Germanium is a group 4A semiconductor. The addition of a dopant atom (group 3A element) that has fewer valence electrons than the host atom results in a p-type semiconductor. The addition of a dopant atom (group 5A elements) that has more valence electrons than the host atom results in an n-type semiconductor. Which of the following elements when used for doping germanium will yield p-type semiconductors, and which elements when used for doping germanium will yield n-type semiconductors?

Answer: check explanation

Explanation:

The balanced equation of reaction is given below;

HCl + NH3 -----------------------> NH4Cl.

We are given the volume in milliliters, let us convert them into Litres;

= 38.30 × 10^-3 Litres.

Here, we have an incomplete question (one parameter is missing- the volume of ammonia,NH3). Therefore, we assume that the volume of Ammonia, NH3 is 10mL(10× 10^-3 Litres).

Step one: we need to calculate the number of moles of HCl.

Number of moles of HCl= molarity × volume.

Number of moles of HCl= 0.507 M × 38.30× 10^-3 L.

Number of moles of HCl= 0.0194181 moles.

From the equation of reaction above, we have that one mole of ammonia is reacting with one mole of Hydrogen Chloride, HCl. Hence, the number of moles of ammonia is equal to the number of moles of Hydrogen Chloride,HCl.

Step two: calculate the molarity of Ammonia, NH3.

The molarity of ammonia= number of moles of ammonia/ volume of Ammonia,NH3.

Molarity of Ammonia= 0.0194181/10× 10^-3 moles NH3.

Molarity of Ammonia= 0.00000194181.

Molarity of Ammonia = 1.94181× 10^-6 M.

The molarity of ammonia after the 2nd addition can be calculated by equating the moles of HCl to the moles of ammonia, as the molar ratio in the reaction equation is 1:1. The equilibrium molarity of ammonia is the moles of ammonia divided by the total volume of the solution (which isn't stated in the question).

In this titration analysis, you've used 0.507 M HCl and 38.30 mL (or 0.03830 L) to neutralize the ammonia. To calculate the molarity of the ammonia, you will be using the reaction equation NH3 + HCl -> NH4Cl. The molar ratio is 1:1, meaning 1 mole of HCl reacts with 1 mole of NH3.

First, we need to find the moles of HCl, which can be obtained by multiplying the HCl molarity by the volume (in liters). So the moles of HCl = 0.507 M * 0.0383 L = 0.01942 moles. This is also the moles of ammonia at equilibrium assuming all of it was neutralized by the HCl.

The equilibrium molarity of ammonia, which is the concentration of ammonia after the 2nd addition, can be calculated by dividing the moles of ammonia by the total volume of the solution. However, the total volume of the solution isn't provided in the question. If it had been stated, you could simply plug in that value (in liters) to get the molarity of ammonia.

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Answer:  A. 1.5 g

Explanation:

[tex]N_2+3H_2\rightarrow 2NH_3[/tex]

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

[tex]\text{Number of moles of nitrogen}=\frac{7.0g}{28g/mol}=0.25 moles[/tex]

According to stoichiometry:

1 mole of [tex]N_2[/tex] requires = 3 moles of [tex]H_2[/tex]

Thus 0.25 moles of [tex]N_2[/tex] will require =[tex]\frac{3}{1}\times 0.25=0.75moles[/tex] of [tex]H_2[/tex]

Mass of [tex]H_2[/tex] required =[tex]moles\times {\text {Molar mass}}=0.75mol\times 2g/mol=1.5g[/tex]

The minimum amount of [tex]H_2[/tex] in grams that would be required to completely react with this amount of [tex]N_2[/tex] is 1.5 grams.

Answer:

The correct answer is option A.

Explanation:

[tex]N_2 + 3 H_2\rightarrow 2 NH_3[/tex]

Moles of nitrogen gas = [tex]\frac{7.0 g}{28 g/mol}=0.25 mol[/tex]

According to reaction, 1 mole of nitrogen reacts with 3 moles of hydrogen gas.

Then 0.25 moles of nitrogen gas will react with:

[tex]\frac{3}{1}\times 0.25 mol=0.75 mol[/tex] of hydrogen gas.

Mass of 0.75 moles of hydrogen gas = 0.75 mol × 2 g/mol = 1.5 g

1.5 grams of hydrogen that would be required to completely react with this amount of nitrogen.

Answer:

n=1,2,3,4

Explanation:

The electronic configuration of vandanium is

1s2 2s2 2p6 3s2 3p6 3d3 4s2

There are two electrons in n=1, eight electrons in n=2, eleven electrons in n=3 and two electrons in n=4

look at image attached

Answer:

Solid KOH will corrode the skin but a solution of KOH in triethyleneglycol will not.

The temperature is monitored with an external thermometer also because the sand looses some heat to energy exchange with the surrounding.

Explanation:

KOH is a deliquescent solid which is very corrosive upon contact with skin. Its solution in an organic liquid is not corrosive.

Secondly, when exposed to the surrounding, heat is lost according to the laws of thermodynamics.

Answer:

HCN

Explanation:

Acids are the species which furnish hydrogen ions in the solution or is capable of forming bonds with electron pair species as they are electron deficient species.

When an acid donates a proton, it changes into a base which is known as its conjugate base.

Bases are the species which furnish hydroxide ions in the solution or is capable of forming bonds with electron deficient species as they are electron rich species. When a base accepts a proton, it changes into a acid which is known as its conjugate acid.

