A piece of unknown metal weighs 217 g . When the metal absorbs 1.43 kJ of heat, its temperature increases from 24.5°C to 39.1°C. Determine the specific heat of the metal.

Explanation:

The Lewis structures can be seen in the Figure below.

The ideal bond angle for both H-C-H and F-C-F is 120° because the have a triangular geometry.

The H-C-H has 117.4° angles due to the repulsion caused by the electrons of the double bond and the electrons of the H-C bond. This repulsion brings closer the H and decreases the angle.

On the other hand, in the F-C-F the same phenomenom happens but given that the F have lone pairs of electrons the repulsion is bigger, the F get closer and the angle is smaller than the H-C-H.

Final answer:

The Lewis structure for tetrafluoroethylene (C2F4) shows each carbon atom double-bonded to the other and single-bonded to two fluorine atoms, with an sp2 hybridization. The ideal bond angles for sp2 hybridized carbon are 120 degrees, but the actual angles differ due to variations in electronegativity and substituent size.

Explanation:

Polymerization and Lewis Structures

To draw the Lewis structure for the monomer tetrafluoroethylene (C2F4) used to make the polymer polytetrafluoroethylene (Teflon), we start by considering that carbon and fluorine both follow the octet rule, meaning that they both seek to be surrounded by eight electrons to reach a stable configuration. In C2F4, each carbon atom is double-bonded to the other carbon atom and single-bonded to two fluorine atoms. The hybridization of carbon in C2F4 is sp2 because it has three sigma bonds (one with the other carbon and two with fluorine atoms) and one pi bond (part of the double bond with the other carbon).

The ideal bond angles for an sp2 hybridized atom are 120 degrees. However, in the case of C2H4 and C2F4, the actual bond angles are 117.4 degrees for H-C-H and 112.4 degrees for F-C-F respectively. These deviations from the ideal angles can be attributed to the differences in electronegativity and size of the substituents, causing variations in electron cloud repulsions and thus slightly adjusting the bond angles.

Answer:

The molar mass of the gas is 140.86 g/mol

Explanation:

Step 1: Data given

Mass of the gas = 0.555 grams

Volume of the gas = 117 mL = 0.117 L

Temperature = 85 °C

Pressure = 0.99 atm

Step 2: Calculate the number of moles

p*V = n*R*T

⇒ with p = the pressure of the gas = 0.99 atm

⇒ with V = the volume = 0.117 L

⇒ with n = the number of moles

⇒ with R = the gas constant = 0.08206 L*atm/K*mol

⇒ with T = the temperature of the gas = 85 °C = 358.15 Kelvin

n = (p*V)/(R*T)

n = (0.99 * 0.117)/(0.08206 * 358.15)

n = 0.00394 moles

Step 3: Calculate molar mass of the gas

Molar mass = mass / moles

Molar mass = 0.555 grams / 0.00394 moles

Molar mass = 140.86 g/mol

The molar mass of the gas is 140.86 g/mol

Answer: 0.04 moles

Explanation:

pV = nRT

p=pressure = 0.99atm

v = volume = 117ml = 0.117L

n = number of moles = n

T = temperature = 85 +273 = 358k

R = rate constant  = 0.082 L.atm/kmol

n = pV/RT = 0.99 X 0.117/(0.082 X 358)

    = 0.004 moles

Answer:

There were 63.67 kJ liberated

If 1 mol of toluene was burned, there would be liberated 3906 kJ / mole

Explanation:

Step 1: Data given

Mass of the toluene sample = 1.500 grams

The temperature of the calorimeter rose from 25.000 °C to 26.413 °C

Write the balanced chemical equation for the reaction in the calorimeter.

C7H8(l)  +  9O2(g) ⇒  7CO2(g) + 4H2O(l)

Step 2: Calculate Q of the calorimeter

Q = C * Δt

Q = 45.06 kJ/ °C * ( 26.413 - 25)

Q = 63.67 kJ

Step 3: Calculate moles of toluene

Moles toluene = mass toluene / molar mass toluene

Moles toluene = 1.500 grams /92.14 g/mol

Moles toluene = 0.0163 moles

Step 4:  Calculate energy per mole

energy per mole = 63.67 kJ / 0.0163 moles = 3906 kJ / mole

Final answer:

The balanced chemical equation for the combustion of toluene is 2 C7H8 + 21 O2 -> 14 CO2 + 8 H2O. The heat liberated by the reaction in the calorimeter is 63.30 kJ. If 1.000 mol of toluene is burned, 0.507 kJ of heat will be liberated.

Explanation:

The balanced chemical equation for the combustion of toluene can be written as:

2 C7H8 + 21 O2 -> 14 CO2 + 8 H2O

To calculate the heat liberated by the reaction, we can use the formula:

q = Ccal * ΔT

where q is the heat liberated, Ccal is the heat capacity of the calorimeter, and ΔT is the change in temperature.

Plugging in the values, we get:

q = (45.06 kJ/°C) * (26.413 °C - 25.000 °C) = 63.30 kJ

To calculate the heat liberated when 1.000 mol of toluene is burned, we can use the molar mass of toluene to convert grams to moles.

The molar mass of toluene is 92.14 g/mol, so:

1.500 g * (1 mol / 92.14 g) = 0.016 mol

Using stoichiometry, we know that for every 2 moles of toluene burned, 63.30 kJ of heat is liberated.

Therefore, for 0.016 mol of toluene:

q = (63.30 kJ / 2 mol) * 0.016 mol = 0.507 kJ

Answer:

pH= 11.49

Explanation:

Ethanolamine is an organic chemical compound of the formula; HOCH2CH2NH2. Ethanolamine, HOCH2CH2NH2 is a weak base.

