A reaction has an equilibrium constant of 7.9×10³ at 298 K. At 713 K, the equilibrium constant is 0.77.
Do you predict that the enthalpy of the reaction is positive or negative? Why?

a. Positive, because endothermic reactions shift toward products at higher temperatures
b. Negative, because exothermic reactions shift toward reactants at higher temperatures
c. Positive, because endothermic reactions shift toward reactants at higher temperatures
d. Negative, because exothermic reactions shift toward products at higher temperatures

Answer:

18.75 mL

Explanation:

We must apply the dilute factor formula to solve this:

Diluted volume . Diluted M = Concentrated volume . concentrated M

250 mL . 0.15 M = Concentrated volume . 2 M

(250 mL . 0.15 M) / 2M = Concentrated volume

18.75 mL = Concentrated volume

Answer:

Fe³⁺ and Mn⁺²

B⁻ and C

Explanation:

Isoelectronic are species that have the same number of electrons. A neutral species has the same number of protons and electrons, and the number of protons in the atomic number (Z) found in the periodic table.

A cation (positive ions) lost the numbe of electrons indicate in its charge, and an anion (negative ions) gain the number of electrons indicate in its charge. So, let's identify the number of electrons (e-) in each one the atoms:

First column:

Zn: Z = 30, e- = 30

Fe³⁺: Z = 26, e- = 26 - 3 = 23

Mn⁺²: Z = 25, e- = 25 - 2 = 23

Ar: Z = 18

Isoelectronic: Fe³⁺ and Mn⁺²

Second column:

Ge⁺²: Z = 32, e- = 32-2 = 30

Cl⁻: Z = 17, e- = 17 + 1 = 18

B⁻: Z = 5, e- = 5 + 1 = 6

C: Z = 6, e- = 6

Isoelectronic B⁻ and C

Isoelectronic species are species that contain the same number of electrons.

Each column as we can see is made up of four chemical species which may be ions or neutral atoms. The species that are isoelectronic are those that contain the same number of electrons. We shall now examine the species in each column to determine which ones are isoelectronic.

In the first column, Fe³⁺ and Mn²⁺ are isoelectronic species, they both contain 23 electrons. In the second column, B⁻ and C are isoelectronic species, they both contain six electrons.

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Explanation:

A covalent bond is defined as the occurrence of a bond due to the sharing of electrons between the combining atoms.

Atomic number of hydrogen atom is 1 and its electronic configuration is [tex]1s^{2}[/tex]. So, in order to complete its octet it needs to gain or mutually shares one electron.

A covalent bond is generally formed between non-metal atoms.

Thus, we can conclude that hydrogen has only one electron that will be involved in the formation of a covalent bond.

In a covalent bond, a hydrogen atom contributes its single electron, which pairs with the electron from the other atom, forming a single covalent bond.

When a hydrogen atom forms a covalent bond with another atom, it shares its single electron, resulting in a paired bonding configuration. As hydrogen has one electron in its valence shell, this one electron becomes involved in bonding. When two hydrogen atoms bond, each contributes their one electron to form a molecular orbital, resulting in a single covalent bond. This is represented in the formation of a hydrogen molecule (H₂), where the bond order is calculated as (2-0)/2=1, signifying one stable covalent bond. The electron configuration for each hydrogen atom in the molecule then resembles that of helium, with two electrons completing the 1s subshell.

It seems that the question is lacking hypothesized structures, which are provided in Figure A

Answer:

Option A has bond angles that have the highest strain beyond known tolerance

Explanation:

Note: Please find the complete question in the attachment

The hypothesized structures in the options were the proposed structures of benzene in the mid 19th century by Ladenburg (1869), Thiele (1899), Armstrong (1887), and Kekule (1865) respectively

Option A suggests sp3 hybridization of the carbon atom. sp3 hybridized atoms are oriented at 109.5° to be stable. The structure suggests that bond angles between C-C bonds are 60° and 90°, these bond angles are well beyond the known tolerable angle strain.

Options B and D (currently accepted structures of benzene) are sp2 hybridized and their bonds angle is 120°, which is the optimal angle for sp2 hybridized atom. Option C also suggests sp2 hybridized carbon, so their will also be no ring strain in it. But it is not an accepted structure as it was suggested to contain radical like properties by Armstrong.

The strain in molecular geometries such as 1-cyclohexyne arises from abnormal bond angles that deviate from ideal sp³ hybridization. Benzene, however, remains stable due to electron delocalization and a regular planar hexagonal structure maintaining equal bond lengths and angles.

Understanding Molecular Geometry and Strain in Hypothesized Structures

The concept of strained bond angles beyond known tolerances relates to proposed molecular structures that do not conform to normal, observed geometries. For instance, the 1,3,5-cyclohexatriene structure proposed for benzene in 1866 by Kekule features alternating single and double bonds. This would predict different bond lengths of 1.48 Å for single and 1.34 Å for double bonds. However, due to aromaticity, benzene is actually a regular planar hexagon with equal C-C bond lengths of 1.39 Å and bond angles of 120°, indicating the presence of resonance and delocalized p-orbital electrons.

In contrast, a structure like 1-cyclohexyne would suffer from severe angle strain since it attempts to maintain a planar structure with sp² hybridization that imposes 120° bond angles, which deviates from the tetrahedral angle of 109.5° expected for sp³ hybridized carbons as in cyclohexane. Additionally, cyclooctatetraene, if forced into a planar structure to achieve delocalization, would also experience prohibitive angle strain, as it can not attain a structure with equivalent π bonds between adjacent carbons without distorting natural bond angles.

