Dimensional analysis with shapes

The energy difference of an electron's transition from n = 6 to n = 1 in a hydrogen atom is approximately -2.17 × 10⁻¹⁸ Joules.

To calculate the energy difference (deltaE) of an electron's transition from n = 6 to n = 1 in a hydrogen atom, we can use the following equation:

deltaE = -13.6 * (1/n_final^2 - 1/n_initial^2) eV
where n_initial is the initial energy level (6 in this case), n_final is the final energy level (1 in this case), and -13.6 eV is the ionization energy of hydrogen.

Converting eV to Joules, we get:
1 eV = 1.602 x 10^-19 J
Therefore, deltaE in Joules can be calculated as follows:
deltaE = -13.6 * (1/1^2 - 1/6^2) * 1.602 x 10^-19 J/eV
deltaE = -2.179 x 10^-18 J

Therefore, the energy difference (deltaE) of an electron's transition from n = 6 to n = 1 in a hydrogen atom is -2.179 x 10^-18 J.

To calculate the energy difference (ΔE) for an electron's transition from n = 6 to n = 1 in a hydrogen atom, you can use the following formula:
ΔE = -13.6 eV * (1/nf² - 1/ni²)

Where ΔE is the energy difference in electron volts (eV), nf is the final energy level (1 in this case), and ni is the initial energy level (6 in this case).
ΔE = -13.6 eV * (1/1² - 1/6²)
ΔE ≈ -13.56 eV

Now convert electron volts to Joules:
1 eV = 1.6 × 10⁻¹⁹ J
ΔE ≈ -13.56 eV * 1.6 × 10⁻¹⁹ J/eV
ΔE ≈ -2.17 × 10⁻¹⁸ J

So,  approximately -2.17 × 10⁻¹⁸ Joules is  the energy difference of an electron's transition from n = 6 to n = 1 in a hydrogen atom.

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b. To determine the number of joules of energy for the slice of cake, we need to convert the calories to joules using the conversion factor 1 calorie = 4.184 joules:

Number of joules = 560 calories * 4.184 joules/calorie = 2343.04 joules

Therefore, the slice of cake contains 2343.04 joules of energy.

c. To determine the number of kilojoules of energy for the slice of cake, we can divide the number of joules by 1000:

Number of kilojoules = 2343.04 joules / 1000 = 2.34304 kilojoules

Therefore, the slice of cake contains 2.34304 kilojoules of energy.

To determine the number of calories in 1.5 kilojoules of energy, we need to convert kilojoules to calories using the conversion factor 1 kilojoule = 1000 calories:

Number of calories = 1.5 kilojoules * 1000 calories/kilojoule = 1500 calories

Therefore, 1.5 kilojoules of energy contains 1500 calories.

Explanation:

Answer:

2. a. 560 Calories, b. 2343.04 J, c. 2.34304 kJ

3. 358.508604 Cal

Explanation:

2.

a. The number of calories found in the slice of chocolate cake is 56012.

b. To determine the number of joules of energy for the slice of cake, we can use the conversion factor that 1 calorie is equal to 4.184 joules3. Therefore, the number of joules in the slice of cake is:

560 Cal ⋅ 4.184 J/ 1 Cal = 2343.04J

c. To determine the number of kilojoules of energy for the slice of cake, we can use the conversion factor that 1 kilojoule is equal to 1000 joules4. Therefore, the number of kilojoules in the slice of cake is:

2343.04 J ⋅ 1 kJ/1000 J = 2.34304 kJ

3. To determine the number of calories in 1.5 kilojoules of energy, we can use the conversion factor that 1 kilojoule is equal to 239.005736 calories1. Therefore, the number of calories in 1.5 kilojoules of energy is:

1.5 kJ ⋅ 239.005736 Cal / 1 kJ = 358.508604 Cal

The pressure of the gas when the volume is 1.09 L and the temperature is 308 K is 2.36 atm.

The final pressure of the gas is calculated by applying ideal gas law as follows;

(P₁V₁)/T₁ = (P₂V₂)/T₂

where

P₂ = (P₁V₁ x T₂)/(V₂ x T₁)

P₂ = (1.01 atm x 2.31 L x 308 K) / (1.09 L x 279 K)

P₂ = 2.36 atm

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The partial pressure of the oxygen is 0.236 atm.

The pressure that one gas component in a mixture of gases exerts is known as partial pressure. It is the pressure that the gas would experience if it took up the same amount of space in the mixture at the same temperature on its own.