From the question, the parameters given are; the concentration of ethanolamine which is = 0.30M, pH value= ??, pOH value= ??, kb=3.2 ×10^-5

Using the formula below;

[OH^-]=√(kb×molarity)----------------------------------------------------------------------------------------------------------(1)

[OH^-] =√(3.2×10^-5 × 0.30M)

[OH^-]= √(9.6×10^-6)

[OH^-]=3.0984×10^-3

pOH= -log[OH^-]

pOH= -log 3.1×10^-3

pOH= 3-log 3.1

pH= 14-pOH

pH= 14-(3-log3.1)

pH= 11+log 3.1

pH= 11+ 0.4914

pH= 11.49

The pH of the ethanolamine solution has been 11.5.

The pH of the solution has been defined as the concentration of hydrogen ion in the solution.

The pH has been the difference of pOH from 14.

The ethanolamine has been a weak base. The OH concentration of ethanolamine has been given as:

[tex]\rm OH^-=\sqrt{Kb\;\times\;Molarity}[/tex]

The Kb of sample has been given as [tex]3.2\;\times\;10^-^5[/tex]

The molarity of sample has been, 0.3 M.

Substituting the values for hydroxide ion concentration:

[tex]\rm OH^-=\sqrt{3.2\;\times\;10^-^5\;\times\;0.3}\\OH^-=3.098\;\times\;10^-^3[/tex]

The pOH has been the negative logarithmic values of hydroxide ion concentration. The pOH has been given as:

[tex]\rm pOH=-log\;OH^-\\pOH=-log\;3.098\;\times\;10^-^3\\pOH=2.5[/tex]

The pOH of the sample has been 2.5.

The pH of ethanolamine sample has been given as:

[tex]\rm pH=14-pOH\\pH=14-2.5\\pH=11.5[/tex]

The pH of the ethanolamine solution has been 11.5.

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Answer:

The missing particle is [tex]_{103}^{259}\textrm{Lr}[/tex].

Explanation:

In a nuclear reaction, the total mass and total atomic number remains the same.

For the given fission reaction:

[tex]^{252}_{98}\textrm{Cf}+^{10}_{5}\textrm{B}\rightarrow ^A_Z\textrm{X}+3^1_0\textrm{n}[/tex]

To calculate A:

Total mass on reactant side = total mass on product side

252 +10 = A+ 3

A = 259

To calculate Z:

Total atomic number on reactant side = total atomic number on product side

98 +5 = Z + 0

Z = 103

Hence, the isotopic symbol of unknown element is [tex]_{103}^{259}\textrm{Lr}[/tex]

(a) [tex]\[ \frac{252}{98}\text{Cf} + \frac{10}{5}\text{B} \rightarrow 3 \frac{1}{0}\text{n} + \frac{259}{103}\text{Lr} \][/tex]

(b) [tex]\[ \frac{2}{1}\text{H} + \frac{3}{2}\text{He} \rightarrow \frac{4}{2}\text{He} + \frac{1}{1}\text{H} \][/tex]

(c) [tex]\[ \frac{1}{1}\text{H} + \frac{11}{5}\text{B} \rightarrow 3 \frac{4}{2}\text{He} \][/tex]

To complete and balance the nuclear equations, we need to ensure that both the atomic numbers (proton numbers) and mass numbers (nucleon numbers) are conserved on both sides of each equation.

(a) [tex]\( \frac{252}{98}\text{Cf} + \frac{10}{5}\text{B} \rightarrow 3 \frac{1}{0}\text{n} + ? \)[/tex]

First, determine the total atomic and mass numbers on the left side:

- Cf: Atomic number = 98, Mass number = 252

- B: Atomic number = 5, Mass number = 10

Sum of atomic numbers on the left side: 98 + 5 = 103

Sum of mass numbers on the left side: 252 + 10 = 262

On the right side, we have 3 neutrons:

- Each neutron: Atomic number = 0, Mass number = 1

- Total for neutrons: Atomic number = 0, Mass number = 3

So, the missing particle must balance the equation:

- Atomic number: 103 - 0 = 103

- Mass number: 262 - 3 = 259

Thus, the missing particle is [tex]\( \frac{259}{103}\text{Lr} \)[/tex] (Lawrencium).

The complete equation is:

[tex]\[ \frac{252}{98}\text{Cf} + \frac{10}{5}\text{B} \rightarrow 3 \frac{1}{0}\text{n} + \frac{259}{103}\text{Lr} \][/tex]

(b) [tex]\( \frac{2}{1}\text{H} + \frac{3}{2}\text{He} \rightarrow \frac{4}{2}\text{He} + ? \)[/tex]

First, determine the total atomic and mass numbers on the left side:

- H: Atomic number = 1, Mass number = 2

- He: Atomic number = 2, Mass number = 3

Sum of atomic numbers on the left side: 1 + 2 = 3

Sum of mass numbers on the left side: 2 + 3 = 5

On the right side, we have one helium-4 nucleus:

- He: Atomic number = 2, Mass number = 4

So, the missing particle must balance the equation:

- Atomic number: 3 - 2 = 1

- Mass number: 5 - 4 = 1

Thus, the missing particle is [tex]\( \frac{1}{1}\text{H} \)[/tex] (a proton).

The complete equation is:

[tex]\[ \frac{2}{1}\text{H} + \frac{3}{2}\text{He} \rightarrow \frac{4}{2}\text{He} + \frac{1}{1}\text{H} \][/tex]

(c) [tex]\( \frac{1}{1}\text{H} + \frac{11}{5}\text{B} \rightarrow 3? \)[/tex]

First, determine the total atomic and mass numbers on the left side:

- H: Atomic number = 1, Mass number = 1

- B: Atomic number = 5, Mass number = 11

Sum of atomic numbers on the left side: 1 + 5 = 6

Sum of mass numbers on the left side: 1 + 11 = 12

Let ?  be [tex]\( \frac{4}{2}\text{He} \) (alpha particle)[/tex]. Since we need 3 of these particles:

- Each He: Atomic number = 2, Mass number = 4

- Total for 3 He: Atomic number = 6, Mass number = 12

So, the 3 particles are 3 [tex]\( \frac{4}{2}\text{He} \)[/tex] (3 alpha particles).

