At a particular temperature 8.0 mol NO2 gas is placed in a 1.0-L container. Over time the NO2 decomposes to NO and O2: 2NO2(g) 2NO(g) O2(g) At equilibrium the concentration of NO(g) was found to be 2.9 mol/L. Calculate the value of K for this reaction (using units of mol/L for the concentrations).

Answer:

glucose

Explanation:

Starch -

Starch is generated by the plants , and is a granular , white , organic compound .

The general formula of the starch is (C₆H₁₀O₅)ₙ , the n shows it to be a polymer , and it is composed of small monomers of glucose , that are linked  at the alpha 1 , 4 linkages.

Starch is is colorless , and tasteless powder.

The most simplest form of starch is the linear polymer amylose , and the branched one amylopectin.

Answer:

80 liters

Explanation:

At STP, 1 mole of ideal gas has a volume of 22.4 liters.

Therefore, since liters and moles are directly proportional, we can use stoichiometry directly.

40L CH₄ × (2L H₂ / 1L CH₄) = 80L H₂

Answer:

Question 1

B) Volume decreases

Question 2

A) Higher pressure

Question 3

A and B

Explanation:

Question 1

B) Volume decreases

Charles' Law: The Temperature-Volume Charles Law. It states that at constant pressure, the volume of a given amount of gas is directly proportional to temperature in Kelvin. As the temperature goes down, the volume also goes down, and vice-versa.

ogadro's law states that "equal volumes of all gases, at the same temperature and pressure, have the same number of molecules." For a given mass of an ideal gas, the volume and amount (moles) of

Question 2

A) Higher pressure

Avogadro's law states that "equal volumes of all gases, at the same temperature and pressure, have the same number of molecules." For a given mass of an ideal gas, the amount (moles) and the volume of the gas are directly proportional at constant temperature and pressure

Increasing the number of moles increases the volume to maintain the same volume of the container with less moles the gas has to be compressed, increasing its pressure

Boyle's law is a gas law, that states that at constant temperature, the pressure and volume of a gas are inversely proportional. increasing the volume , decreases the pressure and vice versa.

Question 3

A and B

A) Doubling pressure while keeping temperature constant. B) Doubling absolute temperature while keeping pressure constant.

The volume of a given mass of gas decreases as pressure increases but increases as absolute temperature increases.

In the first question, a helium balloon initially at room temperature is dunked into a bucket of liquid nitrogen, the volume of the immediately decreases because volume and absolute temperature are directly proportional according to Charles law.

In the second question, two separate containers A and B have the same volume and temperature, but container A has more gaseous molecules than B, then container A will have higher pressure because the pressure of a gas is directly proportional to the number of molecules of a gas present.

In the third question, the change that would cause the volume of a gas to double, assuming moles were held constant is doubling absolute temperature while keeping pressure constant. This follows from Charles law where the volume of a given mass of gas is directly proportional to its volume at constant pressure.

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A barium fluoride solution with a concentration of 1.32 grams per liter, equivalent to its solubility, would be considered saturated. This means that no more barium fluoride can be dissolved in the water at the same temperature.

The solubility of barium fluoride in water is 1.32 grams per liter. Therefore, if a barium fluoride solution had a concentration of 1.32 grams per liter, it would be referred to as being saturated. A saturated solution is a solution in which no more solute can be dissolved in the solvent at a given temperature. In this case, the maximum amount of barium fluoride that can be dissolved in water at the given temperature has been reached.

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Answer:

False.

Explanation:

The given statement is false because for hot vacuum filtration, the filter paper should be wet rather than dry when pouring the hot solution into the Buchner funnel. This is because The possible explanation the filter paper needs to be wetted is not only to allow it to adhere to the funnel, but also to promote the solute to filter down readily across its pores of the paper without wetting it (this is true for organic and aqueous solvents).

Option C, which involves adding 0.1M NaOH to quench unreacted anhydride, then adding diethyl ether and separating the layers, is the extraction procedure that will completely separate an amide from the by-product of the reaction between an amine and excess carboxylic acid anhydride.