We know that;

P[tex]CO_{2}[/tex] = 0.285 torr or 0.000375 atm

P[tex]N_{2}[/tex] = 580.502 torr or 0.764 atm

P[tex]O_{2}[/tex] = ?

Total pressure = 1 atm

Then we have that;

PT =P[tex]CO_{2}[/tex]  +P[tex]N_{2}[/tex]+ P[tex]O_{2}[/tex]

P[tex]O_{2}[/tex] = PT - (P[tex]CO_{2}[/tex]  + P[tex]N_{2}[/tex])

P[tex]O_{2}[/tex] = 1 - (0.000375  + 0.764)

P[tex]O_{2}[/tex]= 0.236 atm

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Answer:O has 6, Na has 1, and Sr has 2.

Explanation:

Answer:

Explanation:

Coal deposits have been found in Antarctica, particularly in the Transantarctic Mountains where they are interbedded in sedimentary rocks of the flat-lying Beacon Supergroup. The presence of coal in Antarctica supports Wegener's continental drift hypothesis because it suggests that Antarctica was once part of a larger landmass that had a warm climate suitable for the formation of coal. Coal is formed from ancient plant matter that has been compressed and heated over millions of years. The presence of coal in Antarctica suggests that the continent was once located closer to the equator and had a climate that supported lush vegetation.

20 g of calcium would release the same number of valence electrons as 23 g of sodium.

The atomic number of calcium (Ca) is 20, which means it has 20 electrons in its neutral state. When calcium loses two electrons, it becomes a Ca2+ ion with 18 electrons.

On the other hand, the atomic number of sodium (Na) is 11, which means it has 11 electrons in its neutral state. When sodium loses one electron, it becomes a Na+ ion with 10 electrons.

To release the same number of valence electrons as 23 g of Na, we need to calculate how many moles of Na there are in 23 g:

Molar mass of Na = 23 g/mol

Number of moles of Na = 23 g / 23 g/mol = 1 mol

Since each Na+ ion has lost one electron, 1 mol of Na+ ions has lost 1 mol of valence electrons.

To release the same number of valence electrons as 1 mol of Na+ ions, we need to calculate how many moles of Ca2+ ions are required:

1 mol of Na+ ions = 1 mol of valence electrons

1 mol of Ca2+ ions = 2 mol of valence electrons

Therefore, we need 0.5 mol of Ca2+ ions to release the same number of valence electrons as 1 mol of Na+ ions.

Finally, we can calculate the mass of calcium that would release the same number of valence electrons as 23 g of Na:

Molar mass of Ca = 40 g/mol

Mass of Ca required = 0.5 mol x 40 g/mol = 20 g

Therefore, 20 g of calcium would release the same number of valence electrons as 23 g of sodium.

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Since the ΔH for the given reaction has negative value, the reaction is exothermic reaction.

A reaction that is exothermic is one in which power is given off as heat or light. In contrast to an endothermic process, which draws energy from its surroundings, an exothermic reaction transfers energy into the environment. The alteration in enthalpy (H) during an exothermic reaction will be negative. Since the ΔH for the given reaction has negative value, the reaction is exothermic reaction.

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We can use the combined gas law to determine the pressure of the gas at the final state. The combined gas law relates the pressure, volume, and temperature of a gas:

(P1 x V1) / T1 = (P2 x V2) / T2

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

We are given the initial pressure (P1 = 1.01 atm), volume (V1 = 2.31 L), and temperature (T1 = 279 K) of the gas, and the final volume (V2 = 1.09 L), and temperature (T2 = 308 K) of the gas. We can solve for P2, the final pressure of the gas:

(P1 x V1) / T1 = (P2 x V2) / T2

P2 = (P1 x V1 x T2) / (V2 x T1)

P2 = (1.01 atm x 2.31 L x 308 K) / (1.09 L x 279 K)

P2 = 2.41 atm (rounded to three significant figures)

Therefore, the pressure of the gas when the volume is 1.09 L and the temperature is 308 K is approximately 2.41 atm.

The concentration of the solution is  0.000435 mol/dm³.

The concentration of a solution is calculated as follows;

Concentration = (Absorbance) / (Molar absorptivity x path length)

the path length =  3cm

the molar absorptivity (E) = 400 dm/mol/cm.

if the solution transmits 30% of the light, it absorbs 70% of the incident light.