The complete equation is:

[tex]\[ \frac{1}{1}\text{H} + \frac{11}{5}\text{B} \rightarrow 3 \frac{4}{2}\text{He} \][/tex]

Complete Question:

Complete and balance the following nuclear equations by supplying the missing particle:

(a) 252/98Cf + 10/5B -> 3 1/0n + ?

(b) 2/1H + 3/2He -> 4/2He + ?

(c) 1/1H + 11/5B -> 3?

Answer: The standard enthalpy of the reaction is -248.78 kJ/mol

Explanation:

The heat released by the reaction is absorbed by the calorimeter and the solution.

The chemical equation used to calculate the heat released follows:

[tex]q=c\times \Delta T[/tex]

where,

c = heat capacity of calorimeter = 448 J/K

[tex]\Delta T[/tex] = change in temperature = 21.1 K

Putting values in above equation, we get:

[tex]q=448J/K\times 21.1K=9452.8J[/tex]

Sign convention of heat:

When heat is absorbed, the sign of heat is taken to be positive and when heat is released, the sign of heat is taken to be negative.

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

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

Given mass of zinc = 2.50 g

Molar mass of zinc = 65.4 g/mol

Putting values in above equation, we get:

[tex]\text{Moles of zinc}=\frac{2.50g}{65.4g/mol}=0.038mol[/tex]

To calculate the standard enthalpy of the reaction, we use the equation:

[tex]\Delta H^o_{rxn}=\frac{q}{n}[/tex]

where,

[tex]q[/tex] = amount of heat released = -9452.8 J

n = number of moles = 0.038 moles

[tex]\Delta H^o_{rxn}[/tex] = standard enthalpy of the reaction

Putting values in above equation, we get:

[tex]\Delta H^o_{rxn}=\frac{-9452.8J}{0.038mol}=-248757.9J/mol=-248.78kJ/mol[/tex]

Conversion factor used:  1 kJ = 1000 J

Hence, the standard enthalpy of the reaction is -248.78 kJ/mol

Answer : The correct answer is "No solution".

Explanation :

As we are given that 38.1 % w/w solution of heptane in chloroform. That means 38.1 gram of heptane present in 100 gram of solution.

Now we have to determine the mass of solution for 65.0 grams of hepatne.

As, 38.1 grams of heptane present in 100 grams of solution

So, 65.0 grams of heptane present in [tex]\frac{65.0}{38.1}\times 100=171[/tex] grams of solution

Since, there is only 20.0 grams of solution available. That means, there is not enough solution.

Thus, the correct answer is "No solution".

Final answer:

The student would need 170.6g of the 38.1% w/w heptane solution to obtain 65.0g of heptane. However, the student only has 20.0g of solution, which is not enough to provide the required amount of heptane, so the student does not have a viable solution.

Explanation:

The student needs to calculate the mass of solution to use to obtain 65.0g of heptane from a 38.1% w/w solution of heptane in chloroform. The weight/weight percentage (w/w%) indicates that for every 100g of solution, there are 38.1g of heptane. To find the mass of the solution needed to obtain 65.0g of heptane, the following calculation can be performed:

m = (desired mass of heptane) / (percentage of heptane in solution as a decimal)

m = 65.0g / 0.381

m = 170.6g

The student would need to use 170.6g of the 38.1% w/w heptane solution to obtain the required amount of pure heptane. However, since the student only has 20.0g of solution, there is not enough to obtain 65.0g of heptane without obtaining more solution. Therefore, the student cannot proceed with the experiment as is and must either get more solution or adjust the experiment.

The theoretical yield of sodium chloride formed from the reaction of 3.3 g of hydrochloric acid and 6.7 g of sodium hydroxide is 5.3 g

How to obtain the theoretical yield of sodium chloride?

We shall begin by determining the limiting reactant for the reaction of 3.3 g of hydrochloric acid and 6.7 g of sodium hydroxide. This is shown below:

HCl + NaOH -> NaCl + H₂O

From the balanced equation above,

36.5 g of HCl reacted with 40 g of NaOH

Therefore,

3.3 g of HCl will react with = (3.3 × 40) / 36.5 = 3.6 g of NaOH

Now, we can see that only 3.6 g of NaOH reacted with 3.3 g of HCl

Thus, the limiting reactant is HCl

Finally, we shall calculate the theoretical yield of sodium chloride. Details below:

HCl + NaOH -> NaCl + H₂O

From the balanced equation above,

36.5 g of HCl reacted to produce 58.5 of NaCl

Therefore,

3.3 g of HCl will react to produce = [tex]\frac{3.3\ \times\ 58.5}{36.5}[/tex] = 5.3 g of NaCl

In conclusion, theoretical yield of sodium chloride is 5.3 g

Complete question:

Aqueous hydrochloric acid reacts with solid sodium hydroxide to produce aqueous sodium chloride and liquid water.

What is the theoretical yield of sodium chloride formed from the reaction of 3.3 g of hydrochloric acid and 6.7 g of sodium hydroxide.? Round your answer to significant figure

Answer:

The dipole moment of H-C bond will be smaller than that of an H-I bond.

Explanation:

The electronegativity of iodine is greater than that of hydrogen.As a result the iodine atom tends to attract the bond electron pair of H-I bond towards itself creating a bond dipole which does not occur in case of H-C bond as the electronegativity of carbon and hydrogen are almost same.

   That"s why dipole moment of H-C bond is smaller than that of H-I bond.

To calculate the mole fraction and partial pressure of each gas, we calculate the number of moles of each gas using the given masses and molar masses. The mole fraction is then determined by dividing the moles of each gas by the total moles. The partial pressure of each gas and the total pressure in the tank are found using the mole fraction and the given total pressure.

To calculate the mole fraction and partial pressure of each gas, we first need to calculate the number of moles of each gas. For dinitrogen difluoride (N2F2), we divide the given mass by the molar mass of N2F2 to get the number of moles.

Moles of N2F2 = 5.53 g / (28.01 g/mol) = 0.1977 mol

Similarly, for sulfur hexafluoride (SF6), the moles are calculated as follows:

Moles of SF6 = 17.3 g / (146.06 g/mol) = 0.1185 mol

The mole fraction of each gas is then calculated by dividing the moles of the gas by the total moles:

Mole fraction of N2F2 = 0.1977 mol / (0.1977 mol + 0.1185 mol) = 0.6258

Mole fraction of SF6 = 0.1185 mol / (0.1977 mol + 0.1185 mol) = 0.3742

To calculate the partial pressure of each gas, we use the formula:

Partial pressure = Mole fraction * Total pressure

Given that the total pressure is 26.9 C, we substitute the values to find:

Partial pressure of N2F2 = 0.6258 * 26.9 C = 16.82 C

Partial pressure of SF6 = 0.3742 * 26.9 C = 10.08 C

Finally, the total pressure in the tank is the sum of the partial pressures:

Total pressure = Partial pressure of N2F2 + Partial pressure of SF6

Total pressure = 16.82 C + 10.08 C = 26.9 C

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The mole fraction of N₂F₂ is 0.4141 with a partial pressure of 0.256 atm, while the mole fraction of SF₆ is 0.5859 with a partial pressure of 0.361 atm. The total pressure in the tank is 0.617 atm.

To solve this problem, we need to use the Ideal Gas Law and the concept of partial pressures.

Step 1: Calculate Moles of Each Gas

First, we calculate the number of moles for each gas:

Step 2: Total Moles in the Tank

Step 3: Calculate the Mole Fraction

Step 4: Calculate Partial Pressures

We use the Ideal Gas Law: PV = nRT

First, calculate the total pressure:

Then, calculate the partial pressures:

Summary

Answer:

All of these oxides of manganese have a mass percent of metal greater than 50%

Explanation:

Step 1: Data given

Molar mass of Mn = 54.94 g/mol

Molar mass of O = 16 g/mol

Step 2: Calculate mass % of Mn in MnO

Molar mass of MnO = 54.94 + 16 = 70.94 g/mol

% Mn = (54.94/70.94)*100 % = 77.45 %

Step 3: Calculate % of Mn in MnO2

Molar mass of MnO2 = 54.94 + 2*16 = 86.94 g/mol

% Mn = (54.94/86.94) * 100% = 63.19 %

Step 4: Calculate % of Mn in Mn2O3

Molar mass of Mn2O3 = 2*54.94 + 3*16 = 157.88 g/mol

% Mn = ((2*54.94)/157.88)*100 % = 69.6 %

All of these oxides of manganese have a mass percent of metal  greater than 50%

Answer:

The final pressure of the carbon dioxide gas will 11.84 atm.

Explanation:

Moles of carbonic acid = [tex]\frac{60.0 g}{62 g/mol}=0.9677 mol[/tex]

[tex]H_2CO_3(aq)\rightarrow H_2O(l) + CO_2(g)[/tex]

According to reaction, 1 mol of carbonic acid gives 1 mole of carbon dioxide gas.

Then 0.9677 moles of carbonic acid will give :

[tex]\frac{1}{1}\times 0.9677 mol=0.09677 mol[/tex] of carbon dioxide

Moles of carbon dioxide gas = n = 0.09677 mol

Volume of soda bottle = [tex]V_1=2 L[/tex]

Pressure of the carbon dioxide gas = P

Temperature of the carbon dioxide gas = T = 298 K

[tex]PV=nRT[/tex] (ideal gas law)

[tex]P=\frac{nRT}{V}=\frac{0.9677 mol\times 0.0821 atm L/mol K\times 298 K}{2 L}=11.84 atm[/tex]

The final pressure of the carbon dioxide gas will 11.84 atm.

By using the ideal gas law equation PV = nRT, calculating the number of moles of CO₂ formed from 60.0 grams of decomposed carbonic acid, and inputting the given values for the volume (V) and temperature (T), we calculated that the final pressure of the CO₂ in the soda bottle would be approximately 1.18 atm.

To calculate the final pressure of carbon dioxide in the soda bottle we must utilize the ideal gas law equation, which is PV = nRT. Where P stands for pressure, V is volume, n is the number of moles of gas, R is the gas constant and T is the temperature.

First, we need to determine the number 'n' of moles of CO₂ formed. Knowing that 60.0 grams of carbonic acid decompose completely into water and CO₂, and knowing that the molar mass of carbonic acid H₂CO₃ is approximately 62.03 g/mol, we can find that the number of moles (n) is equal to the mass (60g) divided by the molar mass (62.03 g/mol), so n equals about 0.967 moles.

Because it specifies that we should assume CO₂ to 'have the full 2.00 L in which to expand', the volume V in the ideal gas law equation is 2.00 L. Similarly, the temperature T is given as 298 K. We will now plug these values into the ideal gas law equation, using 0.0821 as the value for R (the ideal gas constant in units of L*atm/(mol*K)). This gives us P = (nRT) / V, so P is about 1.18 atm, which is the final pressure of CO₂.

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Answer:

Less

Explanation:

The bond angle in NH3 is greater than the bond angle in PH3.

According to the valence shell electron pair repulsion theory, the shape of molecules depends on the number of electron pairs that surround the central atom. The electronegativity of the central atom determines the bond angle of the molecule. The greater the electronegativitty of the central atom, the greater the bond angle.

Nitrogen is more electronegative than Phosphorus hence, the bond angle in NH3 is greater than the bond angle in PH3.

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Answer:

Heat of combustion = 5.6 ×10³ kj/mol

Explanation:

Given data:

Mass of sucrose = 1.010 g

Initial temperature = 24.92 °C

Final temperature = 28.33 °C

Heat capacity of calorimeter = 4.90 KJ/°C

Heat of combustion = ?

