On a small farm, the weights of eggs that young hens lay are normally distributed with a mean weight of 51.3 grams and a standard deviation of 4.8 grams. Using the 68-95-99.7 rule, about what percent of eggs weigh between 46.5g and 65.7g.

Answer : The concentration of the NaOH solution is, 0.738 M

Explanation :

To calculate the concentration of base, we use the equation given by neutralization reaction:

[tex]n_1M_1V_1=n_2M_2V_2[/tex]

where,

[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]H_2SO_4[/tex]

[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is NaOH.

We are given:

[tex]n_1=2\\M_1=0.300M\\V_1=24.6mL\\n_2=1\\M_2=?\\V_2=20.0mL[/tex]

Putting values in above equation, we get:

[tex]2\times 0.300M\times 24.6mL=1\times M_2\times 20.0mL[/tex]

[tex]M_2=0.738M[/tex]

Thus, the concentration of the NaOH solution is, 0.738 M

Answer:

The metallic bond of the sodium metal is weaker than the ionic bond of the sodium chloride

Explanation:

The metallic bond of the sodium metal is weaker than the ionic bond of the sodium chloride. This is seen in the fact that the melting point of the sodium metal which is 97.8°C is lower than the melting point of the sodium chloride which is 801°C. This shows that the inter-atomic bonds of the sodium metal are weaker than that of the inter-ionic bonds of the sodium chloride which will require higher energy to break which is shown by its high melting point of 801°C. The melting point of the sodium metal of 97.8 °C which is lower requires less energy to break its bonds. This is shown by its lower melting point.

Answer:

a. Molecular formula of phenylalanine → C₉H₁₁NO₂

b. 234.2 g of C₉H₁₁NO₂

Explanation:

Let's think the reaction:

C₁₄H₁₈N₂O₅  +  2H₂O →  CH₃OH  +  ‎C₉H₁₁NO₂ + C₄H₇NO₄

In order to work with this reaction, we assume that water is on excess.

Let's determine the moles of aspartame → molar mass = 294 g/mol

Moles = mass / molar mass → 378 g / 294 g/mol = 1.28 moles

As ratio is 1:1, 1 mol of aspartame can produce 1 mol of phenylalanine; therefore 1.28 moles of aspartame must produce 1.28 moles of phenylalanine.

Let's convert the moles to mass → molar mass C₉H₁₁NO₂ = 183 g/mol

Moles . molar mass = Mass → 1.28 mol . 183 g/mol = 234.2 g

A. The molecular formula of phenylalanine is C₉H₁₁NO₂

B. The mass of phenylalanine, C₉H₁₁NO₂ produced from 378 g of aspartame is 212.14 g

A. Determination of the molecular formula of phenylalanine.

To obtain the molecular formula of phenylalanine, we shall write the balanced equation. This is given below

C₁₄H₁₈N₂O₅ + 2H₂O → C₄H₇NO₄ + CH₃OH + C₉H₁₁NO₂

Thus, the molecular formula of phenylalanine is C₉H₁₁NO₂

B. Determination of the mass of phenylalanine, C₉H₁₁NO₂ produced from 378 g of aspartame

C₁₄H₁₈N₂O₅ + 2H₂O → C₄H₇NO₄ + CH₃OH + C₉H₁₁NO₂

Molar mass of C₁₄H₁₈N₂O₅ = (14×12) + (18×1) + (14×2) + (16×5) = 294 g/mol

Mass of C₁₄H₁₈N₂O₅ from the balanced equation = 1 × 294 = 294 g

Molar mass of C₉H₁₁NO₂ = (9×12) + (11×1) + 14 + (16×2) = 165 g/mol

Mass of C₉H₁₁NO₂ from the balanced equation = 1 × 165 = 165 g

From the balanced equation above,

294 g of C₁₄H₁₈N₂O₅ reacted to produce 165 g of C₉H₁₁NO₂

Therefore,

378 g of C₁₄H₁₈N₂O₅ will react to produce = (378 × 165) / 294 = 212.14 g of C₉H₁₁NO₂

Thus, 212.14 g of C₉H₁₁NO₂ were obtained from the reaction.

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Answer:

Option a.

0.01 mol of CaCl₂ will have the greatest effect on the colligative properties, because it has the biggest i

Explanation:

To determine which of the solute is going to have a greatest effect on colligative properties we have to consider the Van't Hoff factor (i)

These are the colligative properties:

ΔP = P° . Xm . i  →  Lowering vapor pressure

ΔT = Kb . m . i  → Boiling point elevation

ΔT = Kf . m . i  →  Freezing point depression

π = M . R . T  →  Osmotic pressure

Van't Hoff factor are the numbers of ions dissolved in the solution. For nonelectrolytes, the i values 1.