The concept of electron delocalization is fundamental in understanding why certain hypothesized structures impose a strain on bond angles. For benzene, the planar structure and delocalization lead to stability, while for other cyclic compounds, the need to maintain ideal hybridization geometry can lead to significant strain if forced into non-tetrahedral geometries.

Answer:

a) 37.04%  b) 37.04%  c) 9.63%

Explanation:

The theoretical percent recovery (Tr), is the total percentage of each compound in the sample. Depending on the technique used to recovery the compounds, the percent recovery will be less than the theoretical, because no technique is 100% efficient.

So, to calculate the theoretical, it will be the mass of the compound divided by the mass of the sample multiplied by 100%.

a) Tr = (250 mg)/(675 mg) * 100%

Tr = 37.04%

b) Tr = (250 mg)/(675 mg) * 100%

Tr = 37.04%

c) Tr = (65 mg)/(675 mg) * 100%

Tr = 9.63%

In each of the Excedrin table which consist of 250 mg of aspirin about 250 mg of acetaminophen and about 65mg of caffeine. The percentage of the theoretical components

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Answer:

1.31x10¹¹ g/cm³

Explanation:

The mass of the proton is equal to the mass of the neutron, which is 1.67x10⁻²⁴ g, so the mass of the alpha particle is 4*1.67x10⁻²⁴ = 6.68x10⁻²⁴ g.

1 fm = 1.0x10⁻²³ cm, thus the radius of the alpha particle is 2.3x10⁻¹² cm. If the particle is a sphere, the volume of it is:

V = (4/3)*π*r³, where r is the radius, so:

V = (4/3)*π*(2.3x10⁻¹²)³

V = 5.1x10⁻³⁵ cm³

The density of the particle is the how mass exists per unit of volume, so, it's the mass divided by the volume:

d = 6.68x10⁻²⁴/5.1x10⁻³⁵

d = 1.31x10¹¹ g/cm³

An alpha particle is a helium-4 nucleus consisting of two protons and two neutrons. The density of an alpha particle can be calculated by dividing its mass by its volume. For an alpha particle with a mass of approximately 4 atomic mass units and a radius of 2.6 femtometers, the density is approximately 4.90 x 10⁶ grams per cubic centimeter.

An alpha particle is a helium-4 nucleus, consisting of two protons and two neutrons. The density of an alpha particle can be calculated by dividing its mass by its volume. Since the mass of an alpha particle is approximately 4 atomic mass units (amu) and its radius is given as 2.6 fm (femtometers), we can use the formula for the volume of a sphere to find the volume of the alpha particle.

V = (4/3)πr³ = (4/3)π(2.6 fm)³ = (4/3)π(2.6 x 10⁻¹⁵ m)³ = (4/3)π(2.6 x 10⁻¹⁵ x 10⁻¹⁵ x 10⁻¹⁵) m³ = (4/3)π(1.94 x 10⁻⁴) m³ ≈ 8.15 x 10⁻¹³ m³

To convert this volume from cubic meters to cubic centimeters, we can multiply by the conversion factor 1 m³ = 1 x 10⁶ cm³.

Volume in cm³ = (8.15 x 10⁻¹³ m³) x (1 x 10⁶ cm³/m³) ≈ 8.15 x 10⁻⁷ cm³

To find the density, we divide the mass of the alpha particle (4 amu) by its volume (8.15 x 10⁻⁷ cm³).

Density = mass/volume = (4 amu) / (8.15 x 10⁻⁷ cm³) ≈ 4.90 x 10⁶ g/cm³

Answer: 6 valence electrons

Explanation: the valence electron is the number of electron in the outermost shell of an atom.

Oxygen has an atomic number of 8. The electronic configuration will be 1s2 2s2 2p4 or 2, 6. Meaning the last shell is having 6 electrons and this is equivalent to the valence electron

Oxygen has 6 valence electrons.

Oxygen has an atomic number of 8, which tells us that it has 8 protons in its nucleus. The atomic mass of oxygen is 16, which is the weighted average mass of all the isotopes of oxygen. Since the atomic number represents the number of protons and the number of electrons in an element, oxygen also has 8 electrons.

In the electron configuration of oxygen, the electrons are arranged in energy levels or shells. The valence electrons are the electrons in the outermost energy level, also known as the valence shell. Oxygen's electron configuration is 1s2 2s2 2p4. From this configuration, we can determine that oxygen has 6 valence electrons.

Valence electrons are responsible for the chemical behavior of an atom. In the case of oxygen, its 6 valence electrons allow it to form two covalent bonds. This enables oxygen to react with other elements to form compounds such as water and carbon dioxide.

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Answer:

19.91 J/K

Explanation:

The entropy is a measure of the randomness of the system, and it intends to increase in nature, thus for a spontaneous reaction ΔS > 0.

The entropy variation can be found by:

ΔS = ∑n*S° products - ∑n*S° reactants

Where n is the coefficient of the substance. The value of S° (standard molar entropy) can be found at a thermodynamic table.

