how many grams of oxygen, o2 are produced if 74.0g of potassium chlorate, kclo3, decomposes? (the molar mass of o2

Neon atoms are attracted to each other by London-dispersion forces, which arise due to the temporary polarization of electron density in one atom that induces a dipole in a neighboring atom. This is the correct option.

Neon atoms are not attracted to each other by covalent bonding, metallic bonding, or H-bonding, as these types of bonding require the sharing or transfer of electrons between atoms, and neon is a noble gas with a complete outer electron shell that is highly stable and does not readily form chemical bonds.

Instead, neon atoms are attracted to each other by London-dispersion forces, which are a type of intermolecular force that arises due to the temporary polarization of electron density in one atom that induces a dipole in a neighboring atom.

In the case of neon, the electron cloud surrounding the nucleus is highly symmetric, and there is no permanent dipole moment.

However, due to the motion of electrons, there is a temporary dipole moment in one neon atom that induces a dipole moment in a neighboring neon atom.

This temporary dipole moment can induce similar temporary dipole moments in adjacent atoms, leading to an attractive force between the neon atoms.

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The loss of absorbance observed when adding the glycine buffer before the sulfanilic acid and sodium nitrite in a spectrophotometric assay is due to the neutralization of any acidic or basic components that could interfere with the reaction.

Here are some possible additional explanations in bullet points:

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No, a mixture of KI and Na2SO3 would not perform the same reduction and precipitation of copper(II) as a mixture of NaI and Na2SO3. This is because the reducing power of the two mixtures is different due to the different nature of the cations (K+ vs Na+).

What is Reduction?

Reduction is a chemical process in which a substance gains electrons and undergoes a decrease in oxidation state. It is a key aspect of many chemical reactions, including combustion, corrosion, and the synthesis of organic compounds.

In the reduction of copper(II) to copper(I), iodide ion (I-) acts as the reducing agent, which gets oxidized to iodine (I2). In the presence of sodium sulfite (Na2SO3), the iodine reacts with sulfite ion (SO32-) to form iodide ion and sulfate ion (SO42-), while copper(I) ions get precipitated as copper(I) sulfite.

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In the separatory funnel, the top diethyl ether layer contains naphthalene, and the bottom aqueous layer contains water, sodium benzoate (converted from benzoic acid), and 2M sodium hydroxide.

An illustration of the location of each component in a separatory funnel containing water, diethyl ether, benzoic acid, naphthalene, and 2M sodium hydroxide is as follows :

1. The separatory funnel contains two immiscible layers due to the presence of water and diethyl ether. The top layer is the less dense diethyl ether layer, and the bottom layer is the more dense aqueous (water) layer.

2. Naphthalene, being nonpolar, will dissolve in the nonpolar diethyl ether layer. So, it will be located in the top layer.

3. Benzoic acid reacts with the 2M sodium hydroxide (a strong base) to form water-soluble sodium benzoate. The reaction can be represented as:
  C₆H₅COOH + NaOH → C₆H₅COONa + H₂O
  Therefore, benzoic acid (or its sodium salt, sodium benzoate) will be located in the bottom aqueous layer.

4. 2M sodium hydroxide, being a strong base, is completely ionized in water. So, it will be located in the bottom aqueous layer as well.

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SO[tex]_3[/tex] has dipole-dipole forces. Dipole-dipole interactions are the electrostatic interactions that occur between permanent polar molecules. Therefore, the correct option is option D.

Dipole-dipole forces and dipole-dipole interactions are the electrostatic interactions that occur between permanent polar molecules. Typically, the positive end of one molecule will draw the negative side of another. As a result, the two molecules get closer to one another, boosting the stability of the material. This interaction is not an ordinary ionic or covalent connection because it involves no transfer as well as exchange of electrons.  SO[tex]_3[/tex] has dipole-dipole forces.

Therefore, the correct option is option D.

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The mass of 2.13 x 10^18 atoms of Li is approximately 2.46 x 10^-4 grams.

To find the mass of 2.13 x 10^18 atoms of Li (lithium), we will use the following steps:

Step 1: Find the molar mass of lithium (Li). The molar mass of lithium is approximately 6.94 g/mol.

