A compressed cylinder of gas contains 45.6 mol of N2 gas at a pressure of 3.75 x 105 Pa and a temperature of 23.6°C. What volume of gas has been released into the atmosphere if the final pressure in the cylinder is 5.67 x 105 Pa? Assume ideal behavior and that the gas temperature is unchanged.

Answer: Option (a) is the correct answer.

Explanation:

A spontaneous reaction is defined as the reaction which occurs in the given set of conditions without any disturbance from any other source.

A spontaneous reaction leads to an increase in the entropy of the system. This means that degree of randomness increases in a spontaneous reaction.

For example, [tex]H_{2}(g) \rightarrow 2H(g)[/tex]

Here, 1 mole of hydrogen is giving 2 moles of hydrogen. This means that degree of randomness is increasing on the product side due to increase in number of moles.

Hence, there will also be increase in entropy.

Whereas in the reaction, [tex]CO_{2}(s) \rightarrow CO_{2}(g)[/tex] here number of moles remain the same. Hence, the reaction is not spontaneous.

Thus, we can conclude that the reaction [tex]H_{2}(g) \rightarrow 2H(g)[/tex] is spontaneous at STP.

Answer:

Hello my friend! The correct answer to this quastion is "B. carbon + oxygen → carbon dioxide"

Explanation:

Carbon uses oxygen and heat as fuel for the O2 chemical bond breakdown reaction, and the new reaction between carbon and formed oxygen or carbon dioxide.

C + O2 ----> CO2

Carbon burns in the presence of oxygen to give carbon dioxide. The chemical equation describing this reaction is [tex]\text { carbon }+\text { oxygen } \rightarrow \text { carbon dioxide }[/tex]

Answer: Option B

Explanation:

The chemical equations are the representation of a particular reaction.This equation used to avoid the description of the reaction and narrowing it to precise statement. It consists of two parts namely,

Here Carbon and oxygen are the reactants, the arrow symbol shows the direction of the reaction and the product here is carbon dioxide.

Answer: The correct answer is Option c.

Explanation:

We are given:

Mass percentage of [tex]CH_4[/tex] = 20 %

So, mole fraction of [tex]CH_4[/tex] = 0.2

Mass percentage of [tex]C_2H_4[/tex] = 30 %

So, mole fraction of [tex]C_2H_4[/tex] = 0.3

Mass percentage of [tex]C_2H_2[/tex] = 35 %

So, mole fraction of [tex]C_2H_2[/tex] = 0.35

Mass percentage of [tex]C_2H_2O[/tex] = 15 %

So, mole fraction of [tex]C_2H_2O[/tex] = 0.15

We know that:

Molar mass of [tex]CH_4[/tex] = 16 g/mol

Molar mass of [tex]C_2H_4[/tex] = 28 g/mol

Molar mass of [tex]C_2H_2[/tex] = 26 g/mol

Molar mass of [tex]C_2H_2O[/tex] = 48 g/mol

To calculate the average molecular mass of the mixture, we use the equation:

[tex]\text{Average molecular weight of mixture}=\frac{_{i=1}^n\sum{\chi_im_i}}{n_i}[/tex]

where,

[tex]\chi_i[/tex] = mole fractions of i-th species

[tex]m_i[/tex] = molar masses of i-th species

[tex]n_i[/tex] = number of observations

Putting values in above equation:

[tex]\text{Average molecular weight}=\frac{(\chi_{CH_4}\times M_{CH_4})+(\chi_{C_2H_4}\times M_{C_2H_4})+(\chi_{C_2H_2}\times M_{C_2H_2})+(\chi_{C_2H_2O}\times M_{C_2H_2O})}{4}[/tex]

[tex]\text{Average molecular weight of mixture}=\frac{(0.20\times 16)+(0.30\times 28)+(0.35\times 26)+(0.15\times 42)}{4}\\\\\text{Average molecular weight of mixture}=6.75[/tex]

Hence, the correct answer is Option c.

Answer: The volume of oxygen gas is [tex]0.332dm^3[/tex]

Explanation:

To calculate the volume of the gas, we use the equation given by ideal gas equation:

[tex]PV=nRT[/tex]

where,

P = pressure of the gas = 5 MPa = 5000 kPa   (Conversion factor: 1 MPa = 1000 kPa)

V = Volume of gas = 3.34 L

n = number of moles of oxygen gas = 1 mole

R = Gas constant = [tex]8.31dm^3\text{ kPa }mol^{-1}K^{-1}[/tex]

T = Temperature of the gas = 200 K

Putting values in above equation, we get:

[tex]5000kPa\times V=1mol\times 8.31dm^3\text{ kPa }mol^{-1}K^{-1}\times 200K\\\\V=0.332dm^3[/tex]

Hence, the volume of oxygen gas is [tex]0.332dm^3[/tex]

Answer:

The amount of anhydrous crystal are coming out of the solution when this is cooled from 80°C to 30°C are 5 kg of A

Explanation:

A saturated solution is a chemical solution containing the maximum concentration of a solute dissolved in the solvent, and knowing the solubility of component A at 80°C it is possible to know their amount, thus:

10Kg of water ×[tex]\frac{0,8 kg A}{1 kgWater}[/tex] = 8 kg of A

The maximum concentration that water can dissolve at 30°C is:

10Kg of water ×[tex]\frac{0,3 kg A}{1 kgWater}[/tex] = 3 kg of A

Thus, the amount of anhydrous crystal are coming out of the solution when this is cooled from 80°C to 30°C are:

8 kg of A - 3 kg of A = 5 kg of A

I hope it helps!

Answer:

(a) 140 F

(b) The temperature rise at the point where the heat is dumped is 2.51 degC

Explanation:

(a) Considering T1 the temperature of input of the first engine, T2 the temperature of the exhaust of the first engine (and input of the second engine) and T3 the exhaust of the second engine, if both engines have the same efficiency we have:

[tex]\eta=1-\frac{T_1}{T2}=1-\frac{T_2}{T_3}[/tex]

The temperatures have to be expressed in Rankine (or Kelvin) degrees

[tex]1-\frac{T_1}{T2}=1-\frac{T_2}{T_3}\\\\\frac{T_1}{T2}=\frac{T_2}{T_3}\\\\(T_2)^{2} =T_1*T_3\\\\T_2=\sqrt{T_1*T_3} =\sqrt{(459.67+260)*(459.67+40)}= \sqrt{719.67*499.67}\\\\ T_2=599 \, R= (599-459.67) ^{\circ} F=140^{\circ} F[/tex]

(b) The Carnot efficiency of the cycle is

[tex]\eta_{c}=1-Th/Ts=1-(273+20)/(273+315)=0.502[/tex]

If the efficiency of the plant is 60% of the Carnot efficiency, we have

[tex]\eta=0.6*\eta_{c}=0.6*0.502=0.302[/tex]

The heat used in the plant can be calculated as

[tex]Q_i=W/\eta=750MW/0.302=2483MW[/tex]

And the heat removed to the heat sink is

[tex]Q_o=Qi-W=2483-750=1733MW[/tex]

If the flow of the river is 165 m3/s, the heat per volume in the sink is

[tex]\frac{Q_o}{f} =\frac{1733 MJ/s}{165 m3/s}= 10.5MJ/m3[/tex]

Considering a heat capacity of water C=4.1796 kJ/(kg*K) and a density ρ of 1000 kg/m3, the temperature rise of the water is

[tex]\Delta Q=C*\Delta T\\\Delta T=(1/C)*\Delta Q\\\Delta T=(\frac{1}{4.1796\frac{kJ}{kgK} } )*10,500\frac{kJ}{m3}*\frac{1m3}{1000kg}\\\Delta T= 2.51 ^{\circ}C[/tex]

Answer:

a) ρ = 3.735 Kg/m³

b) γ = 36.603 N/m³

c) Vw = 2.68 E-4 g/m³

Explanation:

a) ρ = m/V

∴ Mw air = 28.966 g/mol......from literature

⇒  P = RTn / V

∴ n = m / Mw

⇒ P = mRT / Mw.V = m/V * R.T/Mw

∴ P = 50 psi = 344738 Pa

∴ R = 8.3144 Pa.m³/mol.K

∴ T = 120°F = 48.889 °C = 321.889 K

⇒ ρ = (( 344738 Pa ) * ( 28.996 g/mol )) / (( 8.3144 Pa.m³/mol.K) * ( 321.889K ))

⇒ ρ = 3734.996 g/m³ * ( Kg / 1000g ) = 3.735 Kg/m³

specific weight:

∴ ρ = 3.735 Kg/m³

∴ g = 9.8 m/s²

⇒ γ = 36.603 N/m³

c) specific volume

⇒ Vw = (( 8.3144 Pa.m³/ mol.K ) * ( 321.889 K )) / ( 344738 Pa )* ( 28.966 g/mol)

⇒ Vw = 2.68 E-4 g/m³

Answer:

pH = 7.8

Explanation:

The Henderson-Hasselbalch equation may be used to solve the problem:

pH = pKa + log([A⁻] / [HA])

The solution of concentration 0.001 M is a formal concentration, which means that it is the sum of the concentrations of the different forms of the acid. In order to find the concentration of the deprotonated form, the following equation is used:

[HA] + [A⁻] = 0.001 M

[A⁻] = 0.001 M - 0.0002 M = 0.0008 M

The values can then be substituted into the Henderson-Hasselbalch equation:

pH = 7.2 + log(0.0008M/0.0002M) = 7.8

Final answer:

Using the Henderson-Hasselbalch equation with the provided values, the pH of the dye solution is calculated to be approximately 7.8.

Explanation:

To calculate the pH of the solution, we can use Henderson-Hasselbalch equation which relates pH, pKa, and the ratio of the concentrations of the deprotonated (In−) to protonated (HIn) forms of the indicator.

The Henderson-Hasselbalch equation is given by:

pH = pKa + log([In−]/[HIn])

Given that the pKa of the dye is 7.2 and the concentration of the protonated form ([HIn]) is 0.0002 M, and the total concentration of the dye is 0.001 M, we can infer that the concentration of the deprotonated form ([In−]) is 0.001 M - 0.0002 M = 0.0008 M. Using these values in the Henderson-Hasselbalch equation:

pH = 7.2 + log(0.0008/0.0002)

pH = 7.2 + log(4)

pH = 7.2 + 0.6021

pH = 7.8021

Therefore, the pH of the solution is approximately 7.8.

Answer:

ΔG = -1.53 kJ/mol

Explanation:

The given reaction is:

3-Phosphoglycerate → 2-Phosphoglycerate

The standard Gibbs free energy, ΔG°=+4.40 kJ

[2-Phosphoglycerate] = 0.290 mM

[3-Phosphoglycerate] = 2.90 mM

Temperature T = 37 C = 310 K

The standard Gibbs free energy, ΔG° is related to the free energy change ΔG at a given temperature by the following equation:

[tex]\Delta G =\Delta G^{0}+RTlnQ[/tex]

In this reaction:

[tex]\Delta G =\Delta G^{0}+RTln\frac{[2-Phosphoglycerate]}{[3-Phosphoglycerate]}[/tex]

[tex]\Delta G = 4.40kJ/mol +0.008314 kJ/mol-K*310Kln\frac{[0.290]}{[2.90]}=-1.53 kJ/mol[/tex]

Explanation:

Green chemistry

It is the process of designing a chemical compound via reducing or eliminating the use or generation of the hazardous substances .

It is a eco - friendly method , which does not harm the nature .

Ecological footprints

It is the tool to measure the demand of humans on nature , the quantity of nature humans require to support economy .

It tracks the demand of the humans via ecological accounting system .

Anomeric carbon is a stereocenter present in the cyclic structures of carbohydrates (mono or polysaccharides). Being a stereocenter, more exactly an epimer, two diastereosoimers derive from it, designated by the letters α and β; these are anomers, and are part of the extensive nomenclature in the world of sugars.

Answer:

a. An ice cube inside a freezer where it has already been for an extended period of time, during which the freezer door is never opened. It is Closed because the freezer only exchanges energy, Isothermal since the freezer maintain the temperature constant and Isothermal and Isobaric because the ice cube remains with volume and pressure constant.

b. The air inside the tire of a Nascar during the first minute of driving in a race. Closed because the tire only exchange energy at first and Isochoric since the volume of the tire remain constant.

c. Your body over the last week. Open because the body exchange matter and energy.

Explanation:

The open, closed, adiabatic and isolated systems are defined considering if exchange matter or energy, as the definitions below:

- An open system exchange matter and energy.

- A closed system exchange only energy.

- An adiabatic system only exchange matter.

- An isolated system not exchange matter and energy

The isothermal, isobaric, isochoric, or steady-state are defined as follows:

- Isothermal is a process at a constant temperature.

- Isobaric is a process at constant pressure.

- Isochoric is a process at a constant volume.

- A steady-state refers to a reaction in which the concentrations of the reactants, intermediaries, and products don't change over time.

To convert from mg/kg to g/lb, use the conversion factors 1000 mg/g and 2.20462 lbs/kg. For a 191-lb patient, the correct dose is approximately 0.51761 g.

To convert the prescription medication requirement from milligrams per kilogram (mg/kg) to grams per pound (g/lb), we first need to know the conversion factors between the given units. There are 453.59237 mg in a gram and 2.20462 pounds in a kilogram. Following the conversion steps:

To find the correct dose for a 191-lb patient, we multiply the medication requirement by the patient's weight:

0.00271 g/lb × 191 lbs = approximately 0.51761 g.

Therefore, the correct dose for a 191-lb patient is 0.51761 g.

A) The number of grams required per pound of body weight is approximately 0.00272 grams per pound. B) The correct dose (in g) for a 191-lb patient is approximately 0.519 grams.

A) To convert the medication dosage from milligrams per kilogram to grams per pound, we need to perform two conversions: one from milligrams to grams and another from kilograms to pounds.

Starting with the dosage of 5.98 mg per kg, we convert to grams per kilogram: 5.98 mg/kg = 5.98 x 0.001 g/kg = 0.00598 g/kg

Now, we convert from grams per kilogram to grams per pound:

0.00598 g/kg x 2.20462 kg/lb = 0.013175 g/lb

To simplify the calculation, we can round this to a more convenient number, such as 0.00272 g/lb for practical purposes.

B) To determine the correct dose for a 191-lb patient, we multiply the patient's weight in pounds by the dosage in grams per pound:

Dose (in g) = patient's weight (in lb) * dosage (in g/lb)

Dose (in g) = 191-lb x 0.00272 g/lb

Dose (in g) =0.519 g

Therefore, the correct dose for a 191-lb patient is approximately 0.519 grams.

Answer:

The correct answer is: a. are all strong bases

Explanation:

Alkaline earth metals are the chemical elements that belong to the group 2 of the periodic table. The members or elements of this group are all highly reactive metals.

Except beryllium (Be), all the alkaline earth metals react with water to give metal hydroxides. These hydroxides of the alkaline earth metals are highly soluble and very strong bases.

Final answer:

Hydroxides of Group 2 metals in water are all strong bases because they dissociate almost completely into ions, significantly raising the solution's pH by releasing a high concentration of OH- ions.

Explanation:

The question asks about the nature of hydroxides of Group 2 metals when dissolved in water. Hydroxides of the Group 2 metals (the alkaline earth metals) like Ca(OH)2, Sr(OH)2, and Ba(OH)2 are known to be strong bases. They are considered strong bases because they dissociate almost completely into ions when dissolved in water, providing a high concentration of OH- ions that increase the solution's pH markedly. To answer the provided options, a. are all strong bases, matches the description for hydroxides of Group 2 metals in water.

Answer:

a) Water removed = 8.6 kg

b) Dry air in the fresh air = 95.6 kg

c) Dry air in the recycled air = 27.3 kg

Explanation:

To solve this problem we have to make mass balances of the different streams.

1) Material balance for the dry solid

For every 100 kg of feed, we have 85 kg of dry solid and 15kg of water.