S°, Cl(g) = 165.20 J/mol.K

S°, O3(g) = 238.93 J/mol.K

S°, O2(g) = 205.138 J/mol.K

So:

ΔS = (1*205.138 + 1*218.9) - (1*165.20 + 1*238.93)

ΔS = 19.91 J/K

This question does not contain the structures of the molecules. The structures in Daylight SMILES format are:

I. C1=CC=CC=C1C(=O)C

II. C1=CC=CC=C1CC=O

III. C1=CC(C)=CC=C1C=O

IV. C1=CC=CC=C1CCC

V. C1=CC=CC=C1C(C)C

The structures are also attached

Answer:

The structure of compound IV is consistent with the information obtained analysis

Proposed structures for the ions with m/z values of 120, 105,77 and 43 are (also attached):

C1=CC=CC=C1C(=[OH0+])C |^1:7|

C1C([CH0+]=O)=CC=CC=1

C1[CH0+]=CC=CC=1

C(#[OH0+])C

respectively

Explanation:

The IR peak at 1687 cm⁻¹ is indicative of an α unsaturated carbonyl carbon. While the 1H NMR singlet is of the methyl group next to carbonyl and the multiplet near 7.1 ppm is a characteristic peak of benzene. This data shows points towards structure I.

Mass spectrum peak at 120 m/z is of molecular ion peak. In the case of carbonyl-containing molecule, this peak is observable. The signal at 105 shows the loss of a methyl group next to the carbonyl. m/z value of 77 is the characteristic cationic peak of benzene, while the peak at 43 infers the formation of acylium ion (RCO+) due to α-cleavage. All this data agrees with the structure of acetophenone (Structure 1)

The question is incomplete, here is the complete question:

A chemistry student needs 15.00 g of 2-bromobutane for an experiment. He has available 220. g of a 30.0 % w/w solution of 2-bromobutane in ethanol. Calculate the mass of solution the student should use. If there's not enough solution, press the "No solution" button. Round your answer to 3 significant digits.

Answer: The mass of solution, the student should use is 50.0 grams

Explanation:

We are given:

30.0 % (w/w) of 2-bromobutane

This means that 30 grams of 2-bromobutane is present in 100 g of solution

Mass of solution given = 220. g

Mass of 2-bromobutane, the student needs = 15.00 g

Calculating the mass of 2-bromobutane in given amount of solution:

[tex]\Rightarrow 220\times \frac{30}{100}=66g[/tex]

To calculate the mass of solution, we use unitary method:

If 66 grams of 2-bromobutane is present in 220 grams of solution

So, 15 grams of 2-bromonutane will be present in [tex]\frac{220}{66}\times 15=50.0g[/tex] of solution

Hence, the mass of solution, the student should use is 50.0 grams

Final answer:

To calculate the mass of solution, convert the mass of -bromobutane into moles and then use the concentration of the solution to determine the amount of solution needed.

Explanation:

To calculate the mass of solution the student should use, we need to first convert the mass of -bromobutane needed into moles. Then, we can determine the amount of solution needed by using the concentration of the w/w solution of -bromobutane in ethanol. Let's go step by step:

Convert the grams of -bromobutane into liters of solution using the concentration: liters of solution = moles / concentration.

Remember to round your answer to the appropriate number of significant digits.

Final answer:

To calculate the number of cans of soda that would be lethal, we need to determine the amount of caffeine in a lethal dose and the amount of caffeine in one can of soda. Given that the lethal dose of caffeine is 10.0 g and there are 12 oz in a can, drinking approximately 10 cans of soda would be lethal.

Explanation:

To calculate the number of cans of soda that would be lethal, we need to determine the amount of caffeine in a lethal dose and the amount of caffeine in one can of soda. Given that the lethal dose of caffeine is 10.0 g and there are 12 oz in a can, we can use the caffeine concentration of the soda (2.85 mg/oz) to calculate the amount of caffeine in one can.

First, convert the caffeine concentration to mg/mL: 2.85 mg/oz = 2.85 mg/mL. Then, multiply the concentration by the volume of one can (12 oz = 354.88 mL): 2.85 mg/mL * 354.88 mL = 1011.84 mg.

To determine the number of cans that would be lethal, divide the lethal dose by the amount of caffeine in one can: 10.0 g / 1011.84 mg = 9.87 cans (rounded to the nearest whole number). Therefore, drinking approximately 10 cans of soda would be lethal.

Answer:

For 1: The value of [tex]K_c[/tex] is 10.24

For 2: The value of [tex]Q_c[/tex] is 9.07

For 3: The new equilibrium concentration of A is 0.220 M

Explanation:

We are given:

Volume of the container = 2.00 L

Equilibrium moles of A = 0.50 moles

Equilibrium moles of B = 0.50 moles

Equilibrium moles of C = 1.60 moles

Equilibrium moles of D = 1.60 moles

We know that:

[tex]\text{Molarity}=\frac{\text{Moles of solute}}{\text{Volume of solution}}[/tex]

For the given chemical reaction:

[tex]A(g)+B(g)\rightleftharpoons C(g)+D(g)[/tex]

The expression of [tex]K_c[/tex] for above equation follows:

[tex]K_c=\frac{[C][D]}{[A][B]}[/tex]

We are given:

[tex][A]_{eq}=\frac{0.50}{2.00}=0.25[/tex]

[tex][B]_{eq}=\frac{0.50}{2.00}=0.25[/tex]

[tex][C]_{eq}=\frac{1.60}{2.00}=0.8[/tex]

[tex][D]_{eq}=\frac{1.60}{2.00}=0.8[/tex]

Putting values in above equation, we get:

[tex]K_c=\frac{0.8\times 0.8}{0.25\times 0.25}\\\\K_c=10.24[/tex]

Hence, the value of [tex]K_c[/tex] is 10.24

Added moles of B = 0.10 moles

Added moles of C = 0.10 moles

[tex]Q_c[/tex] is the quotient of activities of products and reactants at any stage other than equilibrium of a reaction.

[tex]Q_c=\frac{[C][D]}{[A][B]}[/tex]

Now,

[tex][A]=\frac{0.50}{2.00}=0.25[/tex]

[tex][B]=\frac{0.60}{2.00}=0.3[/tex]

[tex][C]=\frac{1.70}{2.00}=0.85[/tex]

[tex][D]=\frac{1.60}{2.00}=0.8[/tex]

Putting values in above equation, we get:

[tex]Q_c=\frac{0.85\times 0.8}{0.25\times 0.3}\\\\Q_c=9.07[/tex]

Hence, the value of [tex]Q_c[/tex] is 9.07

Taking equilibrium constant as 10.24 for calculating the equilibrium concentration of A.

[tex]K_c=10.24[/tex]

[tex][B]_{eq}=\frac{0.60}{2.00}=0.3[/tex]

[tex][C]_{eq}=\frac{1.70}{2.00}=0.85[/tex]

[tex][D]_{eq}=\frac{1.60}{2.00}=0.8[/tex]

Putting values in expression 1, we get:

[tex]10.24=\frac{0.85\times 0.8}{[A]\times 0.3}[/tex]

[tex][A]_{eq}=\frac{0.85\times 0.8}{0.3\times 10.24}=0.220[/tex]

Hence, the new equilibrium concentration of A is 0.220 M

Answer:

Rate = k * [C₄H₆]²

Explanation:

It is possible to write the reaction as:

The differential rate law for a simple second order reaction of the type 2A → B is:

With the above information in mind, the rate law for the reaction of butadiene would be:

Answer: (c) CHCl3

Explanation:

From the rule : a polar solvent will dissolve a polar compound and a non polar solvent will also dissolves a non polar compound.

(a) CH4 is non- polar and water is a polar solvent. Therefore CH4 is not soluble in water

(b) CCl4 is non polar and water is a polar solvent. Therefore CCl4 is not soluble in wtaer

(c) CHCl3 is polar molecule and water is also a polar solvent.

Therefore CHCl3 is expected to be most soluble in water

Among CH4, CCl4, and CHCl3, CHCl3 (chloroform) is expected to be the most soluble in water due to its somewhat polar nature which corresponds with the polarity of water.

The gas expected to be most soluble in water among CH4, CCl4, and CHCl3 is CHCl3. This is based on the principle of 'like dissolves like'. In terms of polarity, water is polar and CHCl3 (chloroform) is also somewhat polar due to the presence of chlorine atoms, which have higher electronegativity. Since polar molecules are more likely to dissolve in another polar substance, CHCl3 would be the most soluble of the given gases in water.

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Answer:

1. Some polar compounds form hydrogen bridge bonds with water.

2. Polar compounds DO NOT interact in the same way with hydrophobic phospholipid tails.

3. YES: interactions are formed between polar compounds and hydrophobic tails.

4. False

Explanation:

Hello!

Different types of intermolecular junctions can be formed in the molecules:

-Bridge hydrogen bond: It is formed between a hydrogen attached to a very electronegative element (such as oxygen in water) and another very electronegative element (such as oxygen, fluorine). Polar molecules that contain electronegative elements in their structure may form this junction with water.

-Dipole-dipole union: it is formed between polar molecules where the zone with positive charge density of one molecule approaches the zone with negative charge density of another.

-Dipole-induced dipole union: it is formed between polar molecules where the zone with positive charge density of a molecule causes a non-polar molecule to partially polarize. It is a weak union but becomes important in long hydrophobic chains. It is the union that is established between polar compounds and hydrophobic tails of phospholipids.

For a polar compound to enter the hydrophobic space of the phospholipid tails, the cell usually uses other transport systems other than passive transport such as transport by specialized proteins. Water, despite being a polar molecule, due to its small size it can pass through the membrane at low speed.

The bonds formed with water do not break in aqueous medium. the "lack of attraction" does not exist between the molecules, there is always attraction although it can be of different intensity.

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The question concerns the behavior of polar compounds with the phospholipid bilayer of cell membranes. Polar compounds can't interact with the hydrophobic core, not because the core repels them, but because they'd need to break their existing bonds with water and wouldn't be able to form new bonds within the hydrophobic core.

The subject is about how polar compounds interact with the phospholipid bilayer, which forms the structure of cell membranes. Polar compounds form hydrogen bonds with the water molecules outside the membrane due to their polar nature. But they cannot form a similar kind of bond with the hydrophobic tails of the phospholipids, which form the core of the cell membrane.

Hydrophobic substances do not have the necessary conditions to form hydrogen bonds with polar compounds, making them unfavorable for such interactions. For a polar compound to pass through the hydrophobic layer, it would have to break the bonds it has with water and then fail to form any new bonds within the hydrophobic core. This would require a lot of energy and is not favorable, thus making it seem like the hydrophobic core 'repels' polar compounds, when in reality, it's a lack of attraction.

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The structure of Dimethyl sulfide is H3C-S-CH3. It is produced naturally by some marine algae.

Explanation:

Answer:

A

Explanation:

The molecule would not have a net dipole moment because the dipole moments cancel out. Being a vector quantity, dipole moment is affected by the direction of the dipole. If the dipoles are such that they are oriented opposite each other, theory simply cancel out. Polarity of individual bonds in a molecule does not automatically imply that the molecule must poses a net dipolev moment. The polarity of the Si-H and Si-Cl bonds cancel out cause they are oriented opposite each other hence the molecule is non polar.

The molecular shape of the SiH2Cl2 molecule is approximately tetrahedral and its dipole moment is zero.

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Answer:

85.82%

Explanation:

Mass before recovery = 141 mg

Mass after recrystallization = 121 mg

Percent recovery of recrystallization = actual mass / mass of crude extract * 100 = 121 / 141 * 100 = 85.82%

The correct answer is "0.000035 M/s". The rate of disappearance of NO₂ in the reaction is 0.000035 M/s, found by dividing the change in concentration of NO₂ (-0.00350 M) by the time period (100 s).

The rate of disappearance of NO₂ in the reaction 2NO₂
ightarrow 2NO₂ + NO₂ can be calculated using the change in concentration of NO₂ over time. We are given that the concentration of NO₂ drops from 0.0100 to 0.00650 M in 100 seconds.

First, we calculate the change in concentration (also known as the concentration difference):
Concentration difference = Final concentration - Initial concentration = 0.00650 M - 0.0100 M = -0.00350 M (a negative sign indicates disappearance).

To find the rate of disappearance, we divide the concentration difference by the time period:
Rate of disappearance = Concentration difference / Time period
Rate of disappearance = -0.00350 M / 100 s = -0.000035 M/s

Since we are interested in the rate of disappearance, we take the absolute value of the result, making it 0.000035 M/s.

The rate of disappearance of NO2 for the given period is [tex]\( 3.5 \times 10^{-5} \) M/s.[/tex]

To find the rate of disappearance of NO2, we need to calculate the change in concentration of NO2 over time. The reaction given is:

[tex]\[ 2NO_2 \rightarrow 2NO + O_2 \][/tex]

From the problem, we have the initial concentration of NO2 as 0.0100 M and the final concentration as 0.00650 M. The time taken for this change in concentration is 100 seconds.

The rate of disappearance of NO2 can be calculated using the formula:

[tex]\[ \text{Rate} = -\frac{\Delta [NO_2]}{\Delta t} \][/tex]

where [tex]\( \Delta [NO_2] \)[/tex] is the change in concentration of NO2, and [tex]\( \Delta t \)[/tex] is the change in time.

The change in concentration of NO2 is:

[tex]\[ \Delta [NO_2] = [NO_2]_{\text{initial}} - [NO_2]_{\text{final}} \][/tex]

[tex]\[ \Delta [NO_2] = 0.0100 \, \text{M} - 0.00650 \, \text{M} = 0.00350 \, \text{M} \][/tex]

The change in time is:

[tex]\[ \Delta t = t_{\text{final}} - t_{\text{initial}} \][/tex]

[tex]\[ \Delta t = 100 \, \text{s} - 0 \, \text{s} = 100 \, \text{s} \][/tex]

Now, we can calculate the rate:

[tex]\[ \text{Rate} = -\frac{0.00350 \, \text{M}}{100 \, \text{s}} \][/tex]

[tex]\[ \text{Rate} = -3.5 \times 10^{-5} \, \text{M/s} \][/tex]

The negative sign indicates that the concentration of NO2 is decreasing over time. However, the rate of disappearance is typically reported as a positive value, so we take the absolute value:

[tex]\[ \text{Rate} = 3.5 \times 10^{-5} \, \text{M/s} \][/tex]

Answer: [tex]7.0\times 10^2kJ[/tex]  will be released when 0.25 moles of glucose in a can of soda is metabolized

Explanation:

Heat of combustion is the amount of heat released on complete combustion of 1 mole of substance.

Given :

Amount of heat released on combustion 1 mole of glucose = 2.8 MJ = [tex]2.8\times 10^3kJ[/tex]      [tex]1MJ=10^3kJ[/tex]

Thus we can say:  

1 mole of glucose on combustion releases = [tex]2.8\times 10^6J[/tex]

Thus 0.25 moles of glucose on combustion releases =[tex]\frac{2.8\times 10^3kJ}{1}\times 0.25=7.0\times 10^2kJ[/tex]

Thus [tex]7.0\times 10^2kJ[/tex]  will be released when 0.25 moles of glucose in a can of soda is metabolized

Answer:

The temperature of water will be 60 degree Celsius.

Explanation:

For every chemical there present a particular solubility temperature. The phenomenon occurs as the positive charge is attracted towards the negative and hence gives rise to a cohesive structure. When the polar compounds as well as the ions gets added to the water, they start getting into smaller parts and hence dissolve and becomes solution. The partial charge of water starts attracting different parts of the compound and makes them soluble in water.

Final answer:

To dissolve 91.6 g of KCl in 200 g of water, a temperature higher than 25°C will likely be necessary, but an exact temperature cannot be provided without specific solubility data for KCl.

Explanation:

To determine the temperature at which 91.6 g of KCl (potassium chloride) will dissolve in 200 g of water, we need to refer to the solubility data for KCl. Solubility tables or graphs provide this information and show the amount of a solute that can dissolve in a solvent at various temperatures. Generally, the solubility of ionic compounds like KCl increases with temperature. Without specific solubility data for KCl at different temperatures, an exact answer cannot be provided. However, it is known that at 25°C (77°F), approximately 34 g of KCl will dissolve in 100 mL of water. Therefore, for 91.6 g of KCl, a temperature higher than 25°C will likely be necessary to dissolve the entire amount in 200 g of water. More accurate determination requires a solubility chart or experimental data for KCl.

Answer: The percent by mass of ethylene in the equilibrium gas mixture is 3.76 %

Explanation:

We are given:

Initial partial pressure or ethane = 24.0 atm

The chemical equation for the dehydration of ethane follows:

                   [tex]C_2H_6(g)\rightleftharpoons C_2H_4(g)+H_2(g)[/tex]

Initial:          24.0

At eqllm:    24-x            x              x

The expression of [tex]K_p[/tex] for above equation follows:

[tex]K_p=\frac{p_{C_2H_4}\times p_{H_2}}{p_{C_2H_6}}[/tex]

We are given:

[tex]K_p=0.040[/tex]

Putting values in above expression, we get:

[tex]0.040=\frac{x\times x}{24-x}\\\\x^2+0.04x-0.96=0\\\\x=0.96,-1[/tex]

Neglecting the value of x = -1 because partial pressure cannot be negative.

So, partial pressure of hydrogen gas at equilibrium = x = 0.96 atm

Partial pressure of ethylene gas at equilibrium = x = 0.96 atm

Partial pressure of ethane gas at equilibrium = (24-x) = (24 - 0.96) atm = 23.04 atm

To calculate the number of moles, we use the equation given by ideal gas, which follows:

[tex]PV=nRT[/tex]          .........(1)

To calculate the mass of a substance, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]           ..........(2)

We are given:

[tex]P=23.04atm\\V=30.0L\\T=800^oC=[800+273]K=1073K\\R=0.0821\text{ L. atm }mol^{-1}K^{-1}[/tex]

Putting values in equation 1, we get:

[tex]23.04atm\times 30.0L=n\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 1073K\\\\n=\frac{23.04\times 30.0}{0.0821\times 1073}=7.85mol[/tex]

We know that:

Molar mass of ethane gas = 30 g/mol

Putting values in equation 2, we get:

[tex]7.85mol=\frac{\text{Mass of ethane gas}}{30g/mol}\\\\\text{Mass of ethane gas}=(7.85mol\times 30g/mol)=235.5g[/tex]

We are given:

[tex]P=0.96atm\\V=30.0L\\T=800^oC=[800+273]K=1073K\\R=0.0821\text{ L. atm }mol^{-1}K^{-1}[/tex]

Putting values in equation 1, we get:

[tex]0.96atm\times 30.0L=n\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 1073K\\\\n=\frac{0.96\times 30.0}{0.0821\times 1073}=0.33mol[/tex]

We know that:

Molar mass of ethylene gas = 28 g/mol

Putting values in equation 2, we get:

[tex]0.33mol=\frac{\text{Mass of ethylene gas}}{28g/mol}\\\\\text{Mass of ethylene gas}=(0.33mol\times 28g/mol)=9.24g[/tex]

We are given:

[tex]P=0.96atm\\V=30.0L\\T=800^oC=[800+273]K=1073K\\R=0.0821\text{ L. atm }mol^{-1}K^{-1}[/tex]

Putting values in equation 1, we get:

[tex]0.96atm\times 30.0L=n\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 1073K\\\\n=\frac{0.96\times 30.0}{0.0821\times 1073}=0.33mol[/tex]

We know that:

Molar mass of hydrogen gas = 2 g/mol

Putting values in equation 2, we get:

[tex]0.33mol=\frac{\text{Mass of hydrogen gas}}{2g/mol}\\\\\text{Mass of hydrogen gas}=(0.33mol\times 2g/mol)=0.66g[/tex]

To calculate the mass percentage of ethylene in equilibrium gas mixture, we use the equation:

[tex]\text{Mass percent of ethylene gas}=\frac{\text{Mass of ethylene gas}}{\text{Mass of equilibrium gas mixture}}\times 100[/tex]

Mass of equilibrium gas mixture = [235.5 + 9.24 + 0.66] = 245.4 g

Mass of ethylene gas = 9.24 g

Putting values in above equation, we get:

[tex]\text{Mass percent of ethylene gas}=\frac{9.24g}{245.5g}\times 100=3.76\%[/tex]

Hence, the percent by mass of ethylene in the equilibrium gas mixture is 3.76 %

Answer:

The spore coat (proteinous coat)

Explanation:

Endospore is usually a position taken by a gram positive bacteria when it lacks nutrient. Endo- means within and the-spore means offspring. This is a complex stage which is a dormant and highly resistive cell wall that allows the bacteria preserve its genetic materials in times of deprivation within this protective cell.

They are also resistant to UV radiation, heat and chemicals. Once the environment become favourable, they reactivate from this vegetative state and absorb nutrients. This vegetative state consists of the following layers:

1. The outer coating called the exosporium which surrounds the spore is the thin layer responsible for covering the spore coat.

2. The proteinous coating which is the spore coat is the next layer which as a sieve. It is resistant to chemical and other toxins.

3. After this proteinous coating is a thick peptidoglycan called the cortex.

3. A cell/core wall which resides under the cortex. This layer will become the cell wall of the bacterium after the endospore germinates.

5. The center/core of the endospore js the next. The core has the spore chromosomal DNA and some cell structures, such as ribosomes and enzymes but is inactive.

Answer:

The molarity of the solution is 9.73 M

Explanation:

Step 1: Data given

Mass of isopropanol = 585 grams

Molar mass of isopropanol = 60.1 g/mol

Volume = 1000 mL = 1 L

Step 2: Calculate moles of isopropanol

Moles isopropanol = mass isopropanol / molar mass isopropanol

Moles isopropanol = 585 grams / 60.1 g/mol

Moles isopropanol = 9.73 moles

Step 3: Calculate molarity of the solution

Molarity = moles / volume

Molarity = 9.73 moles / 1L

Molarity = 9.73 M

The molarity of the solution is 9.73 M

Answer:

a.Carbon :proton number =6

             electron number =6

b. Fluorine: proton number=9

                  electron number =9

c. tin:  proton number=50

           electron number =50

d. nickel:proton number=28

           electron number =28

Explanation:

In the neutral state of an atom the number of protons is always equal to the number of electrons and that whats makes it electrically neutral as the positive charges of protons balances the negative charges of electrons.