The extraction procedure that will completely separate an amide from the by-product of the reaction between an amine and excess carboxylic acid anhydride is option C. First, you would add 0.1M NaOH (aq) to quench unreacted anhydride. Then, you would add diethyl ether and separate the layers. The amide can be obtained from the aqueous layer by neutralizing with HCl (aq).

Answer:

[tex][Fe^{+3}]=0.700 M[/tex]

[tex][NO_{3}^{-}]=2.10 M[/tex]

Explanation:

Here, a solution of Fe(NO₃)₃ is diluted, as the total volume of the solution has increased. The formula for dilution of the compound is mathematically expressed as:

[tex]C_{1}. V_{1}= C_{2}.V_{2}[/tex]

Here, C and V are the concentration and volume respectively. The numbers at the subscript denote the initial and final values. The concentration of Fe(NO₃)₃ is 1.75 M. As ferric nitrate dissociates completely in water, the initial concentration of ferric is also 1.75 M.

Solving for [Fe],

[tex][Fe^{+3}]=\frac{C_{1}.V_{1}}{V_{2} }[/tex]

[tex][Fe^{+3}]=\frac{(1.75).(30.0)}{45.0+30.0 }[/tex]

[tex][Fe^{+3}]=0.700 M[/tex]

For [NO₃⁻],

There are three moles of nitrate is 1 mole of Fe(NO₃)₃. This means that the initial concentration of nitrate ions will be three times the concentration of ferric nitrate i.e., it will be 5.25 M.

[tex][NO_{3}^{-}]=\frac{C_{1}.V_{1}}{V_{2} }[/tex]

[tex][NO_{3}^{-}]=\frac{(5.25)(30.0)}{30.0+45.0 }[/tex]

[tex][NO_{3}^{-}]=2.10 M[/tex]

Answer:

Option 3.

364.1 g of CO₂ are produced by the reaction

Explanation:

Let's verify the balance equation for this reaction:

Fe₃O₄  +   4 CO  →  3 Fe   +   4CO₂

Let's convert the mass of Fe₃O₄ to moles (mass / molar mass)

478.9 g / 231.55 g/mol  = 2.07 moles

Ratio is 1:4, so let's make a rule of three to determine the moles of CO₂ produced and then its mass.

1 mol of Fe₃O₄ is needed to produce 4 moles of CO₂

Then, 2.07 moles of Fe₃O₄ will produce (2.07  .4) /1 = 8.27 moles of CO₂.

Molar mass . moles = mass

44 g/mol . 8.27 mol = 364 g of CO₂

Answer:

The answer to your question is below

Explanation:

Propane is a hydrocarbon that has only single bonds in its structure so it is an alkane.

Propane formula = C₃H₈

Oxygen formula = O₂

Carbon dioxide formula = CO₂

Water formula = H₂O

Reaction:

                         C₃H₈  +   5O₂   ⇒    3CO₂   +   4H₂O

                  Reactants            Elements     Products

                          3                          C                    3

                          8                          H                    8

                         10                         O                   10

Answer:

The answer to your question is 4.07 x 10²² atoms

Explanation:

Process

1.- Get the molecular weight of Cocaine hydrochloride

C = 12 x 17 = 204 g

H = 22 x 1 = 22 g

Cl = 36 x 1 = 36 g

N = 14 x 1 = 14 g

O = 16 x 4 = 64 g

Molecular mass = 340 g

2.- Calculate the moles of the mass given

                             340 g -------------------  1 mol

                               23 g -------------------  x

                              x = (23 x 1) / 340

                              x = 0.068 moles of Cocaine

3.- Calculate the atoms

                            1 mol -------------------- 6 .023 x 10 ²³ atoms

                   0.068 moles -------------  x

                              x = (0.068 x 6.023 x 10²³) / 1

                              x = 4.07 x 10²² atoms

Answer:

The answer to your question is the coefficients are 2, 1, 1, 2

Explanation:

Chemical Reaction                            

                              HCl  +   Na₂CO₃   ⇒    H₂CO₃   +   NaCl

                       Reactants            Elements             Products

                               1                         Cl                          1

                               2                         Na                        1

                                1                         C                          1

                                1                         H                         2

                                3                        O                          3

This reaction is unbalanced

                            2 HCl  +   Na₂CO₃   ⇒    H₂CO₃   +   2 NaCl

                       Reactants            Elements             Products

                               1                         Cl                          1

                               2                         Na                        2

                               2                         C                          2

                               2                         H                          2

                               3                        O                          3

Now, the reaction is balanced.

