Step 6: Measure Pressure and Volume with the Book
and 1 kg of Weight
Total the mass on the syringe. Record it in the
correct row of the data table.
kg
DONE
4
O-
01
Click and drag weights to change the pressure.
Click the syringe to zoom in and see the volume.

We can use the following formula to determine the standard cell potential from standard free energy changes:

ΔG° = -nFE°

where

ΔG° is the standard free energy change,

n is the number of moles of electrons transferred in the balanced chemical equation,

F is Faraday's constant (96,485 C/mol), and

E° is the standard cell potential.

You can find the standard free energy change (G°) for a specified fuel cell in readily available references. These values ​​are listed:

1. Fuel cell for H2/O2: G° = -237.2 kJ/mol

The chemical formula is 2H2 + O2 -> 2H2O.

In this example, n is equal to 4 (transferring 4 moles of electrons).

After entering the values ​​into the formula, we get:

-4 * 96,485 C/mol * E°1 = -237.2 kJ/mol

As we solve for E°1, we get:

E°₁ ≈ 1.23 V

2. G° = -326.7 kJ/mol for a methanol/O2 fuel cell.

The balanced chemical formula is: CO2 + 2H2O = CH3OH + 1.5O2.

In this example, n is equal to 6 (transferring 6 moles of electrons).

After entering the values ​​into the formula, we get:

-6 * 96,485 C/mol * E°2 = -326.7 kJ/mol

As we solve for E°2, we get:

E ° ₂ ≈ 0.54 V

3. Fuel cell for ethanol and oxygen: G° = -329.6 kJ/mol

The chemical formula is: C2H5OH + 3O2 -> 2CO2 + 3H2O.

In this example, n is equal to 12 (12 electron moles are exchanged).

After entering the values ​​into the formula, we get:

-12 * 96,485 C/mol * E°3 = -329.6 kJ/mol

As we solve for E°3, we get:

E°₃ ≈ 0.27 V

4. Fuel cell for glucose and oxygen: G° = 2,840 kJ/mol

The chemical formula is C6H12O6 + 6O2 -> 6CO2 + 6H2O.

Since 24 moles of electrons are transported, n = 24 in this example.

After entering the values ​​into the formula, we get:

-24 * 96,485 C/mol * E°4 = 2,840 kJ/mol.

Solving for E°4, we get:

E°₄ ≈ 0.37 V

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Based on the indicator chart, the pH indicator that would be the best one to use would be Thymol Blue.

The color change would be from light blue to yellow to orange.

Thymol blue would be best because it would show you where your starting point is and then when you reach the desired pH value of 3. 0. Looking at the indicator chart, Thymol blue has a color of light blue between 8. 0 and 9. 0 so you will know you are at 8. 0 when the reaction starts.

As you add more acid, the color would move to yellow to let you know that it is getting more acidic. Once it reaches orange, the titration should stop.

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To name Type 2 ionic compounds such as AuCl₃, you need to use the Stock system or Roman numeral system to indicate the oxidation state of the cation. Some steps are; Identify the cation, Determine the charge, Write the name, and combine two names.

Here are the steps to name AuCl₃; Identify the cation and anion. In this case, the cation is Au³⁺ and the anion is Cl⁻.

Determine the charge on the cation by using the anion's charge and balancing the charges to zero. Since Cl⁻ has a charge of -1 and there are three Cl⁻ ions in the compound, the total negative charge is -3. Therefore, the Au³⁺ ion has a charge of +3.

Write the name of the cation first, followed by the name of the anion with an -ide ending. Since the cation is Au³⁺, we use the name "gold(III)" to indicate the oxidation state of +3. The anion is Cl⁻, so we add the -ide ending to get "chloride".

Combine the two names to get the compound's name: "gold(III) chloride".

Therefore, the name of the Type 2 ionic compound AuCl₃ is "gold(III) chloride".

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Answer:

The chemical elements are arranged from left to right and top to bottom in order of increasing atomic number, or the number of protons in an atom's nucleus.

