In a titration, 25mL of 0.20M NaOH neutralize 5mL of HCI, what is the acid molarity?

To calculate the pH of the resulting solution, we need to first determine the moles of HNO2 and RbOH that are present in each solution.

Moles of HNO2 = volume (in liters) x concentration

Moles of HNO2 = 0.162 L x 1.860 mol/L

Moles of HNO2 = 0.30372 mol

Moles of RbOH = volume (in liters) x concentration

Moles of RbOH = 0.121 L x 1.090 mol/L

Moles of RbOH = 0.13169 mol

Since HNO2 is a weak acid and RbOH is a strong base, they will undergo a neutralization reaction to form a salt (RbNO2) and water. The balanced chemical equation for this reaction is:

HNO2 + RbOH → RbNO2 + H2O

To determine the amount of HNO2 and RbOH that react with each other, we need to use the stoichiometry of the reaction. Since the reaction is a 1:1 ratio between HNO2 and RbOH, we can say that the amount of HNO2 that reacts is equal to the amount of RbOH added. Therefore, the remaining amount of HNO2 and RbOH that are not used in the reaction can be calculated as follows:

Moles of HNO2 remaining = 0.30372 mol - 0.13169 mol = 0.17203 mol

Moles of RbOH remaining = 0.13169 mol - 0.13169 mol = 0 mol

The 0.13169 mol of RbOH reacts with 0.13169 mol of HNO2 to form 0.13169 mol of RbNO2 and 0.13169 mol of H2O.

Now we need to determine the concentration of HNO2 and RbNO2 in the resulting solution.

Concentration of HNO2 = moles remaining / total volume

Concentration of HNO2 = 0.17203 mol / (0.162 L + 0.121 L)

Concentration of HNO2 = 1.048 M

Concentration of RbNO2 = moles of RbNO2 / total volume

Concentration of RbNO2 = 0.13169 mol / (0.162 L + 0.121 L)

Concentration of RbNO2 = 0.805 M

Finally, we can use the acid dissociation constant (Ka) of HNO2 to calculate the pH of the resulting solution.

Ka = [H+][NO2-] / [HNO2]

5.62 x 10^-4 = [H+]^2 / 1.048

[H+] = 0.00749 M

pH = -log[H+]

pH = -log(0.00749)

pH = 2.13

Therefore, the pH of the resulting solution is approximately 2.13.

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(complete question)

Determine The PH Of The Solution Resulting From Mixing A Solution Of 162 ML Of HNO2 (K, = 5.62e - 04) At A 1.860 M concentration, with 121mL of a 1.090M solution of RbOH?

The heat released during the reaction is -212 kJ. Heat is a fundamental concept in physics, chemistry, and engineering and plays a critical role in many natural phenomena, such as thermodynamics, phase transitions, and thermal radiation.

What is Heat?

Heat is a form of energy that can be transferred from one object to another as a result of a difference in temperature. Heat flows from hotter objects to colder objects until they reach thermal equilibrium, meaning that their temperatures become equal. The amount of heat transferred is typically measured in joules (J) or calories (cal) and is related to the mass of the object, the specific heat capacity of the material, and the temperature change experienced.

The ΔH for the reaction is -847 kJ.

The stoichiometric coefficient of Al in the balanced equation is 2. This means that 2 moles of Al are required to produce 2 moles of Fe and 1 mole of [tex]Al_{2} O^{3}[/tex].

Since the reaction uses 4.0 mol of Al, it will produce 2.0 mol of Fe and 1.0 mol of [tex]Al_{2} O^{3}[/tex].

The amount of heat released during the reaction can be calculated using the equation:

ΔH = q/n

where ΔH is the enthalpy change, q is the heat released, and n is the number of moles of the limiting reactant (in this case, Al).

Substituting the values gives:

ΔH = (-847 kJ) / 4.0 mol = -212 kJ/mol

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Answer:

3. collide with proper energy and proper orientation

Explanation:

Option 3, i.e., "collide with proper energy and proper orientation," is the correct answer.

Chemical reactions involve the rearrangement of atoms to form new molecules or compounds. In order for this rearrangement to occur, the reactant particles must collide with each other in a specific way. The collision must be energetic enough to break the existing chemical bonds and form new ones. Additionally, the particles must collide with each other in a specific orientation that allows the atoms to line up properly and form new chemical bonds.