Absorbance = log (1/Transmittance)

Absorbance  = log (1/0.3)

Absorbance  = 0.523

Concentration = (0.523) / (400 dm/mol/cm x 3 cm)

= 0.000435 mol/dm³

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The ΔH°vap of water at the critical point is approximately 0.04614 kJ/mol.

To calculate the ΔH°vap (enthalpy of vaporization) of water at the critical point, we can use the Clausius-Clapeyron equation;

ln(P₂/P₁) = ΔH°vap/R [1/T₁ - 1/T₂]

where P₁ and T₁ are the pressure and temperature at which the enthalpy of vaporization is known (usually at standard conditions of 1 atm and 100 °C), P₂ and T₂ are the pressure and temperature at the critical point, ΔH°vap is the enthalpy of vaporization, and R is the gas constant (8.314 J/mol∙K).

Using the given values, we can plug them into the equation and solve for ΔH°vap;

ln(3.2 atm / 1 atm) = ΔH°vap / R [1/373 K - 1/275 K]

Simplifying;

ln(3.2) = ΔH°vap / R [0.0026819]

ΔH°vap / R = ln(3.2) / 0.0026819

ΔH°vap / R = 5.552

Multiplying both sides by R:

ΔH°vap = 5.552 x R

ΔH°vap = 5.552 x 8.314 J/mol∙K

ΔH°vap = 46.14 J/mol

Converting to kJ/mol;

ΔH°vap = 0.04614 kJ/mol

Therefore, the ΔH°vap of water is 0.04614 kJ/mol.

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There are 3 bonding pairs and 7 lone pairs of electrons in N2.

An atom's nucleus is orbited by an electron, a subatomic particle with a negative charge. Along with protons and neutrons, it is one of the elementary particles that make up matter. The mass of an electron is exceedingly small, it is roughly 1/1836 that of a proton.

To determine the number of lone pairs of electrons in N2, we need to subtract the number of bonding pairs from the total number of valence electrons:

Number of lone pairs = Total number of valence electrons - Number of bonding pairs

Number of lone pairs = 10 - 3

Number of lone pairs = 7

Therefore, there are 3 bonding pairs and 7 lone pairs of electrons in N2.

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The amount of zinc chloride that would be formed if 77.1 grams of zinc reacts is approximately 160.77 grams

To determine how many grams of zinc chloride ([tex]ZnCl_2[/tex]) would be formed if 77.1 grams of zinc (Zn) reacts, we'll use stoichiometry.

First, we need the molar masses of the substances involved:

Zn: 65.38 g/mol
[tex]ZnCl_2[/tex] : 136.29 g/mol
Now, we'll convert grams of Zn to moles:
77.1 g Zn × (1 mol Zn / 65.38 g Zn) = 1.179 moles Zn
According to the balanced chemical equation, 1 mole of Zn reacts to form 1 mole of [tex]ZnCl_2[/tex]:
1.179 moles Zn × (1 mol ZnCl₂ / 1 mol Zn) = 1.179 moles ZnCl₂

Finally, convert moles of ZnCl₂ to grams:
1.179 moles ZnCl₂ × (136.29 g ZnCl₂ / 1 mol ZnCl₂) ≈ 160.77 g ZnCl₂
So, approximately 160.77 grams of zinc chloride would be formed if 77.1 grams of zinc reacts.

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The ΔG value for the reaction A (aq) + B (aq) → C(aq) at 25 °C and the given concentrations is -8.35 kJ/mol.

The relationship between ΔG and K is given by the following equation:

ΔG = -RTln(K)

where R is the gas constant (8.314 J/(mol·K)), T is the temperature in Kelvin (25 °C = 298.15 K), and ln denotes the natural logarithm.

To calculate K, we need to use the equilibrium expression and the given concentrations:

[tex]K = [C]/([A][B])[/tex]

[tex]K = (5.00 * 10^{-5} M)/((1.50 M)(1.00 M))[/tex]

[tex]K = 3.33 x 10^{-5}[/tex]

Now we can substitute the values for R, T, and K into the equation for ΔG:

ΔG = -RTln(K)

ΔG = [tex]-(8.314 J/(mol.K))(298.15 K)ln(3.33 x 10^{-5})[/tex]

ΔG = -8.35 kJ/mol

Therefore, the ΔG value for the reaction A (aq) + B (aq) → C(aq) at 25 °C and the given concentrations is -8.35 kJ/mol.

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The mass of carbon dioxide formed when 24 kg of carbon is burnt completely in oxygen is 88 kg, and the heat released is 811.2 MJ.

What is Molar Mass?