Explanation:

The law of conservation of mass states that in a chemical reaction, the total mass of the reactants is equal to the total mass of the products. To provide evidence for this law, we can perform an experiment in which calcium carbonate ([tex]CaCO_3[/tex]) is decomposed to produce calcium oxide (CaO) and carbon dioxide ([tex]CO_2[/tex] ), and then measure the masses of the reactants and products.

Method:

Weigh a sample of [tex]CaCO_3[/tex] using a balance.

Heat the [tex]CaCO_3[/tex] in a crucible until it decomposes to CaO and [tex]CO_2[/tex]. The [tex]CO_2[/tex] gas will escape, leaving only CaO in the crucible.

Allow the crucible to cool and then weigh it again to determine the mass of the CaO produced.

Collect the [tex]CO_2[/tex] gas that is released during the reaction in a gas syringe or other collection device. Measure the volume of [tex]CO_2[/tex] gas produced, and calculate its mass using its molecular weight.

Which measurements should be taken:

The following measurements should be taken:

The mass of the [tex]CaCO_3[/tex] used as a reactant.

The mass of the CaO produced as a product.

The volume of [tex]CO_2[/tex] gas produced during the reaction.

The temperature and pressure of the [tex]CO_2[/tex] gas to allow for the calculation of its mass.

How the student could show evidence for the conservation of mass:

To show evidence for the law of conservation of mass, the student can compare the mass of the [tex]CaCO_3[/tex] used as a reactant to the total mass of the products, which includes the mass of CaO produced and the mass of [tex]CO_2[/tex] gas released.

The sum of the masses of CaO and [tex]CO_2[/tex] should be equal to the mass of the [tex]CaCO_3[/tex] used as a reactant, within experimental error. This will provide evidence that the mass of the reactants is conserved and equals the mass of the products, as required by the law of conservation of mass.

Additionally, the student could calculate the theoretical yield of CaO and CO2 based on the balanced equation for the reaction, and compare this to the actual yield obtained from the experiment. Any difference between the theoretical and actual yields could be due to experimental error, but the comparison can still provide additional evidence for the conservation of mass.

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The energy associated with the formation of 2.55 g of 4He by the fusion of 3H and 1H is -2.982 x 10⁻¹⁰ J.

The given masses of the isotopes can be converted to kilograms using the conversion factor: 1 u = 1.661 x 10⁻²⁷ kg.

Mass of 4He = 2.55 g = 2.55 x 10⁻³ kg

Mass of 3H = 3.01605 u = 3.01605 x 1.661 x 10⁻²⁷ kg/u

= 5.0099 x 10⁻²⁷ kg

Mass of 1H = 1.00783 u = 1.00783 x 1.661 x 10⁻²⁷ kg/u

= 1.6737 x 10⁻²⁷ kg

The balanced equation for the fusion reaction is;

3H + 1H → 4He

The molar mass of 4He is 4.0026 g/mol, which can be converted to kg/mol using the conversion factor: 1 g/mol = 1 x 10⁻³ kg/mol.

Molar mass of 4He = 4.0026 g/mol = 4.0026 x 10⁻³ kg/mol

The number of moles of 4He formed can be calculated from its mass;

n(4He) = m(4He) / M(4He)

= 2.55 x 10⁻³ kg / 4.0026 x 10⁻³ kg/mol

= 0.638 mol

From the balanced equation, 3 moles of H atoms react with 1 mole of He atoms to form 1 mole of He atoms. Therefore, the number of moles of H atoms required for the reaction is;

n(H) = 3/4 x n(4He)

= 3/4 x 0.638 mol

= 0.479 mol

The energy released in the reaction can be calculated using the mass-energy equivalence equation;

E = Δm c²

where Δm is change in mass, c is the speed of light.

The change in mass is;

Δm = [3H + 1H - 4He] = [5.0099 x 10⁻²⁷ kg + 1.6737 x 10⁻²⁷kg - 4.0026 x 10⁻³ kg]

= -3.315 x 10⁻²⁷ kg (negative because mass is lost in the reaction)

The energy released is;

E = (-3.315 x 10⁻²⁷ kg) c²

= (-3.315 x 10⁻²⁷ kg) (2.998 x 10⁸ m/s)²

= -2.982 x 10⁻¹⁰ J

The negative sign indicates that energy is released in the reaction (exothermic reaction).