Options 1 and 2 are incorrect as they do not play any role in chemical reactions. The distance between particles and the presence of attractive forces between them do not affect whether a chemical reaction will occur or not. Option 4 is incorrect as chemical reactions do not involve the conversion of chemical energy into nuclear energy.

Hopes this helps

The organic molecule CH2OCH2CHOH is a diol, which means it contains two hydroxyl (-OH) groups. It is also known as ethylene glycol, and it has the molecular formula C2H6O2.

Ethylene glycol is a colorless, odorless, and sweet-tasting liquid that is commonly used as a solvent, antifreeze, and in the production of polyester fibers and resins.

It is also highly toxic if ingested, as it can cause kidney failure and other serious health problems.

Therefore, it is important to handle ethylene glycol with care and follow proper safety precautions when using it in laboratory or industrial settings.

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The rate at which the sun is currently converting hydrogen into helium is about 600 million tons per second. This process, known as nuclear fusion, takes place in the sun's core and produces vast amounts of energy, which is then radiated out into space in the form of sunlight.

The Sun is made up of about 74% hydrogen and 25% helium, with trace amounts of other elements such as oxygen, carbon, and iron. This composition is based on the Sun's observable surface layer, also known as the photosphere.

The Sun produces energy through a process known as nuclear fusion, in which hydrogen atoms combine to form helium. This process releases a tremendous amount of energy in the form of heat and light, which is radiated out into space as sunlight.

The energy produced by the Sun is what allows life to exist on Earth, as it provides the heat and light necessary for plants to grow and animals to survive.

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The given statement "at low temperatures, intermolecular forces become important and pressure of the gas will be higher than predicted by ideal gas law" is false. Because, at low temperatures, intermolecular forces become more significant and cause gas molecules to come closer together, which reduces the volume of the gas.

As a result, the pressure of the gas will be lower than predicted by the ideal gas law, which assumes that gas molecules have no volume and do not interact with each other.

The ideal gas law is only accurate for gases that behave like ideal gases, meaning that they have negligible intermolecular forces and occupy no volume. Real gases, on the other hand, deviate from the ideal gas law at high pressures and low temperatures, where intermolecular forces become more significant.

At these conditions, the volume of the gas becomes significant compared to the volume of the container, and the gas behaves less ideally. In this case, we need to use more complex equations of state to accurately predict the behavior of the gas.

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Answer: 0.5 atm

Explanation: To calculate the partial pressure of carbon dioxide, we need to use the mole fraction of carbon dioxide and the total pressure of the container.

The mole fraction of carbon dioxide is:

X(CO2) = n(CO2) / n(total)

X(CO2) = 0.5 moles / (0.5 moles + 0.32 moles + 0.2 moles)

X(CO2) = 0.4762

The partial pressure of carbon dioxide is:

P(CO2) = X(CO2) * P(total)

P(CO2) = 0.4762 * 1.05 atm

P(CO2) = 0.5 atm (rounded to one decimal place)

Therefore, the partial pressure of carbon dioxide in the container is 0.5 atm.

Answer:

0.58 atm

Explanation:

When we add more carbon in the atmosphere it warms the planet and helps the plants on land grow more. Excess of carbon can makes the water more acidic.

When carbon dioxide is added to atmosphere it reacts with water to form carbonic acid from which hydrogen ions dissociate and results in increasing the acidity of the system. So we can say that in addition to any greenhouse effect the anthropogenic carbon dioxide emissions to the atmosphere can increase the acidity of the atmosphere and precipitation of the atmosphere. Most of the plant species show higher rates of photosynthesis which cause increased growth and the decreased water use and lowered tissue concentrations of nitrogen and protein of the water.

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The correct question is,

What happens when too much CO2 enters the atmosphere?

THF is a poor solvent for Friedel-Crafts reactions using aluminum chloride or boron trifluoride because THF forms stable adducts with these compounds, THF is a strong Lewis base, and THF has a very low boiling point. Here options B, C, and D are correct.

THF is a polar aprotic solvent that is commonly used for many organic reactions, but it is a poor solvent for Friedel-Crafts reactions that use aluminum chloride or boron trifluoride. This is because THF can form stable adducts with these compounds, which can reduce their reactivity and prevent them from effectively catalyzing the reaction. Therefore, option B is correct.

In addition, THF is a strong Lewis base, which means that it can coordinate with Lewis acids like aluminum chloride or boron trifluoride and form adducts. This further reduces the reactivity of the Lewis acid, making the reaction less effective. Therefore, option C is also correct.

Finally, the boiling point of THF is relatively low (66°C), which can make it difficult to maintain the reaction at the desired temperature. This can affect the reaction kinetics and the yield of the product. Therefore, option E is also correct.

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Complete question:

Why is THF (tetrahydrofuran) a poor solvent for Friedel-Crafts reactions that use aluminum chloride or boron trifluoride?

A. THF is unstable in the presence of these compounds

B. THF forms stable adducts with these compounds

C. THF is a strong Lewis base

D. THF is a strong Lewis acid

E. The boiling point of THF is too low.

1. The mass of KClO₃ that decomposed to produce 17.4 g of O₂ is 44.4 g

2. The mass of KClO₃ that formed 23.7 g of KCl is 39.0 g

3. The mole of O₂ that is formed from 1.45 moles of KClO₃ is 2.18 moles

4. The mole of KClO₃ that decomposed to produced 167.7 g of KCl is 0.22 mole

The mass of KClO₃ that decomposed to produce 17.4 g of O₂ can be obtain as follow:

2KClO₃ -> 2KCl + 3O₂

From the balanced equation above,

96 g of O₂ were obtained from 245 g of KClO₃

Therefore,

17.4 g of O₂ will be obtain from = (17.4 × 245) / 96 = 44.4 g of KClO₃

Thus, the mass of KClO₃ that decomposed is 44.4 g

The mass of KClO₃ that formed 23.7 g of KCl can be obtain as follow:

2KClO₃ -> 2KCl + 3O₂

From the balanced equation above,

149 g of KCl were obtained from 245 g of KClO₃

Therefore,

23.7 g of KCl will be obtain from = (23.7 × 245) / 149 = 39.0 g of KClO₃

Thus, the mass of KClO₃ is 39.0 g

The mole of O₂ that is formed from 1.45 moles of KClO₃ can be obtained as follow:

2KClO₃ -> 2KCl + 3O₂

From the balanced equation above,

2 mole of KClO₃ reacted to produce 3 moles of O₂

Therefore,

1.45 moles of KClO₃ will react to produce = (1.45 × 3) / 2 = 2.18 moles of O₂

Thus, the mole of O₂ formed is 2.18 moles

First, we shall obtain the mole of 16.7 g of KCl. Details below:

Mole = mass / molar mass

Mole of KCl = 16.7 / 74.5

Mole of KCl = 0.22 mole

Finally, we shall determine the mole of KClO₃ that decomposed. Details below:

2KClO₃ -> 2KCl + 3O₂

From the balanced equation above,

2 moles of KCl were obtained from 2 moles of KClO₃

Therefore,

0.22 mole of KCl will also be obtain from 0.22 mole of KClO₃

Thus, the mole of KClO₃ that decomposed is 0.22 mole

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The fossil is approximately 17,100 years old.

The age of a fossil can be estimated by using the half-life of carbon-14. The half-life of carbon-14 is approximately 5,700 years, which means that after this time, half of the carbon-14 in a sample will have decayed into nitrogen-14.

If a fossil contains 1 part carbon-14 to 7 parts nitrogen-14, this means that the carbon-14 has decayed to one-eighth (1/8) of its original amount. Since we know the half-life of carbon-14, we can use this information to estimate the age of the fossil.

If we assume that the fossil originally contained only carbon-14, then the number of half-lives that have passed can be calculated as follows:

1/2n = 1/8

where n is the number of half-lives that have passed.

Simplifying this equation, we get:

2n = 8

n = 3

Therefore, the fossil has undergone 3 half-lives of carbon-14 decay. Since the half-life of carbon-14 is approximately 5,700 years, we can estimate the age of the fossil by multiplying the half-life by the number of half-lives:

age = 5,700 years/half-life * 3 half-lives

age = 17,100 years.

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The chemical equation for the reaction between sulfur and carbon to produce carbon disulfide is:

S2(g) + C(s) → CS2(g)

To determine the number of grams of CS2(g) that can be prepared, we need to use the stoichiometry of the reaction and the ideal gas law.

We need to calculate the number of moles of CS2(g) that can be produced by 9.80 mol of S2(g). From the equation, we can see that one mole of S2(g) reacts with one mole of C(s) to produce one mole of CS2(g). Therefore, the number of moles of CS2(g) produced will be equal to 9.80 mol.

We need to use the ideal gas law to calculate the volume of the CS2(g) produced. We know that the reaction vessel has a volume of 5.75 L and is held at 900 K. Assuming the pressure remains constant, we can use the ideal gas law: PV = nRT

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

We can rearrange this equation to solve for n: n = PV/RT

Substituting the values we have:

n = (1 atm)(5.75 L)/(0.08206 L·atm/mol·K)(900 K)

n = 0.318 mol

Therefore, the mass of CS2(g) produced will be:

mass = n × M

where M is the molar mass of CS2(g) which is 76.14 g/mol.

mass = 0.318 mol × 76.14 g/mol

mass = 24.2 g

Therefore, 24.2 grams of CS2(g) can be prepared by heating 9.80 mol S2(g) with excess carbon in a 5.75 L reaction vessel held at 900 K until equilibrium is attained.

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If you use Copper (Cu) as the metal , Oxygen (O) and Carbon (C) as the c, then you can make 2 new compounds in five minutes.

The two possible compounds are:

Copper oxide (CuO) is a chemical compound composed of copper and oxygen. It is a black, solid material with a high melting point and is commonly referred to as cupric oxide or black oxide of copper.

CuO can be produced by heating copper in the presence of oxygen, or by reacting copper sulfate with sodium hydroxide. It is commonly used as a raw material in the production of copper salts, in the manufacturing of ceramics, as a pigment, and as a catalyst in various chemical reactions.

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When a gas expands from 271 ml to 903 ml at a constant temperature, the work done (in joules) by the gas if it expands in vacuum is O J

According to the given data:

Initial volume occupied by the gas, V₁=271 ml =0.271 L

Final volume occupied by the gas, V₂= 903 ml =0.903 L

To calculate the work done (in joules) by the gas if it expands against a vacuum, We can use the following expression.

w = -P × ΔV

Since the gas expands against a vacuum, pressure will be equal to zero

P = 0.

Thus, w = 0 J

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a. Balanced chemical reaction: 4Li + O₂ → 2Li₂O

b. Balanced Chemical reaction: Mg(ClO₃)₂→ MgCl₂ + 3O₂

c. Balanced Chemical reaction: 2HNO₃ + Ca(OH)₂ → Ca(NO₃)₂ + 2H₂O

d. Balanced Chemical reaction: C₅H₁₂+ 8O₂→ 5CO₂ + 6H₂O

Lets understand these reaction types:

1. Synthesis: Chemical synthesis is the process in which chemical reactions are performed with the idea of converting a reactant into a product or multiple products.

For example: 4Li + O₂ → 2Li₂O

2. Decomposition: In these reactions chemical species break up into simpler parts.

For example: Mg(ClO₃)₂→ MgCl₂ + 3O₂

3. Double displacement: is a type of chemical reaction where two compounds react, and positive ions and the negative ions of the two reactants switch places, forming two new compounds or products.

For example:  2HNO₃ + Ca(OH)₂ → Ca(NO₃)₂ + 2H₂O

4. Combustion: These reactions occur when oxygen reacts with another substance and gives off heat and light.

For example: C₅H₁₂+ 8O₂→ 5CO₂ + 6H₂O

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Correct form of question

For each of the following reactions, identify the missing reactant(s) or products(s) and then balance the resulting equation. Note that each empty slot may require one or more substances.

a. synthesis: ___ ⟶Li2O

b. decomposition: Mg(ClO3)2⟶___,

c. double displacement: HNO3+Ca(OH)2→___,

d. combustion: C5H12+O2⟶___

5 to 10 ml urine should be collected for clean-catch procedure. Thus Option A is the correct answer.

In the clean-catch procedure, 5 to 10 ml is adequate amount of urine because more than this quantity will cause a difficulty in sampling the solution. For this process to work there is a set of criteria that needs to be followed to ensure that the given sample is tested with readiness with accuracy.

