Question 7 of 10
Which variable is unknown until the experiment is performed?
O A. A responding variable
OB. A mathematical variable
OC. A controlled variable
OD. A manipulated variable
SUBI

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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Interpreting Data:

Predicting:

Joshua's brother has a 50% chance of being a carrier for cystic fibrosis (Nn) and a 50% chance of having the normal genotype (NN). This is because Joshua's parents are both carriers, so each of their children has a 50% chance of inheriting the recessive allele.

Communicating:

As a genetic counselor, information about many generations of a family is needed to draw conclusions about a hereditary condition because traits are passed down from generation to generation. By studying the family history, we can identify patterns of inheritance and determine the likelihood that a person has inherited a certain trait. This information can help us make informed decisions about genetic testing and treatment options.

Additionally, knowing the family history can help us determine the risk of passing on a genetic condition to future generations.

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Image transcribed and complete question:

Family Puzzle

Problem

A husband and wife want to understand the probability that their children might inherit cystic fibrosis. How can you use the information in the labeled Case Study to predict the probability?

Skills Focus

interpreting data, predicting

Materials

12 index cards

scissors marker

Procedure

1. Read the Case Study. In your notebook, draw a pedigree that shows all the family members. Use circles to represent the females, and squares to represent the males. Shade in the circles or squares representing the individuals who have cystic fibrosis.

2. You know that cystic fibrosis is controlled by a recessive allele. To help you figure out Joshua and Bella's family pattern, create a set of cards to represent the alleles. Cut each of six index cards into four smaller cards. On 12 of the small cards, write N to represent the dominant normal allele. On the other 12 small cards, write n for the recessive allele.

Case Study:

Joshua and Bella

• Joshua and Bella have a son named lan. lan has been diagnosed with cystic fibrosis.

⚫ Joshua and Bella are both healthy.

• Bella's parents are both healthy.

• Joshua's parents are both healthy. • Joshua's sister, Sara, has cystic fibrosis.

Analyze and Conclude

Write your answers in the spaces provided.

1. Interpreting Data What were the genotypes of Joshua's parents? What were the genotypes of Bella's parents?

2. Predicting Joshua also has a brother. What is the probability that he has cystic fibrosis? Explain.

3. Communicating Imagine that you are a genetic counselor. A couple asks why you need information about many generations of their families to draw conclusions about a hereditary condition. Write an explanation you can give to them.

More to Explore

Review the pedigree that you just studied. What data suggest that the traits are not sex-linked? Explain.

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 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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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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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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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: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.

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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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 molality of the solution is 0.0256 m.

The mole fraction of glycerol is  0.00046

The percent by mass concentration of glycerol is 0.23%

The ppm concentration is 2300 ppm

Molality is a measure of the concentration of a solute in a solution. It is defined as the number of moles of solute per kilogram of solvent.

The formula for molality is:

molality = moles of solute / mass of solvent in kilograms

1) Density of water = mass/volume

Mass of water = Density * volume of water

Mass =[tex]0.9982 g/mL * 998.9 mL[/tex]

Mass =0.997 Kg of water

Number of moles of the glycerol =  [tex]2.550* 10^-2 M * 1 L[/tex]

= [tex]2.550*10^-2[/tex] moles

Molality of the solution = [tex]2.550*10^-2[/tex]  moles/0.997 Kg

= 0.0256 m

Number of moles of water = 998.9/18 g/mol

= 55.5 mole

Mole fraction of glycerol = [tex]2.550*10^-2[/tex] /[tex]2.550*10^-2[/tex]  + 55.5

= 0.00046

By percent by mass;

2.3/1001.2 * 100/1

= 0.23%

Mass of glycerol = 2.3 g

Volume of solution = 1 L

Thus we have concentration in ppm as;

[tex]2.3 * 10^3[/tex] mg/ 1 L =2300 ppm

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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 mass of sodium sulfate [tex](Na_2SO_4)[/tex] formed by the complete reaction of 137.3 grams of sodium hydroxide [tex](NaOH)[/tex] is 486.74 grams.

The balanced chemical equation for the reaction between sodium hydroxide [tex](NaOH)[/tex] and sulfuric acid [tex](H_2SO_4)[/tex] is:

[tex]NaOH[/tex] + [tex](H_2SO_4)[/tex] → [tex]Na_2SO_4[/tex] + [tex]2H_2O[/tex]

From the balanced equation, we can see that 1 mole of [tex]NaOH[/tex] reacts with 1 mole of [tex](H_2SO_4)[/tex] to produce 1 mole of [tex]Na_2SO_4[/tex] and 2 moles of water [tex](H_2O).[/tex]

The molar mass of [tex]NaOH[/tex] is 40.00 g/mol, which means that 137.3 grams of [tex]NaOH[/tex] is equal to 137.3 g / 40.00 g/mol = 3.4325 moles of [tex]NaOH[/tex].