Solution:

ΔT = 28.33 °C -  24.92 °C = 3.41 °C

Q = - c. ΔT

Q = 4.90 KJ/°C . 3.41 °C

Q = - 16.7 kj

Number of moles of sucrose :

Number of moles of sucrose = mass/ molar mass

Number of moles of sucrose = 1.010 g / 342.3 g/mol

Number of moles of sucrose = 0.003 mol

Heat of combustion:

Heat of combustion = Q/n

Heat of combustion = - 16.7 kj/0.003 mol

Heat of combustion = -5.6 ×10³ kj/mol

The combustion of sucrose in a bomb calorimeter can be calculated using the formula for heat transfer: q = C * ΔT. Given the heat capacity is 4.90 kJ/°C and the change in temperature is 3.41 °C, the energy released per gram is 16.709 kJ. By multiplying this by the molecular weight of sucrose, over 342.3 g/mol, yields approximately -5.66 x 103 kJ/mole.

The combustion of sucrose in a bomb calorimeter can be determined through use of the formula for heat transfer, q = C x ΔT. Here, C represents the heat capacity of the calorimeter, and ΔT is the change in temperature. Given the heat capacity (C) is 4.90 kJ/°C, and the change in temperature (ΔT) is 28.33 °C - 24.92 °C = 3.41 °C, we plug these values into the equation yielding q = 4.90 kJ/°C x 3.41° C = 16.709 kJ. This is the energy released per gram. To find the energy released per mole, we need to multiply this value by the molecular weight of sucrose which is around 342.3 g/mole. This gives us an answer of -5.66 x 103 kJ/mole, and the negative sign indicates this is an exothermic reaction, meaning heat is released.

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In the ATP hydrolysis reaction, the O-P bond in ATP is broken, requiring a lot of energy. The OH-P bond formed in the reaction is strong and does not release energy.

In the ATP hydrolysis reaction, the O-P bond in ATP is broken, which requires a lot of energy. This breaking of the bond releases energy that is stored in the bond. Therefore, it takes a lot of energy to break the O-P bond in ATP.

The OH-P bond that is formed in the reaction is a strong bond. The formation of this bond is not responsible for releasing energy in the reaction.

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Final answer:

The breaking of the high-energy phosphate bond in ATP during hydrolysis releases energy. The OH-P bond formed in the reaction is strong, but its formation is not the source of energy release. The correct statement is 1) "The OH-P bond that is formed in the reaction is a strong bond."

Explanation:

ATP hydrolysis is an exothermic reaction where energy is released from the breaking of the high-energy phosphate bond in ATP, producing ADP and inorganic phosphate (Pi). The bond between the phosphates in ATP (phosphoanhydride bonds) is considered high-energy, not because it takes a lot of energy to break, but because the products of the reaction (ADP and Pi) have considerably lower free energy than ATP and a water molecule. Hence, the breaking of the O-P bond in ATP releases the energy that was stored in the bond.

The OH-P bond that forms when Pi is created during ATP hydrolysis is actually a strong bond, and the formation of this bond is not responsible for the release of energy, rather it is the breaking of the high-energy bond in ATP that releases energy.

Answer: [tex]2.2326\times 10^{-3}[/tex] moles

Explanation:

We are given:

Moles of electron = 1 mole

According to mole concept:

1 mole of an atom contains [tex]6.022\times 10^{23}[/tex] number of particles.

We know that:

Charge on 1 electron = [tex]1.6\times 10^{-19}C[/tex]

Charge on 1 mole of electrons = [tex]1.6\times 10^{-19}\times 6.022\times 10^{23}=96500C[/tex]

The metal being plated has a +4 charge, thus the equation will be:

[tex]M^{4+}+4e^-\rightarrow M[/tex]

[tex]4\times 96500C[/tex] of electricity deposits = 1 mole of metal

Thus 861.8 C of electricity deposits =[tex]\frac{1}{4\times 96500}\times 861.8=2.2326\times 10^{-3}[/tex] moles of metal

Thus [tex]2.2326\times 10^{-3}[/tex] moles of metal should be plated

The number of moles of the metal deposited is 0.0022 moles of metal.

The term electroplating refers to the use of one metal to cover the surface of another metal. Let the metal in question be M, the equation of the reaction is; M^4+(aq) + 4e -----> M(s).

1 mole of the metal is deposited by 4( 96,500) C

x moles will be  deposited by 861.8 C

x = 1 mole * 861.8 C/ 4( 96,500) C

x = 0.0022 moles of metal

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Answer:

A. 433 years and 63 days

B. Iodine

C. 0.125mg

D. 1.00mg

Explanation:

A. Americium

The formula of a radioactive decay constant half life is t = 0.693/k

Where k is the decay constant.

For americium, k = 0.0016

t = 0.693/0.0016 = 433.125 apprx 433 years

For iodine, k = 0.011

Half life = 0.693/0.011 = 63 days.

B. Iodine decays at a faster rate.

C. After three half lives

For both, first half life yields a mass of 0.5mg, next yields 0.25mg, next yield 0.125mg

D. 1.00mg still remains.

Half life is high for both so the decay after one day is insignificant.

a) The half-lives of Americium-241 and Iodine-125 are 433 years and 63 days, respectively.

b) Iodine-125 decays faster.

c) After three half-lives, 0.125 mg remains.

d) After 4 days, nearly 1.00 mg of Americium-241 and 0.957 mg of Iodine-125 remain.

To answer the given questions regarding the radioactive decay of Americium-241 and Iodine-125:

(a) Half-lives

For a first-order reaction, the half-life (t1/2) can be calculated using the equation:

t1/2 = 0.693 / k

For Americium-241:
k = 1.6 × 10⁻³ yr⁻¹
t1/2 = 0.693 / (1.6 × 10⁻³) = 433 years

For Iodine-125:
k = 0.011 day⁻¹
t1/2 = 0.693 / 0.011 = 63 days

(b) Decay Rate

Since the half-life of Iodine-125 is shorter (63 days) compared to Americium-241 (433 years), Iodine-125 decays at a faster rate.

(c) Remaining Sample After Three Half-lives

After three half-lives, the remaining amount of a radioactive isotope can be calculated using:

Remaining Amount = Initial Amount / (23)
Remaining Amount = 1.00 mg / 8 = 0.125 mg

Therefore, for both isotopes, 0.125 mg remains after three half-lives.

(d) Remaining Sample After 4 Days

For Americium-241 (with t1/2 about 433 years), 4 days is negligible, so nearly the entire 1.00 mg remains.

For Iodine-125, using the first-order decay formula:

N = N₀ e-kt

t = 4 days, k = 0.011 day⁻¹

N = 1.00 mg × e-(0.011 × 4) ≈ 1.00 mg × e-0.044 ≈ 1.00 mg × 0.957 = 0.957 mg

Thus, after 4 days, 0.957 mg of Iodine-125 remains.

Answer:

The correct option is: a. The internal energy depends upon its temperature.

Explanation:

Ideal gas is a hypothetical gas that obeys the ideal gas law. The equation for the ideal gas law:

P·V=n·R·T

Here, V- volume of gas, P - total pressure of gas, n- total mass or number of moles of gas, T - absolute temperature of gas and R- the gas constant

Also, according to the Joule's second law, the internal energy (U) of the given amount of ideal gas depends on the absolute temperature (T) of the gas only, by the equation:

[tex]U = c_{V}nRT[/tex]

Here, [tex]c_{V}[/tex] is the specific heat capacity at constant volume

The internal energy of an ideal gas depends on its temperature and is independent of its volume and pressure. For non-ideal systems, internal energy can be influenced by volume and pressure but only through their effect on temperature.

The internal energy of an ideal gas depends primarily on its temperature. This is because, for an ideal gas, the internal energy is a function of state and is determined by the kinetic energy of its particles, which is directly related to temperature. The volume and pressure of an ideal gas do not directly determine its internal energy, although changes in these properties can indirectly affect temperature and thus internal energy. Option A

In the case of an ideal gas, the internal energy does not change with volume or pressure shifts alone; it is the temperature that directly impacts internal energy. However, for other types of systems or under non-ideal conditions, both volume and pressure can have more complex relationships with internal energy.

The volume of KOH must be added to 500 ml of the acidic solution to neutralize all of the acids is equal to 968 ml.

A neutralization reaction can be described as a chemical reaction in which an acid and base react with each other to produce salt and water.

The neutralization reaction between HCl and NaOH:

HCl    +   KOH   →  KCl   +   H₂O

The concentration of HCl solution = 0.100M

The volume of the solution =  500 ml = 0.5 L

The number of moles of HCl = 0.100 × 0.5 = 0.05 mol

The neutralization reaction between H₂SO₄ and KOH:

H₂SO₄    +  KOH   →   K₂SO₄   +   2H₂O

The concentration of H₂SO₄ solution = 0.100M

The volume of the acidic solution =  500 ml = 0.5 L

The number of moles of H₂SO₄ = 0.210 × 0.5 = 0.105 mol

Total number of moles needed to neutralize = 0.05 + 0.105 = 0.155 mol

The concentration of KOH solution = 0.150 M

The volume of the KOH solution = M/n

V = 0.150/0.155

V = 0.968 L

V = 968 ml

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Answer:

There are initially 0.1 millimoles of acid (HCl) in the beaker.

Explanation:

Step 1: Data given

volume of 0.010 M HCl = 10.0 mL = 0.01 L

Step 2: Calculate the number of moles HCl

Number of moles =  Molarity HCl * volume of solution

Number of moles HCl = 0.010 M * 0.01 L = 0.0001 moles HCl = 0.1 *10^-3 mol = 0.1 mmol HCl

There are initially 0.1 millimoles of acid (HCl) in the beaker.

The initial beaker contains 0.1 millimoles of HCl. The process of adding NaOH to HCl is called titration, which helps measure the concentration of a solution and involves achieving an equivalence point where all reactant molecules have reacted.

The question is asking about the quantity of acid initially present in a beaker before the addition of sodium hydroxide. The millimoles of acid can be calculated using the formula: Molarity (M) = Moles/Liter. For this experiment, you have an initial volume of 10.0 mL (or 0.010 L) of a 0.010 M HCl solution. Therefore, you can calculate the millimoles of acid in the beaker by multiplying the molarity by the volume in liters: 0.010 moles/L * 0.010 L = 0.0001 moles or 0.1 millimoles of HCl.

Titration, mentioned in the references, is a technique used to find the concentration of a solution, where a reaction is created that can easily be observed when the amount of reactant in a solution is exactly sufficient to react with another compound. In your experiment, NaOH is used to titrate HCl. The point when all the molecules of one reactant have reacted with another is called the equivalence point, which you can determine from a titration curve.

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Answer:

The Ka for this acid is 1.4 * 10^-4 (option C)

Explanation:

Step 1: Data given

In a 0.20M aqueous solution, lactic acid is 2.6% dissociated.

Step 2: The equation for the dissociation of HAc is:

HAc ⇌ H+ + Ac¯

The Ka expression is:

Ka = ([H+] [Ac¯]) / [HAc]

1) [H+] using the concentration and the percent dissociation:

(0.026) (0.2) = 0.0052 M

2) Calculate [Ac-] and [H+]

[Ac¯] = [H+] = x = 0.0052 M

3) Calculate [HAc]

[HAc] = 0.20M - x ( since x << 0.20, we can assume [HAc]  = 0.20 M

4) Calculate Ka

Ka= [( 0.0052) ( 0.0052)] / 0.20

Ka = 0.0001352 = 1.4 * 10^-4

The Ka for this acid is 1.4 * 10^-4

Final answer:

The value of Ka for lactic acid, given that a 0.20 M solution is 2.6% dissociated, can be calculated using the concentration of dissociated ions; the closest value to our calculation is 1.4x10^-4.

Explanation:

When you exercise, the burning sensation in your muscles is often due to the buildup of lactic acid. To find the acid dissociation constant (Ka) for lactic acid given a 0.20 M solution that is 2.6% dissociated, we use the dissociation equation of a weak acid:

HC₃H₅O₃ → H⁺ + C₃H₅O₃⁻

The concentration of H⁺ and C₃H₅O₃⁻ will also be 0.0052 M, since it's a one-to-one dissociation. Thus, the expression for Ka becomes:

The closest answer to our calculated Ka value is C.) 1.4x10^-4.