The heat of reaction, ΔH°rxn, for the production of methane, CH4, can be calculated using the enthalpy change of the reaction and the stoichiometric coefficients of the reactants and products.

The heat of reaction, ΔH°rxn, for the production of methane, CH4, can be calculated using the enthalpy change of the reaction and the stoichiometric coefficients of the reactants and products. In this case, the reaction is:

C(s) + O₂(g) → CO₂(g) ΔH° = -393.5 kJ

To calculate the heat of reaction, we can multiply the enthalpy change by the stoichiometric coefficient of methane in the balanced equation:

ΔH°rxn = -393.5 kJ/mol × 1 mol = -393.5 kJ

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The heat of reaction for the production of methane (CH₄) from CO₂ and H₂O is calculated to be -890.4 kJ. Using Hess's Law and the standard enthalpies of formation, the standard enthalpy of formation for CH4 is found to be -74.7 kJ/mol.

To calculate the heat of reaction for the production of methane (CH4), we will use the given thermochemical equation:

Step-by-Step Calculation:

The equation states that 1 mole of CH₄ reacts with 2 moles of O₂ to produce 1 mole of CO₂ and 2 moles of H₂O, releasing 890.4 kJ of energy. This reaction is exothermic, so ΔΗ is negative.

The standard enthalpy of formation [tex](\Delta H^\circ _f)[/tex] of a compound is the change in enthalpy when one mole of a substance is formed from its elements in their standard states.

We know the standard enthalpies of formation at 298 K for CO₂(g) and H₂O(l), which are:
[tex]\Delta H_f^\circ(\text{CO}_2(\text{g})) = -393.5 \, \text{kJ/mol}[/tex]
[tex]\Delta H_f^\circ(\text{H}_2\text{O}(\text{l})) = -285.8 \, \text{kJ/mol}[/tex]

Using Hess's Law: [tex]\Delta H^\circ_{\text{rxn}} = \left[ \sum \Delta H_f^\circ (\text{products}) \right] - \left[ \sum \Delta H_f^\circ (\text{reactants}) \right][/tex]

Insert values into the equation:

[tex]\Delta H^\circ_{\text{rxn}} = \left[ (-393.5 \, \text{kJ/mol}) + 2(-285.8 \, \text{kJ/mol}) \right] - \left[ \Delta H_f^\circ(\text{CH}_4(\text{g})) + 2(0 \, \text{kJ/mol for O}_2) \right][/tex]

[tex]\Delta H^\circ_{\text{rxn}} = [-393.5 - 571.6] - [\Delta H_f^\circ(\text{CH}_4(\text{g})) + 0][/tex]

[tex]\Delta H^\circ_{\text{rxn}} = -965.1 \, \text{kJ} - \Delta H_f^\circ(\text{CH}_4(\text{g}))[/tex]

Given [tex]\Delta H^\circ_{\text{rxn}}[/tex] = -890.4 kJ, solve for [tex]\Delta H_f^\circ(\text{CH}_4(\text{g})): -965.1 \, \text{kJ} - \Delta H_f^\circ(\text{CH}_4(\text{g})) = -890.4 \, \text{kJ}[/tex]

ΔHf°(CH₄(g)) = -965.1 kJ + 890.4 kJ

ΔHf°CH₄(g)) = -74.7 kJ/mol

Therefore, the standard enthalpy of formation for methane, CH₄, is -74.7 kJ/mol.

The percent by mass of acetyl bromide in the given solution can be calculated by dividing its mass by the total mass of the solution and then multiplying by 100%, leading to a result of 43.5%.

To calculate the percent by mass of each component in the solution, we first need to find the total mass of the solution. The total mass is the sum of the mass of chloroform, acetone, and acetyl bromide, which is 96.2 g + 31.2 g + 98.1 g = 225.5 g.

Next, we calculate the percent by mass for each substance by dividing the mass of each substance by the total mass and then multiplying it by 100%.

For acetyl bromide, it's (98.1 g / 225.5 g) * 100% = 43.5%.

Please note that the number of significant digits in the answer should match the lowest number of significant digits in the input, that's why it's rounded off to three significant digits.

Learn more about Percent by Mass here:

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Answer: some other substance must be reduced.

Explanation:

Oxidation-reduction reaction or redox reaction is defined as the reaction in which oxidation and reduction reactions occur simultaneously.

Oxidation reaction is defined as the reaction in which a substance looses its electrons. The oxidation state of the substance increases.

[tex]M\rightarrow M^{n+}+ne^-[/tex]

The substance which gets oxidized itself reduces others and thus is called as reducing agent.

Reduction reaction is defined as the reaction in which a substance gains electrons. The oxidation state of the substance gets reduced.

[tex]M^{n+}+ne^-\rightarrow M[/tex]

The substance which gets itself reduced , oxidise others and thus is called as oxidising agent.

Oxidation is the loss of electrons and is usually coupled with the reduction of another substance, known as the oxidizing agent. The production of hydronium ions is not a core characteristic of all oxidation processes.