The acid and the base which is only differ by absence or presence of the proton are known as acid conjugate base pair.

Also, the strongest acid leads to the weakest conjugate base and vice versa.

Thus, Out of HI, HCN, [tex]HNO_3[/tex], [tex]HClO_4[/tex] and HCl , the weakest acid is:- HCN

Thus, HCN corresponds to the strongest conjugate base.

Final answer:

The initial rate of formation of P2 in the dimerization reaction of a protein with a concentration of 1.6 × 10^-4 M is approximately 1.59 × 10^-10 M/s.

Explanation:

To calculate the initial rate of formation of P2 in the dimerization reaction of a protein, we can use the rate equation:

rate = k[P]^2

where k is the rate constant and [P] is the concentration of the protein.

Plugging in the given values:

rate = (6.2 × 10^-3 M^-1s^-1)(1.6 × 10^-4 M)^2

Simplifying:

rate = 6.2 × 10^-3 M^-1s^-1 * (1.6 × 10^-4 M)^2

rate = 6.2 × 10^-3 M^-1s^-1 * 2.56 × 10^-8 M^2

rate ≈ 1.59 × 10^-10 M/s

Therefore, the initial rate of formation of P2 is approximately 1.59 × 10^-10 M/s.

Answer:

a) H2O2 is the limiting reactant

b) There will remain 0.450 moles of N2H4

c) There will be produced 0.250 moles of N2 and 1 mol of H2O

Explanation:

Step 1: Data given

Number of moles of N2H4 = 0.7 moles

Number of moles H2O2 = 0.500 moles

Molar mass of N2H4 = 32.05 g/mol

Molar mass of H2O2 = 34.01 g/mol

Step 2: The balanced equation

N2H4 + 2H2O2 → N2 + 4H2O

Step 3: Calculate the limiting reactant

For 1 mol of N2H4 we need 2 moles of H2O2 to produce 1 mol of N2 and 4 moles of H2O

H2O2 is the limiting reactant. It will completely be consumed. (0.500 moles).

N2H4 is in excess. There will react 0.500/2 = 0.250 moles of N2H4

There will remain 0.700 - 0.250 moles = 0.450 moles of N2H4

Step 4: Calculate moles of products

For 1 mol of N2H4 we need 2 moles of H2O2 to produce 1 mol of N2 and 4 moles of H2O

For 0.500 moles of H2O2. we'll have 0.250 moles of N2 and 1 mol of H2O

Answer:

Explanation:

(a). Hydrogen peroxide is the limiting reactant, because from the rxn the mole ratio is 1:2, therefore, 0.7mol of N2H4 is supposed to react with 1.4mol of H2O2.

(b). From the mole ratio, 0.5mol of H2O2 will react with 0.25mol of N2H4. Therefore, the unused mol of N2H4 will be (0.7-0.25)mol

=0.45mol

(c). 0.25mol of N2 and 1.0mol of H20 are the products formed based on the mole ratio from the rxn.

The intermolecular forces between a formaldehyde molecule and a hydrogen sulfide molecule are dipole-dipole forces and London dispersion forces. Both forces occur due to the polar nature of these molecules and the temporary creation of dipoles.

The types of intermolecular forces that act between a formaldehyde (H2CO) molecule and a hydrogen sulfide (H2S) molecule are dipole-dipole forces and London dispersion forces. In this case, both molecules are polar, meaning they have uneven charge distribution. Therefore, the positive end of one molecule is attracted to the negative end of the other, resulting in dipole-dipole forces. Both molecules, being non-ideal gases, exhibit London dispersion forces, which are weak, temporary attractions occurring when the electrons in two adjacent atoms occupy positions that make the atoms form temporary dipoles.

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Answer:

1) ΔG°r(298 K) = - 28.619 KJ/mol

2) ΔG°r will decrease with decreasing temperature

Explanation:

1) ΔG°r = ∑νiΔG°f,i

⇒ ΔG°r(298 K) = ΔG°CO2(g) + ΔG°H2(g) - ΔG°H2O(g) - ΔG°CO(g)

from literature, T = 298 K:

∴ ΔG°CO2(g) = - 394.359 KJ/mol

∴ ΔG°CO(g) = - 137.152 KJ/mol

∴ ΔG°H2(g) = 0 KJ/mol........pure substance

∴ ΔG°H2O(g) = - 228.588 KJ/mol

⇒ ΔG°r(298 K) = - 394.359 KJ/mol + 0 KJ/mol - ( - 228.588 KJ/mol ) - ( - 137.152 KJ7mol )

⇒ ΔG°r(298 K) = - 28.619 KJ/mol

2) K = e∧(-ΔG°/RT)

∴ R = 8.314 E-3 KJ/K.mol

∴ T = 298 K

⇒ K = e∧(-28.619/(8.314 E-3)(298) = 9.624 E-6

⇒ ΔG°r = - RTLnK

If T (↓) ⇒ ΔG°r (↓)

assuming T = 200 K

⇒ ΔG°r(200 K) = - (8.314 E-3)(200)Ln(9.624E-3)

⇒ ΔG°r (200K) = - 19.207 KJ/mol < ΔG°r(298 K) = - 28.619 KJ/mol

Final answer:

The ΔG° at 298 K for the given reaction is 130.0 kJ/mol. Lowering the temperature will decrease ΔG° because the reaction is exothermic.

Explanation:

Delta G (ΔG°) at 298 K can be calculated using the equation ΔG° = ΔH° - TΔS°, where ΔH° is the standard enthalpy change, T is the temperature in Kelvin, and ΔS° is the standard entropy change of the reaction. Substitute the given values into the equation to obtain ΔG°. For the given reaction, CO(g) + H2O(g) → H2(g) + CO2(g), ΔG° = 177.8 kJ/mol - (298 K * 0.1605 kJ/K mol) = 130.0 kJ/mol.

The effect of lowering the temperature on ΔG° can be determined by understanding how changes in temperature affect the equilibrium constant (K) of the reaction. According to Le Chatelier's principle, if the reaction is exothermic (negative ΔH°), a decrease in temperature will cause the equilibrium to shift towards the products, leading to a decrease in ΔG°. On the other hand, if the reaction is endothermic (positive ΔH°), a decrease in temperature will cause the equilibrium to shift towards the reactants, leading to an increase in ΔG°. In this case, since the reaction is exothermic, ΔG° will decrease with decreasing temperature.

The energy required to break the π bond in 2-butene is approximately 4.42 x 10^-19 joules per molecule.

The energy required to break only the π bond of a C=C double bond can be calculated by subtracting the energy of a C-C single bond from the energy of a C=C double bond. So, we need to subtract the bond energy of a single bond (348 kJ/mol) from that of a double bond (614 kJ/mol). Therefore, the π bond energy is 614 kJ/mol - 348 kJ/mol = 266 kJ/mol. However, as the student asked for the energy required to break the π bond in 2-butene, we must convert this energy into energy per molecule by using Avogadro's number (6.022 x 10^23 molecules/mol). Hence, the energy required would be 266 kJ/mol x 10^3 J/kJ / (6.022 x 10^23 molecules/mol) = approximately 4.42 x 10^19 joules per molecule.

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To estimate the energy required to break the π bond in 2-butene, subtract the energy of a single C-C bond from that of a C=C double bond and divide by Avogadro's number, yielding approximately 4.42 x 10^-19 J/molecule.

The student asked how to estimate the energy needed to break only the π bond of the double bond in 2-butene, expressed in joules per molecule. Given that the energy for a C=C double bond is 614 kJ/mol and the energy for a C-C single bond is 348 kJ/mol, we can calculate the energy associated with the π bond. First, we find the difference between the double bond and the single bond energies: 614 kJ/mol - 348 kJ/mol = 266 kJ/mol. This difference represents the π bond energy per mole.

Now, convert this energy to joules (since 1 kJ = 1000 J), we have 266 kJ/mol x 1000 J/kJ = 266,000 J/mol. To find the energy per molecule, we divide by Avogadro's number (approximately 6.022 x 1023 mol-1):

266,000 J/mol ÷ 6.022 x 1023 mol-1 = 4.42 x 10-19 J/molecule.

Note that this is an estimate based on average bond energies and actual energies may vary based on the molecular environment.

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Answer:

[tex]T_{C}[/tex] = -4.2°C

[tex]T_{H}[/tex] = 49.4°C

Explanation:

A Carnot cycle is known as an ideal cycle in thermodynamic. Therefore, in theory, we have:

|[tex]\frac{Q_{C} }{Q_{H} }[/tex]| = [tex]\frac{T_{C} }{T_{H} }[/tex]

Similarly,

|[tex]Q_{H}[/tex]| = |[tex]Q_{C}[/tex]| + |[tex]W_{S}[/tex]|

During winter, the value of |[tex]T_{H}[/tex]| = 20°C = 273.15 + 20 = 293.15 K and |[tex]W_{S}[/tex]| = 1.5 kW. Therefore,

|[tex]Q_{H}[/tex]| = 0.75([tex]T_{H}[/tex] -  [tex]T_{C}[/tex])

Similarly,

|[tex]\frac{W_{S} }{Q_{H} }[/tex]| = 1 - [tex]\frac{T_{C} }{T_{H} }[/tex]

1.5/0.75*(293.15-[tex]T_{C}[/tex]) = 1 - ([tex]T_{C}[/tex]/293.15

Further simplification,

[tex]T_{C}[/tex] = -4.2°C

During summer, [tex]T_{C}[/tex] = 25°C = 273.15+25 = 298.15 K, and |[tex]W_{S}[/tex]| = 1.5 kW. Therefore,

|[tex]Q_{C}[/tex]| = 0.75([tex]T_{H}[/tex] -  [tex]T_{C}[/tex])

Similarly,

|[tex]\frac{W_{S} }{Q_{C} }[/tex]| = [tex]\frac{T_{H} }{T_{C} }[/tex] - 1

1.5/0.75*([tex]T_{H}[/tex] - 298.15) = ([tex]T_{H}[/tex]/298.15

Further simplification,

[tex]T_{H}[/tex] = 49.4°C

A) The minimum outside temperature for which the house can be maintained at 20°C is; T_L = -4.21 °C

B) The maximum outside temperature for which the house can be maintained at 20°C is; T_H = 37.21 °C

A) Heat transfer rate; Q'_H = 0.75 kJ/s/°C

Rating of heat pump motor; W'_in = 1.5 kW = 1.5 kJ/s

Temperature inside house; T_H = 20 °C = 293.15 K

Formula to find the minimum outside temperature is;

[tex]\frac{Q_{H}}{W'_{in}} = \frac{T_{H}}{T_{H} - T_{L}}[/tex]

Where [tex]T_{L}[/tex] is the minimum outside temperature.

[tex]{Q_{H}} = {Q'_{H}}({T_{H} - T_{L}})[/tex]

Thus, plugging in the relevant values gives;

(0.75/1.5)(293.15 - [tex]T_{L}[/tex]) = 293.15/(293.15 - [tex]T_{L}[/tex])

0.5(293.15 - [tex]T_{L}[/tex]) = 293.15/(293.15 - [tex]T_{L}[/tex])

(293.15 - [tex]T_{L}[/tex])² = 293.15/0.5

(293.15 - [tex]T_{L}[/tex]) = √586.3

[tex]T_{L}[/tex] = 293.15 - 24.21

[tex]T_{L}[/tex] = 268.94 K

Converting to °C gives

[tex]T_{L}[/tex] = -4.21°C

B) Now, we want to find the maximum outside temperature when T_L = 25°C = 298.15 K

Thus, we will use the formula;

[tex]\frac{Q_{H}}{W'_{in}} = \frac{T_{L}}{T_{H} - T_{L}}[/tex]

(0.75/1.5) * ([tex]T_{H} - T_{L}[/tex]) = [tex]\frac{T_{L}}{T_{H} - T_{L}}[/tex]

0.5 * ([tex]T_{H} - 298.15[/tex]) = 298.15/([tex]T_{H} - 298.15[/tex])

149.075 = ([tex]T_{H} - 298.15[/tex])²

√149.075 = ([tex]T_{H} - 298.15[/tex])

12.21 + 298.15 = [tex]T_{H}[/tex]

T_H = 310.36 K

Converting to °C gives;

T_H = 37.21 °C

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Answer:

The heat energy transfer required to raise the temperature of 13.0 kg liquid ammonia from -50.0 °C to 0.0 °C is 36,896.44 kJ.

Explanation:

The process involved in this problem are :

[tex](1):NH_3(l)(-50^oC)\rightarrow NH_3(l)(-33.4^oC)\\\\(2):NH_3(l)(-33.4^oC)\rightarrow NH_3(g)(-33.4^oC)[/tex]

[tex](3):NH_3(g)(-33.4^oC)\rightarrow NH_3(g)(-0.0^oC)[/tex]

Now we have to calculate the amount of heat released or absorbed in both processes.

For process 1 :

[tex]Q_1=m\times c_{1}\times (T_{final}-T_{initial})[/tex]

where,

[tex]Q_1[/tex] = amount of heat absorbed = ?

m = mass of ammonia = 13000 g

[tex]c_1[/tex] = specific heat of liquid ammonia  = [tex]4.7J/g^oC[/tex]

[tex]T_1[/tex] = initial temperature = [tex]-50.0^oC[/tex]

[tex]T_2[/tex] = final temperature = [tex]-33.4^oC[/tex]

Now put all the given values in [tex]Q_3[/tex], we get:

[tex]Q_1=13000 g\times 4.7 J/g^oC\times ((-33.4)-(-50.0))^oC[/tex]

[tex]Q_1=1,014,260 J=1.1014.260 kJ[/tex]

For process 2 :

[tex]Q_2=n\times \Delta H_{fusion}[/tex]

where,

[tex]Q_2[/tex] = amount of heat absorbed = ?

m = mass of solid ammonia = 13.0 Kg  = 13000 g

n = Moles of ammonia = [tex]\frac{13000 g}{17 g/mol}=764.71 mol[/tex]

[tex]\Delta H_{vap}[/tex] = enthalpy change for vaporization=23.5 kJ/mol

Now put all the given values in [tex]Q_1[/tex], we get:

[tex]Q_2=764.71 mol\times 23.5 kJ/mol=17,970.6 kJ[/tex]

For process 3 :

[tex]Q_3=m\times c_{p,l}\times (T_{final}-T_{initial})[/tex]

where,

[tex]Q_3[/tex] = amount of heat absorbed = ?

m = mass of ammonia = 13000 g

[tex]c_2[/tex] = specific heat of gaseous ammonia  = [tex]2.2J/g^oC[/tex]

[tex]T_1[/tex] = initial temperature = [tex]-33.4^oC[/tex]

[tex]T_2[/tex] = final temperature = [tex]0.0^oC[/tex]

Now put all the given values in [tex]Q_3[/tex], we get:

[tex]Q_3=13000 g\times 2.2J/g^oC\times (0.0-(-33.4))^oC[/tex]

[tex]Q_3=955,240 J=955.240 kJ[/tex]

The heat energy transfer required to raise the temperature of 13.0 kg liquid ammonia from -50.0 °C to 0.0 °C  = Q

[tex]Q=Q_1+Q_2+Q_3=17,970.6 kJ+17,970.6 kJ+955.240 kJ[/tex]

Q = 36,896.44 kJ

Answer:

6.00986

Explanation:

Answer: Option (4) is the correct answer.

Explanation:

Evaporation is defined as a process in which molecules of a substance absorb energy and hence, they gain kinetic energy. As a result, state of the substance changes from liquid phase to vapor phase.

For example, when we boil water then heat is absorbed by molecules of the liquid and hence, they change into vapor state.

Endothermic processes are defined as the process in which heat energy is absorbed by the reactant molecules.

The weak intermolecular forces which can arise either between nucleus and electrons or between electron-electron are known as dispersion forces. These forces are also known as London dispersion forces and these are temporary in nature.

Dispersion forces are present in between both polar and non-polar molecules.

Therefore, we can conclude that evaporation is an endothermic process is the correct statement.

Of the presented statements, the correct one is that evaporation is an endothermic process, as it involves the absorption of heat to overcome intermolecular forces.

The correct statement among the options provided is that evaporation is an endothermic process. This means that when a liquid turns to a gas, it absorbs heat from the surroundings, which is necessary to overcome the intermolecular forces that hold the molecules together in the liquid phase. Dispersion forces, also known as London dispersion forces, are a type of intermolecular force that exist between all atoms and molecules, whether they are polar or nonpolar. These forces arise from temporary dipoles caused by the fluctuating positions of electrons in atoms or molecules, so it is incorrect to say that only nonpolar, only molecules that can hydrogen bond, or only polar molecules have dispersion forces. Therefore, statements 1, 2, 3, and 5 are incorrect. Statement 6 is not applicable as statement 4 is correct.

Answer:

B

Explanation:

That electron is found in the 3d orbital.

ml=-2,-1,0,1,2

n=3

l=2

ml=0

ms=-1/2

Since it must be of opposite spin according to Pauli exclusion principle.

Answer:

8

Explanation:

Oxidation:

[tex]Fe^{2+} -->Fe^{3+}+e^{-}[/tex]

Reduction:

[tex]Cr_{2}O_{7}^{2-}+6e^{-} -->2Cr^{3+}[/tex]

We have to equalise the number  of moles of electrons gained and lost in a redox reaction in order to get a balanced reaction.

Hence we have to multiply the oxidation reaction throughout by 6.

and adding the two half-reactions we obtain:

[tex]6Fe^{2+}+Cr_{2}O_{7}^{2-} -->6Fe^{3+}+2Cr^{3+}[/tex]

Still the total charge and number of oxygen is not balanced.

Since the reaction takes place in acidic conditions, we will add required number of H+ to the appropriate side to balance the charge and add half the amount of H2O to balance the hydrogen atoms.

We add 14 H+ on LHS and 7H2O on RHS to obtain:

[tex]6Fe^{2+}+Cr_{2}O_{7}^{2-}+14H^{+} -->6Fe^{3+}+2Cr^{3+}+7H_{2}O[/tex]

Sum of coefficients of product cations = 6+2 = 8

Answer: The molarity of NaOH solution is 0.1046 M.

Explanation:

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 KHP = 0.5516 g

Molar mass of KHP = 204.22 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of KHP}=\frac{0.5516g}{204.22g/mol}=0.0027mol[/tex]

The chemical reaction for the reaction of KHP and NaOH follows

[tex]KHC_8H_4O_4(aq.)+NaOH\rightarrow KNaC_8H_4O_4(aq.)+H_2O(l)[/tex]

By Stoichiometry of the reaction:

1 mole of KHP reacts with 1 mole of NaOH.

So, 0.0027 moles of KHP will react with = [tex]\frac{1}{1}\times 0.0027=0.0027mol[/tex] of NaOH.

To calculate the molarity of NaOH, we use the equation:

[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]

We are given:

Moles of NaOH = 0.0027 moles

Volume of solution = 25.82 mL  = 0.02582L      (Conversion factor:  1L = 1000 mL)

Putting values in above equation, we get:

[tex]\text{Molarity of NaOH }=\frac{0.0027mol}{0.02582L}=0.1046M[/tex]

Hence, the molarity of NaOH solution is 0.1046 M.

Based on the data provided, the balanced equation of the reaction is:

The equation of the reaction between KHP and NaOH is given below as:

From the equation of reaction, mole ratio of KHP and NaOH is 1 : 1

Moles of KHP = mass/molar mass

molar mass of KHP = 204 g/mol

Moles of KHP = 0.5516/204

Moles of KHP = 0.00271 moles

Moles of NaOH = molarity × volume

Volume of NaOH = 25.82 mL = 0.02582 L

0.00271 = molarity × 0.02582

Molarity = 0.00271/0.02582

Molarity of KOH = 0.105 M

Hence, the molarity of NaOH solution is 0.105 M.

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Answer : The value of [tex]\Delta G_{rxn}[/tex] is -24.9 kJ/mol

Explanation :

First we have to calculate the value of 'Q'.

The given balanced chemical reaction is,

[tex]Fe_2O_3(s)+3CO(g)\rightarrow 2Fe(s)+3CO_2(g)[/tex]

The expression for reaction quotient will be :

[tex]Q=\frac{(p_{CO_2})^3}{(p_{CO})^3}[/tex]

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Now put all the given values in this expression, we get

[tex]Q=\frac{(2.1)^3}{(1.4)^2}[/tex]

[tex]Q=3.375[/tex]

Now we have to calculate the value of [tex]\Delta G_{rxn}[/tex].

The formula used for [tex]\Delta G_{rxn}[/tex] is:

[tex]\Delta G_{rxn}=\Delta G^o+RT\ln Q[/tex]    ............(1)

where,

[tex]\Delta G_{rxn}[/tex] = Gibbs free energy for the reaction  = ?

[tex]\Delta G_^o[/tex] =  standard Gibbs free energy  = -28.0 kJ/mol

R = gas constant = [tex]8.314\times 10^{-3}kJ/mole.K[/tex]

T = temperature = 298 K

Q = reaction quotient = 3.375

Now put all the given values in the above formula 1, we get:

[tex]\Delta G_{rxn}=(-28.0kJ/mol)+[(8.314\times 10^{-3}kJ/mole.K)\times (298K)\times \ln (3.375)[/tex]

[tex]\Delta G_{rxn}=-24.9kJ/mol[/tex]

Therefore, the value of [tex]\Delta G_{rxn}[/tex] is -24.9 kJ/mol

The standard free energy change of the reaction is -25kJ/mol.

The perform the task we must first calculate Kp from the data provided as follows;

P(CO) = 1.4 atm

P(CO2) = 2.1 atm

Kp = (p.CO2)^3/(p.CO)^3

Kp = (2.1)^3/(1.4)^3

Kp = 9.3/2.7

Kp = 3.4

ΔG°rxn =ΔG° + RTlnKp

Where;

R = 8.314 J/Kmol

T =  298 K

ΔG°rxn = -28.0 kJ + (8.314 * 298 * ln 3.4) * 10^-3

ΔG°rxn = -25kJ/mol

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Answer:

OH⁻ and H₂O

Explanation:

From the equation given in question ,

NH₃ (aq) + H₂O (l) ⇌ NH₄⁺ (aq) + OH⁻ (aq)

Acids are the species which furnish hydrogen ions in the solution or is capable of forming bonds with electron pair species as they are electron deficient species.

When an acid donates a proton, it changes into a base which is known as its conjugate base.

Bases are the species which furnish hydroxide ions in the solution or is capable of forming bonds with electron deficient species as they are electron rich species. When a base accepts a proton, it changes into a acid which is known as its conjugate acid.

The acid and the base which is only differ by absence or presence of the proton are known as acid conjugate base pair.

Hence , from the equation given in the question ,

The conjugate acid - base pair is -

OH⁻ and H₂O

Answer:

Their particles are arranged randomly

Explanation:

The random arrangement of molecules cause the translational motion of the liquid and Brownian motion of the gas

Answer:

h = 0.346 m

Explanation:

mercury barometer:

∴ Pa = 0.455 atm = 46102.875 Pa = 46102.875 Kg/ms²

∴ ρ = 13600 Kg/m³

∴ g = 9.80 m/s²

⇒ h = (46102.875 Kg/ms²) / (13600 Kg/m³ )(9.80 m/s²)

⇒ h = 0.346 m

The height of the mercury column in a barometer on top of a mountain, where the atmospheric pressure is 0.455 atm, would be around 33 centimeters.

To compute the height of the mercury column, we use the formula h = P/(gρ). Here, h represents height, P is the atmospheric pressure, g is the acceleration due to gravity and ρ is the density of the fluid (in this case, mercury). Given that P = 0.455 atm, ρ = 13.6*10^3 kg/m^3, and the standard value for g is approximated as 9.81 m/s^2, we then convert atmospheres to pascal by multiplying by 101325, since 1 atm = 101325 Pa.

This gives us P = 0.455 atm * 101325 Pa/atm = 46107.875 Pa. Using these values in our formula, we get h = 46107.875 Pa /(9.81 m/s^2 x 13.6*10^3 kg/m^3) which simplifies to approximately 0.33 meters or 33 centimeters. Hence, on top of a high mountain, where the atmospheric pressure is 0.455 atm, the height of the mercury column would be around 33 cm.

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The reaction would progress faster and the activation energy would be lowered when the enzyme gets added.

Enzymes are proteins that basically speed up the chemical reaction without being used. Enzymes are usually specific for a particular substrate. The substrate in the reaction bind to the active site of the enzyme which is present on the surface of the enzymes forming the enzyme-substrate complex.

Performing the enzyme-substrate complex the enzyme changes the shape slightly so that the substrate can fit tightly to its active site. Then this enzyme-substrate Complex undergoes a reaction to form a product. Enzymes lower the activation energy of a reaction i.e the required amount of energy needed for a reaction to occur.They do this by binding to a substrate and holding it in a way that allows the reaction to happen more efficiently.

Answer: b)298 K converts to 24.9 °C, so the water would remain in its liquid state

Explanation: Kelvin is an absolute temperature scale, that means that 0K (zero Kelvin) is the lowest temperature possible and compare to Celsius scale the change of 1 degree is the same, but 0°C is the same as 273,1K

So, in order to convert Kelvin to Celsius you have to subtract 273,1

In this case, 298K - 273,1 = 24,9°C

This temperature is room temperature, so water is in liquid state.

Answer:  Tetrahedral geometry

Explanation: The structure of the acetic acid is shown in the image below.

The left carbon atom is the one which is attached to three hydrogen atoms through single bonds and to one carbon atom through single bond as well.

Thus the orbitals which are used in the process of the formation of the chemical bond between these 4 are sp3 orbitals. And these orbitals results in the formation of the tetrahedral geometry.

The right carbon atom that is attached to a  carbon atom and an oxygen atom through single bonds and a second oxygen atom through a double bond has trigonal planar geometry which involves the sp2 orbitals for the formation of the bond.

Final answer:

The left carbon atom in acetic acid has a tetrahedral molecular geometry with sp3 hybridization.

Explanation:

The molecular geometry of the left carbon atom in acetic acid, which is attached to three hydrogen atoms and one carbon atom, is tetrahedral. This carbon atom has a sp3 hybridization, which means it forms four sp3 hybrid orbitals.

Each of these orbitals forms a single bond with either a hydrogen atom or the other carbon atom, resulting in a tetrahedral shape, with bond angles close to 109.5 degrees.

Acetic acid has two distinct carbon-oxygen bonds because the right carbon atom is part of a carboxylic acid group. In acetic acid, this carbon is double-bonded to one oxygen atom (forming a carbonyl group) and single-bonded to the oxygen atom in the hydroxyl group (OH). This leads to different bond lengths and strengths.

However, when acetic acid loses a hydrogen ion and becomes the acetate ion, the negative charge is delocalized over the two oxygen atoms, resulting in equivalent resonance structures where both carbon-oxygen bonds have partial double-bond character, making them appear as one type of bond.

The increase in entropy when the partition is removed is calculated using the entropy change formula ΔS=nRln(Vf/Vi). The total entropy change is the sum of the entropy changes for each gas. For the first case, the total entropy increase is 11.52 J/K; for the 2 moles of gas A case, the increase is 17.29 J/K; and for the case where both compartments contain gas A, the increase is also 11.52 J/K.

For this, we will use the formula for the entropy change, ΔS, which is given by ΔS=nRln(Vf/Vi), where n is the number of moles, R is the gas constant, and Vf and Vi are the final and initial volumes respectively. In the first case, the total entropy change when the partition is removed is the sum of the entropy changes for both gas A and gas B, since both volumes double, so ΔS=(1mol)(8.31 J/mol K)ln(2)+(1mol)(8.31 J/mol K)ln(2)=11.52 J/K.

If there are 2 moles of gas A in the first compartment, the entropy change is ΔS=(2mol)(8.31 J/mol K)ln(2)+(1mol)(8.31 J/mol K)ln(2)=17.29 J/K. For the final case where both compartments initially contain gas A, when the partition is removed, the total volume available to the gas doubles, so each gas experiences an entropy change of ΔS=(1mol)(8.31 J/mol K)ln(2), summed over both gases, gives us again 11.52 J/K.

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Case 1 - For Ideal Gas A and B the value of ΔStotal is 11.52 J/K.

Case 2 - For 2 Moles of Ideal Gas A the value of ΔS is 11.52 J/K.

Case 3 - Both Compartments with Ideal Gas A has the value of ΔS as 11.52 J/K.

Case 1: Ideal Gas A and B

For the first case, where there is 1 mole of ideal gas A in one compartment and 1 mole of ideal gas B in the other, both at 1 atm:

Case 2: 2 Moles of Ideal Gas A

For the second case, with 2 moles of ideal gas A in one compartment:

Case 3: Both Compartments with Ideal Gas A

Now, let’s consider the scenarios in which both compartments contain the same type of gas A:

If one compartment had 2 moles of gas A and the other also had 2 moles, the calculation is the same, but for each set of 2 moles:
ΔS = 2 * 8.314 * ln(2) ≈ 11.52 J/K

The key idea is that the type of molecules and their identical properties simplify calculations using the entropy formula for ideal gases.

Answer:

D) Oxygen is oxidized and hydrogen is reduced.

Explanation:

In the electrolysis of water, an electric current passes through an electrolytic solution (e.g. aqueous NaCl), leading to the following redox reaction.

H₂O(l) → H₂(g) + 1/2 O₂(g)

The corresponding half-reactions are:

Reduction:  2 H₂O(l) + 2 e⁻ → H₂(g) + 2 OH⁻

Oxidation:  2 H₂O(l) → O₂(g) + 4 H⁺(aq) + 4 e⁻

As we can see, H in water is reduced (its oxidation number decreases from 1 to 0), while O in water is oxidized (its oxidation number increases from -2 to 0).

Answer:

There will be 15.05 Joules of heat released.

Explanation:

Step 1: Data given

Mass of mercury = 5.00 grams

The specific heat of mercury = 0.14 J/g*K = 0.14 J/g°C

Raise of temperature from 15.0°C to 36.5 °C

Step 2: Calculate heat released

q = m*c*ΔT

⇒ with q = The heat released in Joules

⇒ with m = the mass of mercury = 5.00 grams

⇒ with c = The specific heat of mercury = 0.14 J/g°C

⇒ with ΔT = The change of temperature = T2 - T1 = 36.5°C - 15.0°C = 21.5 °C

q = 5.00g * 0.14 J/g°C * 21.5 °C = 15.05 J

There will be 15.05 Joules of heat released.

The quantity of heat needed to raise 5g of mercury  from 15.0°C to 36.5°C is  15.05 Joules

Given Data

We know that the expression for the amount of heat is given as

Q =. mcΔT

Substituting our given data into the expression we have

Q = 5*0.14*(36.5 - 15)

Q = 0.7*21.5

Q = 15.05 Joules

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Answer:

B) The term "inert" was dropped because it no longer described all the group 8A elements.

Explanation:

Inert elements in chemistry simply refers to elements that are chemically inactive and are not expected to form any compounds. this is the general belief for the group 8 elements as they all have complete duplet/octet configurations (and ideally, they ought to be very stable with no tendency to form compounds by participating in the loss and gain of electrons). However the discovery of compounds like xenon tetrafluoride (XeF4) proved this to be wrong.

Again, the reason the term - inert gses was droppedis beacause this term is not strictly accurate because several of them do take part in chemical reactions.

After dropping the term - Inert gases, they are now referred to as noble gases.

The term 'inert gases' was dropped because it was found that these gases can participate in chemical reactions under certain circumstances. Consequently, they were not entirely 'inert'. Additionally, to avoid confusion with other gases referred to as 'inert' in organic chemistry, the term was replaced with 'noble gases'.

The term 'inert gases' was primarily used to describe the group 8A elements in the periodic table, now commonly referred to as the noble gases. Initially, these gases were called inert due to their low reactivity. However, as science progressed, it was discovered that these gases could indeed take part in chemical reactions under certain conditions - hence, they weren't completely 'inert'. This led to the term being dropped in favor of 'noble gases'. Additionally, the term 'inert' began to cause confusion as it was also used in organic chemistry to refer to gases like nitrogen and methane that do not readily react.

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Answer:

[tex]k_{sp}=4.7 \times 10^{-13}.[/tex]

Explanation:

Now equation of tqo halves are:

Oxidation : [tex]Ag(s)-->Ag^+(aq)+e^-[/tex]

Reduction : [tex]AgBr(s)+e^--->Ag(s)+Br^-(aq)[/tex]

We know,

[tex]E^o_{cell}=0.071-(0.8)=-0.729\ V.[/tex]

[tex]\Delta G^o=-n\times F \times E^o= -R\times T\times ln(k_s_p)\\-1\times 96485\times (-0.729)=-8.314\times 298\times ln(k_s_p)\\k_s_p=4.7\times 10^{-13}.[/tex]

[tex]k_{sp}=4.7 \times 10^{-13}.[/tex]

Hence, this is the required solution.

Final answer:

To calculate Ksp for AgBr, use the given standard reduction potentials and the Nernst equation. The equilibrium constant, K, for the half-reaction can be found using the Nernst equation. The equilibrium constant Ksp can then be calculated from K.

Explanation:

To calculate Ksp for AgBr, we can use the given standard reduction potentials and the Nernst equation. The balanced half-reaction for the reduction of AgBr(s) to Ag(s) is:

AgBr(s) + e- → Ag(s) + Br-(aq)

The reduction potential for this half-reaction is +0.071 V. Using the Nernst equation, we can find the equilibrium constant, K, for this reaction:

K = [Ag(s)][Br-(aq)] / [AgBr(s)] = 10^(nE°/0.0592) = 10^((1)(0.071)/0.0592) = 1.44

Since the stoichiometry of the reaction is 1:1, the equilibrium constant Ksp for AgBr can be directly calculated from K:

Ksp = [Ag+(aq)][Br-(aq)] = 1.441

Answer:

The amount of energy required to transport hydrogen ions from a cell into the stomach is 37.26KJ/mol.

Explanation:

The free change for the process can be written in terms of its equilibrium constant as:

ΔG° = [tex]-RTInK_(eq)[/tex]

where:

R= universal gas constant

T= temperature

[tex]K_eq[/tex]= equilibrum constant for the process

Similarly, free energy change and cell potentia; are related to each other as follows;

ΔG= -nFE°

from above;

F = faraday's constant

n = number of electrons exchanged in the process; and  

E = standard cell potential

∴ The amount of energy required for transport of hydrogen ions from a cell into stomach lumen can be calculated as:

ΔG° = [tex]-RTInK_(eq)[/tex]

where;

[texK_eq[/tex]=[tex]\frac{[H^+]_(cell)}{[H^+(stomach lumen)]}[/tex]

For transport of ions to an internal pH of 7.4, the transport taking place can be given as:

[tex]H^+_{inside}[/tex] ⇒ [tex]H^+_{outside}[/tex]

Equilibrum constant for the transport is given as:

[tex]K_{eq}=\frac{[H^+]_{outside}}{[H^+]_{inside}}[/tex]

[tex]=\frac{[H^+]_{cell}}{[H^+]_{stomach lumen}}[/tex]

[tex][H^+]_{cell}[/tex]= 10⁻⁷⁴

=3.98 * 10⁻⁸M

[tex][H^+]_{stomach lumen}[/tex] = 10⁻²¹

=7.94 * 10⁻³M

Hence;

[tex]K_{eq}=\frac{[H^+]_{cell}}{[H^+]_{stomachlumen}}[/tex]

=[tex]\frac{3.98*10^{-8}}{7.94*10{-3}}[/tex]

= 5.012 × 10⁻⁶

Furthermore, free energy change for this reaction is related to the equilibrium concentration given as:

ΔG° = [tex]-RTInK_(eq)[/tex]

If temperature T= 37° C ; in kelvin

=37° C + 273.15K

=310.15K; and

R-= 8.314 j/mol/k

substituting the values into the equation we have;

ΔG₁ = [tex]-(8.314J/mol/K)(310.15)TIn(5.0126*10^{-6})[/tex]

= 31467.93Jmol⁻¹

≅ 31.47KJmol⁻¹

If the potential difference across the cell membrane= 60.0mV.

Energy required to cross the cell membrane will be:

ΔG₂ = [tex]-nFE°_{membrane}[/tex]

ΔG₂ = [tex]-(1 mol)(96.5KJ/mol/V)(60*10^{-3})[/tex]

= 5.79KJ

Therefore, for one mole of electron transfer across the membrane; the energy required is 5.79KJmol⁻¹

Now, we  can calculate the total amount of energyy required to transport H⁺ ions across the membrane:

Δ [tex]G_{total} = G_{1}+G_{2}[/tex]

= (31.47+5.79) KJmol⁻¹

= 37.26KJmol⁻¹

We can therefore conclude that;

   The amount of energy required to transport ions from cell to stomach lumen is 37.26KJmol⁻¹