Answer:

The coefficients are: 2,1,1,2 (Option 3)

Explanation:

Step 1: Unbalanced equation

HCl + Na2CO3 → H2CO3 + NaCl

Step 2 : Balancing the equation

On the right side we have 2x H (in H2CO3), on the left side we have 1x H (in HCl). To balance the amount of H, we have to multiply HCl, on the left side, by 2.

2 HCl  + Na2CO3  →  H2CO3 + NaCl

On the left side we have 2x Na (in Na2CO3), on the right side, we have 1x Na (in NaCl). To balance the amount of Na, we have to multiply NaCl, on the right side, by 2. Now the equation is balanced.

2 HCl  + Na2CO3  →  H2CO3 + 2NaCl

The coefficients are: 2,1,1,2 (Option 3)

Answer:

The answer to your question is 0.3 g of H₂

Explanation:

Data

N₂ (g) = 1.4 g

H₂ (g) = ?

Balanced Reaction

                            N₂(g)  + 3H₂ (g)   ⇒ 2NH₃ (g)

Process

1.- Calculate the atomic mass of nitrogen and hydrogen.

N₂  = 14 x 2 = 28 g

H₂ = 1 x 6 = 6 g

2.- Use proportions and cross multiplication to solve it

                             28 g of N₂  ------------------- 6 g of H₂

                              1.4 g of N₂ -------------------- x

                             x = (1.4 x 6) / 28

                            x = 0.3 g of Hydrogen

Answer:

0.3758moles

Explanation:

moles of kcl = mass of kcl/ molar mass of kcl = 28/74.5 = 0.3758moles

Answer:

We have 0.376 moles in 28.0 grams of KCl

Option C is correct.

Explanation:

Step 1: Data given

Mass KCl = 28.00 grams

Molar mass KCl = 74.55 g/mol

Step 2: Calculate moles KCl

Moles KCl = mass KCl / moalr mass KCl

Moles KCl = 28.0 grams / 74.55 g/mol

Moles KCl = 0.376 moles KCl

We have 0.376 moles in 28.0 grams of KCl

Option C is correct.

The Rydberg formula can calculate the wavelengths of light emitted by hydrogen atoms. However, with some modifications, it can also be used to calculate the wavelengths of light emitted by atoms of other elements. Hence, the initial energy level is n = 3

Rydberg equation:

1/λ = f/c = R(1/m² - 1/n²)

where c is the speed of light, f is frequency, λ is the wavelength, R is Rydberg constant, and n and m are the quantum numbers of the energy levels.

The initial energy level has quantum number n.

The final energy level is a ground state with the quantum number m = 1.

2.924 x 1015/2.998 x 108 = 1.097 x 107 x (1/12 - 1/n²)

(1/12 - 1/n²) = 0.8891

1/n² = 0.1109

=> n²

= 9

=> n = 3

Thus, the initial energy level is n = 3

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The electron in the hydrogen atom was in the 9th energy level (n = 9) before emitting the photon.

When a hydrogen atom transitions from an excited state to the ground state, it emits a photon with a certain frequency. In this case, the frequency is given as ν = 3.084 x 10^15 s^-1. We can use the equation ν = E/h, where E is the energy of the photon and h is Planck's constant, to find the energy of the emitted photon. Once we have the energy, we can determine the energy level the electron was in before the photon was emitted.