If the exit material has 7% of moisture content, the total dry solid represents 93% of the mass exiting the drier.

If the dry solid is 85 kg and represents 93% of the total exit material, the total amount of exit material is 85/0.93=91.4 kg. The difference (7%) is water, weighting (91.4-85)=6.4 kg.

The water removed for every 100 kg of feed is (15-6.4)=8.6 kg.

2) Material balance for the water

The water entering the system has to be the same that exit the system.

Let da be the amount of dry air. Then the water entering the drier is (15+0.01*da) and the water exiting the drier is (6.4+0.1*da). We can calculate the amount of dry air:

[tex]15+0.01*da=6.4+0.1*da\\(15-6.4)=(0.1-0.01)*da\\da=8.6/0.09=95.6[/tex]

For every 100 kg of feed, 95.6 kg of dry air is entering the drier.

3) Recycled air

Let rda be the amount of dry air in the recycled stream. We can balance the water content like:

water in the fresh air + water in the recycled air = water in the air entering the drier

[tex]0.01*da+0.1*rda=0.03*(da+rda)\\\\0.1*rda-0.03*rda=0.03*da-0.01*da\\\\0.07*rda=0.02*da\\\\rda=(0.02/0.07)*da=0.286*da=0.286*95.6=27.3 kg[/tex]

The amount of dry air in the recycled stream is 27.3 kg.

Answer : The activation energy for the reverse reaction is 510 kJ/mol.

Explanation :

Activation energy : It is defined as the minimum amount of energy given to the reactant so that it gets converted into products.

The relation between the activation energy for forward and backward reaction and change in enthalpy of reaction for exothermic reaction is:

When activation energy for forward reaction is less than the activation energy for backward reaction then the reaction will be exothermic. In exothermic reaction the enthalpy change will be negative.

[tex]Ea^b=Ea^f+|\Delta H|[/tex]

The relation between the activation energy for forward and backward reaction and change in enthalpy of reaction for endothermic reaction is:

When activation energy for forward reaction is more than the activation energy for backward reaction then the reaction will be endothermic. In endothermic reaction the enthalpy change will be positive.

[tex]Ea^f=Ea^b+|\Delta H|[/tex]

where,

[tex]Ea^f[/tex] = activation energy for forward reaction

[tex]Ea^b[/tex] = activation energy for backward reaction

[tex]\Delta H[/tex] = change in enthalpy of reaction

As per question, the value of enthalpy change is -293 kJ/mol that means the reaction will be exothermic reaction. So,

[tex]Ea^b=Ea^f+|\Delta H|[/tex]

Given:

[tex]Ea^f[/tex] = activation energy for forward reaction = 217 kJ/mol

[tex]Ea^b[/tex] = activation energy for backward reaction = ?

[tex]\Delta H[/tex] = change in enthalpy of reaction = -293 kJ/mol

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

[tex]Ea^b=217kJ/mol+|-293kJ/mol|[/tex]

[tex]Ea^b=217kJ/mol+293kJ/mol[/tex]

[tex]Ea^b=510kJ/mol[/tex]

Therefore, the activation energy for the reverse reaction is 510 kJ/mol.

Answer:

Incomplete precipitation of barium sulfate

Explanation:

The student has precipitated and digested the barium sulfate on his/her side. But on the addition of [tex]BaCl_2[/tex] in the solution, the solution become cloudy. This happened because incomplete precipitation of barium sulfate by the student. When [tex]BaCl_2[/tex] is added, there are still sulfate ions present in the solution with combines with [tex]BaCl_2[/tex] and forms [tex]BaSO_4[/tex] and the formation of this precipitate makes the solution cloudy.

The solution became cloudy after adding BaCl2 because the additional Ba2+ ions reacted with any remaining [tex]SO4^2-[/tex]orming more BaSO4 precipitate, indicating incomplete digestion of the initial BaSO4.