The given criteria are the sample must be stored away in an compact space, the sample must be free from any impurities, the sample must be fresh for the process to run smoothly. Furthermore, this set of process is crucial for finding the cause of the disease that are caused by bacteria.

This type of process is adequate for all age groups ranging from infants to adults.

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The molarity of the sample of HNO3 (aq.) is neutralized by 22.1 milliliters of 0.250 m KOH(aq.) is 0.193M

Molarity is also called as molar concentration. Molarity can be defined as the measure of the concentration of a chemical specie of the reaction in particular of a solute in a solution which is in terms of amount of substance per unit volume of solution.

The molarity can be calculated as,

M1 V1 = M2 V2

Hehe, M1 is the molarity of the acid and V1 is the volume  of the an acid. M2 is the molarity of the base and V2 is the volume of the base of the solution.

When 35.0-milliliter sample of HNO3 (aq.) is neutralized by 22.1 milliliters of 0.250 m KOH(aq.)

Putting all the values in the expression we get,

M1=M2V2/V1

    =(0.150) (32.1)/25.0

    = 0.193M

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Answer:

The main places you would find glaciers on in today's world are places like Alaska, Antarctica, and Greenland. Places that are cold either year-round or most of the year and are in an ocean area are typically where glaciers form.

Explanation:

These places typically are cold year-round and are the perfect area for glaciers to form, but not everywhere forms glaciers as they need certain weather conditions to form.

a. The fermi level assuming two electrons per atom 3.37 eV.

b. The density of states as a function of electron energy is, D(E) = 2 / [6.626 x 10^-34 J.s * (π x 0.15 x 10^-9 m)^2] * √(2 x 9.11 x 10^-31 kg) * √(E - 3.37 eV)

c. The behavior of the density of states near the Fermi energy can be used to check if it is correct.

The number of atoms in the one-dimensional crystal is given by:

N = L/a = (20 x 10^-6 m) / (0.15 x 10^-9 m) = 1.33 x 10^5 atoms

The number of electrons in the crystal, assuming two electrons per atom, is, 2 x N = 2 x 1.33 x 10^5 = 2.66 x 10^5 electrons

The Fermi level is the energy level at which the probability of finding an electron is 0.5. For a system of non-interacting electrons, the Fermi energy can be calculated using the equation:

EF = (h^2/8m)(3π^2n)^(2/3)

where h is Planck's constant, m is the mass of an electron, and n is the electron density. Plugging in the values, we get:

EF = (6.626 x 10^-34 J.s)^2 / (8 x 9.11 x 10^-31 kg) x (3π^2 x 2.66 x 10^5 m^-1)^(2/3) = 3.37 eV

The density of states is a function of the energy and can be calculated using the following equation,

D(E) = 2 / [h * (πa)^2] * √(2m) * √(E - EF)

where h is Planck's constant, a is the lattice spacing, m is the mass of an electron, and EF is the Fermi energy. Plugging in the values, we get:

D(E) = 2 / [6.626 x 10^-34 J.s * (π x 0.15 x 10^-9 m)^2] * √(2 x 9.11 x 10^-31 kg) * √(E - 3.37 eV)

In particular, the density of states should approach zero as the energy approaches the Fermi energy, because all available energy states have been filled. If the density of states near the Fermi energy does not exhibit this behavior, it may indicate that the model used to calculate the density of states is incorrect or that the system is not well-described by the model.

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--The complete question is, consider a one-dimensional crystal (similar to a nanowire) with length 20 um and lattice spacing 0.15 nm.

a. what is the fermi level assuming two electrons per atom?

b. what is the density of states as a function of electron energy?

c. how do you know if the density of states is correct--

Plug in the values of pKa, C₁, and C₂_new into the equation to find the pH of the resulting buffer solution.

To determine the pH of the resulting buffer solution, we need to first identify the acid and its conjugate base present in the mixture. Since the exact substances are not provided in the student question, I will use a generic example of a weak acid (HA) and its conjugate base (A-).
Let's assume that we have V₁ volume of HA with a concentration of C₁ and V₂ volume of A- with a concentration of C₂.
Step 1: Calculate the total volume of the buffer solution.
V_total = V₁ + V₂
Step 2: Calculate the moles of HA and A- in the mixture.
moles_HA = C₁ * V₁
moles_A- = C₂ * V₂
Step 3: Calculate the new concentrations of HA and A- in the buffer solution.
C₁_new = moles_HA / V_total
C₂_new = moles_A- / V_total
Step 4: Use the Henderson-Hasselbalch equation to determine the pH of the buffer solution.
pH = pKa + log ([A-] / [HA])
Here, pKa is the acid dissociation constant of HA, [A-] represents the concentration of A- (C₂_new), and [HA] represents the concentration of HA (C₁_new).