Molar mass is the mass of one mole of a substance and is expressed in grams per mole (g/mol). It is calculated by adding up the atomic masses of all the atoms in a molecule or formula unit of a compound. The molar mass is used in stoichiometry calculations to convert between mass and moles of a substance.

The balanced equation for the combustion of carbon is:

C + O₂ → CO₂

From the equation, we can see that one mole of carbon reacts with one mole of oxygen to produce one mole of carbon dioxide. The molar mass of carbon dioxide is 12 + (2 × 16) = 44 g/mol.

First, let's find the number of moles of carbon in 24 kg:

n(C) = m/M = 24000 g / 12 g/mol = 2000 mol

Therefore, 2000 mol of CO₂ will be produced.

The mass of CO₂ produced can be calculated as:

m(CO₂) = n(CO₂) × M(CO₂) = 2000 mol × 44 g/mol = 88,000 g = 88 kg

Now, let's calculate the heat released during combustion:

Heat released = 33.8 MJ/kg × 24 kg = 811.2 MJ

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The hydrocarbon's molecular structure is [tex]C_9H_20[/tex].The hydrocarbon's empirical formula is [tex]C_9[/tex]/4H5.

To solve this problem, we need to use stoichiometry to relate the amount of [tex]CO__2[/tex] and [tex]H_2O[/tex] produced to the amount of [tex]CxHy[/tex] burned.

(a) To find the molecular formula of the hydrocarbon, we need to first calculate the number of moles of [tex]CO__2[/tex] and [tex]H_2O[/tex] produced. From the ideal gas law, we know that 1 mole of gas at STP (standard temperature and pressure) occupies 22.4 L. Therefore, 4.032 L of [tex]CO__2[/tex] at STP corresponds to:

4.032 L / 22.4 L/mol = 0.180 mol [tex]CO__2[/tex]

Similarly, the mass of H2O produced corresponds to:
3.602 g / 18.02 g/mol = 0.200 mol [tex]H_2O[/tex]

Since the hydrocarbon undergoes complete combustion, it reacts with oxygen to form [tex]CO__2[/tex] and [tex]H_2O[/tex] according to the balanced chemical equation:

[tex]CxHy[/tex] + (x + (y/4))O2 → [tex]CO__2[/tex] + (y/2)[tex]H_2O[/tex]
where x and y are the coefficients of the balanced equation. We can use the stoichiometric ratios to set up two equations:

0.180 mol [tex]CO__2[/tex] = x mol [tex]CxHy[/tex] → x = 0.180 mol / 0.0200 mol = 9
0.200 mol [tex]H_2O[/tex] = (y/2) mol [tex]CxHy[/tex] → y = 0.400 mol / 0.0200 mol = 20

Therefore, the molecular formula of the hydrocarbon is [tex]C_9H_20[/tex].
(b) To find the empirical formula of the hydrocarbon, we need to divide the subscripts by their greatest common factor. In this case, both subscripts are divisible by 4, so we get:

[tex]C_9H_20[/tex] → C9/4H5
Therefore, the empirical formula of the hydrocarbon is C9/4H5.

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Answer: Mixtures can be physically separated by using methods that use differences in physical properties to separate the components of the mixture, such as

Explanation:

Step-by-step Explanation:

8.51 moles is to 5.12 L  as  'x'  moles is to (13.3-5.12) L

8.51 moles / 5.12 L    =   x / ( 13.3-5.12)

x = 13.6 moles

Answer:

Explanation:

Both 2CH4 and C2H8 have the same number and kind of elements. But practically, 2CH4 will be existing but C2H8 cannot exist.

The concentration of the ammonia from the calculation is  [tex]1.1 * 10^-15[/tex]M.

The ratio of the concentrations of products to reactants in chemical equilibrium, for a given chemical reaction at a given temperature, is described by the equilibrium constant, abbreviated as Kc or Keq.

We can see that;

[tex]Keq = [NH_{3} ]^2/[N_{2} ] [ H_{2}]^3\\Keq[N_{2} ] [ H_{2}]^3 = [NH_{3}]^2\\\\N_{2} ] = [NH_{3}]^2/Keq[ H_{2}]^3[/tex]

=[tex](0.000123 )^2/ 0.00659 * (0.0275)^3= 1.1 * 10^-15 M[/tex]

Thus we would have the nitrogen concentration as [tex]1.1 * 10^-15[/tex] M

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Answer:

Explanation:

a) The balanced equation for the double replacement reaction between lead (II) nitrate and potassium bromide is:

Pb(NO₃)₂(aq) + 2KBr(aq) → PbBr₂(s) + 2KNO₃(aq)

In this reaction, lead (II) bromide (PbBr₂) will precipitate, while potassium nitrate (KNO₃) will remain in solution.

b) To determine the amount of precipitate produced, we need to first determine the limiting reactant. We can do this by calculating the number of moles of each reactant and comparing it to the stoichiometry of the balanced equation.