Therefore, the energy associated is -2.982 x 10⁻¹⁰ J.

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Answer:

Energy

4p ⇵ ⇵ ⇵

3d ⇵ ⇵ ⇵ ⇵ ⇵

4s ⇵

3p ⇵ ⇵ ⇵

3s ⇵

2p ⇵ ⇵ ⇵

2s ⇵

1s  ⇵

The gas law that is calculated with the pressure and volume variables at a constant temperature is Boyle's Law. Boyle's Law states that the pressure (P) of a gas is inversely proportional to its volume (V) when temperature (T) is held constant.

Mathematically, it is expressed as P₁V₁ = P₂V₂, where P₁ and V₁ represent the initial pressure and volume, and P₂ and V₂ represent the final pressure and volume.According to Boyle's Law, if the volume of a gas is reduced while keeping the temperature constant, the pressure will increase proportionally.

Similarly, if the volume is increased, the pressure will decrease. This relationship holds as long as the temperature remains constant throughout the process. Boyle's Law is one of the fundamental gas laws and provides insights into the behavior of gases under changing pressure and volume conditions at a constant temperature.

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a) The half-reaction that occurs at the anode is;

Cu(s) ----> Cu^2+(aq) + 2e

b) The Keq of the reaction is 3.75 * 10^20

Electrode potential is a measure of the tendency of an electrode to gain or lose electrons when it is in contact with a solution containing ions.

For the oxidation half reaction;

E° = Ecathode - Eanode

Eanode =  Ecathode - E°

Eanode = 0.951 -  0.609

= 0.342 V

Note that;

E° = 0.0592/nlogKeq

logKeq = E° * n/ 0.0592

= 0.609 * 2/0.0592

Keq =3.75 * 10^20

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The system will react by changing the concentrations to restore equilibrium when the CO and CO2 concentrations in the chemical equation 2CO + O2 2CO2 are raised. The system will try to reduce the rise in CO and CO2 by moving the reaction towards the products side, in accordance with Le Chatelier's concept.

The reaction will go forward as the CO concentration rises, eating part of the extra CO and turning it into CO2. This change lowers the excess CO concentration and aids in reestablishing equilibrium.

However, since CO2 is already a product, its concentration does not have a direct impact on the reaction. However, to keep the stoichiometric balance, it can result in a somewhat greater concentration of CO.

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If 3 moles of cl reacts with 3 moles oxygen, there is no limiting reactant or excess reactant because the reactants are in stoichiometric proportions.

To determine the limiting reactant and excess reactant, we need to compare the stoichiometry of the reaction to the given amounts of each reactant.

The balanced chemical equation for the reaction between chlorine (Cl2) and oxygen (O2) can be represented as follows:

2Cl2 + O2 → 2Cl2O

According to the balanced equation, it requires 2 moles of chlorine (Cl2) to react with 1 mole of oxygen (O2) to produce 2 moles of chlorine oxide (Cl2O).

Given that we have 3 moles of chlorine (Cl2) and 3 moles of oxygen (O2), we can determine the limiting reactant by comparing the ratio of moles between the two reactants.

The ratio of Cl2 to O2 required for complete reaction is 2:1. However, since we have equal amounts of Cl2 and O2 (both 3 moles), neither reactant is present in excess.

Therefore, in this scenario, there is no limiting reactant or excess reactant because the reactants are in stoichiometric proportions. All of the chlorine and oxygen will be consumed in the reaction, resulting in the complete conversion to chlorine oxide (Cl2O).

It's important to note that if the amounts of Cl2 and O2 were different, the reactant present in lesser quantity would be the limiting reactant, and the reactant in greater quantity would be the excess reactant.

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The temperature of the oxygen gas sample is 609 K, which is approximately 336°C or 637°F. The answer is A.