Using the energy-frequency relationship and Planck's constant, we have:

E = hν = (6.626 x 10^-34 J s)(3.084 x 10^15 s^-1) ≈ 2.041 x 10^-18 J

From this energy, we need to find the corresponding energy level. The energy levels of hydrogen are given by the formula: E = -13.6 eV/n^2, where E is the energy of the level and n is the principal quantum number. Rearranging the formula, we have:

n^2 = -13.6 eV/E ≈ -13.6 eV/(2.041 x 10^-18 J) ≈ -6.662 x 10^17

Taking the square root of both sides, we find:

n ≈ -8.159 x 10^8

Since n must be a positive integer, we can conclude that the electron was in the 9th level (n = 9) before emitting the photon.

Answer:

Option A, The Rutherford experiment proved the Thomson "plum-pudding" model of the atom to be essentially correct.

Explanation:

Thomson's plum pudding model:

Plum pudding model was proposed by J.J Thomson. In Thomson's model, atoms are proposed as sea of positively charge in which electrons are distributed through out.

Result of Rutherford experiment:

As per Rutherford's experiment:

Most of the space inside the atom is empty.

Positively charge of the atom are concentrated in the centre of the atom known as nucleus.

Electrons are present outside the nucleus and revolve around it.

As it is clear that, result of Rutherford experiment did not supported the Thomson model.

The Rutherford experiment did not prove the Thomson 'plum-pudding' model of the atom to be essentially correct. Instead, it disproved this model by showing that atoms have a small, dense, positively charged nucleus.

The statement 'The Rutherford experiment proved the Thomson 'plum-pudding' model of the atom to be essentially correct.' (option A) is not accurate. In fact, the Rutherford experiment disproved the Thomson 'plum-pudding' model of the atom. Rutherford's experiment revealed that atoms have a small, dense, positively charged nucleus, contradicting Thomson's model which proposed that positive charge was spread evenly throughout the atom. All other listed experiments (options B, C, and D) did indeed provide the results described.

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Answer:

Basic

Explanation:

The titration of a strong base with a weak acid would always result in a solution whose pH is greater than 7,i.e a basic solution. For this reason its advisable to use an indicator whose color change will occur in that pH range.

For other cases, the titration of a stong acid and strong will result in a neutral solution and the titration of a weak base versus a strong acid would result in an acidic solution.

Final answer:

At the equivalence point in a titration between a weak acid and a strong base, the solution will be basic due to the formation of a conjugate base from the weak acid which then increases the OH- concentration.

Explanation:

When performing a titration involving a weak acid and a strong base, the resulting solution at the equivalence point will typically be basic. This is due to the fact that the weak acid has a less complete dissociation in water compared to a strong base. At the equivalence point, a weak acid has been completely neutralized by the strong base, which means it has converted to its conjugate base by losing a proton. The conjugate base may then react with water to form OH- (hydroxide ions), making the solution basic.

Indicators are chosen according to the expected pH at the equivalence point. When titrating a weak acid with a strong base, an indicator that changes color in the basic pH range would be suitable, such as phenolphthalein. It is importantly noted that in a titration of a strong acid with a strong base, the equivalence point occurs at a neutral pH of 7.00; however, this is not the case for a titration involving a weak acid.

Answer:

The answer to your question is the original dose of Ba-142 was 1184 μg

Explanation:

Data

Total time = 1.25 hours

The Final amount of Ba = 9.25 μg

The Half-life of Ba = 10.6 minutes

Process

1.- Convert total time to minutes

                      1 h ----------------- 60 min

                       1.25 h ------------ x

                        x = 75 min

2.- Draw a table of this process

                   Final amount of Ba               Time

                           9.25                                 75 min

                           18.5                                  64.4 min

                           37                                     53.8 min

                           74                                     43.2 min

                          148                                     32.6 min

                          296                                    22  min

                          592                                    11.4 min

                         1184                                      0.8 min        

Answer:

108.25 ºC

Explanation:

The boiling point elevation for a given solute in water is given by the expression:

ΔTb = i Kbm

where ΔTb is the boiling point elevation

           i is the van´t Hoff factor

           Kb the boiling constant which for water is 0.512 ºC/molal

           m is the molality of the solution

The molality of the solution is the number of moles per kilogram of solvent. Here we run into a problem since we are not given the identity of the coolant, but a search in the literature tells you that the most typical are ethylene glycol and propylene glycol. The most common is ethylene glycol and this it the one we will using in this question.