CaCl₂ and KNO₃ are two ionic solutes. They dissociate as this:

CaCl₂  → Ca²⁺ + 2Cl⁻

We have 1 mol of Ca²⁺ and 2 chlorides, so 3 moles of ions → i = 3

KNO₃ → K⁺ + NO₃⁻

We have 1 mol of K⁺ and 1 mol of nitrate, so 2 moles of ions → i = 2

Option a, is the best.

0.01 mol of CaCl2 has the greatest effect on colligative properties because it dissociates into three ions per formula unit, resulting in a higher number of dissolved particles compared to the same amount of KNO3, which dissociates into two ions, and CO(NH2)2, which does not dissociate.

The solute that has the greatest effect on the colligative properties for a given mass of pure water from the options provided would be 0.01 mol of CaCl2 (an electrolyte). This is because the colligative properties such as freezing point depression, boiling point elevation, and osmotic pressure depend on the total number of dissolved particles in solution.

When CaCl2 dissolves in water, it dissociates into three ions: one Ca2+ ion and two Cl- ions. This means for every mole of CaCl2 we get three moles of particles. In contrast, 0.01 mol of KNO3, another electrolyte, produces two ions per formula unit, and 0.01 mol of CO(NH2)2 (urea, a nonelectrolyte) does not dissociate into ions when dissolved, hence producing only one mole of particles per mole of solute.

Therefore, for the given mass of 0.01 mol, CaCl2 produces the greatest number of particles and therefore has the greatest effect on colligative properties when compared to KNO3 and urea.

Answer:

It'll be lower.

Explanation:

Water is a universal solvent, which means it can dissolve virtually anything except oils and non-polar substance because of its polarity and ability to form hydrogen bond .Water molecules are also attracted to other polar molecules and to ions. A charged or polar substance that interacts with and dissolves in water is said to be hydrophilic: hydro means "water," and philic means "loving." In contrast, nonpolar molecules like oils and fats do not interact well with water

Answer:

Vital

Explanation:

Hello! The factors mentioned previously are essential substances for the life and health of our bodies. Not only in humans, but also for animals. They are factors that are linked and contribute to physical and mental energy, growth and life. Are essentials because humans can't live without that.

Thanks for your question! Feel free to ask more!

Calcium carbonate

Explanation:

The chemical most commonly used to counteract the effects of acid precipitation on aquatic ecosystems is calcium carbonate.

Acid rain or acid deposition or acid precipitation is any form of precipitation with an elevated level of hydrogen ion concentration in them.

To nullify this acidic precipitation in aquatic ecosystem, we need to use an environmentally friendly alkaline agent.

The most desired is calcium carbonate. The carbonate neutralizes the acid by producing carbon dioxide, water and calcium salts.

Answer:

The answer to your question is None of your answers is correct, maybe the data are wrong.

Explanation:

Data

Concentration 1 = C1 = 1 M

Volume 2 = 5 ml

Concentration 2 = 0.05 M

Volume 1 = x

To solve this problem use the dilution formula

             Concentration 1 x Volume 1 = Concentration 2 x Volume 2

Solve for Volume 1

              Volume 1 = (Concentration 2 x Volume 2)/ Concentration 1

Substitution

              Volume 1 = (0.05 x 5) / 1

Simplification

              Volume 1 = 0.25/1

Result

               Volume 1 = 0.25 ml

Answer:

Explanation:

You must find how many grams of a 14.0% solution contains 35.0 g of thiophene (solute) and then evaluate if the amount available (1.0 kg) is enough.

Formula:

Substitute and solve for the mass of solution:

        [tex]14.0 =(35.0g/x)\times 100\\\\14.0/100=35.0g/x\\\\x=35.0g\times100/14\\\\x=250g[/tex]

Hence, the student should use 250g of the 14.0% w/w solution of thiophene in ethanol. Since, 1.0 kg is 1,000g there is enough available.

Final answer:

To acquire 35.0 grams of thiophene from a 14.0% w/w thiophene solution, the student should measure out 250 grams of the solution. The 1.0 kg of available solution is more than sufficient for the student's needs.

Explanation:

To calculate the mass of the thiophene solution that the student needs, we first need to understand the percentage concentration of the solution. A 14.0% w/w solution of thiophene in ethanol means that there are 14 grams of thiophene for every 100 grams of solution. To find out how many grams of solution contain the required 35.0 grams of thiophene, we use the following formula:

Mass of thiophene = (Percentage/100) × Total mass of solution

Therefore, we can rearrange this to solve for the total mass of solution:

Total mass of solution = Mass of thiophene / (Percentage/100)

Total mass of solution = 35.0 g / (14.0/100) = 250 g

The student should use 250 grams of the solution to obtain 35.0 grams of thiophene. Since the available solution is 1.0 kg, which is 1000 grams, there is enough solution for the experiment.

Answer:

The sun is the large astronomical body that is located at the center of the solar system. The interior structure of the sun are as follows-

Answer:

The corrext answer is E. make; break

Explanation:

In living organisms, the metabolism is either anabolic or catabolic where anabolic metabolism is energy consuming and catabolic metabolism is eneegy releasesing. It should however be noted that anabolic reaction builds or biosynthesize new mollecular structures while catabolic reaction breaks down complex structure bonds into simple structures

The braking down of bonds in catabolic reations realeses energy to sustain the anabolic rection process for the formation of new bonds

Anabolic reactions create or 'make' new bonds while forming complex structures, requiring energy. Catabolic reactions involve 'break' down bonds to create simpler structures, often releasing energy.

The correct answer is E. Anabolic reactions are those that 'make' or form new bonds, they involve the joining of smaller molecules to form larger, more complex ones. This process often requires energy. Conversely, catabolic reactions are those that 'break' or degrade bonds, they involve the breaking down of larger molecules into smaller, simpler ones. This process often releases energy.

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Answer:62.66°C or 235.66K

Explanation:Q=McpT, the energy was given in calories so you first convert to Joules by multiplying the value in calories by 4.184J.

17*4.184=71.128kJ.

71.128kJ=mcpT

71.128kJ=245*4.187*(T-Tm)

Tm is the final temperature of the mixture. The T is the temperature given which should be converted to Kelvin by adding 273...T=32+273=305K.

71128J=245*4.187*(305-Tm)

71128=312873.575-1025.815Tm

1025.815Tm=312873.575-71128

1025.815Tm=241745.58

Tm=241745.58/1025.815

Tm=235.66K

Final answer:

Approximately 34.97 kilocalories of heat energy were absorbed by the water in the velomobile seats.

Explanation:

In this scenario, we need to calculate the amount of heat energy absorbed by the water in the velomobile seats. To do this, we can use the formula:

Where:

Let's calculate the heat energy absorbed:

Converting this to kilocalories:

Therefore, approximately 34.97 kilocalories of heat energy were absorbed by the water in the velomobile seats.

Answer:

a) C₈H₁₈ + (23/2)O₂ -----> 8CO₂ + 9H₂O

b) Mass of CO₂ produced annually from this combustion of isooctane gasoline = 1.12 × 10⁵ Kg

c) CO₂ produced from the combustion of the gasoline in a year will occupy 5.632 × 10⁷ L

d) There needs to be a minimum of 1.752 × 10⁷ moles of air and 3.92 × 10⁸ L of air for the oxygen to be in excess all through the year of gasoline combustion.

Explanation:

a) C₈H₁₈ + (23/2)O₂ -----> 8CO₂ + 9H₂O

b) C₈H₁₈ has a density of 0.792 mg/L.

Since density = mass/volume;

mass = density × volume

Mass of C₈H₁₈ with 4.6 x 10^10 L volume = 0.792 × 4.6 x 10^10 = 3.643 × 10^10 mg = 3.643 × 10⁷ g.

To obtain the mass of CO₂ produced, we need the number of moles of C₈H₁₈ that burned.

Number of moles = mass/molar mass

Molar mass of C₈H₁₈ = (8×12) + 18 = 114g/mol

Number of moles of C₈H₁₈ = (3.643 × 10⁷)/114 = (3.2 × 10⁵) moles.

From the chemical reaction,

1 mole of C₈H₁₈ burns to give 8 moles of CO₂

(3.2 × 10⁵) moles will give 8 × 3.2 × 10⁵ = (2.56 × 10⁶) moles of CO₂

Mass of CO₂ produced = number of moles × Molar mass

Molar mass of CO₂ = 44 g/mol

Mass of CO₂ produced = 2.56 × 10⁶ × 44 = 1.12 × 10⁸ g = 1.12 × 10⁵ kg

c) 1 mole of any gas at stp occupies 22.4L

2.56 × 10⁶ moles of CO₂ will occupy 2.56 × 10⁶ × 22.4 = 5.632 × 10⁷ L

d) 1 mole of C₈H₁₈ requires 23/2 moles of O₂ for complete combustion yearly.

3.2 × 10⁵ moles would require 3.2 × 10⁵ × 23/2 = 3.68 × 10⁶ moles of O₂

O₂ makes up 21% of the air

That is,

0.21 moles of O₂ would be contained in 1 mole of air

3.68 × 10⁶ moles of O₂ would be contained in (3.68 × 10⁶ × 1)/0.21 = 1.752 × 10⁷ moles of air.

1 mole of any gas at stp occupies 22.4L

1.752 × 10⁷ of air will occupy

1.752 × 10⁷ × 22.4/1 = 3.92 × 10⁸ L of air!

Final answer:

The balanced equation for the combustion of isooctane is C8H18(l) + 11O2(g) → 8CO2(g) + 9H2O(g). Assuming that gasoline is 100% isooctane, the mass of CO2 produced each year by the annual US gasoline consumption of 4.6 x 10^10 L is 9.04 x 10^6 kg. The volume of this CO2 at STP is 2.03 x 10^11 L. We use the molar ratios from the balanced equation to determine the number of moles of air necessary for the combustion of 1 mol of isooctane. The volume in L of this air at STP is 1173 L.

Explanation:

The balanced chemical equation for the combustion of isooctane (C8H18) is:

C8H18(l) + 11O2(g) → 8CO2(g) + 9H2O(g)

Assuming that gasoline is 100% isooctane, we can use the balanced equation to calculate the mass of CO2 produced each year by the annual US gasoline consumption of 4.6 x 1010 L. We start by calculating the mass of isooctane consumed, then use the molar ratios from the balanced equation to calculate the mass of CO2 produced. Given that the density of isooctane is 0.792 g/mL, the mass consumed is 4.6 x 1010 L x 0.792 g/mL = 3.64 x 1010 g. The molar mass of CO2 is 44.01 g/mol, so the number of moles of CO2 produced is 28 x 3.64 x 1010 g / 114.22 g/mol = 9.04 x 109 mol. Finally, we convert to kilograms by dividing by 1000, so the CO2 produced is 9.04 x 106 kg.

To calculate the volume of CO2 at STP, we use the ideal gas law. At STP (Standard Temperature and Pressure), 1 mole of gas occupies 22.4 L. So, the volume of CO2 at STP is 9.04 x 109 mol x 22.4 L/mol = 2.03 x 1011 L.

We use the molar ratios from the balanced equation to determine the number of moles of air necessary for the combustion of 1 mol of isooctane. From the equation, we can see that 11 moles of O2 are required for the combustion of 1 mol of isooctane. Since air is 21.0% O2 by volume, the volume of air required is 11 / 0.21 = 52.4 mol. At STP, 1 mole of gas occupies 22.4 L, so the volume of air required is 52.4 mol x 22.4 L/mol = 1173 L.

Answer: 0.75 moles of Phosphorus and 4.5 moles of Hydrogen are produced respectively from 3 moles of Phosporus.

Explanation:

4PH3 --------> P4 + 6H2

From the stochiometry of the reaction,

4 moles of Phosphine gives 1 mole of Phosphorus

3 moles of Phosphine will give (3×1)/4 moles of Phosphorus.

Therefore, 0.75 moles of Phosphorus is produced.

Similarly, 4 moles of Phosphine gives 6 moles of Hydrogen

3 moles of Phosphine will give (3×6)/4 moles of Hydrogen.

Therefore, 4.5 moles of Hydrogen is produced.

QED!

Equation of reaction

NaHCO3 +  HCl  --------->  NaCl  +  H2O  +  CO2(g)

1)The mass(actual yield) of CO2 can be gotten by isolating it from other products and getting its mass.

Assume 1.0g of CO2 was gotten as a product of the experiment

2) For the theoretical yield

The mass of NaHCO3 was not stated, but for the purpose of this solution, I would assume 2g of NaHCO3 was used for the experiment. You can substitute any parameter by following the steps I follow

Number of mole of NaHCO3 = Mass of NaHCO3/ Molar Mass of NaHCO3

Number of mole of NaHCO3 = 2/84.01 = 0.0238 moles

1 mole of NaHCO3 yielded 1 mole of CO2

0.0238 moles of NaHCO3 will yield 0.0238 moles of CO2

Mass of CO2 = Number of moles * Molar Mass

Mass of CO2 = 0.0238 * 44 = 1.0472g

Theoretical yield of CO2 = 1.0472 grams

3) Percentage yield of CO2 = Actual yield/Theoretical yield *100%

Percentage yield of CO2 = 1.0/1.0472 *100

Percentage yield of CO2 = 95.49%

Answer:

1) The mass of [tex]CO_{2}[/tex] produced is 1.29g

2) The theoretical yield for [tex]CO_{2}[/tex] is 1.311g

3) The percent yield is 98.4%

Explanation:

Step 1: in an experiment let's allow sodium bicarbonate (baking soda) to react with hydrochloric acid HCl to obtain high yield [tex]CO_{2}[/tex].

[tex]NaHCO_{3}+HCl[/tex] ⇒ [tex]NaCl + H_{2} O + CO_{2}[/tex]

from the balanced equation it can be seen that 1 mole of [tex]NaHCO_{3[/tex] produces 1 mole of [tex]CO_{2}[/tex].