We can use the ideal gas law equation, PV = nRT, to solve for the temperature of the oxygen gas sample.

First, we need to calculate the number of moles of oxygen gas present in the sample using its mass and molar mass:

n = m/M

where:

n = number of moles

m = mass (in grams)

M = molar mass (in g/mol)

The molar mass of oxygen gas (O2) is 32.00 g/mol.

n = 48 g / 32.00 g/mol = 1.50 mol

Next, we can rearrange the ideal gas law equation to solve for temperature (T):

T = (PV) / (nR)

where:

T = temperature (in Kelvin)

P = pressure (in atm)

V = volume (in liters)

n = number of moles

R = gas constant (0.0821 L atm/mol K)

Plugging in the given values, we get:

T = (3.0 atm x 25 L) / (1.50 mol x 0.0821 L atm/mol K)

T = 609 K

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Schemes for the formation of bonds:

The schemes for the formation of bonds between the atoms of the following elements are:

Carbon and phosphorus:

C + P → CP

This is an example of a covalent bond, which is a type of bond that is formed by the sharing of electrons. In this case, the carbon atom shares one electron with the phosphorus atom, forming a single covalent bond.

Sulfur:

S + S → S₈

This is an example of a molecular bond, which is a type of bond that is formed by the sharing of electrons between multiple atoms of the same element. In this case, the sulfur atoms share two electrons each, forming a double bond.

Magnesium and silicon:

Mg + Si → Mg₂Si

This is an example of an ionic bond, which is a type of bond that is formed by the transfer of electrons from one atom to another. In this case, the magnesium atom gives up two electrons to the silicon atom, forming a magnesium ion with a charge of +2 and a silicon ion with a charge of -4. These ions are then attracted to each other by the opposite charges.

2. The types of bonds and crystal lattices for the following elements:

CaO: ionic bond, ionic lattice

C: covalent bond, diamond lattice

SiO₂: covalent bond, tetrahedral lattice

Fe: metallic bond, body-centered cubic lattice

K₃N: ionic bond, cubic lattice

3. The processes depicted by the following diagrams, along with the corresponding electronic balances:

Cu0 → Cu⁺²: oxidation

Cu0 → Cu⁺² + 2e⁻

S0⁻ → S⁻²: reduction

S0⁻- + 2e⁻ → S⁻²

Fe⁺³ → Fe⁺²: reduction

Fe⁺³ + 1e⁻ → Fe⁺²

4. The redox reactions and the coefficients arranged by the electronic balance method:

H₂O + CO₂ → HCL + O₂

2H⁺ + ¹/₂O₂ → HCL

2e⁻ + 2H⁺ → H₂

¹/₂O₂ + 2e⁻ → O₂⁻

Fe₂O₃ + H₂ → Fe + H₂O

Fe₂O₃ + 3H₂ → 2Fe + 3H₂O

3Fe⁺³ + 6e⁻ + 6H⁺ → 2Fe + 6H₂O

2O₂⁻ + 6H⁺ → 4H₂O

H₂SO₄ + S → SO₂ + H₂O

2H⁺ + SO₄²⁻ → SO₂ + H₂O

2e⁻ + 2H+ → H₂

S → S²⁻ + 2e⁻

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Answer: The values you provided show that as the current (measured in amps) and voltage (measured in volts) increase, the resistance (measured in ohms) remains relatively constant.

Looking at the values, it appears that the current (in amps) divided by the voltage (in volts) yields resistance values (in ohms) that are relatively close to each other. The calculated resistance values range from approximately 27 ohms to 33 ohms, with some variation in between.

The price of the ingot in South African Rand (SAR) is 19,533,171.507 R.

To calculate the price of the ingot in South African Rand (SAR), we need to follow these steps:

Convert the dimensions of the ingot to cm³:

Length = 23.7 cm

Width = 75.5 mm = 7.55 cm

Height = 10.9 cm

The volume of the ingot is calculated by multiplying these dimensions:

Volume = Length × Width × Height

= 23.7 cm × 7.55 cm × 10.9 cm

= 1830.0475 cm³

Calculate the weight of the ingot in grams:

Since 1 cm³ of gold weighs 19.30 g, we can multiply the volume of the ingot by this conversion factor to obtain the weight:

Weight = Volume × 19.30 g

= 1830.0475 cm³ × 19.30 g

= 35,380.1375 g

Convert the weight of the ingot to ounces:

Since 1 ounce is equal to 28.35 g, we can divide the weight of the ingot by this conversion factor:

Weight in ounces = Weight / 28.35 g

= 35,380.1375 g / 28.35 g

= 1247.0461 ounces

Calculate the price of the ingot in USD:

The current price of gold is $1629.8 per ounce, so we can multiply the weight of the ingot in ounces by this price:

Price in USD = Weight in ounces × Price per ounce

= 1247.0461 ounces × $1629.8/ounce

= $2,032,881.72278

Convert the price from USD to SAR:

The exchange rate is 9.6 R/$, so we can multiply the price in USD by this exchange rate:

Price in ZAR = Price in USD × Exchange rate = $2,032,881.72278 × 9.6 R/$ = 19,533,171.507 R

Therefore, the price of the ingot in South African Rand (SAR) is approximately 19,533,171.507 R.

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As Earth makes its way around the sun (earth orbit) along a tilted path relative to its own axis we experience staggered views of different stars and constellations depending on where we find ourselves across our annual trajectory.

The cycle of our planet traveling along an elliptical orbit around the sun provides us with what we know as a year or roughly 365.25 days of time measurement; however, this orbital path isn't uniformed and appears like a flattened circle with one focus positioned near our star - The Sun.

This causes changes in distance between both objects throughout different periods of time during its annual journey; for instance, early January denotes when we reach perihelion or our shortest distance from Sun while we hit aphelion on early July - marking our farthest measured distance from it.

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In forming the Al³⁺ ion, the Al atom loses the electrons in its outermost energy level, causing a decrease in the atomic radius, hence option A is correct.

The number of protons in the nucleus of AlandAl3+ AlandAl3+ is the same, however there are differing numbers of electrons in the final shell. Al³⁺ is smaller than Al because it has fewer electrons.

The Al atom will become an Al³⁺ ion when it loses its third electron and develops a tri-positive charge on it. In forming the Al³⁺ ion, the Al atom loses the electrons in its outermost energy level, causing a decrease in the atomic radius.

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The kinetic energy of a single molecule of nitrogen (N₂) with a velocity of 400.0 m/s is approximately 3.72 x 10^-21 joules.

To calculate the kinetic energy of a single molecule of nitrogen (N₂) with a velocity of 400.0 m/s, we can use the equation for kinetic energy: KE = 1/2 * m * v², where KE represents the kinetic energy, m is the mass of the molecule, and v is the velocity.

(a) Unit conversions:

The mass of a nitrogen molecule (N₂) is approximately 28 atomic mass units (u) or 4.65 x 10^-26 kilograms (kg).

The given velocity is already in meters per second (m/s), so no further conversion is needed.

(b) Calculating the kinetic energy:

Plugging the values into the kinetic energy equation, we have:

KE = 1/2 * (4.65 x 10^-26 kg) * (400.0 m/s)²

Evaluating the expression:

KE = 1/2 * 4.65 x 10^-26 kg * 160000 m²/s²

= 3.72 x 10^-21 joules (J)

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Answer:

To determine the number of moles of oxygen consumed to produce 68.1g of water, we need to know the amount of hydrogen in the water. Since water has two hydrogen atoms and one oxygen atom, the molar mass of water is 18.015 g/mol.

Using the molar mass of water, we can calculate the number of moles of water produced:

68.1 g ÷ 18.015 g/mol = 3.78 mol

Since there are two hydrogen atoms in each molecule of water, there must be twice as many moles of hydrogen as there are moles of water:

3.78 mol × 2 = 7.56 mol of hydrogen

Finally, we can use the balanced chemical equation for the reaction of hydrogen and oxygen to form water to determine the number of moles of oxygen consumed:

2H2 + O2 -> 2H2O

For every 1 mole of oxygen consumed, 2 moles of water are produced. Therefore, the number of moles of oxygen consumed is:

7.56 mol H2O ÷ 2 = 3.78 mol O2

So, 3.78 moles of oxygen would be consumed to produce 68.1g of water.