Now the i factor in the equation above is 1 for non ionizable compounds such as ethylene glycol.

Our equation is then:

ΔTb =  Kbm

So lets calculate the molality and then ΔTb:

m = moles ethylene glycol / Kg solvent

Converting gal to L

4.90 g x 3.785 L/gal = 18.55 L

in a 50/50 blend by volume we have 9.27 L of ethylene glycol, and 9.27 L of water.

We need to convert this 9.27 l of ethylene glycol to grams assuming a solution density of 1 g/cm³

9.27 L x 1000 cm³ / L = 9273.25 cm³

mass ethylene glycol = 9273.25 cm³ x 1 g/cm³ = 9273.25 g

mol ethylene glycol = 9273.25 g/ M.W ethylene glycol

                                 = 9273.25 g / 62.07 g/mol =149.4mol

molality solution =  149.4 mol / 9.27 Kg H₂O = 16.12 m

( density of water 1 kg/L )

Finally we can calculate ΔTb:

ΔTb =  Kbm = 0.512  ºC/molal x 16.12 molal = 8.25 ºC

boiling point = 100 º C +8.25 ºC = 108.25 ºC

( You could try to solve for propylene glycol the other popular coolant which should give around 106.7 ºC )

Answer:

Option c → 8:1

Explanation:

This is the reaction:

C₅H₁₂ + 8O₂ → 10CO₂ + 6H₂O

1 mol of pentane needs 8 moles of oxygen to be combusted and this combustion produces 10 mol of carbon dioxide and 6 moles of water.

To determine the ratio, look the stoichiometry.

For every 8 moles of oxygen, I need 1 mole of pentane gas.

Answer:

FALSE                            

Explanation:

The incident of Muese Valley occured in 1930 due to air pollution.

Muese Valley lies along the river Muese which is situated Huy and Liege, Belgium. This region was crowded with industries including steel manufacturers, glass manufacturers, explosives plants, zinc smelter, etc.

The increase number of industries and population lead to the sources of pollution. Also increase in burning of domestic coal increased pollution surrounding the area.

Air pollution became so severe at this region that people have severe  respiratory problems. Residents suffered from vomiting, retrosternal pain, coughing fits and several experienced nausea. There were fog and smog all over and many people died.

Hence the answer is FALSE.

Answer:

For 0.353 moles AgNO3, we'll have 0.353 moles AgCl

Explanation:

How many moles of AgCl will be produced from 60.0g AgNO3 assuming NaCl is available in excess.

Step 1: Data given

Mass of AgNO3 = 60.0 grams

Molar mass AgNO3 = 169.87 g/mol

NaCl is in excess, so AgNO3 is the limiting reactant

Step 2: The balanced equation

AgNO3 + NaCl → AgCl + NaNO3

Step 3: Calculate moles AgNO3

Moles AgNO3 = mass AgNO3 / molar mass AgNO3

Moles AgNO3 = 60.0 grams / 169.87 g/mol

Moles AgNO3 = 0.353 moles

Step 4: Calculate moles AgCl

For 1 mol AgNO3 we need 1 mol NaCl to produce 1 mol AgCl and 1 mol NaNO3

For 0.353 moles AgNO3, we'll have 0.353 moles AgCl

Final answer:

To find the moles of AgCl produced from a reaction, calculate using stoichiometry and molar mass.