The molar mass of lead (II) nitrate is 331.21 g/mol and the molar mass of potassium bromide is 119.00 g/mol.

The number of moles of lead (II) nitrate is 32.5 g / 331.21 g/mol = 0.0981 mol The number of moles of potassium bromide is 38.75 g / 119.00 g/mol = 0.3256 mol

According to the balanced equation, one mole of lead (II) nitrate reacts with two moles of potassium bromide to produce one mole of lead (II) bromide. This means that if all the lead (II) nitrate were to react, it would require 0.0981 mol * 2 = 0.1962 mol of potassium bromide.

Since we have more than enough potassium bromide (0.3256 mol > 0.1962 mol), lead (II) nitrate is the limiting reactant.

The number of moles of lead (II) bromide produced will be equal to the number of moles of lead (II) nitrate consumed, which is 0.0981 mol.

The molar mass of lead (II) bromide is 367.01 g/mol, so the mass of lead (II) bromide produced will be 0.0981 mol * 367.01 g/mol = 36.0 g.

c) To determine the amount of excess reactant remaining, we need to subtract the amount consumed from the initial amount.

The number of moles of potassium bromide consumed is half the number of moles of lead (II) nitrate consumed, which is 0.0981 mol / 2 = 0.04905 mol.

The mass of potassium bromide consumed is 0.04905 mol * 119.00 g/mol = 5.84 g.

The mass of potassium bromide remaining is 38.75 g - 5.84 g = 32.91 g.

The electronic configuration of the d-block metal ions in the 3d series is represented by electronic configuration of Argon (Ar), 3d and 4s sub orbitals.

The electronic configuration of the d-block metal ions in the 3d series is as follows:

Scandium (Sc): [Ar] 3d¹ 4s²

Titanium (Ti): [Ar] 3d² 4s²

Vanadium (V): [Ar] 3d³ 4s²

Chromium (Cr): [Ar] 3d⁵ 4s¹

Manganese (Mn): [Ar] 3d⁵ 4s²

Iron (Fe): [Ar] 3d⁶ 4s²

Cobalt (Co): [Ar] 3d⁷ 4s²

Nickel (Ni): [Ar] 3d⁸ 4s²

Copper (Cu): [Ar] 3d¹⁰ 4s¹

Zinc (Zn): [Ar] 3d¹⁰ 4s²

Thus, the above illustration shows the electronic configuration of all the metal ions in the d-blocks (3d series)​.

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The final temperature of the copper and water after they come to thermal equilibrium is 109.8°C.

Temperature is a measure of the amount of thermal energy present in a substance or object. It is measured in degrees on a scale such as Fahrenheit, Celsius, or Kelvin. Temperature is important in determining the physical and chemical properties of a substance, such as its melting point, boiling point, and specific gravity. Temperature also affects the rate of a chemical reaction and the speed of diffusion.

The change in temperature of the copper can be calculated using the equation
ΔT = (Q/mc), where Q is the heat transferred, m is the mass of the copper, and c is the specific heat of copper.
Q = mcΔT = (192 g)(0.385 J/g °C)(100°C) = 74080 J
The heat transferred from the copper must equal the heat transferred to the water. Therefore,
(74080 J) = (0.85 L)(4.184 J/g°C)(ΔT)
ΔT = (74080 J)/[(0.85 L)(4.184 J/g°C)] = 109.8°C
The final temperature of the copper and water after they come to thermal equilibrium is 109.8°C.