Whenever a substance is oxidized, it means that it has lost electrons. This process is usually coupled with the reduction of another substance, which gains the electrons. Thus, statement 1 is incorrect, and statement 2 is correct: when a substance is oxidized, another substance must be reduced.

Statement 3 is also correct: a substance that is oxidized and causes the reduction of another substance is called the oxidizing agent. Statement 4 is incorrect: the production of hydronium ions is not a general characteristic of oxidation reactions, it’s specific to certain reactions that involve acids.

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Explanation:

According to the Einstein law, it is known that

            [tex]h \times \nu = \phi + \frac{1}{2} mv^{2}[/tex]

where,   h = energy of light

           [tex]\phi[/tex] = work function

           [tex]m v^{2}[/tex] = kinetic energy of electron

It is given that the value of [tex]h \nu[/tex] is [tex]7.2 \times 10^{-19} J[/tex]. And,

               1 eV = [tex]1.6 \times 10^{-19} J[/tex]

Here,  [tex]\phi[/tex] for titanium is 4.33 eV

          = [tex](4.33 \times 1.6 \times 10^{-19})[/tex] J  

          = [tex]6.928 \times 10^{-19}[/tex] J

(a)   First of all, kinetic energy will be calculated as follows.

             [tex]\frac{1}{2}mv^{2} = h \nu - \phi[/tex]

                         = [tex](7.2 \times 10^{-19} - 6.92 \times 10^{-19})[/tex] J

                         = [tex]0.272 \times 10^{-19}[/tex] J

It is known that mass of electrons is equal to [tex]9.109 \times 10^{-31} kg[/tex].

Therefore, [tex]mv^{2} = 0.544 \times 10^{-19} J[/tex]

and,     [tex](mv)^{2} = 9.109 \times 0.544 \times 10^{-19} \times 10^{-31}[/tex]

                       = [tex]4.955 \times 10^{-50}[/tex]

                mv = [tex]2.225 \times 10^{-25}[/tex]

Now, the relation between wavelength and mv is as follows.

      [tex]\lambda = \frac{6.626 \times 10^{-34}}{2.225 \times 10^{-25}}[/tex]

                   = [tex]2.98 \times 10^{-9} m[/tex]

Therefore, the wavelength of the ejected electrons is [tex]2.98 \times 10^{-9} m[/tex].

(b)   It is known that relation between energy and wavelength is as follows.

               E = [tex]h \nu = \frac{hc}{\lambda}[/tex]

                  [tex]\lambda = \frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{7.2 \times 10^{-19}}[/tex]

                     = [tex]\frac{6.626 \times 3 \times 10^{-26}}{7.2 \times 10^{-19}}[/tex]

                    = [tex]2.76 \times 10^{-7} m[/tex]

Hence, the wavelength of the ejected electrons is [tex]2.76 \times 10^{-7} m[/tex].

(c)   For iron surface, [tex]\phi = 4.7 eV[/tex]

                                       = [tex](4.7 \times 1.6 \times 10^{-19})[/tex] J

                                       = [tex]7.52 \times 10^{-19}[/tex] J

Here, the value of [tex]\phi[/tex] is more than the value of UV light source. Hence, we need a shorter wavelength light as we know that,

                  [tex]E \propto \frac{1}{\lambda}[/tex]

Therefore, lesser will be the wavelength higher will be the energy.

Final answer:

In a photoelectric effect experiment, the wavelength of the ejected electrons can be calculated using the equation E = h / λ, where E is the energy of the electron, h is Planck's constant, and λ is the wavelength. The wavelength of the ejected electrons is found to be 13.8 femtometers. The wavelength of the incident UV light can be calculated using the same equation and is found to be 9.2 picometers. For an iron surface with a higher work function, a longer wavelength of light would be required to eject electrons with the same wavelength.

Explanation:

(a) To calculate the wavelength of the ejected electrons, we can use the equation E = h / λ, where E is the energy of the electron, h is Planck's constant (6.626 x 10^-34 J*s), and λ is the wavelength.

We can rearrange the equation to solve for λ: λ = h / E.

When we substitute the energy of the ejected electrons as given (3 eV = 4.8 x 10^-19 J), we get λ = (6.626 x 10^-34 J*s) / (4.8 x 10^-19 J) = 13.8 x 10^-15 m, or 13.8 femtometers.

(b) To calculate the wavelength of the incident UV light, we can use the same equation E = h / λ, but this time substitute the energy of the incident light (7.2 x 10^-19 J) and solve for λ. We get λ = (6.626 x 10^-34 J*s) / (7.2 x 10^-19 J) = 9.2 x 10^-16 m, or 9.2 picometers.

(c) The work function of an iron surface is provided (4.7 eV). Since the work function represents the minimum energy required to eject an electron, a longer wavelength of light would be required to eject electrons with the same calculated wavelength in part (a). This is because longer wavelength light corresponds to lower energy photons.

Answer:

206 mL

Explanation:

In the annexed picture you can see your same question, just in another format.

First we calculate the total moles of CuF₂ that are required in the working solution:

1967 μM ⇒ 1967 / 10⁶ = 1.967 x10⁻³M

1.967 x10⁻³M * 0.275 L = 5.409x10⁻⁴ mol

Now we divide those moles by the concentration of the stock solution, to calculate the volume:

5.409x10⁻⁴ mol ⇒ 5.409x10⁻⁴ * 1000 = 0.5409 mmol

0.5409 mmol ÷ (2.63 mmol/L) = 0.206 L

0.206 L ⇒ 0.206 * 1000 = 206 mL