Explanation:

To determine the moles of AgCl produced:

Step 1: Data given

Mass of AgNO3 = 60.0 grams

Molar mass AgNO3 = 169.87 g/mol

NaCl is in excess, so AgNO3 is the limiting reactant

Step 2: The balanced equation

AgNO3 + NaCl → AgCl + NaNO3

Step 3: Calculate moles AgNO3

Moles AgNO3 = mass AgNO3 / molar mass AgNO3

Moles AgNO3 = 60.0 grams / 169.87 g/mol

Moles AgNO3 = 0.353 moles

Step 4: Calculate moles AgCl

For 1 mol AgNO3 we need 1 mol NaCl to produce 1 mol AgCl and 1 mol NaNO3

For 0.353 moles AgNO3, we'll have 0.353 moles AgCl

Explanation:

The mass of carbon and hydrogen is calculated from the mass of their oxides ([tex]CO_{2}[/tex] and [tex]H_{2}O[/tex]) as follows.

    Mass of C = [tex]3.00 g CO_{2} \times \frac{12.01 g C}{44.0 g CO_{2}}[/tex]

                      = 0.818 g C

    Mass of H = [tex]1.23 g CO_{2} \times \frac{2.016 g H}{18.02 g H_{2}O}[/tex]

                      = 0.137 g H

So, the mass of C + mass of H is as follows.

                        0.818 g + 0.137 g = 0.955 g

This mass is actually less than the mass of sample. And, the missing mass must be caused by O. Hence, the mass of O will be calculated as follows.

            Mass of O = 1.50 g - 0.955 g

                              = 0.545 g

Now, we convert masses to moles and find their moles and ratio as follows.

  Element     Mass/g      Moles    Ratio    [tex]\times 2[/tex]     Integers

       C             0.818       0.068       1        2              2

       H             0.137        0.068      1        2               2

       O             0.545       0.0340    0.5   1                1

Thus, we can conclude that simplest formula for the given compound is [tex]C_{2}H_{2}O[/tex].

Answer:

C2H4O (ut quest)

Answer:

Catenation

Explanation:

Catenation is one the most important bonding characteristics of carbon. It is the ability of carbon atoms to join with themselves to form chains of different chain length.

This is the principal reason why there are a lot of organic compounds. The chain forming capability of carbon, otherwise known as catenation is responsible for this

The versatility of carbon in bonding with other atoms and forming important compounds is primarily attributed to its tetravalent nature, stable covalent bonds with other carbon atoms, and the formation of double and triple bonds.

The most significant feature of carbon that contributes to its versatility in bonding with other atoms and forming important compounds is its tetravalent nature. Carbon has four valence electrons in its outermost energy level, allowing it to form up to four covalent bonds with other atoms. This ability to form multiple bonds makes carbon the building block of a wide variety of organic molecules.

Another important feature of carbon is its ability to form stable covalent bonds with other carbon atoms, resulting in the formation of long chains, branched structures, and rings. This property gives rise to the diversity and complexity of organic compounds found in nature.

Furthermore, carbon is capable of forming double and triple bonds with other atoms, such as oxygen and nitrogen. These multiple bonds contribute to the unique properties and reactivity of specific carbon compounds, such as aldehydes, ketones, and aromatic compounds.

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To calculate the molar concentration of HNO3 in the solution, use the balanced chemical equation to determine the mole ratio and then calculate the moles of HNO3 used. Finally, divide the moles of HNO3 by the volume of the solution to get its molar concentration.

To calculate the molar concentration of HNO3, we first need to use the balanced chemical equation to determine the mole ratio between HNO3 and LiOH. The balanced equation is 2HNO3 + 2LiOH → 2LiNO3 + H2O. From the equation, we can see that 2 moles of HNO3 react with 2 moles of LiOH. Since we know the molar concentration of LiOH and the volume used, we can calculate the moles of LiOH used: (12.5 mL)(1.0x10^-4 M) = 0.00125 moles of LiOH. Since the mole ratio is 1:2, the moles of HNO3 used would be half of that, which is 0.000625 moles of HNO3. Now, we can use the volume of the HNO3 solution to calculate its molar concentration:

Molar concentration of HNO3 = (0.000625 moles) / (25.0 mL) = 0.025 M HNO3

Final answer:

The molar concentration of HNO3 is 5.00x10-5 M.