Answer:

The true stress required = 379 MPa

Explanation:

True Stress is the ratio of the internal resistive force to the instantaneous cross-sectional area of the specimen. True Strain is the natural log to the extended length after which load applied to the original length. The cold working stress – strain curve relation is as follows,

σ(t) = K (ε(t))ⁿ, σ(t) is the true stress, ε(t) is the true strain, K is the strength coefficient and n is the strain hardening exponent

True strain is given  by

Epsilon t =㏑ (l/l₀)

Substitute㏑(l/l₀) for ε(t)

σ(t) = K(㏑(l/l₀))ⁿ

Given values l₀ = 49.7mm, l =51.7mm , n =0.2 , σ(t) =379Mpa

379 x 10⁶ = K (㏑(51.7/49.7))^0.2

K = 379 x 10⁶/(㏑(51.7/49.7))^0.2

K = 723.48 MPa

Knowing the constant value would be same as the same material is being used in the second test, we can find out the true stress using the above formula replacing the value of the constant.

σ(t) = K(㏑(l/l₀))ⁿ

l₀ = 49.7mm, l = 51.7mm, n = 0.2, K = 723.48Mpa

σ(t) = 723.48 x 106 x (㏑(51.7/49.7))^0.2

σ(t) = 379 MPa

The true stress necessary to plastically elongate the specimen is 379 MPa.

Answer:i ea

Explanation:

Answer:

-) (2-chloroethyl)cyclopropane

-) (1-chloroethyl)cyclopropane

-) 1-chloro-2-ethylcyclopropane

-) 1-chloro-1-ethylcyclopropane

Explanation:

The monochlorination are products in which Cl is added only 1 time. Now, in the ethylcyclopropane we will have 5 carbons, so we will have 5 options for monochlorination (see figure 1). But, for carbons 2 and 3 we will have equivalent structures, therefore at the end, we will have only 4 possible structures. In carbon 5, carbon 4, carbon 3 and carbon 1 (see figure 2).

Final answer:

Monochlorination of ethylcyclopropane can produce several isomers, based on the position where the chlorine atom is added - either on the cyclopropane ring or the ethyl group attached to it.

Explanation:

The subject of the question is the monochlorination products from the free-radical chlorination of ethylcyclopropane. During the free-radical chlorination, chlorine radicals can react with the different types of hydrogens in ethylcyclopropane - those on the cyclopropane ring as well as those on the ethyl group. Chlorination at the ring will yield two isomers due to two different types of hydrogens (geminal and vicinal), while chlorination on the ethyl group could occur at either one of the methylene hydrogens or the methyl group hydrogens. When considering the ethyl group, monochlorination would give a 1-chloroethylcyclopropane or a 2-chloroethylcyclopropane, depending on which hydrogen is substituted. With regard to the cyclopropane ring, substitution could lead to two different chlorocyclopropanes due to the different positions of hydrogen on the ring.

Answer:

The water lost is 36% of the total mass of the hydrate

Explanation:

Step 1: Data given

Molar mass of CuSO4*5H2O = 250 g/mol

Molar mass of CuSO4 = 160 g/mol

Step 2: Calculate mass of water lost

Mass of water lost = 250 - 160 = 90 grams

Step 3: Calculate % water

% water = (mass water / total mass of hydrate)*100 %

% water = (90 grams / 250 grams )*100% = 36 %

We can control this by the following equation

The hydrate has 5 moles of H2O

5*18. = 90 grams

(90/250)*100% = 36%

(160/250)*100% = 64 %

The water lost is 36% of the total mass of the hydrate

Answer:

The heat produced is -15,1kJ

Explanation:

For the reaction:

2SO₂+O₂ → 2SO₃

The enthalpy of reaction is:

ΔHr = 2ΔHf SO₃ - 2ΔHf SO₂

As ΔHf SO₃ = -395,7kJ and ΔHf SO₂ = -296,8kJ

ΔHr = -197,8kJ

Using n=PV/RT, the moles of reaction are:

[tex]n = \frac{1,00atm*3,75L}{0,082atmL/molK*298,15K}[/tex] = 0,153 moles of reaction

As 2 moles of reaction produce -197,8kJ of heat, 0,153moles produce:

0,153mol×[tex]\frac{-197,8kJ}{2mol}[/tex] = -15,1kJ

I hope it helps!

14.8 KJ/mol of heat energy is produced when a volume of 3.75 L of SO2(g) is converted to 3.75 L of SO3(g).

The equation of the reaction is;

2SO2(g) + O2(g) ⇄ 2SO3(g)

We have the following information;

ΔHf SO2(g) = -296.8 KJ/mol

ΔHf O2(g) = 0 KJ/mol

ΔHf SO3(g) =  -395.7 kJ/mol

We can calculate the heat of reaction ΔHrxn from;

ΔHrxn = [ΔHf (products) - ΔHf (reactants)]

ΔHrxn = [2( -395.7 kJ/mol)] - [2(-296.8 KJ/mol) + 0 KJ/mol]

ΔHrxn = (-791.4  kJ/mol) + 593.6 KJ/mol

ΔHrxn = -197.8 KJ/mol

We can find the number of moles of SO2 reacted using the ideal gas equation;

P = 1.00 atm

T = 25.0°C + 273 = 298 K

n = ?

V = 3.75 L

R = 0.082 atmLK-1mol-1

So,

PV = nRT

n = PV/RT

n = 1.00 atm × 3.75 L/0.082 atmLK-1mol-1 ×   298 K

n = 0.15 moles

If 2 moles of SO2 produced -197.8 KJ/mol

0.15 moles of SO2 will produce  0.15 moles  × (-197.8 KJ/mol)/ 2 moles

= -14.8 KJ/mol

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Answer:

A mixture of 100. mL of 0.1 M HC3H5O3 and 50. mL of NaOH

Explanation:

The pH of a buffer solution is calculated using following relation

[tex]pH=pKa+log(\frac{salt}{acid} )[/tex]

Thus the pH of buffer solution will be near to the pKa of the acid used in making the buffer solution.