Step 2: Calculate the number of moles using Avogadro's number (6.022 x 10^23 atoms/mol). To do this, divide the given number of atoms (2.13 x 10^18) by Avogadro's number:

Number of moles = (2.13 x 10^18 atoms) / (6.022 x 10^23 atoms/mol) = 3.54 x 10^-6 moles

Step 3: Multiply the number of moles by the molar mass to find the mass of Li:

Mass of Li = (3.54 x 10^-6 moles) x (6.94 g/mol) = 0.0246 g

So, the mass of 2.13 x 10^18 atoms of Li is approximately 0.0246 grams.

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The correct answer is C) —COO− + —NH2+ → —COOH + —NH2. This is because pK2 refers to the second dissociation constant of the amino acid valine, which is the dissociation of the carboxyl group (—COOH) from the amino group (—NH2+).

During titration with a strong base like NaOH, the base will react with the acidic proton (H+) on the carboxyl group, resulting in the formation of the carboxylate ion (—COO−) and water (H2O). However, once all the carboxyl groups have been neutralized, the excess base will react with the amino group (—NH2+) and remove the proton (H+) from it, resulting in the formation of the amino ion (—NH2) and water (H2O). This is the point at which pK2 is reached, and the reaction is represented by the equation C) —COO− + —NH2+ → —COOH + —NH2. The other answer choices are not relevant to the titration of valine with a strong base.

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The ratio of copper ions to chlorine anions in the unit cell is 4:4, which can be simplified to 1:1. Therefore, the empirical formula of the compound is CuCl.

Step 1: Identify the number of atoms in the ccp array
In a cubic close-packed array of anions (chlorine in this case), there are 4 anion atoms present per unit cell.

Step 2: Determine the number of tetrahedral holes in the ccp array
There are two tetrahedral holes per anion in a ccp structure. So, for 4 anion atoms, there will be 4 * 2 = 8 tetrahedral holes.

Step 3: Calculate the number of occupied tetrahedral holes by copper ions
As copper ions occupy one-half of the tetrahedral holes, the number of occupied holes will be 8 * 0.5 = 4 copper ions.

Step 4: Determine the empirical formula of the compound
The ratio of copper ions to chlorine anions in the unit cell is 4:4, which can be simplified to 1:1. Therefore, the empirical formula of the compound is CuCl.

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Here are the chemical formulas, coordination numbers, and information on ligand dentate, isomerism, and enantiomers for each of the complexes you listed.

1. Hexaaquacobalt(III): [Co(H2O)6]3+. Coordination number: 6. Ligand: monodentate water molecules. No cis-trans isomers or enantiomers due to the symmetry of the octahedral shape.

2. Trioxalatocobaltate(III): [Co(C2O4)3]3-. Coordination number: 6. Ligand: polydentate oxalate ions. There are cis-trans isomers due to the presence of three bidentate oxalate ions in the complex.

3. Triglycinatocobaltate(III): [Co(gly)3]3-. Coordination number: 6. Ligand: tridentate glycinate ions. No cis-trans isomers or enantiomers due to the symmetry of the octahedral shape.

4. Tricarbonatocobaltate(III): [Co(CO3)3]3-. Coordination number: 6. Ligand: polydentate carbonate ions. There are no cis-trans isomers or enantiomers due to the symmetry of the octahedral shape.

5. Tris(1,10-phenanthroline)cobalt(III): [Co(phen)3]3+. Coordination number: 6. Ligand: bidentate phenanthroline molecules. There are cis-trans isomers due to the presence of three bidentate phenanthroline ligands in the complex, and there are enantiomers due to the chiral nature of the phenanthroline ligands.

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In a chemical reaction, the rate is the speed at which reactants are consumed and products are formed. The initial rate refers to the rate at the beginning of the reaction, when the concentrations of reactants are highest and the rate is typically fastest.

The rate law describes the relationship between the concentrations of reactants and the rate of the reaction. It can be expressed mathematically using the terms you mentioned:

A) Rate = k[A]^B - This rate law states that the rate is directly proportional to the concentration of reactant A raised to the power of B. This is known as a "simple" or "unimolecular" reaction.

B) Rate = K[A] - This rate law states that the rate is directly proportional to the concentration of reactant A. This is known as a "first-order" reaction.