The ideal gas law can be used to determine the pressure of mercury gas inside a fluorescent bulb:

PV = nRT

where

R is the gas constant,

n is the number of moles,

P is the pressure, V is the volume, and T is the temperature in Kelvin.

As we solve for n, we get:

n = pv / rt

Given the low pressure and moderate temperature, we can assume that mercury gas behaves better.

Therefore, the amount of mercury gas in the bulb is expressed as:

n = (1.06 Pa)(1.62 L)/(8.31 J/K/mol)(40 + 273.15 K) = [tex]7.71 * 10^-^4[/tex] mol

The molar mass of mercury can be used to determine how much liquid mercury is needed to produce this amount of gas:

m = nM

where M is the molar mass of mercury and m is its mass in mass units.

M = 200.59 g/mol

m = (7.71 x 10^-4 mol)(200.59 g/mol) = 0.154 g

Therefore, to achieve a pressure of 1.06 Pa, 154 mg of liquid mercury must evaporate at a temperature of 40 °C for a 1.62-L fluorescent light bulb.

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Answer:

In a quantum model of the atom, the shape of an electron orbital represents the probability distribution of finding an electron in a particular region of space around the nucleus of an atom. In other words, an electron orbital is a three-dimensional space around the nucleus where there is a high probability of finding an electron with a given energy level.

Different orbitals can have different shapes, such as spherical, hourglass-shaped, or more complex shapes.

Electron Orientations:

s = 1 orbital orientation

p = 3 orientations ([tex]6e^-[/tex])

d = 5 orientations ([tex]10e^-[/tex])

f = 7 orientations ([tex]14e^-[/tex])

The crystal I chose is sodium chloride crystals and it is beneficial for man as it is used in the preservation of food as well as in seasoning of food.

A solid whose components are arranged in a highly ordered microscopic structure to form an all-pervasive crystal lattice is referred to as a crystal.

Sodium chloride also referred to as common salt is an ionic compound that has the chemical formula NaCl.

Sodium chloride is an essential nutrient employed in healthcare. It is employed as a spice to improve flavor and as a food preservative. Additionally, sodium chloride is employed in the production of plastics and other goods and is applied to de-ice sidewalks and roadways

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A) Keep temperature constant and increase the pressure of the reaction; reaction rate will increase.

The temperature and pressure of a reaction will affect the rate of reaction.

Pressure

Pressure and the rate of reaction have a direct relationship. This means that as one increases, so does the other. So, if the pressure increases. then the rate of reaction will also increase. This is due to the number of collisions. As pressure increases, the number of collisions between molecules increases. This causes the reaction to occur faster. Thus, the rate of reaction increases.

Temperature

Kinetic energy and temperature are proportional. This means that as temperature increases, so does kinetic energy. This leads to temperature and rate of reaction also having a direct relationship. So, temperature and rate of reaction increase and decrease together. This is due to the fact that when temperature increases, the energy of the molecules increases. This leads to an increased number of collisions. As stated above, more collisions lead to a faster reaction.

An exothermic reaction releases 325 kJ. 1359.8 kJ energy is this in calories.

An exothermic reaction is a chemical reaction that releases heat energy into the surroundings. During an exothermic reaction, the products of the reaction have less potential energy than the reactants, and the excess energy is released in the form of heat.

One calorie is defined as the amount of energy required to raise the temperature of one gram of water by one degree Celsius.

One joule is defined as the amount of energy required to apply a force of one newton over a distance of one meter.

1 cal = 4.184 J

1 cal = 0.004184 kJ

325000 cal = x kJ

0.004184/ 1 = x / 325000

x = 1359.8 kJ

Thus, 1359.8 kJ energy is this in calories.

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The atomic mass of Aluminum is 23, which means it has a total of 23 particles in its nucleus, including protons and neutrons.

Aluminum has an atomic number of 13, which means it has 13 protons in its nucleus.

To find the number of neutrons, we subtract the atomic number from the atomic mass. So, Aluminum has 23 - 13 = 10 neutrons in its nucleus. Electrons are the negatively charged particles that orbit around the nucleus of an atom.

Aluminum, being a neutral atom, has an equal number of electrons to the number of protons in its nucleus, which is 13. These electrons are distributed in different energy levels or shells around the nucleus.

Aluminum is a widely used metal in different applications due to its unique properties such as low density, high strength, and resistance to corrosion. It is used in the manufacturing of cans, foils, and aircraft parts. The number of protons and electrons determines the atomic number and chemical properties of an element. The number of neutrons affects the stability and isotopes of an element.

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The mass (in grams) of carbon dioxide, CO₂ produced, given that 360 g of water vapor was produced from the reaction is 586.67 g

First, we shall obtain the mass of C₂H₆ that reacted to produce 360 g of water vapor, H₂O. This is obtain as follow:

2C₂H₆ + 7O₂ -> 4CO₂ + 6H₂O

From the balanced equation above,

108 g of H₂O were obtained from 60 g of C₂H₆

Therefore,

360 g of H₂O will be obtain from = (360 × 60) / 108 = 200 g of C₂H₆

Thus, we can mass of C₂H₆ that reacted is 200 g

Finally, we shall obtain the mass of carbon dioxide, CO₂ produced. This is shown below:

2C₂H₆ + 7O₂ -> 4CO₂ + 6H₂O

From the balanced equation above,

60 g of C₂H₆ reacted to produce 176 g of CO₂

Therefore,

200 g of C₂H₆ will react to produce = (200 × 176) / 60 = 586.67 g of CO₂

Thus, the mass of carbon dioxide, CO₂ produced is 586.67 g

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Answer:

Explanation:

balance

2 C2H6 + 7  O2  --> 4 CO2 + 6 H2O

given 360 g H20 (g)

required =586.67 g CO2

360 g H20 x (1mole/18 g H20) X (4 mole CO2/6 moles H20) X (44g CO2/1mole) =586.67 g CO2

Answer:

Explanation:

constant = 33.43 ml  charles law; formula is v1/t1=v2/t2

129.77 mmhg= boyles law;  = p1v1=p2v2

659.51 K ; gay-lussac law; P1/T1=P2/T2

20.79 mL; combined gas law; p1v1/t1=p2v2/t2

90.67 mmHg; combined gas law; P1V1/T1=P2V2/T2

Answer:

Explanation:

H+ = 1 X 10^-7.41 = 3.89 X 10^ -8

POH = 14-7.41 = 6.59

OH- = 1 x 10 ^-6.59 = 2.57 X 10^ -7

The [H+] and [OH−] concentrations of the solution are approximately 2.38 × 10^(-7) M, and the pOH is 6.59.

The pH of a solution is a measure of the concentration of hydrogen ions ([H+]) in the solution. The pH scale ranges from 0 to 14, with a pH of 7 considered neutral. A pH of 7.41 indicates that the solution is slightly basic. To calculate the [H+], [OH−], and pOH of the solution, we can use the relationship:

pH + pOH = 14

Given that the pH is 7.41, we can subtract it from 14 to find the pOH:

pOH = 14 - 7.41 = 6.59

Since pH + pOH = 14, we can also determine the [OH−] by taking the antilogarithm of the pOH value:

[OH−] = 10^(-pOH)

[OH−] = 10^(-6.59)

[OH−] ≈ 2.38 × 10^(-7) M

Since the solution is neutral, the concentration of [H+] will be equal to the concentration of [OH−]:

[H+] = [OH−] ≈ 2.38 × 10^(-7) M

Therefore, the [H+] and [OH−] concentrations of the solution are approximately 2.38 × 10^(-7) M, and the pOH is 6.59.

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