Please provide a synthesis for the following molecules with the provided starting material.

Answer:

Adsorption is the adhesion of atoms, ions or molecules from a gas, liquid or dissolved solid to a surface. This process creates a film of the adsorbate on the surface of the adsorbent. This process differs from absorption, in which a fluid is dissolved by or permeates a liquid or solid.

while ,

Absorption is a physical or chemical phenomenon or a process in which atoms, molecules or ions enter some bulk phase – liquid or solid material. This is a different process from adsorption, since molecules undergoing absorption are taken up by the volume, not by the surface

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The law of conservation of mass say that, in a chemical reaction, the mass of the reagents will always be equal to the mass of the products.

If we have 29.5 g of HBr, we must have 100 g of H2 and Br2 together, so:

Answer:

Answer choices A, B, and C all contain a certain level (or amount, if you will) of Calcium Carbonate, but B. chalk contains the most amount out of all three.

Explanation:

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Answer: A. The particles move faster

Explanation:

Increasing the temperature increases the kinetic energy of the particles, causing them to move faster.

The ratio of diffusion rates of ¹H₂ to ²H₂ is 1.4 while the ratio of diffusion rates of O₂ to O₃ is 1.2.

Graham’s law states can be described as the rate of diffusion of a gas is inversely proportional to the square root of its molecular weight of the gas.

According to this Law, molecules with lower molecular mass will diffuse faster than those with higher molecular mass molecules at constant pressure and temperature.

The mathematical formula is utilized to compare the rates of diffusion of two different gases at pressures and temperatures.

[tex]{\displaystyle \frac{R_1}{R_2} =\sqrt{\frac{M_2}{M_1} }[/tex]

Where M₁ and M₂ represent the molar mass of gas 1 and gas 2.

Given, the molar mass of  ¹H₂  = 2.0 g/mol

The molar mass of to ²H₂ = 4.0 g/mol

The ratio of the rate of diffusion of ¹H₂ to ²H₂:

[tex]{\displaystyle \frac{R_{^1H_2}}{R_{^2H_2}} =\sqrt{ \frac{M_{^2H_2}}{M_{^1H_2}} }[/tex]

[tex]{\displaystyle \frac{R_{^1H_2}}{R_{^2H_2}} =\sqrt{ \frac{4}{2} }[/tex]

[tex]{\displaystyle \frac{R_{^1H_2}}{R_{^2H_2}} =1.4[/tex]

Given, the molar mass of  O₂  = 32.0 g/mol

The molar mass of to O₃ = 48.0 g/mol

The ratio of the rate of diffusion of O₂ to O₃:

[tex]{\displaystyle \frac{R_{O_2}}{R_{O_3}} =\sqrt{ \frac{M_{O_3}}{M_{O_2}} }[/tex]

[tex]{\displaystyle \frac{R_{O_2}}{R_{O_3}} =\sqrt{ \frac{48}{32} }[/tex]

[tex]{\displaystyle \frac{R_{O_2}}{R_{O_3}} =1.2[/tex]

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C. The number of moles of H in 0.109 mole of N₂H₄ is 0.436 mole

D. The number of moles of H in 34 moles of C₁₀H₂₂ is 748 moles

1 mole of N₂H₄ contains 4 moles of H

Therefore,

0.109 mole of N₂H₄ will contain = 0.109 × 4 = 0.436 mole of H

1 mole of C₁₀H₂₂ contains 22 moles of H

Therefore,

34 mole of C₁₀H₂₂ will contain = 34 × 22 = 748 mole of H

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The amount of water that will be produced is 50.36 grams

The metabolism of glucose is represented by the following equation:

[tex]C_6H_1_2O_6(s)+6O_2(g)--- > 6CO_2(g)+6H_2O(g)[/tex]

The mole ratio of glucose metabolized to the water produced is 1:6.

Mole of 84.0 g glucose = 84/180.156 = 0.4662 moles

Equivalent mole of water = 0.4662 x 6 = 2.7975 moles

Mass of 2.7975 moles water = 2.7975 x 18 = 50.36 grams

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20.24 grams of [tex]N_2O[/tex] is formed if 0.46 mole of [tex]NH_4NO_3[/tex] is used in the reaction.

A mole is defined as 6.02214076 × [tex]10^{23}[/tex]of some chemical unit, be it atoms, molecules, ions, or others. The mole is a convenient unit to use because of the great number of atoms, molecules, or others in any substance.

Balanced equation:

[tex]NH_4NO_3[/tex] → [tex]N_2O + 2H_2O[/tex].

1 mole [tex]NH_4NO_3[/tex] gives 1 mole of [tex]N_2O.[/tex]

0.46 mole of [tex]NH_4NO_3[/tex] gives [tex]\frac{ 1 mole}{ 1 mole}[/tex] x 0.46 mole [tex]N_2O[/tex].

0.46 mole [tex]N_2O[/tex] =[tex]\frac{mass}{44.013 g/mol}[/tex]

Mass= 0.46 x 44.013

Mass=20.24gram.

Hence, 20.24 grams of [tex]N_2O[/tex] is formed if 0.46 mole of [tex]NH_4NO_3[/tex] is used in the reaction.

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Each section would be 230 centimeters or 2.30 meters long.

The original wood pole is 12 meters long.

12 meters = 12 x 100 = 1200 centimeters

It is to be cut into 5 equal sections: 1200/5 = 240 centimeters

But each cut wastes 10 cm of the pole.

          240 - 10 = 230 centimeters.

Thus, each section would be 230 centimeters or 2.30 meters long.

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Answer:

100 KJ

Explanation:

yep yep cool is that correct

Wastewater can be treated using series of eight steps which would be clarified below.

Wastewater is the type of water that is used for four different purposes in homes,offices, industries and factories.

These wastewater contain human waste, food scraps, oils, soaps and chemicals which can be treated and reused through the following means:

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Thorium 234 is the daughter product of Uranium-238 decays whereas protactinium 234 is the daughter product of Thorium -228.

If an atom of Uranium-238 decays via alpha emission, a nucleus of uranium 238 decays by alpha emission to form a daughter nucleus, thorium 234 while If an atom of Thorium -228 decayed via beta emission, the daughter isotope is protactinium 234.

So we can conclude that thorium 234 is the daughter product of Uranium-238 decays whereas protactinium 234 is the daughter product of Thorium -228.

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Answer:

The bright side of the moon will become less and less and will eventually be covered in complete darkness :)

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Answer:

electrolysis is the process by which electric current is passed through a substance to effect a chemical change

The volume of [tex]Ba(OH)_2[/tex] required will be 83.33 mL

The equation of the reaction is as follows:

[tex]Ba(OH)_2 + 2HBr --- > 2H_2O + BaBr_2[/tex]

Mole ratio of reactants = 1:2

Mole of 50 mL, 0.05 M HBr = 0.05 x 0.05 = 0.0025 moles

Equivalent mole of  [tex]Ba(OH)_2[/tex] = 0.0025/2 = 0.00125 moles

Volume of 0.015 M, 0.00125 moles  [tex]Ba(OH)_2[/tex] = 0.00125/0.015 = 0.0833 L

0.0833 L = 83.33 mL

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The molecular mass of a 2.66g sample of gas at a temperature of pressure of 845torr and occupying a volume of 0.70L is 84.16g/mol.

The molecular mass can be calculated by first calculating the number of moles using the following formula:

PV = nRT

Where;

1.112 × 0.7 = n × 0.0821 × 300

0.7784 = 24.63n

n = 0.032mol

molecular mass = 2.66g ÷ 0.032mol

molecular mass = 84.16g/mol

Therefore, the molecular mass of a 2.66g sample of gas at a temperature of pressure of 845torr and occupying a volume of 0.70L is 84.16g/mol.

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Answer:

24.0 g C₃H₈

Explanation:

To find the mass of C₃H₈, you need to (1) convert grams CO/H₂ to moles CO/H₂ (via molar mass), then (2) convert moles CO/H₂ to moles C₃H₈ (via mole-to-mole ratio from reaction coefficients), and then (3) convert moles C₃H₈ to grams C₃H₈ (via molar mass). It is important to arrange the conversions/ratios in a way that allows for the cancellation of units (the desired unit should be in the numerator). The final answer should have 3 sig figs to reflect the sig figs in the given values.

Molar Mass (CO): 12.011 g/mol + 15.998 g/mol

Molar Mass (CO): 28.009 g/mol

Molar Mass (H₂): 2(1.008 g/mol)

Molar Mass (H₂): 2.016 g/mol

Molar Mass (C₃H₈): 3(12.011 g/mol) + 8(1.008 g/mol)

Molar Mass (C₃H₈): 44.097 g/mol

3 CO + 7 H₂ ----> 1 C₃H₈ + 3 H₂O
^           ^              ^

45.8 g CO           1 mole              1 mole C₃H₈          44.097 g
-----------------  x  ------------------  x  --------------------  x  ------------------  =
                           28.009 g           3 moles CO            1 mole

=  24.0 g C₃H₈

87.3 g H₂           1 mole           1 mole C₃H₈          44.097 g
----------------  x  ---------------  x  ---------------------  x  -----------------  =  
                          2.016 g           7 moles H₂             1 mole

=  273 g C₃H₈

It was necessary to find the mass of the products from both of the reactants because you did not know which one was the limiting reagent. The limiting reagent is the reactant which is completely used up first. Because CO produced the smaller amount of product, it must be the limiting reagent. Therefore, the actual amount of C₃H₈ produced is 24.0 grams.

The amount of Al required will be 15.77 grams

First, the equation of the reaction:

[tex]2Al +3 FeO -- > Al_2O_3 + 3Fe[/tex]

The mole ratio is 2:3.

Mole of 63.0 g of FeO = 63/71.84 = 0.8769 moles

Equivalent moles of Al = 0.8769 x 2/3 = 0.5846 moles

Mass of 0.5846 moles Al = 0.5846 x 26.98 = 15.77 grams

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Taking into account the reaction stoichiometry, 13.28 grams of Al is required to react with 63 grams of FeO.

In first place, the balanced reaction is:

2 Al + 3 FeO → 3 Fe + Al₂O₃

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

The molar mass of the compounds is:

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

The following rule of three can be applied: If by reaction stoichiometry 215.55 grams of FeO react with 54 grams of Al, 53 grams of FeO react with how much mass of Al?

[tex]mass of Al= \frac{53 grams of FeOx54 grams of Al}{215.55 grams of FeO}[/tex]

mass of Al= 13.28 grams

Finally, 13.28 grams of Al is required to react with 63 grams of FeO.

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A cofactor is an organic or inorganic molecule or ion necessary for proper functioning of some biological catalysts (option E).

A cofactor in biochemistry is a molecule that binds to and regulates the activity of a protein.

A cofactor is a metal or coenzyme responsible for the functioning of an enzyme and must be present.

Therefore, a cofactor is an organic or inorganic molecule or ion necessary for proper functioning of some biological catalysts.

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[tex]K_2CO_3 + CaBr_2[/tex] → [tex]KBr + CaCO_3[/tex]  is the net ionic equation for this reaction.

An ionic equation is a synthetic equation where electrolytes are composed as separated particles.

[tex]K_2CO_3 + CaBr_2[/tex] → [tex]KBr + CaCO_3[/tex] is the net ionic equation for this reaction.

A balanced ionic equation. shows the reacting ions. in a chemical reaction. These equations can be used to represent what happens in precipitation reactions.

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Answer:

C- both!

Explanation:

In a chemical reaction, the total mass and total charge has to be balanced on both sides of the "equation", therefore both are conserved

Answer:

Technology is the application of scientific knowledge for practical purposes especially in industry.

Considering the combined law, the volume of the sample is 81.639 L at 423 K and 1.00 atm.

Charles's law establishes the relationship between the temperature and the volume of a gas when the pressure is constant. This law says that the volume is directly proportional to the temperature of the gas: if the temperature increases, the volume of the gas increases while if the temperature of the gas decreases, the volume decreases.

Mathematically, Charles's law states:

[tex]\frac{V}{T}=k[/tex]

Boyle's law states that the pressure of a gas in a closed container is inversely proportional to the volume of the container, when the temperature is constant. That is, if the pressure increases, the volume decreases while if the pressure decreases, the volume increases.

Mathematically, this law is expressed as:

P×V= k

Gay-Lussac's law establishes the relationship between the temperature and pressure of a gas when the volume is constant. This law says that the pressure of a gas is directly proportional to its temperature: if the temperature increases, the pressure will increase, while if the temperature decreases, the pressure will decrease.

Mathematically, Gay-Lussac's law states:

[tex]\frac{P}{T}=k[/tex]

Combined law equation is the combination of three gas laws called Boyle's, Charlie's and Gay-Lusac's law:

[tex]\frac{VxP}{T}=k[/tex]

Considering an initial state 1 and a final state 2, it is fulfilled:

[tex]\frac{V1xP1}{T1}=\frac{V2xP2}{T2}[/tex]

In this case, you know:

Replacing in combined law equation:

[tex]\frac{57.9 Lx1 atm}{300 K}=\frac{V2x1 atm}{423 K}[/tex]

Solving:

[tex]V2= \frac{423 K}{1 atm} \frac{57.9 Lx1 atm}{300 K}[/tex]

V2= 81.639 L

Finally, the volume of the sample is 81.639 L at 423 K and 1.00 atm.

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The number of molecules will depend on the molar mass of the compound.

According to Avogadro, 1 mole of any substance contains 6.022 x [tex]10^{23}[/tex].

To find the number of moles in a substance:

Mole = mass/molar mass.

In this case, we do not know the compound but we know the mass. Hence, we cannot know the molar mass.

Assuming the molar mass is X.

Mole = 84.7/X

Number of molecules in 84.7 grams of the compound =

              84.7/X x 6.022 x [tex]10^{23}[/tex] molecules.

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Whether a living or non-living object, everyone has a particular life cycle. It starts right from the birth and ends upon the death of that object. Just like others, stars also have their own life cycle and that has different stages.

Nebula ⇒ Protostar ⇒ Supergiant ⇒ Supernova ⇒ Neutron Star

In the initial stage, a star is in the form of a nebula. At this stage, it is a huge cloud of gas and dust. The changes occur in the star and the final size totally depends upon the size of that star.

In a massive star, the fuel is used very quickly because of its life span of a few hundred thousand years. The second stage is the Protostar.

The next stage is the Supergiant. At this stage, the nuclear reactions that occur because of the hydrogen will gradually run out and the star will enter the final phase of its life span. The bigger star will have more massiveness.

Following this, the star will enter the next stage, i.e., supernova, and after that, neutron star. In a supernova, the star will have a very energetic and violent end. In the last stage, when the dust particles clear, a neutron star will be visible.

For better understanding, an image is attached at the end.

Thus, the life cycle of a high-mass star is arranged from its birth to its death.

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The stages of a high-mass star's life cycle in the correct order, from a star's birth to its death are-

Nebula

Protostar

Supergiant

Supernova

Neutron star

The life cycle of a high-mass star begins with a nebula, a cloud of gas and dust in space. As gravity pulls the nebula together, it forms a protostar—a young, hot, and dense object. The protostar continues to gather mass and becomes a main-sequence star, burning hydrogen in its core.

Once the hydrogen is exhausted, the star expands into a supergiant and starts fusing helium into heavier elements. Eventually, it undergoes a supernova explosion, releasing an enormous amount of energy and leaving behind a dense core known as a neutron star. This process showcases the incredible transformations that high-mass stars undergo throughout their life cycle.

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Based on the electron configuration of the two atoms predict the ratio of metal cationic (+) atom to nonmetal anionic (-) atom in the compound hence the ratio is 1:1.

The Octet Rule states that the atoms of the elements bond to each other in an attempt to complete their valence shell (the last shell of the electrosphere).

Basically, they obey the octet rule which states that the capacity of atoms (e.g. here Magnesium and Sulfur) to prefer to have 8 (eight) electrons (e-) in their respective valence shells. In this case,  Magnesium is Mg (+2) and Sulfur is (-2) hence the ratio is 1:1.

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The molecular formula of fructose is C₆H₁₂O₆.

The molecular formula is derived in the following steps:

Step 1. the molality of the solution is determined first

Molality = Change in boiling point/Kb

Molality = (78.59 - 78.35)/1.2 = 0.2 molal

Step 2. Moles of solute is determined.

Moles of solute = molality * mass of solvent (kg)

moles = 0.2 * 0.325 = 0.065

Step 3. Molar mass of fructose = mass/moles

Molar mass of fructose = 11.7/0.065 = 180 g/mol

Step 4: Mole ratio of constituent elements

Carbon = 40% * 180/12 = 6

Hydrogen = 6.7% * 180/1 = 12

Oxygen = 53.3% * 180/16 = 6

Therefore, the molecular formula of fructose is C₆H₁₂O₆.

In conclusion, the molecular formula of fructose is derived from the molality of the solution.

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The mass in kg of the water on the roof after the rainstorm is mathematically given as

The mass of the water on the roof after a rainstorm is 1.1 *10^{10}g  or 1.1*10^{7}kg

Generally, the equation for the Area of the roof is  mathematically given as

[tex]A=0.20 \times 10^{10} \mathrm{~cm}^{2}[/tex]

height of the rain =5.5cm

the volume of the rain on the roof =1.1 * 10^{10} CC

Generally, the equation for mass is  mathematically given as

mass=volume*density

[tex]&=\left(1.1 \times 10^{10} \mathrm{cC} \times 1 \mathrm{~g} / \mathrm{cc}\right) \\&=1.1 \times 10^{10} \mathrm{~g}[/tex]

In conclusion, the Mass of the water on the roof after a rainstorm is 1.1 *10^{10}  or $1.1*10^{7}

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