C) Rate = K[A][B] - This rate law states that the rate is directly proportional to the concentrations of both reactants A and B. This is known as a "second-order" reaction.

D) Rate = k[A]*[B] - This rate law states that the rate is directly proportional to the product of the concentrations of reactants A and B. This is known as a "bimolecular" reaction.

E) Rate = kAI*B) - This rate law states that the rate is directly proportional to the concentrations of reactants A and B, as well as the concentration of a catalyst I. This is known as a "catalyzed" reaction.

By measuring the initial rate of a reaction under different conditions, scientists can determine the rate law and reaction order. This information is important for understanding the mechanism of the reaction and predicting how changes in reactant concentrations or reaction conditions will affect the rate.

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The value of Kp for this reaction at 25°C is 6.17 × 10^9 (approx.)

To find the value of Kp for the reaction C(s) + H₂(g) ↔ CH₄(g), which converts coal to methane at equilibrium, we can use the equation:

ΔG° = -RT ln(Kp)

Given ΔG° = -50.79 kJ/mole and the temperature T = 25°C (298.15 K), we can solve for Kp. First, let's convert ΔG° to J/mole:

ΔG° = -50.79 kJ/mole * 1000 J/kJ = -50790 J/mole

Now, we can rearrange the equation and solve for Kp:

ln(Kp) = -ΔG° / (RT)
Kp = e^(-ΔG° / (RT))

Using R = 8.314 J/(mole·K):

Kp = e^(-(-50790 J/mole) / (8.314 J/(mole·K) * 298.15 K))
Kp ≈ 6.17 × 10^9

Therefore, by calculating we can say that the value of Kp for this reaction at 25°C is approximately 6.17 × 10^9. Given the high value of Kp, the reaction heavily favors the formation of methane (CH₄) at equilibrium, suggesting that studying this reaction as a means of methane production is worth pursuing.

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Compare the protons in 127^I and 131^I.

Both isotopes are of iodine, which is represented by the symbol 'I'. The atomic number of iodine is 53, which indicates the number of protons in the nucleus.

Therefore, the comparison of protons in 127^I and 131^I is as follows:
A. 127^I has more protons than 131^I - Incorrect
B. 131^I has more protons than 127^I - Incorrect
C. 131^I has the same number of protons as 127^I - Correct

Your answer: C. 131^I has the same number of protons as 127^I

Both 127^I and 131^I have 53 protons, as they are both isotopes of iodine. The difference between them is in the number of neutrons, not protons.

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The formula used to find the concentration of a solution after dilution is: C1V1 = C2V2,

The dilution formula relates the initial concentration of a solution to its final concentration after it has been diluted. To find the concentration of a solution after dilution, you can use the dilution formula:

C1V1 = C2V2

Where:
C1 = initial concentration of the solution
V1 = initial volume of the solution
C2 = final concentration of the solution (which you want to find)
V2 = final volume of the solution (after dilution)

By using this formula, you can calculate the concentration of the solution after dilution (C2) by rearranging the formula as follows:
C2 = (C1V1) / V2

Simply plug in the values for C1, V1, and V2, and you can find the concentration of the solution after dilution.

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During the electrolysis of an aqueous NaCl solution, H2, and OH- are produced at the cathode, and O2 and H+ are produced at the anode.

In the electrolysis of an aqueous NaCl solution, the products at the cathode and anode can be determined by comparing the standard reduction potentials (Eo) of the half-reactions involved.

At the cathode, reduction occurs, and we have two possible reactions:
1. Na+ + e- → Na; Eo = -2.7 V
2. 2H2O + 2e- → H2 + 2OH-; Eo = -0.4 V

Since a more positive Eo is favored, the second reaction (Eo = -0.4 V) will occur at the cathode, producing H2 and OH-.

At the anode, oxidation takes place, and we have two potential reactions:
1. Cl2 + 2e- → 2Cl-; Eo = 1.4 V
2. O2 + 4H+ + 4e- → 2H2O; Eo = -0.82 V

For oxidation, we need to reverse the reactions and change the sign of the Eo values:
1. 2Cl- → Cl2 + 2e-; Eo = -1.4 V
2. 2H2O → O2 + 4H+ + 4e-; Eo = 0.82 V

The more positive Eo is still favored, so the second reaction (Eo = 0.82 V) will occur at the anode, producing O2 and H+.