Since 1 mole of [tex]NaOH[/tex] reacts with 1 mole of [tex](H_2SO_4)[/tex] to produce 1 mole of [tex](Na_2SO_4)[/tex] , we can say that 3.4325 moles of [tex]NaOH[/tex] will react with 3.4325 moles of [tex]H_2SO_4[/tex] to produce 3.4325 moles of [tex]Na_2SO_4[/tex].

The molar mass of [tex](Na_2SO_4)[/tex] is 142.04 g/mol, which means that 1 mole of [tex](Na_2SO_4)[/tex] has a mass of 142.04 g.

Therefore, 3.4325 moles of [tex](Na_2SO_4)[/tex] has a mass of 3.4325 moles x 142.04 g/mol = 486.74 grams.

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A potential problem associated with travel to another planet is : the extended time for humans to be in space.

It is highly likely that humans will travel to another planet, and Mars is currently considered the most viable option for human exploration. However, there are many potential problems associated with this endeavor, and one of the major issues is the extended time that humans would need to spend in space.

Traveling to Mars would take several months, and once there, astronauts would need to spend significant amount of time on planet before returning to Earth. This means that they would be exposed to high levels of radiation and would need to find ways to survive in harsh and unforgiving environment.

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At equivalence point, the reaction is seen to consume approximately 0.0024973 moles of KHP and then 0.0024973 moles of  NaOH

The primary aim in this problem is to standardize a solution of the sodium hydroxide, NaOH, with the aid of potassium hydrogen phthalate, KHP.

The beginning point in this problem is the balanced chemical equation with respect to this neutralization reaction

KHP (aq] + NaOH(aq] → KNaP(aq] + H₂O(l]

The important thing that we are going to observe is that there is a ratio of 1:1 mole ratio between the two reactants. This suggests to us that the equivalence point can be attained by getting equal number of moles of  KHP and of NaOH to react with each other.

We will begin with 0.5100 g of KHP. To obtain the molar amount of acid utilized for the experiment, we will make use of its molar mass of 0.5100g⋅

molar mass of KHP

1 mole KHP 204.22g = 0.0024973 moles KHP

Thus, at equivalence point, the reaction is seen to consume approximately 0.0024973 moles of KHP and then 0.0024973 moles of  NaOH, due to the fact that it's what the  1:1 mole ratio suggests to us.

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

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

Surface area = 2lw + 2lh + 2wh

Substituting the given values, we get:

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

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

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

1 m² / 10,000 cm²

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

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

Surface area = 0.034 m²

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

The minimum concentration of fluoride ions needed is 2.726 x 10⁻⁴ M.

To find the minimum concentration of fluoride ions needed to precipitate CaF₂, we'll use the solubility product constant (Ksp) and the calcium ion concentration.

Ksp = [Ca²⁺][F⁻]²

Given: [Ca²⁺] = 5.25 x 10⁻³ M, Ksp = 3.9 x 10⁻¹¹

3.9 x 10⁻¹¹ = (5.25 x 10⁻³)[F⁻]²

Solve for [F⁻]:

[F⁻]² = (3.9 x 10⁻¹¹) / (5.25 x 10⁻³)

[F⁻]² = 7.4286 x 10⁻⁹

[F⁻] = 2.726 x 10⁻⁴ M

The minimum concentration of fluoride ions needed is 2.726 x 10⁻⁴ M.

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PbS, sometimes referred to as galena, is made up of lead (Pb) and sulfur (S) atoms and has a straightforward structure. Each lead atom forms a tetrahedral link with four sulfur atoms, while each sulfur atom forms a covalent bond with two lead atoms.

Each sulfur atom forms an octahedral link with six iron atoms, while each iron atom forms a covalent bond with two sulfur atoms. FeS2's crystal structure is a cubic, tightly packed lattice.

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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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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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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 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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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

Answer:

Cvapor = 2.03 J/(g·°C)heu

The solution has a total volume of 300 mL and is composed of 100 mL of ethanol (the solute) and 200 mL of water (the solvent).

Less than 100 ml will result from mixing 50 ml each of ethanol and water in an equal ratio. This happens because ethanol molecules, which are smaller than those of water, may fit inside big water molecules. As a result, the alcohol content in a 250 mL mix of water and alcohol is 60%.

The combined volumes of the ethanol and water make up the total volume of the solution, which is:

Total volume = 100 mL + 200 mL = 300 mL

Therefore, the solution is made up of 100 mL of ethanol (the solute) and 200 mL of water (the solvent), with a total volume of 300 mL.

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

Answer: They have eight outer electrons.

Explanation: They don't need any more electrons to be added to them and can't give out any electrons to other groups. They have a complete outer shell.

Answer:

as the frequency goes down, the speed goes down by the same factor, and so the wavelength doesn't change.

Explanation:

Answer: as the frequency goes down, the speed goes down by the same factor, and so the wavelength doesn't change.

Explanation: