describe how these nutrients and elements such as carbon, oxygen, and nitrogen are cycled through ecosystems.

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:

[tex]-5.226\times 10^{-7} C[/tex] is the charge on each sphere.

Explanation:

Coulomb's law is given as ;

[tex]F=K\times \frac{q_1\times q_2}{r^2}[/tex]

[tex]q_1,q_2[/tex] = Charges on both charges

r = distance between the charges

K = Coulomb constant =[tex]9\times 10^{9} N m^2/C^2[/tex]

We have ;

Charge of ion =[tex]q_1=-q[/tex]

Charge of electron =[tex]q_2=-q[/tex]

[tex]r=26.55 cm =0.2655 m[/tex]

Force between the charges  at r distance will be :  F

F = 0.03500 N

[tex]0.03500 N=9\times 10^{9} N m^2/C^2\times \frac{(-q)\times (-q)}{(0.2655 m)^2}[/tex]

[tex]q=5.226\times 10^{-7} C[/tex]

[tex]-5.226\times 10^{-7} C[/tex] is the charge on each sphere.

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 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:

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.

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:

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:

1. 8.7moles of H2

2. 2.25moles of O2

Explanation:

1. 2NH3 —> N2 + 3H2

From the equation,

2moles of NH3 produce 3 moles of H2.

Therefore, 5.8moles of NH3 will produce Xmol of H2 i.e

Xmol of H2 = (5.8x3)/2 = 8.7moles

2. C3H8 + 5O2 —> 3CO2 + 4H2O

From the equation,

5moles of O2 produced 4moles of H2O.

Therefore, Xmol of O2 will produce 1.8mol of H2O i.e

Xmol of O2 = (5x1.8)/4 = 2.25moles

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!

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!

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:

option d

Explanation:

Molecular sizes of gaseous molecules are very less. Volume occupied by the all the molecules of the gases are very less or negligible as compared to the container in which it is kept. Therefore, most of the volume occupied by gaseous molecules are negligible.

Volume occupied by the gaseous molecules are actually the volume of the container and its does not depend upon the amount, molecular mass or dipole moment of the gaseous molecules.

Therefore, the correct option is d ‘Because most of the volume occupied by the substance is empty space.’

Answer:

weaker and longer

Explanation:

Since there are 3 bonds in ethyne in comparision with the 2 bonds of ethyne between carbon atoms, they are attracted more to each other → the bond gets shorter . And since there are one more bond that supports the union → the bond gets stronger

thus the carbon-carbon double bond in ethene is weaker and longer than the carbon-carbon triple bond in ethyne

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:

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:

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

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:

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:

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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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:

The answer is 18.12KJ is required to vaporise 48.7 g of dichloromethane at its boiling point

Explanation:

To solve the above question we have the given variable as follows

ΔHvap = heat of vaporisation of dichloromethane per mole = 31.6KJ/mole

However since the heat of vaporisation is the heat to vaporise one mole of dichloromethane, then, for 48.7 grams of dichloromethane, we have.

The number of moles of dichloromethane present = 48.7/84.93 = 0.573 moles

Therefore, the amount of heat required to vaporise 48.7 grams of dichloromethane at its boiling point is 31.6KJ/mole×0.573moles =18.12KJ

Answer:

[tex]\beta=63.85^{\circ}[/tex] from the vertical OR [tex]16.15^{\circ}[/tex] from horizontal.

Explanation:

Given:

Thrust of launching the rocket, [tex]F=5\times 10^6\ N[/tex]

angle of launch from the horizontal, [tex]\theta=37^{\circ}[/tex]

mass of the rocket, [tex]m=200000\ kg[/tex]

Now the direction of acceleration due to the thrust force is in the direction of force:

[tex]a=\frac{F}{m}[/tex]

[tex]a=\frac{5000000}{200000}[/tex]

[tex]a=25\ m.s^{-2}[/tex]

And the acceleration due to gravity is always directed towards the center of the earth i.e. vertically downwards.

Since acceleration is a vector quantity we, approach accordingly:

[tex]\tan\beta=\frac{a\cos\theta}{g}[/tex]

[tex]\tan\beta=\frac{25\cos37^{\circ}}{9.8}[/tex]

[tex]\beta=63.85^{\circ}[/tex] from the vertical OR [tex]16.15^{\circ}[/tex] from horizontal.

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 rate at which [tex]P_4[/tex] is being produced is 0.0228 M/s.

The rate at which [tex]PH_3[/tex] is being consumed is 0.0912 M/s.

Explanation:

[tex]4PH_3\rightarrow P_4(g)+6H_2(g)[/tex]

Rate of the reaction : R

[tex]R=\frac{-1}{4}\frac{d[PH_3]}{dt}=\frac{1}{6}\frac{d[H_2]}{dt}=\frac{1}{1}\frac{d[P_4]}{dt}[/tex]

The rate at which hydrogen is being formed = [tex]\frac{d[H_2]}{dt}=0.137 M/s[/tex]

[tex]R=\frac{1}{6}\frac{d[H_2]}{dt}[/tex]

[tex]R=\frac{1}{6}\times 0.137 M/s=0.0228 M/s[/tex]

The rate at which [tex]P_4[/tex] is being produced:

[tex]R=\frac{1}{1}\frac{d[P_4]}{dt}[/tex]

[tex]0.0228 M/s=\frac{1}{1}\frac{d[P_4]}{dt}[/tex]

The rate at which [tex]PH_3[/tex] is being consumed :

[tex]R=\frac{-1}{4}\frac{d[PH_3]}{dt}[/tex]

[tex]0.0228 M/s\times 4=\frac{-1}{1}\frac{d[PH_3]}{dt}[/tex]

[tex]\frac{-1}{1}\frac{d[PH_3]}{dt}=0.912 M/s[/tex]

The rate at which P4 is being produced is 0.034 M/s and the rate at which PH3 is being consumed is 0.2055 M/s.

The given reaction is 4PH3(g) → P4(g) + 6H2(g). We are given the rate at which molecular hydrogen is being formed, which is 0.137 M/s. To find the rate at which P4 is being produced, we need to consider the stoichiometry of the reaction. From the balanced equation, we can see that for every 4 moles of PH3 consumed, 1 mole of P4 is produced. Therefore, the rate at which P4 is being produced is 0.137/4 or 0.034 M/s.

Similarly, to find the rate at which PH3 is being consumed, we can use the stoichiometry of the reaction. From the balanced equation, we can see that for every 4 moles of PH3 consumed, 6 moles of H2 is produced. Therefore, the rate at which PH3 is being consumed is (6/4) * 0.137 or 0.2055 M/s.

Learn more about reaction rates here:

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Answer: The correct statement is, It behaves as the electron donor.

Explanation:

According to the Lewis concept:

A Lewis-acid is defined as a substance that accepts electron pairs.

A Lewis-base is defined as a substance which donates electron pairs.

For example : Acid + Base ⇄ Acid-base adduct

[tex]H^++NH_3\rightleftharpoons NH_4^+[/tex]

As per question, the characteristic of a Lewis base is that it behaves as the electron donor.

Hence, the correct statement is, It behaves as the electron donor.

Answer:

A

Explanation:

It behaves as the electron donor

Answer: carbonyl group C=O

Explanation:

Formaldehyde is an organic compound, it is the simplest form of Aldehydes. It formula is CH2O and has a carbonyl functional group, C=O. The general formula for adehydes is R-COH. The carbon atom is bonded to oxygen with a double bond and one of the two remaining bonds is occupied by hydrogen, and the other by an alkyl group.

One of the properties of adehydes is their solubility in water. The lower members (up to 4 carbons) of aldehydes are soluble in water due to H-bonding. Ofcourse the the higher members are not soluble in water because their hydrophobic long chains.

Aldehydes contain carbonyl group, therefore they undergo reactions like nucleophilic addition reactions, oxidation, reduction, halogenation.

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

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