Step 2: Let assume 2.5g of [tex]NaHCO_{3[/tex] were used

mole of [tex]NaHCO_{3[/tex] = 2.5g/molecular mass of [tex]NaHCO_{3[/tex]

molecular mass of[tex]NaHCO_{3[/tex] = 23+1+12+(16×2) = 84g/mol

converting mass of [tex]NaHCO_{3[/tex] into mole we have:

           [tex]\frac{2.5g}{84gmol^{-1} }[/tex]= 0.0298 moles

Step 3: since [tex]NaHCO_{3[/tex] : [tex]CO_{2}[/tex] mole is ratio 1:1, then 0.0298 mole of [tex]NaHCO_{3[/tex] will produce 0.0298 mole of [tex]CO_{2}[/tex]

The mass of this amount of [tex]CO_{2}[/tex] = 0.0298 × molecular mass of [tex]CO_{2}[/tex]

                = 0.0298 mol × ( 12+(16×2))g/mol = 1.3111g of [tex]CO_{2}[/tex]

 therefore; the theoretical yield expected is 1.311g

Step 4: Let assume for the experiment that we obtained 1.29g  of [tex]CO_{2}[/tex] as the actual yield

then, percent yield for [tex]CO_{2}[/tex] = [tex]\frac{Actual yield obtained}{theoretical yield that should have been obtained}[/tex] × 100%

                                             = 1.2g/1.311g × 100% =98.4%

Answer:

Option (4)

Explanation:

The primitive atmosphere was comprised of lesser or no amount of oxygen. With the progressive time, primitive organisms appeared such as the primitive aquatic plants, algae, and cyanobacteria, that carried out the process of photosynthesis in the presence of sunlight, water, and CO₂, and in return produced the food and oxygen. This is how the rapid photosynthesis process led to the increasing amount of oxygen in the atmosphere that facilitates the growth and evolution of different life forms on earth.

Thus, the correct answer is option (4).

Answer:

Archaebacteria belong to the kingdom 'Archaea'. They resemble bacteria when observed under a microscope.They are single-cell organisms. However,they differ from prokaryotes in many ways.

•They have a completely different cell membrane so that they can survive and thrive in harsh environments

•Unlike bacteria,their cell wall and membrane can be stiff and gives a specific shape such as flat,rod shaped or cubic.

•the absence of peptidoglycan in cell wall,instead contain lipid and protein to give them strength and resistance against chemical.

•The cell membrane of archae are made up of phospholipid bilayer but unlike bacteria and eukaryotes,the bilayers have ether bonds.These ether bonds have the ability to resist chemical and helps them to survive in very extreme environments that might otherwise kill them.

•They have multiple RNA polymerase that contain multiple polypeptide

•They have initiator tRNA with unmodified methionine(unlike bacteria).

•They have the ability to survive in very high temperatures,in acidic environment and alkaline environments and even have tolerance for high salt content.

Answer:

C. Running bond

Explanation:

The common bond is a structural bonding pattern for bricks with a single wythe.

The structural bonding pattern for bricks that features a single wythe is the common bond.

Answer:

Full disk encryption

Explanation:

Disk encryption is a protection technique used in securing information by converting it into unreadable codes that cannot decrypt in other to prevent unauthorized persons from accessing the information.

When cryptography is applied to entire disks, it is termed Full disk encryption.

Answer:

the electric field at Z = 12 cm is E =   9.68 × 10³ N/C = 9.68 kN/C

Explanation:

Given: radius of disk, R = 2.0 cm = 2 × 10⁻² cm, surface charge density,σ = 6.3 μC/m² = 6.3 × 10⁻⁶ C/m², distance on central axis, z = 12 cm = 12 × 10⁻² cm.

The electric field, E at a point on the central axis of a charged disk is given by E = σ/ε₀([tex]1 - \frac{z}{\sqrt{z^{2} + R^{2} } }[/tex])

Substituting the values into the equation, it becomes

E = σ/ε₀([tex]1 - \frac{z}{\sqrt{z^{2} + R^{2} } }[/tex]) = 6.3 × 10⁻⁶/8.854 × 10⁻¹²([tex]1 - \frac{0.12}{\sqrt{0.12^{2} + 0.02^{2} } }[/tex]) = 7.12 × 10⁵([tex]1 - \frac{0.12}{0.1216}[/tex]) = 7.12 × 10⁵(1 - 0.9864) = 7.12 × 10⁵ × 0.0136 = 0.0968 × 10⁵ = 9.68 × 10³ N/C = 9.68 kN/C

Therefore, the electric field at Z = 12 cm is E =   9.68 × 10³ N/C = 9.68 kN/C

Answer:FeS

Explanation:

Mass of sulphur=5.95-3.78=2.17g

Fe S

3.78/56. 2.17/32

0.0675/0.0675. 0.0678/0.0675

1:1

Empirical formula=Fe1S1=FeS

An empirical formula is a compound's chemical formula that only specifies the ratios of the elements it contains and not the precise number or arrangement of atoms. The empirical formula is FeS.

The formula of a material expressed with the smallest integer subscript is referred to as an empirical formula for a compound. The empirical formula provides details regarding the ratio of atom counts in the molecule. A compound's empirical formula is directly related to its % content.