Explanation:

To calculate the molar concentration of HNO3, we first need to determine the number of moles of LiOH used in the titration. The molarity of the LiOH solution is given as 1.0x10-4 M, and the volume used is 12.5 mL (0.0125 L).

moles of LiOH = Molarity x Volume = (1.0x10-4 M) x (0.0125 L) = 1.25x10-6 mol

Since the balanced chemical equation shows a 1:1 ratio between LiOH and HNO3, the number of moles of HNO3 is also 1.25x10-6 mol.

To find the molar concentration of HNO3, we use the volume of the HNO3 solution, which is 25.0 mL (0.025 L).

Molarity of HNO3 = moles of HNO3 / Volume = (1.25x10-6 mol) / (0.025 L) = 5.00x10-5 M HNO3

Answer:

5290.7 kg of Al₂O₃ are produced in the reaction of 2800 kg of Al

Explanation:

The reaction to produce alumina is:

4 Al  +  3O₂  →  2 Al₂O₃

So, let's convert the mass of Al to moles (mass / molar mass)

2800 kg = 2800000 g (Then, 2.8 ×10⁶ g)

2.8 ×10⁶ g / 26.98 g/mol = 103780.5 moles of Al

Ratio is 4:2, so with the moles I have, I will produce the half of moles of alumina.

103780.5 mol / 2 = 51890.25 moles of Al₂O₃

Now we can convert these moles to mass ( molar mass . mol )

101.96 g/mol . 51890.25 mol = 5290729.8 g

5290729.8 g →  5290.7 kg

The mass of alumina Al₂O₃ formed from the reaction between Aluminum and oxygen is 5290.73 kg

The equation for the reaction that leads to the formation of alumina(Al2O3) can be expressed as:

[tex]\mathbf{2Al + \dfrac{3}{2}O_2 \to Al_2O_3}[/tex]

Given that:

the mass of aluminium Al = (2800 kg = 2800 × 1000) grams

= 2800000 grams

The molar mass of Aluminium Al = 26.98 g/mol

Number of moles of Al = 2800000 g/ 26.98 g/mol

Number of moles of Al = 103780.5782 moles

Since the ratio of the aluminium in the reactant and product is 2:1  

There will be half moles of alumina;

∴

Moles of alumina will be:

[tex]\mathbf{\dfrac{1}{2} \times 103780.5782 \ moles}[/tex]

= 51890.2891 moles of Al₂O₃

Mass = number of moles × molar mass

The molar mass of Al₂O₃  = 101.96 g/mol

Mass of  Al₂O₃= 51890.2891 g × 101.96 g/mol

Mass of  Al₂O₃ = 5290733.877 grams

Mass of  Al₂O₃ = (5290733.877/1000 ) kg

Mass of  Al₂O₃ = 5290.73 kg

Learn more about the mass of a substance here:

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Answer:

57.478atm

Explanation:

T = 400k

n = 5mol

v = 2.00

a = [tex]6.49L^{2}[/tex]

b = 0.0652L/mol

R = 0.08206

Formula

P =  [tex]\frac{RT}{(\frac{v}{n} )-b} - \frac{a}{(\frac{v}{n} )^{2} }[/tex]

[tex]P = \frac{0.08206 * 400}{(\frac{2.00}{5.00} )-0.0652} - \frac{6.49}{(\frac{2.00}{5.00} )^{2} }[/tex]

[tex]P = \frac{32.824}{0.3348} - \frac{6.49}{0.16}[/tex]  

[tex]P = 98.041 - 40.563[/tex]

[tex]P = 57.478atm[/tex]

The van der Waals equation, which accounts for the actual volume of gas molecules and the attractions between them, can be used to find the pressure of a 5.00 mol sample of Cl₂ at 400.0 K in a 2.00 L container.

The pressure of a 5.00 mol sample of Cl₂ at 400.0 K in a 2.00 L container can be found using the van der Waals equation. The van der Waals equation is a modification of the ideal gas law that takes into account the volume of the gas molecules themselves (represented by the parameter 'b') and the attractions between gas molecules (represented by 'a').