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1. The volume of 0.350 M NH₄I solution is required is 286 mL

2. The mole of PbI₂ formed is 0.0498

First, we shall determine the mole in 415 mL of 0.120 M Pb(NO₃)₂. Details below:

Mole = molarity × volume

Mole of Pb(NO₃)₂ = 0.120 × 0.415

Mole of Pb(NO₃)₂ = 0.0498 mole

Next, we shall determine the mole of NH₄I that reacted. Details below:

Pb(NO₃)₂(aq) + 2NH₄I(aq) ⟶ PbI₂(s) + 2NH₄NO₃(aq)

From the balanced equation above,

1 moles of Pb(NO₃)₂ reacted with 2 moles of NH₄I

Therefore,

0.0498 mole of Pb(NO₃)₂ will react with = 0.0498 × 2 = 0.1 mole of NH₄I

Finally, we shall determine the volume of NH₄I required for the reaction. Details below:

Volume = mole / molarity

Volume of NH₄I = 0.1 / 0.350

Volume of NH₄I = 0.286 L

Multiply by 1000 to express in mL

Volume of NH₄I = 0.286  × 1000

Volume of NH₄I = 286 mL

The mole of PbI₂ formed can be obtain as follow:

Pb(NO₃)₂(aq) + 2NH₄I(aq) ⟶ PbI₂(s) + 2NH₄NO₃(aq)

From the balanced equation above,

1 mole of Pb(NO₃)₂ reacted to produced 1 mole of PbI₂

Therefore,

0.0498 mole of Pb(NO₃)₂ will also react to produce 0.0498 mole of PbI₂

Thus, the number of mole of PbI₂ formed is 0.0498 mole

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The volume of oxygen gas, O₂ collected at 1.05 atm pressure and 44.0 °C when 42.5 g of KClO₃ decomposed is 13.01 L

We shall begin by obtaining the mole in 42.5 g of KClO₃. Details below:

Mole = mass / molar mass

Mole of CaC₂ = 42.5 / 122.5

Mole of CaC₂ = 0.35 mole

Next, we shall determine the mole of oxygen gas, O₂. produced. Details below:

2KClO₃ -> 2KCl + 3O₂

From the balanced equation above,

2 moles of KClO₃ decomposed to produced 3 mole of O₂

Therefore,

0.35 mole of KClO₃ will decompose to produce = (0.35 × 3) / 2 = 0.525 mole O₂

Finally, we shall determine the volume of oxygen gas, O₂ collected. Details below:

PV = nRT

1.05 × V = 0.525 × 0.0821 × 317

Divide both sides by 1.05

V = (0.525 × 0.0821 × 317) / 1.05

Volume of oxygen gas = 13.01 L

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To determine the enthalpy of reaction for the given reaction, use Hess's Law, which states that the total enthalpy change of a reaction is independent of the pathway between the initial and final states.

Break the given reaction into a series of steps for which the enthalpy changes are known or can be measured experimentally.

Add the enthalpy changes of each step to determine the overall enthalpy change for the reaction.

For example, determine the enthalpy of formation for A₂X₄(l), X₂(g), and AX₃(g) and use them to calculate the enthalpy of reaction for the given equation.

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Answer: 4 moles i think this is right im not sure

1. To find the volume and units of a solution that contains 0.50 moles of NaOH dissolved in enough distilled water to make 3.0 M NaOH solution:

We first need to use the formula:

moles = concentration (in moles/L) x volume (in L)

Rearranging the formula to solve for volume, we get:

volume = moles / concentration

Substituting the given values, we get:

volume = 0.50 moles / 3.0 M = 0.17 L

Since the volume is given in liters, the units of the solution are L. Therefore, the solution contains 0.50 moles of NaOH dissolved in 0.17 L of distilled water, which makes a 3.0 M NaOH solution.

2. To find the molarity of a solution that contains 60.0 g of Ca(OH)2 dissolved in 150 mL of solution:

We first need to convert the mass of Ca(OH)2 to moles using the molar mass:

molar mass of Ca(OH)2 = 40.08 g/mol + 2 x 16.00 g/mol + 2 x 1.01 g/mol = 74.10 g/mol

moles of Ca(OH)2 = 60.0 g / 74.10 g/mol = 0.810 moles

Next, we need to convert the volume of the solution from milliliters to liters:

volume of solution = 150 mL / 1000 mL/L = 0.150 L

Finally, we can use the formula:

molarity = moles / volume

Substituting the given values, we get:

molarity = 0.810 moles / 0.150 L = 5.4 M

Therefore, the molarity of the solution is 5.4 M.

The amount of heat required to vaporize 181.20 g of ethanol would be 170.1 kJ.

Using the formula:

Q = n ∆Hvap

where:

Moles of 181.20 g of ethanol = 181.20 g / 46.07 g/mol = 3.933 mol

Substituting the values:

Q = 3.933 mol x 43.3 kJ/mol = 170.1 kJ

In other words, the amount of heat required to vaporize 181.20 g of ethanol is 170.1 kJ.

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