The pKa value of HC₃H₅O₃ acid is more closer to required pH = 4 than CH₃NH₃⁺ acid.

pKa = -log [Ka]

For HC₃H₅O₃

pKa = 3.1

For CH₃NH₃⁺

pKa = 10.64

pKb = 14-10.64 = 3.36 [Thus the pKb of this acid is also near to required pH value)

A mixture of 100. mL of 0.1 M HC3H5O3 and 50. mL of NaOH

Half of the acid will get neutralized by the given base and thus will result in equal concentration of both the weak acid and the salt making the pH just equal to the pKa value.

The branch of science that deals with the chemicals and the bond are called chemistry. there are two types of solution out of which one is acidic and the other is basic

The correct answer is C.

The equation used to solve the question is as follows:-

[tex]pH = pKa +log\frac{salt}{acid}[/tex]

Thus the pH of the buffer solution will be near to the pKa of the acid used in making the buffer solution because the salt is used in the question with respect to the acid.

The pKa value of [tex]HC_3H_5O_3[/tex]acid is closer to the required pH = 4 than [tex]CH_3NH_3^+[/tex]acid. [tex]pKa = -log [Ka]HC_3H_5O_3[/tex] , the pKa is 3.1

[tex]CH_3NH_3^+[/tex]the pKa is 10.64

The total pKb value is

= 14-10.6

= 3.36

Hence, the correct answer is C which is A mixture of 100 mL of 0.1 M HC3H5O3 and 50. mL of NaOH

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Answer: A

Explanation:

Please check the attachment for the answer.

Answer:

a. Isopropyl pentanoate.

Explanation:

Hello,

In this case, such esterification is:

[tex]CH_3CH_2CH_2CH_2COOH+OHCH(CH_3)_2\rightarrow CH_3CH_2CH_2CH_2COOCH(CH_3)_2+H_2O[/tex]

Wherein the first reactant is the pentanoic acid, the second one the isopropanol and the first product is the result, which is a. isopropyl pentanoate. This is because the isopropyl get separated from the hydroxile and bonds with the carboxile group of the pentanoic acid by also releasing water as the byproduct.

Best regards.

Answer:

The statements are definitions to chromatography terms which have been highlighted below.

Explanation:

Match the chromatography term with its definition.

Volumetric Flow Rate = The volume of solvent traveling through the column per unit time.

Retention time = The elapsed time between sample injection and detection.

Adjusted Retention Time = The time required by a retained solute to travel through the column beyond the time required by the un -retained solvent.

Linear Flow Rate = The distance traveled by the solvent per unit time.

Retention factor = Describes the amount of time that a sample spends in the stationary phase relative to the mobile phase. It is sometimes also called the capacity factor or capacity ratio.

Relative Volume = Volume of the mobile phase required to elute a solute from the column.

Relative Retention = Ratio of the adjusted retention times or retention factors of two solutes. It is sometimes also called the separation factor.

Partition coefficient = The ratio of the solute concentrations in the mobile and stationary phases.

Chromatography is a laboratory technique used to separate and analyze the components of a mixture based on their different affinities for a stationary phase and a mobile phase. The different terms in chromatography can be defined as given below-

The volume of solvent traveling through the column per unit time - Flow rate

The elapsed time between sample injection and detection - Retention time

The time required by a retained solute to travel through the column beyond the time required by the unretained solvent - Adjusted retention time

The distance traveled by the solvent per unit time - Velocity

Describes the amount of time that a sample spends in the stationary phase relative to the mobile phase. It is sometimes also called the capacity factor or capacity ratio - Retention factor

Volume of the mobile phase required to elute a solute from the column - Elution volume

Ratio of the adjusted retention times or retention factors of two solutes. It is sometimes also called the separation factor - Selectivity factor

The ratio of the solute concentrations in the mobile and stationary phases - Distribution coefficient

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Answer:

kf = 1.16 x 10¹⁸

Explanation:

Step 1: [Ni(H₂O)₆]²⁺  + 1en → [Ni(H₂O)₄(en)]²⁺  ΔG°1 = -42.9 kJmol⁻¹

Step 2: [Ni(H₂O)₄(en)]²⁺  + 1en → [Ni(H₂O)₂(en)₂]²⁺  ΔG°2 = -35.8 kJmol⁻¹

Step 3: [Ni(H₂O)₂(en)₂]²⁺ + 1en →  [Ni(en)₃]²⁺  ΔG°3 = -24.3 kJmol⁻¹

________________________________________________________

Overall reaction: [Ni(H₂O)₆]²⁺  + 3en → [Ni(en)₃]²⁺  ΔG°r

ΔG°r = ΔG°1 + ΔG°2 + ΔG°3

ΔG°r = -42.9 - 35.8 - 24.3

ΔG°r = -103.0 kJmol⁻¹

ΔG°r = -RTlnKf

-103,000 Jmol⁻¹ =  - 8.31 J.K⁻¹mol⁻¹ x 298 K x lnKf

kf = e ^(-103,000/-8.31x298)

kf = e ^41.59

kf = 1.16 x 10¹⁸

Answer:

60.0 years must pass to reduce a 24 mg of cesium 237 to 6.0 mg

Explanation:

For radioactive decay of a radioactive nuclide-

         [tex]N_{t}=N_{0}(\frac{1}{2})^{(\frac{t}{t_{\frac{1}{2}}})}[/tex]

Where, [tex]N_{t}[/tex] is amount of radioactive nuclide after "t" time , N_{0} is initial amount of radioactive nuclide and [tex]t_{\frac{1}{2}}[/tex] is half-life of radioactive nuclide

Here N_{0} = 24 mg, N_{t} = 6.0 mg and [tex]t_{\frac{1}{2}}[/tex] = 30.0 years

So, [tex]6.0mg=24mg\times (\frac{1}{2})^{(\frac{t}{30.0years})}[/tex]

or, [tex]t=60.0years[/tex]

So 60.0 years must pass to reduce a 24 mg of cesium 237 to 6.0 mg