Mass of sulphur = 5.95 - 3.78 = 2.17g

Moles of Fe and S are:

Fe = 3.78/56

n = 0.0675

S = 2.17/32

n = 0.0678

Dividing with the smallest number:

0.0675/0.0675 = 1

0.0678/0.0675 = 11

The ratio is:

1:1

Empirical formula = FeS

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Answer:

[tex]\rho=7.15\ g/cm^3[/tex]

Explanation:

The expression for density is:

[tex]\rho=\frac {Z\times M}{N_a\times {{(Edge\ length)}^3}}[/tex]

[tex]N_a=6.023\times 10^{23}\ {mol}^{-1}[/tex]

M is molar mass of Chromium = 51.9961 g/mol

For body-centered cubic unit cell , Z= 2

[tex]\rho[/tex] is the density  

Radius = 125 pm = [tex]1.25\times 10^{-8}\ cm[/tex]

Also, for BCC, [tex]Edge\ length=\frac{4}{\sqrt{3}}\times radius=\frac{4}{\sqrt{3}}\times 1.25\times 10^{-8}\ cm=2.89\times 10^{-8}\ cm[/tex]

Thus,  

[tex]\rho=\frac{2\times \:51.9961}{6.023\times \:10^{23}\times \left(2.89\times 10^{-8}\right)^3}\ g/cm^3[/tex]

[tex]\rho=7.15\ g/cm^3[/tex]

The density of solid crystalline chromium will be "7.15 g/cm³".

According to the question,

Radius,

Molar mass of Chromium,

For body-centered,

For BCC,

The edge length will be:

= [tex]\frac{4}{\sqrt{3} }\times radius[/tex]

= [tex]\frac{4}{\sqrt{3} }\times 1.25\times 10^{-8}[/tex]

= [tex]2.89\times 10^{-8} \ cm[/tex]

hence

The expression for density will be:

→  [tex]\rho = \frac{Z\times M}{N_a(Edge \ length)^3}[/tex]

By substituting the values, we get

      [tex]= \frac{2\times 51.9961}{6.023\times 10^{23}\times (2.89\times 10^{-8})}[/tex]

      [tex]= 7.15 \ g/cm^3[/tex]

Thus the above approach is right.

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Answer:

10mol

Explanation:

3H2 + N2 -> 2NH3

Stoichiometry is a tool that chemists can use to find the amount of substance present in any part of a reaction. The arrow (->) suggests that the reaction goes to completion (100%), so assume that left side = right side.

3H2

15 mol

You can divide the amount of moles by the coefficient to find the number of moles when you have a coefficient of 1. This number can then be used to find the value of moles for the rest of the products/reactants:

15/3=5mol

NH3 has a coefficient of 2, so we have to multiply the value we got (5mol) by 2. This results in having 10mol of ammonia as the end result.

Final answer:

The production of ammonia from hydrogen and excess nitrogen follows a 3:2 mole ratio, according to the balanced equation N2(g) + 3H2(g) → 2NH3(g). From 15 moles of hydrogen, 10 moles of ammonia are produced.

Explanation:

The question asks about the production of ammonia (NH3) from hydrogen (H2) in the presence of excess nitrogen (N2) based on the chemical reaction provided. According to the stoichiometry of the balanced equation, N2(g) + 3H2(g) → 2NH3(g), there is a 3:2 mole ratio between hydrogen and ammonia. Therefore, for every 3 moles of hydrogen, 2 moles of ammonia are produced.

To calculate the amount of ammonia produced from 15 mol of hydrogen, we use the mole ratio from the balanced equation. Since 3 moles of hydrogen produce 2 moles of ammonia, the amount of ammonia produced from 15 moles of hydrogen can be found using cross-multiplication:

(2 mol NH3) / (3 mol H2) = (x mol NH3) / (15 mol H2)

x = (15 mol H2 × 2 mol NH3) / 3 mol H2

x = 10 mol NH3

The answer is that 10 moles of ammonia can be produced from 15 moles of hydrogen reacting with excess nitrogen.

Answer:

0.1976 atm is the total pressure in the container at equilibrium.

Explanation:

The value of [tex]K_p[/tex] for the given reaction = 0.842

Partial pressure of the hydrogen sulfide = 0.104 atm

Partial pressure of the hydrogen sulfide at equilibrium = (0.104-x) atm

Partial pressure of the hydrogen gas at equilibrium = x

Partial pressure of the sulfur gas at equilibrium = x

[tex]H_2S(g)\rightleftharpoons 2H_2(g)+S(g)[/tex]

Initially

0.104 atm                      0  0

At equilibrium

(0.104 -2x)                      2x   x

The expression of  [tex]K_p[/tex] is given by ;

[tex]K_p=\frac{x\times x}{(0.104 -x) }[/tex]

[tex]0.842 =\frac{x^2}{(0.104-x)}[/tex]

Solving for x:

x = 0.0936 atm

Partial pressure of the hydrogen sulfide at equilibrium = (0.104-x) atm

= (0.104-0.0936 ) atm = 0.0104 atm

Partial pressure of the hydrogen gas at equilibrium = x = 0.0936 atm

Partial pressure of the sulfur gas at equilibrium = x = 0.0936 atm

Total pressure in the container will be sum of all the partial pressure of the hydrogen sulfide gas, hydrogen gas and sulfur gas at the equilibrium.