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Approximately 597.2 kilowatt-hours (kWh) of electrical energy are produced by the wind turbine per month.

We may use the following conversion factors to change the energy produced by the wind turbine from joules to kilowatt-hours:

1 kilowatt-hour (kWh) = 3.6 x 10^6 joulesGiven:

Energy generated by wind turbine = 2.15 x 10^12 J

Now that we know how much energy the wind turbine produces in kWh, we can compute it as follows:

Kilowatt-hours (kWh) of energy are equal to (2.15 x 10^12 J) / (3.6 x 10^6 J/kWh)

Energy in kilowatt-hours (kWh) = 597.2 kWh (rounded to one decimal place)

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So, Na (Z=11) has the largest atomic radius among the given elements in the third period.

The element with the largest atomic radius among the given compounds is Na (Z=11) which is a group 1 element and has only one valence electron, leading to a larger atomic radius compared to the other elements in the third period.

1. List the elements given: Al (Z=13), Na (Z=11), Si (Z=14), and Cl (Z=17).
2. Understand that atomic radius generally decreases across a period from left to right due to an increase in effective nuclear charge.
3. Since Na is the farthest left element among the options provided, it has the largest atomic radius.

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A nickel-metal hydride battery has the following general features: it is commonly found in power tools, the anode reaction involves the oxidation of hydrogen atoms absorbed within a metal alloy, and the cathode reaction is the reduction of nickel(III) to nickel(II).

The general features of a nickel-metal hydride battery include:

1. Common usage in power tools: Nickel-metal hydride batteries are often used in power tools due to their high energy density and ability to handle high discharge rates.

2. Anode reaction: The anode reaction in a nickel-metal hydride battery involves the oxidation of hydrogen atoms absorbed within a metal alloy. This process releases electrons, which then flow through the external circuit, providing power to the device.

3. Cathode reaction: The cathode reaction in a nickel-metal hydride battery is the reduction of nickel(III) to nickel(II). This occurs as the electrons return to the battery through the external circuit and combine with the nickel(III) ions at the cathode, reducing them to nickel(II) ions.

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The Ksp for Ag2S is 6.81 × 10-63, which corresponds to option a).

To calculate the Ksp for Ag2S given its molar solubility, we need to follow these steps:

1. Write the balanced chemical equation for the dissolution of Ag2S:
Ag2S (s) ⇌ 2Ag+ (aq) + S2- (aq)

2. Use the molar solubility (1.26 × 10-16 M) to determine the concentration of each ion at equilibrium:
[Ag+] = 2 × (1.26 × 10-16 M) = 2.52 × 10-16 M
[S2-] = 1.26 × 10-16 M

3. Write the expression for the Ksp of Ag2S:
Ksp = [Ag+]^2 × [S2-]

4. Substitute the equilibrium concentrations into the Ksp expression:
Ksp = (2.52 × 10-16)^2 × (1.26 × 10-16)

5. Calculate the Ksp:
Ksp = 6.81 × 10-63

So, the Ksp for Ag2S is 6.81 × 10-63.

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Two-Dimensional Gel Electrophoresis (2DGE) is a gel electrophoresis method that separates proteins on the basis of their relative contents of acidic and basic residues. When a protein is at its pI (isoelectric point), it has a net charge of zero.

The primary technique for proteomics research is two-dimensional gel electrophoresis, or 2D-PAGE. It isolates a complex combination of samples by using two different protein characteristics. Proteins can be separated in the first dimension by their pI value, and in the second dimension by their relative molecular weight.  Proteins were initially visualised using ³²P or ³⁵S labelling. This is now being superseded by more sensitive approaches like SYBRO Ruby. Advances have been made at several phases of the 2D-PAGE technology, which can separate up to a staggering 10,000 proteins in just one single gel. A 2D-PAGE gel picture is recorded and analysed to determine the amount of proteins expressed in a certain tissue.

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Carbon tetrachloride should have the highest surface tension at a given temperature. Surface tension is a measure of the force required to stretch or break the surface of a liquid. It is dependent on the intermolecular forces between the molecules of the liquid.

The greater the strength of these intermolecular forces, the higher the surface tension.