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Based on the chemical equation provided, the missing reactant in the reaction is an alkene.

Tertiary alcohols are alcohols in which the carbon atom attached to the hydroxyl group is bonded to three other carbon atoms. They can be formed by the hydration of tertiary alkenes, which are alkenes in which the carbon atom at the site of the double bond is bonded to three other carbon atoms.

In the given reaction, the hydrogen molecule is added to the double bond of a tertiary alkene to form a new carbon-carbon single bond, while the two hydrogen atoms are added to the two carbon atoms of the double bond. The resulting compound is a tertiary alcohol. Therefore, the missing reactant in the reaction is a tertiary alkene that would undergo the addition of H₂ to form the given tertiary alcohol product.

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--The complete reaction is, What is the missing reactant in this organic reaction?--

Balanced chemical equation for the reaction between hydrobromic acid (HBr) and aqueous ammonium carbonate is: 2HBr (aq) + (NH₄)2CO₃ (aq) → 2NH₄Br (aq) + CO₂ (g) + H₂O (l).

Process that results in chemical transformation of one set of chemical substances to another is called chemical reaction.

In this reaction, hydrobromic acid (HBr) reacts with aqueous ammonium carbonate [ (NH₄)2CO₃] to form ammonium bromide (NH₄Br), carbon dioxide gas (CO₂ ), and water (H₂O).

It's important to note that this reaction should be carried out with caution as hydrobromic acid is a strong acid and can be corrosive and dangerous. Proper safety precautions and equipment should be used when conducting this reaction.

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The value of the equilibrium constant (Kc) at 853 K for the given reaction is approximately [tex]8.81e^{-5}[/tex].

The quantitative representation of a chemical reaction's state at equilibrium is the equilibrium constant, abbreviated as K. It is described as the ratio of the reactant and product concentrations (or partial pressures for gas-phase reactions), each concentration being raised to the power of the corresponding stoichiometric coefficient in the balanced chemical equation.

The equilibrium constant, denoted as Kc, is a measure of the extent of a chemical reaction at equilibrium. For the given reaction:

[tex]2 HI(g)[/tex]  ⇌ [tex]H_2(g) + I_2(g)[/tex]

The equilibrium constant expression is:

[tex]K_c = \frac{[H_2][I_2]}{[HI]^2}[/tex]

Where [tex][H_2][/tex], [tex][I_2][/tex], and [tex][HI][/tex] are the molar concentrations of H₂, I₂, and HI, respectively, at equilibrium.

Given data:

[tex][HI] = 0.182 M[/tex]

[tex][H_2] = 2.53e^{-2} M[/tex]

[tex][I_2] = 3.28e^{-2} M[/tex]

Plugging these values into the equilibrium constant expression, we get:

[tex]K_c = \frac{(2.53e^{-2}) * (3.28e^{-2})}{(0.182)^2}[/tex]

[tex]K_c = 8.81e^{-5}[/tex]

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To determine the empirical formula of the lead oxide, we need to find the mole ratios of lead and oxygen in the product.

First, we need to determine the moles of each element present in the product:

Mass of lead = 1.154 g - x (where x is the mass of oxygen in the product)

Mass of oxygen = x

Next, we need to convert the masses of lead and oxygen to moles:

moles of lead = (1.154 g - x) / 207.2 g/mol

moles of oxygen = x / 16.00 g/mol

We can set up a ratio of moles of lead to moles of oxygen:

(1.154 g - x) / 207.2 g/mol : x / 16.00 g/mol

Cross-multiplying and simplifying:

(1.154 g - x) x 16.00 g/mol = 207.2 g/mol x

18.464 g - 16.00 x = 207.2 x

191.736 g = 223.2 x

x = 0.859 g

So, the mass of oxygen in the product is 0.859 g. The mass of lead in the product is 1.154 g - 0.859 g = 0.295 g.

Now we can calculate the moles of each element in the product:

moles of lead = 0.295 g / 207.2 g/mol = 0.001422 mol

moles of oxygen = 0.859 g / 16.00 g/mol = 0.0537 mol

To find the empirical formula, we need to divide the moles of each element by the smallest number of moles:

0.001422 mol / 0.001422 mol = 1

0.0537 mol / 0.001422 mol = 37.7

Rounding to the nearest whole number, we get the empirical formula PbO38.

The limiting reagent is nitrogen. The number of molecules of ammonia formed are 8.97. The number of hydrogen molecules in excess is 3.

You might have learnt that the formation of ammonia can be represented by:

N₂ + 3H₂ → 2NH₃

From the image provided by you, you have 6 molecules of hydrogen and 3 molecules of nitrogen. To determine the limiting reagent, you'll need to compare the number of moles of each reactant with the stoichiometric ratio in the balanced equation.

For nitrogen, you have:

3 molecules N₂ × 1 mole N₂/6.022 × 10²³ molecules N₂ = 4.98 × 10⁻²⁴ moles N₂

For hydrogen:

6 molecules H₂ × 1 mole H₂/6.022 × 10²³ molecules H₂ = 9.96 × 10⁻²⁴ moles H₂

Using the stoichiometric ratio from the balanced equation, you'll see that 1 mole of N₂ reacts with 3 moles of H₂ to produce 2 moles of NH₃.

Based on the calculations above, you can deduce easily that the amount of nitrogen is limiting because you only have 4.98 × 10⁻²⁴ moles of N₂, which is less than the amount of hydrogen required to react with it.

To find out how many molecules of NH₃ are formed, you'll have to use the amount of limiting reagent in your calculation. In this case, you're with 3 moles of NH₃ produced for every 1 mole of N₂, so calculate:

4.98 × 10⁻²⁴ moles N₂ × 3 moles NH₃/1 mole N₂ = 1.49 × 10⁻²³ moles NH₃

To convert this to molecules of NH₃, can use Avogadro's number:

1.49 × 10⁻²³ moles NH₃ × 6.022 × 10²³ molecules/mole = 8.97  molecules NH₃

So, 8.97 molecules of NH₃ are formed.

To determine the excess reactant, you first calculate the amount of hydrogen remaining after the reaction:

Amount of H₂ consumed = 3 moles NH₃ × 3 moles H₂/1 mole NH₃ = 9 moles H₂

Amount of H₂ remaining = 6 moles H₂ - 9 moles H₂ = -3 moles H₂

The negative value for the amount of hydrogen remaining indicates that there is an excess of hydrogen by 3 moles.

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The measured Kc is different from the calculated Kc, the system is not at equilibrium. There is either too much hydrogen iodide and/or too little hydrogen and/or iodine in the cylinder.

To determine whether the system is at equilibrium, we can use the equilibrium constant expression for the reaction;

H₂(g) + I₂(g) ⇌ 2HI(g)

The equilibrium constant expression for this reaction is;

Kc = [HI]² / [H₂] [I₂]

We are given the partial pressures of each component in the cylinder, so we can calculate the concentrations using the ideal gas law.

[P] = n/V = (m/M) / V

where P is the partial pressure, n is the number of moles, V is the volume, m is the mass, M is the molar mass, and the square brackets denote concentration.

For hydrogen (H₂)

[P] = n/V = (m/M) / V

[m/M] = [P] x V

[m/M] = 0.127 atm x V / R x T

For iodine (I₂)

[P] = n/V = (m/M) / V

[m/M] = [P] x V

[m/M] = 0.134 atm x V / R x T

For hydrogen iodide (HI)

[P] = n/V = (m/M) / V

[m/M] = [P] x V

[m/M] = 1.055 atm x V / R x T

Switching these expressions into the equilibrium constant expression, we get;

Kc = ([HI]/V)² / ([H₂]/V) x ([I2]/V)

Kc = ([HI]² / V²) / ([H₂] x [I2] / V²)

Put in the values we get:

Kc = [(1.055 atm / V)² / (0.127 atm / V) x (0.134 atm / V)]

Kc = 8.37 atm² / V²

If the system is at equilibrium, the measured concentrations should give the same value for Kc as calculated above. If the measured Kc is different, then the system is not at equilibrium.

Therefore, we can calculate the measured Kc as;

Kc = ([HI]² / V²) / ([H₂] x [I2] / V²)

Kc = [(1.055 atm / V)² / (0.127 atm / V) x (0.134 atm / V)]

Kc = 8.87 atm² / V²

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The higher temperature will witness more volume in gas when the amount of reactants is same.

The Charle's Law states that the volume is directly proportional to the temperature when other parameters are kept constant. Thus, the increase in temperature will accompany increase in volume.

The reason for such phenomenon is expansion of gases on heating due to excess absorption of energy. It is available in the form of kinetic energy which subsequently is evident through square of velocity.

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Answer:

≈15

Explanation:

1.43-.96=.57

.57/1.43 (change / original) * 25 ≈ 9.96

25 - 9.96 ≈ 15

Answer:   12.55 C

Explanation:

The relationship between gas volume and temperature is described by 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 ideal gas constant, and T is the temperature in Kelvin.

P1V1 = nRT1 (initial conditions)

P2V2 = nRT2 (final conditions)

Since the number of moles is constant, we can set nRT1 equal to nRT2:

P1V1 = P2V2

We can rearrange this equation to solve for the final temperature:

T2 = (P1V1/T1) * V2/P2

Substituting the given values:

T2 = (1 atm * 0.675 L / 308.15 K) * (0.635 L / 1 atm)

where we converted the initial temperature of 35 C to Kelvin by adding 273.15 K.

Simplifying and solving for T2:

T2 = 285.7 K - 273.15 K

T2 = 12.55 C

Therefore, the temperature of the room where the gas has a volume of 0.635 L at 1 atm is approximately 12.55 C.

The pH of the solution after 50.0 mL of 0.500 M HCl has been added is 4.74.

The balanced chemical equation for the reaction between NH₃ and HCl is:

NH₃ + HCl → NH₄Cl

This reaction is a neutralization reaction, in which an acid (HCl) reacts with a base (NH₃) to form a salt (NH₄Cl) and water.

Since NH₃ is a weak base and HCl is a strong acid, the reaction will go to completion, and all of the NH₃ will react with the HCl. Before any HCl is added, the solution contains only NH₃, so its pH is basic. As HCl is added, it reacts with the NH₃ to form NH₄Cl, which is a neutral salt. The pH of the solution will decrease as more HCl is added, until all of the NH₃ has reacted and the solution becomes acidic.

To find the pH of the solution after 50.0 mL of 0.500 M HCl has been added to 100.0 mL of 0.500 M NH₃, we need to calculate the moles of NH₃ and HCl that have reacted.

Moles of NH₃ = (100.0 mL)(0.500 M) = 0.0500 moles

Moles of HCl = (50.0 mL)(0.500 M) = 0.0250 moles

Since NH₃ and HCl react in a 1:1 ratio, all of the HCl has reacted with half of the NH₃, leaving 0.0250 moles of NH₃ unreacted.

To calculate the concentration of NH₄⁺ ions in the solution, we need to consider the equilibrium reaction between NH₃  and NH₄⁺ :

NH₃ + H₂O ⇌ NH₄⁺ + OH⁻

The equilibrium constant for this reaction is:

Kb = [NH₄⁺][OH⁻] / [NH₃]

At equilibrium, the concentration of NH₃ will be 0.0250 moles / 0.150 L = 0.167 M (since the total volume of the solution is 150.0 mL). The concentration of OH- ions can be calculated using the Kb value for NH₃, which is 1.8 × 10⁻⁵:

Kb = [NH₃⁺][OH⁻] / [NH₃]

1.8 × 10⁻⁵ = [NH₄⁺][OH⁻] / 0.167

[NH₄⁺][OH⁻] = 3.006 × 10⁻⁶

Since the solution is neutral at equilibrium, the concentration of H+ ions is equal to the concentration of OH- ions. Therefore, the pH of the solution is:

pH = -㏒[H⁺]

[H⁺] = [OH⁻] = √(Kw / [NH₄⁺])

                     = √(1.0 × 10⁻¹⁴/ 3.006 × 10⁻⁶) = 1.83 × 10⁻⁵ M

               pH = -㏒(1.83 × 10⁻⁵) = 4.74

As a result, the pH of the solution after adding 50.0 mL of 0.500 M HCl is 4.74.

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