P = 0.0104 atm + 0.0936 atm + 0.0936 atm = 0.1683 atm

0.1976 atm is the total pressure in the container at equilibrium.

The total pressure in the system at equilibrium is calculated using the equilibrium constant Kp and the initial pressure of the system. By setting up a equilibrium constant expression and solving the quadratic equation for 'x', the change in pressure, we find that the total pressure at equilibrium in the system is ~0.101 atm.

The decomposition reaction of H2S into H2 and S is a chemical equilibrium problem in which the equilibrium constant is given in terms of pressure, or Kp. We know that at the beginning, only H2S is present at a pressure of 0.104 atm, and at equilibrium, the pressure of H2S decreases by 'x', being 'x' the pressure of H2 and S at equilibrium. According to the expression of the equilibrium constant Kp, Kp = (pH2* pS) / p_H2S = x^2 / (0.104-x).

From here, if we solve for 'x' using the quadratic formula and the given Kp = 0.842, we find two solutions, ~0.0967 atm and ~0.0072 atm. We discard the result 0.0967 because it cannot be higher than the initial pressure of the system, so the correct 'x' is ~0.0072 atm. Therefore, the total pressure will be the sum of the individual pressures at equilibrium. Total Pressure = pH2 + pS + p_H2S = 2x + (0.104-x) = 2*0.0072 + (0.104-0.0072) = ~0.101 atm.

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Answer:

130 g of O₂

Explanation:

Let's apply the Ideal Gases Law, to solve this.

Pressure . volume = moles . R . T° (K)

First of all, we need to convert the volume in mL to L, because the units of R

10000 mL . 1L/1000mL = 10L

Let's replace → 10 atm . 10L = n . 0.082L.atm/mol.K . 300K

(100 atm.L) / (0.082L.atm/mol.K . 300K) = n

4.06 moles = n

Let's convert the moles to mass → 4.06 mol . 32 g/1mol = 130 g

Answer: The molar composition of nitrogen gas is 0.6 and that of oxygen gas is 0.4

Explanation:

To calculate the molarity of solution, we use the equation:

[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}[/tex]

Given mass of nitrogen gas = 51.3 mg = 0.0513 g     (Conversion factor:  1 g = 1000 mg)

Molar mass of nitrogen gas = 28 g/mol

Volume of solution = 2 L

Putting values in above equation, we get:

[tex]\text{Molarity of nitrogen gas}=\frac{0.0513g}{28g/mol\times 2L}\\\\\text{Molarity of nitrogen gas}=9.16\times 10^{-4}mol/L[/tex]

To calculate the partial pressure, we use the equation given by Henry's law, which is:

[tex]C_{N_2}=K_H\times p_{N_2}[/tex]

where,

[tex]K_H[/tex] = Henry's constant = [tex]6.1\times 10^{-4}mol/L.atm[/tex]

[tex]C_{N_2}[/tex] = molar solubility of nitrogen gas = [tex]9.16\times 10^{-4}mol/L[/tex]

Putting values in above equation, we get:

[tex]9.16\times 10^{-4}mol/L=6.1\times 10^{-4}mol/L.atm\times p_{N_2}\\\\p_{N_2}=\frac{9.16\times 10^{-4}mol/L}{6.1\times 10^{-4}mol/L.atm}=1.50atm[/tex]

We are given:

Total pressure of the mixture = 2.50 atm

Partial pressure of oxygen gas = 2.50 - 1.50 = 1.00 atm

To calculate the mole fraction of a substance at 25°C, we use the equation given by Raoult's law, which is:

[tex]p_{A}=p_T\times \chi_{A}[/tex]       ......(1)

where,

[tex]p_A[/tex] = partial pressure

[tex]p_T[/tex] = total pressure

[tex]\chi_A[/tex] = mole fraction

We are given:

[tex]p_{N_2}=1.50atm\\p_T=2.50atm[/tex]

Putting values in equation 1, we get:

[tex]1.50atm=2.50\times \chi_{N_2}\\\\\chi_{N_2}=\frac{1.50}{2.50}=0.6[/tex]

We are given:

[tex]p_{O_2}=1.00atm\\p_T=2.50atm[/tex]

Putting values in equation 1, we get:

[tex]1.00atm=2.50\times \chi_{O_2}\\\\\chi_{O_2}=\frac{1.00}{2.50}=0.4[/tex]

Hence, the molar composition of nitrogen gas is 0.6 and that of oxygen gas is 0.4

The mole fraction of nitrogen and oxygen is 0.6 and 0.4 respectively.

Data Given;

The law states that the mass of a dissolved gas in a given volume of solvent at equilibrium will be proportional to the partial pressure of the gas.

Mathematically;

C = KP

The number of moles Nitrogen dissolved is

[tex]n = mass/ molar mass\\ n = 0.0513/28\\ n = 0.00183127 moles[/tex]

The concentration of Nitrogen in water is

[tex]\frac{0.00183127}{2} *1 = 0.0009156M[/tex]

Applying Henry's law,

[tex]0.0009156 = 6.1*10^-^4 * P\\ P = 1.5atm[/tex]

The partial pressure of nitrogen in the mixture is 1.5atm

The total pressure of the gas is 2.50atm

Partial Pressure of oxygen = total pressure - partial pressure of nitrogen

partial pressure of oxygen = 2.50 - 1.50 = 1

The pressure fraction of the gas is the ratio between the partial pressure to the total pressure

pressure fraction of nitrogen = 1.5/2.5 = 0.6

pressure fraction of oxygen = 1/2.5 = 0.4

But partial pressure is equal to molar fraction.

This makes the mole fraction of nitrogen equals 0.6 and mole fraction of oxygen equals 0.4.

Learn more about Henry's law here;

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Answer:

Unstable Nuclei. ... Too many neutrons or protons upset this balance disrupting the binding energy from the strong nuclear forces making the nucleus unstable. An unstable nucleus tries to achieve a balanced state by given off a neutron or proton and this is done via radioactive decay.

Explanation:

The volume change during Iron's BCC to FCC transformation is computed via cubing the atomic radii and applying them to a percent change formula. The increasing radius hints at a likely volume increase, however, the exact percent change requires numerical calculation.

To answer your question on whether the volume increases or decreases during the allotropic transformation of Iron (Fe) from a BCC phase (with atomic radius rBCC = 0.12584 nm) to an FCC phase (rFCC = 0.12894 nm), we need to recognize that the volume of an atom in a crystal structure can be determined by cubing its atomic radius, and the volume change can be obtained by taking the difference between the initial and final volumes.

The initial volume (of the BCC phase) is (0.12584 nm)³, while the final volume (of the FCC phase) is (0.12894 nm)³. The percent volume change can then be computed by [(final volume - initial volume)/initial volume] x 100%.

If this calculation yields a positive value, this would mean the volume increases; if the result is a negative value, the volume decreases. While it's evident that the radius increases during this transformation, due to the cubic relationship between radius and volume, it's likely that the volume also increases, although the actual percent volume change would require numerical computation.

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Final answer:

The percent volume change during the allotropic transformation of iron from BCC to FCC can be calculated using the given atomic radii and results in an increase in volume.

Explanation:

To compute the percent volume change associated with the allotropic transformation of iron (Fe) from a BCC (body-centered cubic) structure to an FCC (face-centered cubic) structure at 912°C, we can use their respective atomic radii and the equation for the volume of a sphere, V = (4/3)πr3. Given the atomic radii for BCC as rBCC = 0.12584 nm and for FCC as rFCC = 0.12894 nm, the respective volumes can be calculated.

After obtaining the volumes, we calculate the percent volume change with the formula: Volume Change (%) = (VFCC - VBCC)/VBCC x 100%. Using the given radii, we find that the volume of FCC iron is larger than BCC iron, indicating that the volume increases during the transformation. The actual numerical percent volume change can be computed using the given radii values and the volume equation stated above.

Answer:

The retrieved bullet is normally marked at the tip or base with the initials of the investigator while ensuring that no markings are placed on the sides of the retrieved bullet. It is required to ensure that the markings made on the bullets does not go over or obscure striations or markings already present on the bullet.

Where the retrieved bullet is mutilated whereby it is impossible to engrave the required markings on it, it should be placed in a marked envelope, container or pill box

Explanation:

It is possible to trace retrieved casings and bullets from crime scene back to the gun from which it was fired or to the suspect's gun. The retrieved casings and bullet, when scrutinized at the crime lab, can reveal the gun model and make from which the casing or bullet was fired. The retrieved bullet or casing could also be traced back to the lot or batch of ammunition in possession of the suspect.

When a bullet is retrieved for identification purposes, it is marked with a unique identification number and other relevant information. Care must be taken to avoid damaging the bullet's markings and altering its shape or surface. Proper documentation and careful handling are crucial for accurate identification of the bullet.

When a bullet is retrieved for identification purposes, it is marked using a variety of methods. One common method is to assign a  unique identification number to the bullet, typically by inscribing or engraving it on the base or side of the bullet. This number can be used to match the bullet to a specific firearm. Additionally, the bullet may be photographed and cataloged, and any unique characteristics such as rifling marks or imperfections can be documented and compared to the firearm that fired it.

When marking a bullet for identification, it is important to avoid damaging the bullet's crucial markings or altering its shape in any way that could affect its comparison to a firearm. The use of permanent markers or corrosive substances should be avoided, as they can damage the bullet surface. Careful handling and proper documentation are critical to preserving the integrity of the bullet and ensuring accurate identification.

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