Carbon tetrachloride has four chlorine atoms, which are highly electronegative and hence, can form strong dipole-dipole interactions with neighboring molecules. These intermolecular forces result in a strong cohesive force between the molecules of carbon tetrachloride, leading to higher surface tension.

In contrast, methane has weak van der Waals forces between its molecules, which result in lower surface tension. Carbon tetrafluoride, carbon tetrabromide, and carbon tetraiodide also have weaker intermolecular forces than carbon tetrachloride, which leads to lower surface tension values.

Therefore, among the given options, carbon tetrachloride should have the highest surface tension at a given temperature.

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The formation of H2 (g) and Cl2 (g) during hydrolysis in the presence of NaCl is due to the difference in reduction potentials of the half-cell equations and the selective reduction and oxidation of the ions present in the solution.

The best explanation for the formation of H2 (g) and Cl2 (g) during the hydrolysis of water in the presence of NaCl lies in the standard reduction potentials of the half-cell equations given in the accompanying table.

It can be observed that the reduction potential of Cl2 is much higher than that of O2, and the reduction potential of H2 is much lower than that of Na. This means that Cl2 has a stronger tendency to be reduced than O2, and H2 has a weaker tendency to be oxidized than Na.

During hydrolysis, the water molecule is split into H+ and OH- ions. The H+ ions are reduced by the Cl- ions to form HCl, which then reacts with more H+ ions to form H2 gas.

The OH- ions are oxidized by water molecules to form O2 gas and more OH- ions. The Na+ ions, however, have a higher tendency to be oxidized than the water molecule, so they do not participate in the reaction.

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The conservation laws for nuclear reactions state that the total number of protons and neutrons must remain the same before and after the reaction. However, the identity of the elements can change as a result of the reaction.

In the case of reactions starting with isotopes of hydrogen, such as deuterium or tritium, the atoms produced may contain a different number of protons and neutrons, leading to the formation of different elements. For example, the fusion of deuterium and tritium can result in the production of helium and a neutron.

Therefore, the conservation laws for nuclear reactions are still upheld, but the identity of the elements can change as a result of the reaction.

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The  type of the weapon delivery is used with the B61 weapon is Ballistic missile.

The B61 nuclear bomb is the variable yield dual which is use the tactical and the strategic bomb equipped with the Full Fuzing Option that is FUFO and it designed for the external carriage by the high-speed aircraft.

The Ballistic missile is the missiles that is use in the ballistic trajectory normally in the deliver the warhead over the horizon, and at the distances of the thousands of the kilometers, that is in the case of the "intercontinental ballistic missiles" and the "submarine-launched ballistic missiles".

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The new volume of the gas is 3.7 liters.

Volume is the measure of the amount of a space which is occupied by an object or substance. It is commonly used to describe the three-dimensional size or the capacity of an object or a region.

To solve this problem, we can use the combined gas law, which relates the pressure, volume, as well as temperature of a gas. The formula is;

P₁V₁/T₁ = P₂V₂/T₂

where P₁, V₁, and T₁ is the initial pressure, volume, and temperature, respectively, and P₂, V₂, and T₂ is the final pressure, volume, and the temperature, respectively.

Substituting the given values into the formula, we have;

P₁V₁/T₁ = P₂V₂/T₂

Pressure will be constant, so we can eliminate it

V₂ = V₁T₂/T₁

V₂ = 3.5 L304 K/288 K

V₂ = 3.7 L

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You have 0.1032 mol of PbSO4 in 31.3 g, and 68.96 g of PbSO4 in 0.2275 mol.

The formula weight of PbSO4 is calculated by adding the atomic weights of its constituent elements: Pb (207.2 g/mol), S (32.07 g/mol), and O (16.00 g/mol x 4).

Formula weight = 207.2 + 32.07 + (16.00 x 4) = 303.27 g/mol

To find the moles of PbSO4 in 31.3 g, use the formula:

moles = mass / formula weight
moles = 31.3 g / 303.27 g/mol = 0.1032 mol

To find the grams of PbSO4 in 0.2275 mol, use the formula:

mass = moles x formula weight
mass = 0.2275 mol x 303.27 g/mol = 68.96 g

So, you have 0.1032 mol of PbSO4 in 31.3 g, and 68.96 g of PbSO4 in 0.2275 mol.