In this problem, the values of 'a' and 'b' for Cl₂ are given to be 6.49 L²･atm/mol² and 0.0652 L/mol respectively. The van der Waals equation is given by:

[P + a(n/V)²] (V/n - b) = RT

where P is the pressure we are trying to find, n is the number of moles of gas, V is the volume of the container, R is the gas constant (0.0821 L･atm/mol･K), and T is the temperature in kelvins. Substituting given values into this equation and simplifying it, we get the value of the pressure. Please note, this type of problem often requires some algebraic manipulation.

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Answer:

The explanation is given below

Explanation:

Toxicity is a measure of how harmful a substance is on a living organism (causing illness, injury or even death). There are several factors that must be considered when evaluating toxicity:

- Dose of exposure: the amount of substance to which the organism has been exposed to;

- Frequency of exposure: how many times (and for how long) the individual has been exposed to the substance;

- Age and health condition of the individual exposed;

- Genetic makeup: the genetic backgroun of an organism will determine its response and degree of sensitivity to a given substance.

Moreover, there are five main characteristics of substances that determine its toxicity: Solubility (hidrophilic or lipophilic substances), persistence (for how long the substance remains the same and cause the same damage), bioaccumulation (when the substance is progresivelly incorporated in tissues), biomagnification (when the amount of substance rise along trophic levels) and other chemical interactions (different interactions with other chemicals can increase the degree of damage) .  

The question is incomplete, here is the complete question:

What is the concentration, in parts per million, of a solution prepared by dissolving 0.00040 mol HCl in 2.2 L [tex]H_2O[/tex] ? Assume that the volume of the solution does not change when the HCl is added.  Assume the density of the solution is the same as that for pure water (1.00 g/mL)

Answer: The concentration of HCl in the solution is 6.64 ppm

Explanation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]

We are given:

Moles of HCl = 0.00040 moles

Molar mass of HCl = 36.5 g/mol

Putting values in above equation, we get:

[tex]0.00040mol=\frac{\text{Mass of HCl}}{36.5g/mol}\\\\\text{Mass of HCl}=(0.00040mol\times 36.5g/mol)=0.0146g[/tex]

[tex]\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}[/tex]

Density of water = 1.00 g/mL

Volume of water = 2.2 L  = 2200 mL      (Conversion factor:  1 L = 1000 mL)

Putting values in above equation, we get:

[tex]1.00g/mL=\frac{\text{Mass of water}}{2200mL}\\\\\text{Mass of water}=(1.00g/mL\times 2200mL)=2200g[/tex]

ppm is the amount of solute (in milligrams) present in kilogram of a solvent. It is also known as parts-per million.

To calculate the ppm of HCl in the solution, we use the equation:

[tex]\text{ppm}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 10^6[/tex]

Both the masses are in grams.

We are given:

Mass of HCl = 0.0146 g

Mass of solution = 2200 g

Putting values in above equation, we get:

[tex]\text{ppm of HCl in solution}=\frac{0.0146g}{2200g}\times 10^6\\\\\text{ppm of HCl in solution}=6.64[/tex]

Hence, the concentration of HCl in the solution is 6.64 ppm

Answer:

6.6

Explanation:

Answer:

The answer is B. False

Explanation:

The ratio of sizes between the ionic radii of cations and anions in a cell influences the manner of packing for that cell thereby predicting the possible cation/anion coordination number in any compound and establishing the structure of ionic solids.

Answer:

We have to add 2RbNO3 (option A)

10 Rb +2RbNO3 → 6 Rb2O + N2

Explanation:

Step 1: The equation:

10 Rb + _____ —> 6 Rb2O + N2

Step 2: Balancing the equation

On the right side we have 6x2 = 12 Rb atoms.

On the left side we have 10x Rb

This means we need to add 2x Rb on the left side.

On the right side  we have 2x N, On the left side 0x N.

This means we need to add 2x N on the left side.

On the right side we jave 6x O, on the left side we have 0x O.

This means we need to add 6x O on the left side.

We add this by adding RbNO3

This means we have to add 2x RbNO3 (option A)

10 Rb +2RbNO3 → 6 Rb2O + N2