Phenolphthalein has a pKa of 9.7 and is colorless in its acid form and pink in its basic form. calculate [In-}/{HIn} for the following pH

4.9

2.1

7.8

11.5

The percentage of I2 converted = 78.6 %

Explanation:

Write down the given values

Kp= 54.4

The percentage of I2 will be converted to HI is 0.200 moled each of H2 and I2.

It should come to an equilibrium in 1.00 lit container .

Change in I2 (iodine) and H2 (hydrogen)  = x in each

Change in HI = 2x

total ni (nickel)

number of moles = 0.2 -x + 0.2 -x + 2x

=0.4 moles

Mole fractions :

I2 = 0.2-x / 0.4 H2

=0.2-x  / 0.4 HI

= 2x /0.4

Kp = HI ^2 / H2* I2

= (2x) ^2 / (0.2-x) ^2 = 54.4

by  taking square root:

2x / 0.2-x = 7.375

= x=0.157

percentage of I2 converted = 78.6 %

Answer:

a. NaHCO₃ + HCl → NaCl + H₂O + CO₂

b. 39.14 g is the mass of NaHCO₃ required to produce 20.5 moles of CO₂

Explanation:

A possible reaction for NaHCO₃ to make dioxide is this one, when it reacts with hydrochloric to produce the mentioned gas.

NaHCO₃ + HCl → NaCl + H₂O + CO₂

Ratio in this reaction is 1:1

So 1 mol of baking soda, produce 1 mol of CO₂

Let's calculate the moles

20.5 g CO₂ / 44 g/m = 0.466 moles

This moles of gas came from the same moles of salt.

Molar mass baking soda = 84 g/m

Molar mass . moles = mass

84 g/m .  0.466 moles = 39.14 g

Answer:

The boiling point of water at 550 torr will be 91 °C or 364 Kelvin

Explanation:

Step 1: Data given

Pressure = 550 torr

The heat of vaporization of water is 40.7 kJ/mol.

Step 2: Calculate boiling point

⇒ We'll use the Clausius-Clapeyron equation

ln(P2/P1) = (ΔHvap/R)*(1/T1-1/T2)

ln(P2/P1) = (40.7*10^3 / 8.314)*(1/T1 - 1/T2)

⇒ with P1 = 760 torr = 1 atm

⇒ with P2 = 550 torr

⇒ with T1 = the boiling point of water at 760 torr = 373.15 Kelvin

⇒ with T2 = the boiling point of water at 550 torr = TO BE DETERMINED

ln(550/760) = 4895.4*(1/373.15 - 1/T2)

-0.3234 = 13.119 - 4895.4/T2

-13.4424= -4895.4/T2

T2 = 364.2 Kelvin = 91 °C

The boiling point of water at 550 torr will be 91 °C or 364 Kelvin

The boiling point of water decreases with altitude due to reduced atmospheric pressure. At an atmospheric pressure of 550 torr, which one might find on a peak in the Rocky Mountain National Park, the boiling point of water would be slightly higher than 90°C, but less than 100°C.

The boiling point of water is impacted by the atmospheric pressure. At higher altitudes, like on a peak in Rocky Mountain National Park where the pressure is 550 torr, the atmospheric pressure is lower and results in a lower boiling point compared to sea level. This can be confirmed with the use of a vapor pressure curve.

Deciphering the vapor pressure curve: Usually, 500 torr corresponds to a temperature around 80-90°C. So, it can be inferred that the boiling point of water at 550 torr would be slightly higher than 90°C, but still less than 100°C, which is the boiling point of water at sea level at standard atmospheric pressure.

Summarizing the concept, the boiling point of a liquid increases as pressure increases and decreases as pressure decreases. In other words, the boiling point of water decreases with altitude as the atmospheric pressure decreases.

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Answer:

a) Representation - (in attachment)

b) Tetrahedral geometry and

c) [tex]sp^{3}[/tex] hybrid orbitals invovle sigma bonding.

[tex]\pi[/tex] orbitals are formed by the overlapping of d-orbitals of sulfur with p-orbitals of oxygen.

Explanation:

a)

Representation in attachment.

The overall charge in the both structures are -2. The structure (b) is favored by the resonance structures since the formal charge with in the species are remained.

Therefore, structure (b) is the better representation of sulfate ion.

b)

In sulfate ion, sulfur atom is attached with four different oxygen atoms. According to the VSEPR theory , the sufate ion has tetrahedral geometry.

There is four sigma bonds and zero lone pairs present on the central metal atom.

Hence, the hybridization of the sulfur atom is [tex]sp^{3}[/tex]

c)

The s and p orbitals of sulfur invovles hybridization and form sigma bonds.The orbitals of sulfur involves [tex]\pi[/tex] bonding.

Therefore, [tex]\pi[/tex] bonds are formed by the overlapping of d-orbitals of sulfur with  p-orbitals of oxygen.

The sulfate ion is best represented with four S―O single bonds maintaining a tetrahedral geometry and sulfurs' sp³ hybridization. The pi bonds are formed due to the overlap of sulfur's 3d and oxygen's 2p orbitals.

The sulfate ion can be represented better with four S―O single bonds. This equalizes the formal charges, since every oxygen atom carries a charge of -2 while sulfur carries a charge of +6.

The geometry shape of the sulfate (SO4²-) ion is tetrahedral, and the hybridization of the sulfur (S) atom is sp³, based on the fact that there are four regions of electron density around sulfur in the sulfate ion. These regions come from the four sigma bonds formed between sulfur and oxygen.

In view of the tetrahedral shape of the sulfate ion, the sulfur atom uses its 3p orbitals and 3s orbital to form sp³ hybrid orbitals for sigma bonds with the oxygen atom's 2p orbitals. The double bonds observed in some resonance structures of sulfate ion arise due to the overlap of the 3d orbitals of sulfur with the 2p orbitals of oxygen to form pi bonds.

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Answer:C

(CH3CH2CH2)2CuLi

Explanation:

The reaction between acyl halides and Grignard reagents or acyl halides and organolithium compounds does not form ketones because the reagents ( Grignard reagent and organo lithium compounds) are too reactive hence the ketone intermediate reacts further to form tertiary alcohols hence the ketone cannot be isolated.

Answer:

If 25 grams of NaI is mixed with an excess amount of Pb(NO3)2 we get 38.45 grams of  PbI2 and 14.18 grams of NaNO3

Explanation:

Step 1: Data given

Mass of NaI = 25.00 grams

Pb(NO3)2 is in excess

Step 2: The balanced equation

2NaI + Pb(NO3)2 → PbI2 + 2NaNO3

Step 3: Calculate moles of NaI

Moles NaI = mass NaI/ molar mass NaI

Moles NaI = 25.00/ 149.89 g/mol

Moles NaI = 0.1668 moles

Step 4: Calculate moles of PbI2 and NaNO3

For 2 moles of NaI we need 1 mol of Pb(NO3)2 to produce 1 mol of PbI2 and 2 moles of NaNO3

For 0.1668 moles of of NaI we will have 0.1668/2 = 0.0834 moles of PbI2 and 0.1668 moles of NaNO3

Step 5: Calculate mass of the products

Mass PbI2 = 0.0834 *461.01 g/mol

Mass PbI2 = 38.45 grams

Mass NaNO3 = 0.1668 mol * 84.99 g/mol

Mass NaNO3 = 14.18 grams

If 25 grams of NaI is mixed with an excess amount of Pb(NO3)2 we get 38.45 grams of  PbI2 and 14.18 grams of NaNO3

Answer:

Reduction (cathode): Cu²⁺(⁺aq) + 2 e⁻ → Cu(s)  

Oxidation (anode): Zn(s) → Zn²⁺(⁺aq) + 2 e⁻        

Cu²⁺(⁺aq) + Zn(s) → Cu(s) + Zn²⁺(⁺aq)

E°cell = 1.10 V

Explanation:

The half-reactions are missing, but I will propose some to show you the general procedure and then you can apply it to your equations.

Suppose we have the following half-reactions.

Cu²⁺(⁺aq) + 2 e⁻ → Cu(s)   E°red = 0.34 V

Zn²⁺(⁺aq) + 2 e⁻ → Zn(s)    E°red = -0.76 V

To identify how to make a spontaneous cell, we need to consider the standard reduction potentials (E°red). The half-reaction with the higher E°red will occur as a reduction (in the cathode), whereas the one with the lower E°red will occur as an oxidation (in the anode).

Reduction (cathode): Cu²⁺(⁺aq) + 2 e⁻ → Cu(s)   E°red = 0.34 V

Oxidation (anode): Zn(s) → Zn²⁺(⁺aq) + 2 e⁻        E°red = -0.76 V

To get the overall equation we add both half-reactions.

Cu²⁺(⁺aq) + Zn(s) → Cu(s) + Zn²⁺(⁺aq)

The standard cell potential (E°cell) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.

E°cell = E°red, cat - E°red, an

E°cell = 0.34 V - (-0.76 V) = 1.10 V

Since E°cell > 0, the reaction is spontaneous.

With the provided half-reactions information, one could write the reduction and oxidation reactions at the cathode and anode, respectively, and determine the overall balanced equation for the spontaneous redox reaction in a galvanic cell. The positive and negative electrodes, identifying the cathode and anode, play a key role in this determination. However, without the specific details of the half-reactions, it's not possible to write the balanced equations or calculate the cell voltage.

To answer questions about the galvanic cell, we require specifics about the half-reactions involved. The half-reaction that occurs at the cathode is the reduction process, and the half-reaction that occurs at the anode is the oxidation process. However, without the actual half-reactions or the substances involved, it is impossible to write the precise equations.

In general terms, if the half-reactions and their respective standard reduction potentials are known, we can write the reactions as follows:

The electrode at which reduction occurs, the cathode, is the positive electrode, while the electrode at which oxidation occurs, the anode, is the negative electrode. For the overall balanced equation that powers the cell, simply combine the two half-reactions to get the full-cell reaction.

To calculate the cell voltage under standard conditions, the difference in standard electrode potentials between the cathode and anode is taken (E°cathode - E°anode). The overall reaction should be spontaneous if the cell voltage is positive.

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Answer:

K = 0.0450, T > 1501 K

Explanation:

We may apply the Gibbs free energy equation relating the Gibbs free energy to an equilibrium constant of a reaction. The relationship is described by the following equation:

[tex]\Delta G^o = -RT ln(K)[/tex]

Rearrange the equation for the equilibrium constant:

[tex]ln(K) = -\frac{\Delta G^o}{RT}\therefore K = e^{-\frac{\Delta G^o}{RT}}[/tex]

Given the temperature [tex]T = 1280 K[/tex] and the ideal gas law constant  [tex]R = 8.314 \frac{J}{K mol}[/tex], we obtain:

[tex]K = e^{-\frac{33\cdot 10^3 J}{8.314 \frac{J}{K mol}\cdot 1280 K}} = 0.0450[/tex]

Now notice if [tex]K > 1[/tex], then [tex]ln(K) > 1[/tex] and [tex]\Delta G^o < 0[/tex].

We may firstly solve for the entropy change of this reaction using the following equation:

[tex]\Delta G^o = \Delta H^o - T\Delta S^o \therefore \Delta S^o = \frac{\Delta H^o - \Delta G^o}{T} = \frac{224\cdot 10^3 J - 33\cdot 10^3 J}{1280 K} = 149.2 \frac{J}{K}[/tex]

Using the same equation, solve when the change in the Gibbs free energy is negative:

[tex]\Delta H^o - T\Delta S^o < 0\therefore T > \frac{\Delta H^o}{\Delta S^o} = \frac{224\cdot 10^3 J}{149.2 \frac{J}{K}} = 1501 K[/tex]

Final answer:

The equilibrium constant (K) at 1280 K for the reaction Zn(s) + H2O(g) → ZnO(s) + H2(g) is calculated using the standard reaction free energy and the relationship ΔG0 = -RTln(K). K is found to be e⁻³.152, which is less than 1. The temperature at which K becomes greater than 1 is approximately 1501 K.

Explanation:

The student's question is about calculating the equilibrium constant (K) at 1280 K for the reaction Zn(s) + H2O(g) → ZnO(s) + H2(g) given the standard reaction enthalpy (ΔHR0 = 224 kJ) and the standard reaction free energy (ΔG0 = +33 kJ) at this temperature. To calculate K, we can use the relationship between free energy and the equilibrium constant given by the equation ΔG0 = -RTln(K), where R is the gas constant (8.314 J/mol·K) and T is the temperature in kelvins. Rearranging and solving for K yields:

K = e(-ΔG0)/(RT)

At 1280 K:

K = e(-33000 J/mol)/(8.314 J/mol·K × 1280 K)

K ≈ e-3.152

The equilibrium constant K at 1280 K is approximately e-3.152, which is less than 1.

To find the temperature at which K becomes greater than 1, we need to find the temperature at which ΔG0 becomes negative. Since the relationship between ΔH0, ΔG0, and ΔS0 (the standard entropy change) is given by ΔG0 = ΔH0 - TΔS0, we can set ΔG0 to 0 and solve for T, assuming ΔS0 remains constant.

0 = ΔH0 - TΔS0

T = ΔH0 / ΔS0

We don't have ΔS0, but using the provided ΔH0 and ΔG0 values at a known temperature, we can calculate it:

ΔS0 = (ΔH0 - ΔG0) / T

ΔS0 = (224000 J/mol - 33000 J/mol) / 1280 K

ΔS0 = 149.22 J/(mol·K)

Now, substituting ΔS0 back into the original equation to solve for T when K becomes greater than 1:

0 = ΔH0 - TΔS0

T = ΔH0 / ΔS0

T = 224000 J/mol / 149.22 J/(mol·K)

T ≈ 1501 K

Therefore, the equilibrium constant becomes greater than 1 at a temperature of approximately 1501 K.

The artifact has an approximate age of 6146 years.

To determine the age of the artifact, we can use the formula for radioactive decay. The decay rate of the artifact is given as 2.75 dis/min·gC, and the decay rate of a living organism is 15.3 dis/min·gC. Using the formula:

t = (ln(N0/N))/(k)

where t is the time in years, N0 is the initial amount of carbon-14, N is the current amount of carbon-14, and k is the decay constant, we can calculate the age of the artifact.

Substituting the given values:

t = (ln(15.3/2.75))/(k)

where k = 0.693/t1/2

Using the half-life of carbon-14 (5730 years), we can calculate k:

t = (ln(15.3/2.75))/(0.693/5730)

Calculating this expression gives us an answer of approximately 6146 years, so the artifact is approximately 6146 years old.

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Answer:

Pressure and number of moles.

Explanation:

According to Charles law,

For an ideal gas, volume is directly proportional to temperature at constant pressure for a fixed amount of gas.

V ∝ T

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

From the law we can say that the two conditions are:

The condition that remains constant in Charles's law is Pressure and number of moles.

Charles's law states that the volume occupied by a fixed amount of gas is directly proportional to its absolute temperature, if the pressure remains constant.

The volume of a gas is directly proportional to it's number of moles.

Therefore, the condition that remains constant in Charles's law is Pressure and number of moles.

Learn more about Charles's law here:

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Answer:

i: answer is in attachment.

ii: 24 valance electron in overall structure

iii: carbon having positive charge while oxygen having negative

iv: a bond angle of approximately 104.5 degrees.

v:  CH3 having sp3 hybridization  and both CH having sp2 hybridization

vi: yes. resonance occur between CH=CH and oxygen.

Explanation:

iii: oxygen is electronegative so attraction of electrons occur at oxygen results in negative charge.

iv: a bond angle is about 104.5 degrees because the oxygen has 2 lone pairs and 2 bond pairs of electrons so lone pairs repulsion occur more than bonding pairs that's why the angle is less than that of an atom with tetrahedral or pyramidal geometry.

The Lewis structure for CH3CHCHOH comprises a chain of three carbon atoms with associated hydrogens and an alcohol group, having 22 valence electrons in total. The formal charges for the carbon and oxygen atoms are zero, the ideal C-O-H bond angle is approximately 109.5 degrees, the first and third carbon atoms are sp3 hybridized, and the second carbon is sp2 hybridized. There are no resonance structures for CH3CHCHOH.

The Lewis structure for CH3CHCHOH can be drawn by considering that carbon typically forms four bonds, hydrogen forms one bond, and oxygen forms two bonds. Here's a step-by-step process: Place the carbons in a row since they are connected to each other, attach the three hydrogens to the first carbon, add a double bond between the second and third carbon, and complete the octets for each carbon. Attach the hydroxyl group (-OH) to the third carbon. After drawing this, complete the structure by adding lone pairs to the oxygen atom.

Total number of valence electrons: Carbon has 4 valence electrons, Hydrogen has 1, and Oxygen has 6. The compound has three carbons (3x4=12), four hydrogens (4x1=4), and one oxygen (1x6=6), totaling 22 valence electrons.

Formal charges on each atom are calculated by using the formula: Valence electrons - (Nonbonding electrons + 1/2 Bonding electrons). After calculating, you will find that the formal charges for the carbons and the oxygen in CH3CHCHOH are all zero.

The ideal bond angle for the C-O-H atoms in an alcohol group is approximately 109.5 degrees, based on a tetrahedral arrangement.

The hybridization for the carbon atoms depends on the number of sigma bonds and lone pairs around each carbon. For CH3CHCHOH, we have: first carbon (C1) is sp3 hybridized, second carbon (C2) is sp2 hybridized, and the third carbon (C3, which is part of the alcohol group) is sp3 hybridized.

Regarding the presence of resonance structures, there are no resonance structures for CH3CHCHOH since there are no delocalizable pi electrons or lone pairs adjacent to the pi bonds that could result in resonance.

Answer:

The copper cube has a mass of 11.68 grams

Explanation:

Step 1: Data given

Mass of the copper cube = 100 grams

Temperature of water = 100°C

Temperature of ice = 0°C

Specific heat of copper = 0.39J/g°C

Enthalpy of fusion of water = 334 J/g

Step 2:  Calculate the energy for cooling the copper

energy in the copper as it cools from 100º to 0º = 0.39 J/g°C * 100g *100°C = 3900 J

Step 3: Calculate the mass of the copper cube

3900 J / 334 J/g = 11.68 grams

The copper cube has a mass of 11.68 grams

M = 3900/334 = 11.68 g

change is 3900 J

The maximum mass of ice that could be melted by the heat exchange is approximately 11.68 grams.

To find the maximum mass of ice that could be melted by the heat exchange, we need to calculate the amount of thermal energy transferred from the copper cube to the ice. We can use the equation Q = m * c * ΔT, where Q is the thermal energy, m is the mass, c is the specific heat, and ΔT is the change in temperature.

First, let's calculate the thermal energy transferred by the copper cube in the boiling water bath. The cube's mass is 100 grams, the specific heat of copper is 0.39 J/g-C, and the temperature change is 100.0 - 0.0 = 100.0 Celsius. Therefore, Q1 = (100g) * (0.39 J/g-C) * (100.0 C) = 3900 J.

Next, let's calculate the thermal energy required to melt the ice. The enthalpy of fusion for water is 334 J/g. Assuming that all the thermal energy is used to melt the ice, we can calculate the maximum mass of ice that could be melted using the equation Q2 = m * ΔHf, where Q2 is the thermal energy, m is the mass, and ΔHf is the enthalpy of fusion. Rearranging the equation, we have m = Q2 / ΔHf.

Since the thermal energy transferred by the copper cube is equal to the thermal energy required to melt the ice, we can equate Q1 and Q2: 3900 J = m * (334 J/g). Solving for m, we get m = 3900 J / 334 J/g = 11.68 g.

Therefore, the maximum mass of ice that could be melted by the heat exchange is approximately 11.68 grams.

The Molecule Geometry of an ABE3 molecule, with three electron groups and no lone pairs, is trigonal planar, with groups arranged 120° apart in a plane.

The Molecule Geometry of a molecule with the designation ABE3, where 'A' represents the central atom and 'E' represents electron groups (or bonding domains) surrounding the central atom, relates to how these groups are spatially distributed around the central atom. In an ABE3 molecule configuration, with three electron groups around the central atom and no lone pairs (indicated by the '3'), all three groups are arranged to be as far apart from one another as possible. This arrangement forms trigonal planar geometry, where the groups adopt the positions at the corners of an equilateral triangle, each 120° apart and all in the same plane.

Answer:

Starting material is Diisopropylamine.

Explanation:

By reacting Diisopropylamine with n-Butyllithium in dry cold conditions with Tetrahydrofuran as solvent, Lithium diisopropylamide is prepared. Please see the attached image for reference.

Answer:

monochlorinated products: 4

dichlorinated products: 12

Explanation:

Chlorination of alkanes is a reaction that takes place when the chlorine is in presence of light. This actually decomposes the chlorine, and one atom of Chlorine substracts an hydrogen from the alkane. Now, this hydrogen substracted comes usually from the most substitued carbon, because it's more stable (A tertiary carbon is more stable than a secondary carbon, and this more stable than primary).

When this happens, the other chlorine atom, goes as electrophyle in that carbon and formed the chlorinated product. Now, although a tertiary carbon is more stable, we can still have (in minor quantities) chlorinated products that comes from a secondary and primary carbon. The first picture shows the general mechanism of the chlorination, and the possible products for a monochlorinated.

The second picture shows the possible dichlorinated products, which are in higher quantities than the monochlorinated basicallu because of the variety of positions the chlorine can be. So, second picture shows all the products.

The chlorination of 3-methylpentane can produce varying products, depending on the number of hydrogen atoms being substituted by chlorine. Monochlorination can produce 4 unique products, while dichlorination can yield 9 unique products.

The chlorination process can result in multiple products due to the number of hydrogen atoms in 3-methylpentane that can be substituted by chlorine. The original compound, 3-methylpentane, has 12 hydrogen atoms. Therefore, monochlorination can produce various constitutional isomers. The unique monochlorinated constitutional isomers that can be obtained when we replace one hydrogen by one chlorine atom are 1-chloro-3-methylpentane, 2-chloro-3-methylpentane, and a pair of 3-chloro-3-methylpentanes, giving a total of four.

For dichlorination (two chlorine substituents), we consider each of the monochlorinated products and determine where the second chlorine atom can be placed. This will give 1,1-dichloro-3-methylpentane, 1,2-dichloro-3-methylpentane and others, totaling up to 9 unique dichlorinated constitutional isomers. Therefore,monochlorination of 3-methylpentane can produce 4 products and dichlorination can produce 9 products.

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Answer: Total ATP yield for the complete oxidation of the 16-carbon unsaturated fatty acid palmitoleic acid (a 16:1-Δ9 fatty acid) = 108 ATP molecules.

Note: Two ATP molecules are used in the activation of palmitoleic acid to palmitoleoyl-CoA. Therefore the net ATP yield is 106 molecules.

Explanation:

Palmitoleic acid is a 16-carbon fatty acid. The complete oxidation of palmioleic acid yields eight  acetyl-CoA molecules and 7FADH2 and &NADH2. The overall equation for the reaction is shown below

Palmiltoleoyl-CoA + 7CoA + 7 FAD + 7NAD+ + 7H2O---> 8 acetyl-CoA + 7FADH2 + 7NADH + 7H+

Each of the eight acetyl-Coa molecules enters the citric acid cycle to yield three NADH and one FADH2 which equals to; 8 * 3NADH = 24NADH and 8 * 1FADH2 = 8FADH2. Also the substrate level phosphoryation by succinyl-CoA synthetase yields 8 GTP molecules which later is converted to 8 ATP molecules.

Total FADH2 = 7 + 8 = 15FADH2

Total NADH = 7 + 24 NADH = 31 NADH

Each FADH2 and NADH enters the electron transport chain to produce 1.5 ATP and 2.5 ATP per molecule respectively.

1.5 * 15 FADH2 = 22.5 ATP molecules

2.5 * 31 NADH = 77.5 ATP molecules

Total ATP yield for the complete oxidation of the 16-carbon unsaturated fatty acid palmitoleic acid (a 16:1-Δ9 fatty acid) = 8 +22.5+77.5 = 108 ATP molecules.

Note: Two ATP molecules are used in the activation of palmitoleic acid to palmitoleoyl-CoA. Therefore the net ATP yield is 106 molecules.

The complete oxidation of palmitoleic acid yields 129 ATPs through several steps, including activation, the production of acetyl-CoA, the citric acid cycle, and reoxidation of reducing equivalents.

The complete oxidation of palmitoleic acid yields 129 ATPs. This is determined by several steps. First, 1 mole of ATP is used for activation. Then, 8 moles of acetyl-CoA are produced, each yielding 10 moles of ATP in the citric acid cycle, resulting in 80 ATPs. The β-oxidation reactions are repeated seven times, producing 7 moles of NADH and 7 moles of FADH2. These reducing equivalents are reoxidized through respiration, yielding 2.5-3 moles of ATP per NADH and 1.5-2 moles of ATP per FADH2. Overall, this results in approximately 49 moles of ATP from NADH and 14 moles of ATP from FADH2. Therefore, the total ATP yield for the complete oxidation of palmitoleic acid is 129 ATPs.

Answer:

a. 7.75 × 10²² molecules

b. 1.59 × 10²⁷ molecules

c. 6.41 × 10¹⁷ molecules

Explanation:

There are 7.11 × 10²⁴ molecules in 100.0 cm³ of a certain substance. We will use this ratio in our conversion fractions.

a. What is the number of molecules in 1.09 cm³ of the substance?

1.09 cm³ × (7.11 × 10²⁴ molecules/100.0 cm³) = 7.75 × 10²² molecules

b. What is the number of molecules in 2.24 × 10⁴ cm³ of the substance?

2.24 x 10⁴ cm³ × (7.11 × 10²⁴ molecules/100.0 cm³) = 1.59 × 10²⁷ molecules

c. What number of molecules would be in 9.01 × 10⁻⁶ cm³?

9.01 × 10⁻⁶ cm³ × (7.11 × 10²⁴ molecules/100.0 cm³) = 6.41 × 10¹⁷ molecules

Calculating the number of particles in a given volume involves using Avogadro's number and proportionality between the volume and the number of particles, assuming constant density.

The question involves concepts from chemistry, specifically dealing with Avogadro's number, the mole, and calculations of the number of molecules in a given volume. Avogadro's number is defined as the number of atoms or molecules in one mole of a substance, which is 6.022 × 10²23. When given the number of molecules in a specific volume, you can calculate the number of molecules in a different volume by setting up a proportional relationship. This is because the number of molecules is directly proportional to the volume if the density remains the same.

For example, if there are 7.11 × 10²24 molecules in 100.0 cm³ of a substance, to find the number of molecules in 1.09 cm³:

Similarly, you would use the same approach for other volumes to determine the number of molecules in 2.24 × 10²4 cm³ and 9.01 × 10²²6 cm³ respectively.

Answer:

ΔHv = 17.04 KJ/mol

Explanation:

T(°C)   T(K)      Pv(atm)         1/T(K)                LnPv

34      307      0.236       0.00325733      - 1.4439

44      317       0.292       0.0031545         - 1.2310

54      327      0.355       0.003058          - 1.0356

Clausius-Clapeyron:

⇒ δLnP = ΔH/R (δT/T²)

∴ δT/T² = δ/δT(- 1/T )

⇒ δLnP/δT = - ΔH/R

Graphing: LnP vs 1/T

we get an ecuation that corresponds to a straight line:

y = - 2049.6x + 5.2331 ...... R² = 1

where the slope of this line is:

y = mx + b

⇒ m = - 2049.6 = - ΔH/R.....Clausius-Clapeyron

⇒ ΔH = (2049.6)(R)

∴ R = 8.314 E-3 KJ/mol.K

⇒ ΔHv = 17.04 KJ/mol

Answer:

T = 3006.976 K

Explanation:

∴ ΔH° = 125 KJ/mol ( 800K - 1500K)

∴ ΔG° = 25 KJ/mol (1150 K)

⇒ T = ? ∴ K > 1

In the equilibrium:

∴ K > 1

If ΔG°/RT = 1

⇒ e∧(1) = 2.72 > 1

∴ ΔG°/RT = 1

⇒ T = ΔG°/R = (25 KJ/mol)/(8.314 E-3 KJ/mol.K)

⇒ T = 3006.976 K

verifying:

⇒ K = e∧(25/((8.314 E-3)(3006.976)))

⇒ K = 1.000000061 > 1

Final answer:

The student is asking about estimating the temperature at which the equilibrium constant becomes greater than 1 based on the given standard enthalpy and Gibbs energy values.

Explanation:

The given question is related to standard enthalpy and Gibbs energy of a reaction, and the estimation of temperature at which the equilibrium constant becomes greater than 1.

Firstly, it is stated that the standard enthalpy of the reaction is approximately constant at +125 kJmol-1 from 800K to 1500K. This information helps us understand the change in enthalpy with respect to temperature.

Secondly, the standard Gibbs energy is given as +25 kJmol-1 at 1150K. This value allows us to calculate the equilibrium constant using the equation: ΔG = -RT lnK.

To estimate the temperature at which the equilibrium constant becomes greater than 1, we need to solve for temperature in the equation K = e-ΔG/T and find the value of T that results in K > 1.

Answer: The mass difference between the two is 7.38 grams.

Explanation:

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

[tex]PV=nRT[/tex]

where,

P = pressure = 125 psi = 8.50 atm    (Conversion factor:  1 atm = 14.7 psi)

V = Volume = 855 mL = 0.855 L    (Conversion factor:  1 L = 1000 mL)

T = Temperature = [tex]25^oC=[25+273]K=298K[/tex]

R = Gas constant = [tex]0.0821\text{ L. atm }mol^{-1}K^{-1}[/tex]

n = number of moles = ?

Putting values in above equation, we get:

[tex]8.50atm\times 0.855L=n\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 298K\\\\n=\frac{8.50\times 0.855}{0.0821\times 298}=0.297mol[/tex]

To calculate the number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]      .....(1)

Moles of air = 0.297 moles

Average molar mass of air = 28.8 g/mol

Putting values in equation 1, we get:

[tex]0.297mol=\frac{\text{Mass of air}}{28.8g/mol}\\\\\text{Mass of air}=(0.297mol\times 28.8g/mol)=8.56g[/tex]

Mass of air, [tex]m_1[/tex] = 8.56 g

Moles of helium = 0.297 moles

Molar mass of helium = 4 g/mol

Putting values in equation 1, we get:

[tex]0.297mol=\frac{\text{Mass of helium}}{4g/mol}\\\\\text{Mass of helium}=(0.297mol\times 4g/mol)=1.18g[/tex]

Mass of helium, [tex]m_2[/tex] = 1.18 g

Calculating the mass difference between the two:

[tex]\Delta m=m_1-m_2[/tex]

[tex]\Delta m=(8.56-1.18)g=7.38g[/tex]

Hence, the mass difference between the two is 7.38 grams.

Final answer:

The mass of air in the tire calculated using the ideal gas law is 8.4752 grams, and the mass of helium is 1.1778 grams; the mass difference is 7.2974 grams, with helium being the lighter gas.

Explanation:

To calculate the mass of air and helium in a bike tire, we will use the ideal gas law which relates pressure (P), volume (V), temperature (T), and the number of moles of gas (n) using the equation PV = nRT where R is the gas constant. For this scenario, we are given that the volume (V) of the tire is 0.855 L (converted from 855 ml), the pressure (P) is 125 psi (which we must convert to atmospheres, since the gas constant uses atmospheres), and the temperature (T) is 25 degrees Celsius (which we must convert to Kelvin).

First, let's convert the pressure to atmospheres. Since 1 atm = 14.7 psi, the pressure in atmospheres is 125 psi / 14.7 psi/atm = 8.5034 atm. Next, let's convert the temperature to Kelvin: T = 25 °C + 273.15 = 298.15 K.

Using the ideal gas law equation PV = nRT, and solving for n (the number of moles), we get n = PV / RT. Substituting the known values for helium (R = 0.0821 L atm/mol K) and air (assuming average R = same as for helium, as there are no significant differences for the purposes of this problem):

For Air: n = (8.5034 atm) * (0.855 L) / (0.0821 L atm/mol K × 298.15 K) = 0.294 moles of air

For Helium: We use the same calculation as for air, since the volume, pressure, and temperature are the same.

Now, using the molar mass, we calculate the mass of each gas:

The mass of air = n (number of moles of air) × molar mass of air = 0.294 moles × 28.8g/mol = 8.4752 g

The molar mass of helium = 4.0026 g/mol, so the mass of helium = n (number of moles of helium) × molar mass of helium = 0.294 moles × 4.0026 g/mol = 1.1778 g

The mass difference between air-filled and helium-filled tires = mass of air - mass of helium = 8.4752 g - 1.1778 g = 7.2974 g. Thus, filling tires with helium instead of air would make them lighter by 7.2974 g.

Answer:

239 L

Explanation:

1. Propane reacts with oxygen to produce carbon dioxide and water:

[tex]C_3H_8 (g) + O_2 (g)\rightarrow CO_2 (g) + H_2O (l)[/tex]

Firstly, 3 carbon atoms are required on the right:

[tex]C_3H_8 (g) + O_2 (g)\rightarrow 3 CO_2 (g) + H_2O (l)[/tex]

Secondly, 8 hydrogens in total (4 water molecules) are required on the right:

[tex]C_3H_8 (g) + O_2 (g)\rightarrow 3 CO_2 (g) + 4 H_2O (l)[/tex]

On the right, we have a total of 10 oxygen atoms, this implies we need 5 oxygen molecules on the left:

[tex]C_3H_8 (g) + 5 O_2 (g)\rightarrow 3 CO_2 (g) + 4 H_2O (l)[/tex]

2. Calculate moles of propane using the the ratio of mass to molar mass:

[tex]n_1 = \frac{m_1}{M_1} = \frac{150 g}{44.1 g/mol} = 3.40 mol[/tex]

According to the stoichiometry, we have 3 times greater amount of carbon dioxide:

[tex]n_2 = 3n_2 = 3\cdot 3.40 mol = 10.2 mol[/tex]

Use the ideal gas law to solve for volume:

[tex]pV = nRT\therefore V = \frac{nRT}{p} = \frac{10.2 mol\cdot 0.08206 \frac{L atm}{mol K}\cdot 285.15 K}{1 atm} = 239 L[/tex]

Final answer:

The reaction of cyclohexene with concentrated KMnO₄ cleads to the formation of adipic acid due to oxidative cleavage of the carbon-carbon double bond.

Explanation:

When cyclohexene is treated with concentrated KMnO₄ (potassium permanganate), the reaction is an oxidative cleavage, and it produces a compound called adipic acid. This is because KMnO₄ is a strong oxidizing agent, which will typically cleave the double bond in the cyclohexene and oxidize the resulting fragments into carboxylic acids. Given that cyclohexene has a six-carbon ring, the oxidative cleavage will produce a six-carbon diacid, which is known as adipic acid.

Therefore, the correct answer to what compound is produced when cyclohexene is treated with concentrated KMnO₄ is B) adipic acid.

Answer:

[tex]\large \boxed{\text{(a) 42 000 yr;  (b) 45 000 yr}}[/tex]  

Explanation:

Two important equations in radioactive decay are

[tex]\ln \dfrac{N_{0} }{N_{t}} = kt\\\\t_{\frac{1}{2}} = \dfrac{\ln2}{k }[/tex]

We use them for carbon dating.

(a) Initial activity = 0.23 Bq

(i) Calculate the rate constant

The half-life of ¹⁴C is 5730 yr.

[tex]\begin{array}{rcl}t_{\frac{1}{2}}& = &\dfrac{\ln2}{k }\\\\k& = &\dfrac{\ln2}{t_{\frac{1}{2}}}\\\\ & = & \dfrac{\ln2}{\text{5730 yr}}\\\\ & = & 1.210 \times 10^{-4}\text{ yr}^{-1}\\\end{array}[/tex]

(ii) Calculate the age of the sample

[tex]\begin{array}{rcl}\ln \dfrac{N_{0} }{N_{t}} & = & kt\\\\\ln \dfrac{0.23 }{0.0015} & = & k\times 1.210 \times 10^{-4}\text{ yr}^{-1}\\\\\ln 153 & = &  1.210 \times 10^{-4}k \text{ yr}^{-1}\\5.03 & = & 1.210 \times 10^{-4}k \text{ yr}^{-1}\\k & = & \dfrac{5.03}{1.210 \times 10^{-4} \text{ yr}^{-1}}\\\\ & = & \textbf{42 000 yr}\\\end{array}\\\text{The age of the sample is $\large \boxed{\textbf{42 000 yr}}$}[/tex]

(b) Initial activity =  45 % larger

N₀ = 1.45 × 0.230 Bq = 0.334 Bq

[tex]\begin{array}{rcl}\ln \dfrac{N_{0} }{N_{t}} & = & kt\\\\\ln \dfrac{0.334 }{0.0015} & = & k\times 1.210 \times 10^{-4}\text{ yr}^{-1}\\\\\ln 222 & = &  1.210 \times 10^{-4}k \text{ yr}^{-1}\\5.40 & = & 1.210 \times 10^{-4}k \text{ yr}^{-1}\\k & = & \dfrac{5.40}{1.210 \times 10^{-4} \text{ yr}^{-1}}\\\\ & = & \textbf{45 000 yr}\\\end{array}\\\text{The age of the sample is $\large \boxed{\textbf{45 000 yr}}$}[/tex]

The molar concentration of phosphate (present as dihydrogen phosphate) in seawater can be calculated by converting the given mass concentration of phosphate (0.07 ppm) into moles using the molar mass of phosphate. After this conversion, the phosphate concentration is approximately 7.14 x 10^-4 mol/L.

The question asks us to calculate the molar concentration of the dihydrogen phosphate ion in seawater. Remember that molar concentration is the amount of a solute (in this case phosphorus as dihydrogen phosphate) divided by the volume of the solution (seawater). We are given that the concentration of phosphorus is 0.07 ppm by mass, which, converted into grams per liter (g/L), yields 0.07 g/L. Additionally, the molar mass of H₂PO₄⁻ is approximately 97.994 g/mol.

Step-by-step calculation:

Therefore, the molar concentration of phosphate in seawater is approximately 7.14 x 10⁻⁴ mol/L.

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Answer:

The C22- ion is stable

Explanation:

The C22- ion is a stable ion having a bond order of three. It has a favourable stabilization energy of 6∆ compared to 4∆ in C2. In the carbide ion carbon firms three bonds rather than two in dicarbon, hence the formation of the carbide ion is preferred. The molecular orbital configuration of the carbide ion is shown in the image attached.

Final answer:

The carbide ion (C2 2-) is formed from acetylene (C2H2) by loss of two protons and has a bonding scheme where the triple bond of acetylene is reduced to a double bond in the ion, yielding a bond order of 2 for the carbide ion compared to 3 for acetylene, according to molecular orbital theory.

Explanation:

The carbide ion (C2 2-), formed when acetylene (C2H2) loses two protons, has an interesting bonding scheme described by molecular orbital theory. In acetylene, there is a triple bond between the two carbon atoms consisting of one sigma (σ) bond and two pi (π) bonds. This triple bond is the result of the overlapping of sp hybrid orbitals for the σ bond and the side-by-side overlap of unhybridized p orbitals for the π bonds. When the two protons are removed to form the carbide ion, two additional electrons are added to the system which occupy the antibonding π orbitals (π*). In molecular orbital theory, the bond order is equal to the difference between the number of bonding electrons and antibonding electrons divided by two. Thus, in the carbide ion, the triple bond of acetylene is reduced to a double bond due to the additional electrons in the antibonding orbitals, which gives a bond order of 2 in C2 2- as compared to a bond order of 3 in C2.

Final answer:

To read beakers accurately, identify the smallest increment, estimate one decimal place beyond it, align your eye level with the meniscus, and report the measurement with the correct number of significant figures and units.

Explanation:

To read beakers to the correct amount of significant figures, you should:

For instance, if you measure a volume that reads between 50 mL and 60 mL and the meniscus is halfway between the two, you would report a volume of 55 mL. If you are using a more precise beaker with 1 mL increments, you might report 55.5 mL where the 5 is an estimate.

It is crucial to ensure that all your reported measurements and calculated results have the proper number of significant figures and proper units, maintaining precision and accuracy in your experiments.

Answer:

Surface runoff and condensation

Explanation:

Let's define each of the given processes in order to understand them better:

All in all, notice that surface runoff keeps water in its liquid state, while all the other three options consider phase change. The only phase change of interest is condensation: we produce liquid water from water vapor and then we can analyze its movement in the liquid state.

The equilibrium concentration of H2 in the given reaction is approximately 0.614 mol/L.

Given that the equilibrium constant (K) is 1.00×10^2 for the reaction H2(g) + F2(g) ⇌ 2HF(g), we can use the stoichiometry of the reaction to determine the equilibrium concentration of H2. Since 1 mol of H2 reacts with 1 mol of F2 to form 2 mol of HF, the concentration of H2 at equilibrium will be half of its initial concentration. Therefore, the equilibrium concentration of H2 in the 1.14-L flask would be 1.40 mol divided by 2 and then divided by 1.14 L, which gives us approximately 0.614 mol/L.

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The equilibrium concentration of H₂ is calculated to be 0.205 M.

To find the equilibrium concentration of H₂, we first need to set up an ICE (Initial, Change, Equilibrium) table:

Initial concentrations:

Changes in concentration:

Equilibrium concentrations:

According to the equilibrium constant expression (K):

[tex]K = [HF]^2/ [H_2][F_2][/tex]

Substitute the equilibrium concentrations into the expression:

[tex]1.00 \times 10^2 = (2x)^2 / (1.23 - x)(1.23 - x)[/tex]

This simplifies to:

[tex]100 = 4x^2 / (1.23 - x)^2[/tex]

Taking the square root of both sides:

10 = 2x / (1.23 - x)

Solve for x:

10(1.23 - x) = 2x

12.3 - 10x = 2x

12.3 = 12x

x = 1.025

The equilibrium concentration of H₂  is:

[H₂ ] = 1.23 - 1.025 = 0.205 M

Answer:

D. Carboxylate ion

Explanation:

The nitriles are hydrolyzed wiyh aqueous soda, under heating, to form carboxilates and ammonia:

H3C-C≡CN + NaOH(aq) → H3C-C=OONa+  + NH3

Upon the complete reaction of 29.6 grams of phosphoric acid with excess potassium hydroxide, approximately 64 grams of potassium phosphate will be produced.

To determine the amount of potassium phosphate that will be produced, we first need to find the molar mass of phosphoric acid (H3PO4), which is approximately 98 g/mol. So, 29.6 grams of phosphoric acid is equal to 0.302 moles (29.6g / 98g/mol).

The balanced equation shows that each mole of phosphoric acid will produce one mole of potassium phosphate (K3PO4) in the reaction. Therefore, 0.302 moles of phosphoric acid will produce 0.302 moles of potassium phosphate.

Next, we need to convert this amount from moles to grams. The molar mass of potassium phosphate is roughly 212 g/mol. So, 0.302 moles of K3PO4 is equal to approximately 64 grams (0.302 moles * 212 g / mol).

Therefore, upon the complete reaction of 29.6 grams of phosphoric acid with excess potassium hydroxide, approximately 64 grams of potassium phosphate will be formed.

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