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The geometry of the hexafluoroaluminate ion (AlF6)^-3 is octahedral. The correct answer is option a.

In this ion, an aluminum (Al) atom is surrounded by six fluoride (F) atoms. The octahedral geometry is a result of the aluminum atom's coordination with six fluoride ions, which distribute themselves symmetrically around the aluminum atom.

In an octahedral arrangement, the central atom forms six bonding pairs with its surrounding ligands, creating six bond angles of 90 degrees between adjacent fluoride atoms. The octahedral shape maximizes the distance between the fluoride ions, minimizing electron repulsion and stabilizing the overall structure of the ion.

This arrangement is distinct from other geometries such as tetrahedral (option b), trigonal bipyramidal (option c), and hexagonal (option d), which involve different numbers of bond pairs and bond angles. The octahedral geometry is a common structure for metal coordination complexes and provides insights into the electronic properties and reactivity of the hexafluoroaluminate ion.

Therefore, option a is correct.

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It is crucial to ensure that the magnesium wire is securely tied to the copper wire to avoid this error and ensure accurate results in the calculation of the molar mass of magnesium.

The common error in this experiment, where the magnesium wire breaks free from the bottom of the buret, can have a significant impact on the final calculated molar mass of magnesium.

This is because the magnesium wire not reacting completely with the acid means that the amount of magnesium consumed is not accurate, which leads to an incorrect value for the molar mass of magnesium.

When the magnesium wire breaks free, it stops reacting with the acid prematurely, resulting in a reduced amount of magnesium being consumed. This means that the calculated molar mass of magnesium will be higher than its actual value, as there is less magnesium consumed than there should be.

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The concentration of X after 155 seconds is 7.6 x 10⁻³ M. The answer is a.

To solve this problem, we can use the second-order rate law, which is given by the equation:

1/[X]t - 1/[X]0 = kt

Where [X]t is the concentration of X at time t, [X]0 is the initial concentration of X, k is the rate constant, and t is the time elapsed.

Substituting the given values into the equation, we get:

1/[X]155 - 1/0.45 = (0.035 M⁻¹s⁻¹) x (155 s)

Solving for [X]155, we get:

[X]155 = 7.6 x 10⁻³ M

Therefore, the concentration of X is 7.6 x 10⁻³ M, which is option A.

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The theoretical yield of H2O that can be obtained from the reaction of 4.5 g H2 and excess O2 is 40. g. The correct answer is option C.

The theoretical yield of H2O that can be obtained from the reaction of 4.5 g H2 and excess O2 can be calculated using stoichiometry. From the balanced equation, we can see that for every 2 moles of H2 that react with 1 mole of O2, 2 moles of H2O are produced. This means that the mole ratio of H2O to H2 is 2:1.

First, we need to convert the mass of H2 to moles using its molar mass:

4.5 g H2 x (1 mol H2 / 2.02 g H2) = 2.23 mol H2

Since the O2 is in excess, it won't be completely consumed and doesn't affect the calculation of theoretical yield. Now, we can use the mole ratio to calculate the theoretical yield of H2O:

2.23 mol H2 x (2 mol H2O / 2 mol H2) x (18.02 g H2O / 1 mol H2O) = 40.4 g H2O

Therefore,  from the reaction of 4.5 g H2 and excess O2 , the theoretical yield of H2O that can be obtained is 40. g.

So, option C  is correct.

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10,000 ppm is 1%. Thus, the right answer is the option (c).

Percent is the number that is expressed by the denominator as 100 or hundred. That is 1% can be expressed as 1/100 or 0.01 in fractions and decimals respectively. It can also be expressed as [tex]10^{-2[/tex].

Ppm is the per part million that is expressed by the denominator as a million or 1,000,000. That is 1 ppm can be expressed as 1/1,000,000 or 0.000001 in fractions and decimals respectively. It can be expressed as [tex]10^{-6[/tex].

Therefore 1 ppm = [tex]10^{-6[/tex]

1% = [tex]10^{-2[/tex].

[tex]10^{-2[/tex]. is expressed as [tex]\frac{10^{-2}}{10^{-6}}[/tex] or 10,000

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