Cu

Next, we use the stoichiometry ratio to determine the number of moles of Ag produced:

0.0551 mol Cu x (2 mol Ag / 1 mol Cu) = 0.1102 mol Ag


In summary, we use stoichiometry to determine the number of moles of Ag produced from 3.50 g of Cu by first converting the mass of Cu to moles, and then using the stoichiometry ratio from the balanced chemical equation.

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

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

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

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

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

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

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

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

Finally, we can calculate the change in entropy:

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

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

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

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

PV = nRT

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

First, we need to convert the temperature from Celsius to Kelvin:
T = 28°C + 273.15 = 301.15 K

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

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

V = nRT / P

Substitute the known values into the equation:

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

V ≈ 55.44 L

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

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According to the question the enolate nucleophile and carbonyl electrophiles are attached in the images below.

A nucleophile is a species (atom, molecule, or ion) that donates an electron pair to form a new covalent bond in a reaction. Nucleophiles are attracted to electron-deficient or positively-charged sites, such as the electrophilic sites of organic molecules or cations. In organic chemistry, nucleophiles are typically Lewis bases, such as amines or other electron-rich molecules. In inorganic chemistry, nucleophiles include anions and neutral molecules containing lone pairs of electrons. In chemical reactions, nucleophiles interact with electrophiles, which are positively-charged species.

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Complete Question:

The cell potential for the given reaction at 25°C is -2.3895 V.

First, we need to balance the equation;

Mg(s) + Fe³⁺(aq) → Mg²⁺(aq) + Fe(s)

Next, we can use the Nernst equation to calculate the cell potential (Ecell) at 25°C;

Ecell = E°cell - (RT/nF)ln(Q)

where; E°cell is the standard cell potential

R is the gas constant (8.314 J/mol·K)

T is the temperature in Kelvin (298 K)

n is number of electrons transferred in balanced reaction

F is the Faraday constant (96,485 C/mol)

Q is the reaction quotient

Since the reaction is not balanced in terms of electrons transferred, we need to balance it and determine the number of electrons transferred:

Mg(s) + Fe³⁺(aq) → Mg²⁺(aq) + Fe(s) + 2e⁻

n = 2

The reaction quotient (Q) will be calculated using concentrations of the reactants and products;

Q = [Mg²⁺][Fe(s)] / [Mg(s)][Fe³⁺]

Substituting the given values, we get;

Q = (0.000612 M)(1) / (1)(1.29 M)

Q = 0.000474

Now, we can calculate the cell potential (Ecell) using the Nernst equation;

Ecell = E°cell - (RT/nF)ln(Q)

= (-2.37 V) - (0.0257 V)log10(0.000474)

= -2.37 V - 0.0195 V

= -2.3895 V

Therefore, the cell potential is -2.3895 V.

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The final pressure of the gas sample is 1.09 atm when heated to 50ºC.

The final pressure of a gas sample initially at 1.00 atm and 25ºC when heated to 50ºC can be calculated using the ideal gas law:

P₁ × V₁ ÷ T₁ = P₂ × V₂ ÷ T₂

where P₁, V₁, and T₁ are the initial pressure, volume, and temperature of the gas, respectively, and P₂, V₂, and T₂ are the final pressure, volume, and temperature of the gas, respectively.

Assuming that the volume of the gas remains constant, V₁ = V₂, and rearranging the ideal gas law, we get:

P₂ = P₁ (T₂ ÷ T₁)

Substituting the values, we get:

P₂ = (1.00 atm) × (323 K) ÷ (298 K) = 1.09 atm

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Yes, two magnets can create magnetic force fields that allows them to interact without touching.

Magnetic forces are non contact forces; they pull or push on objects without touching them. Magnets are only attracted to a few 'magnetic' metals and not all matter. Yes, the investigation did answer the question about whether two magnets create magnetic force fields that allow them to interact without touching.

The investigation provided enough evidence to support the idea that invisible magnetic force fields exist:

The investigation involved observing how two magnets interact with each other without touching. The magnets were brought closer together until they interacted, and then they were moved further apart. This process was repeated several times, and the results were observed and recorded. During the investigation, it was observed that the magnets interacted with each other even when they were not touching. This interaction occurred because the magnets created magnetic force fields that allowed them to interact with each other even when they were not in direct contact. This is because the interaction between the magnets could not be explained by any other means except through the existence of magnetic force fields.

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

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

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

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

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

Now the equation is balanced, and coefficients are:

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

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

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

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

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The new pressure at 273 K, given that the initial pressure was 378 KPa, is 249.9 KPa

The following parameters were obtained from the question:

The new pressure of the gas at 273 K can be obtained as shown below:

P₁ / T₁ = P₂/ T₂

378 / 413 = P₂ / 273

Cross multiply

413 × P₂ = 378 × 273

413 × P₂ = 103194

Divide both sides by 413

P₂ = 103194 / 413

P₂ = 249.9 KPa

Thus, from the above calculation, we can conclude the new pressure at 273 K is 249.9 KPa

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

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

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

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

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Missing parts;

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

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

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

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

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

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

The pH of the apple juice is approximately 3.52.

The pH of ammonia is approximately 11.13.

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

[H₃O⁺] = 10^(-pH)

0.0003 = 10^(-pH)

-pH = ㏒(0.0003)

pH = -㏒(0.0003)

pH = 3.52

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

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

NH₃ + H₂O → NH₄⁺ + OH⁻

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

Kb = [NH4⁺][OH⁻]/[NH₃3

1.8 x 10⁻⁵ = x²/0.05

x = 1.34 x 10⁻³

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

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

pH = 14 - pOH = 14 - 2.87 = 11.13

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

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