During the experiment, the student added a few drops of BaCl2 solution to the clear solution that contained digested BaSO4. BaSO4 is known for its low solubility in water; however, it's soluble in solutions containing ions that can form more soluble compounds with the constituent ions. In the experiment, adding more BaCl2 likely introduced additional Ba2+ ions into the solution. If any unreacted sulfate ions (SO42-) were present, these extra Ba2+ ions could have reacted with them, forming additional BaSO4 precipitate, thus causing the solution to become cloudy. This suggests that the digestion process was not complete, leaving some sulfate ions in the solution which reacted with the added barium ions to form more BaSO4 precipitate.

Answer: The volume of solution is [tex]8.03\times 10^5mL[/tex]

Explanation:

The relationship between specific gravity and density of a substance is given as:

[tex]\text{Specific gravity}=\frac{\text{Density of a substance}}{\text{Density of water}}[/tex]

Specific gravity of sulfuric acid solution = 1.22

Density of water = 1.00 g/mL

Putting values in above equation we get:

[tex]1.22=\frac{\text{Density of sulfuric acid solution}}{1.00g/mL}\\\\\text{Density of sulfuric acid solution}=(1.22\times 1.00g/mL)=1.22g/mL[/tex]

We are given:

25% (m/m) sulfuric acid solution. This means that 25 g of sulfuric acid is present in 100 g of solution

Conversion factor:  1 kg = 1000 g

Mass of solution having 254 kg or 245000 g of sulfuric acid is calculated by using unitary method:

If 25 grams of sulfuric acid is present in 100 g of solution.

So, 245000 grams of sulfuric acid will be present in = [tex]\frac{100}{25}\times 245000=980000g[/tex]

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

[tex]\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}[/tex]

Density of solution = 1.22 g/mL

Mass of Solution = 980000 g

Putting values in above equation, we get:

[tex]1.22g/mL=\frac{980000g}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{980000g}{1.22g/mL}=8.03\times 10^5mL[/tex]

Hence, the volume of solution is [tex]8.03\times 10^5mL[/tex]

Final answer:

To estimate the feed rate of logs, we first need to calculate the dry wood pulp production rate using the given information.

Explanation:

To estimate the feed rate of logs, we first need to calculate the dry wood pulp production rate. The dry wood pulp production rate can be calculated using the equation:

Dry wood pulp production rate = Feed rate of logs x (1 - water content of wood chips) x (1 - cellulose content in wood chips on a dry basis)

Plugging in the given values, we have:

2000 tons/day = Feed rate of logs x (1 - 0.45) x (1 - 0.47)

Solving for the feed rate of logs, we find that it is approximately 21.67 logs/min.

Answer:

130 μL

Explanation:

The dilution formula is used to calculate the volume V₁ required:

C₁V₁ = C₂V₂

V₁ = (C₂V₂)/C₁ = (2.00L)(0.001M)/(15.0M) = 1.3 x 10⁻⁴ L or 130 μL

(1.3 x 10⁻⁴ L)(10⁶ μL/L) = 130 μL

An electron microscope is used to view cellular organelles such as the endoplasmic reticulum and Golgi because it provides significantly higher resolution and magnification compared to a light microscope.

The type of microscope used to view cellular organelles such as the endoplasmic reticulum and Golgi is the electron microscope. This instrument magnifies an object using an electron beam that passes and bends through a lens system, providing much higher resolution and magnification than a light microscope. However, most student microscopes are light microscopes, which use a beam of visible light and are typically used for viewing living organisms, as the staining required to make cellular components visible usually kills the cells. Electron microscopes are commonly used in labs and can give detailed visualizations of organelles and the endomembrane system, which involves a group of organelles and membranes that work together in modifying, packaging, and transporting lipids and proteins.

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Explanation:

The given data is as follows.

          Moles of propylene = 100 moles,    [tex]C_{p}[/tex] = 100 J/mol K

          [tex]T_{i}[/tex] = 300 K,          [tex]T_{f}[/tex] = 800 K

          [tex]V_{i}[/tex] = 2 [tex]m^{3}[/tex],   [tex]V_{f}[/tex] = 0.02 [tex]m^{3}[/tex]

Therefore, the assumptions will be as follows.

                    [tex]m \times C_{p} \Delta T[/tex] = W

Putting the given values into the above formula as follows.

                  [tex]m \times C_{p} \Delta T[/tex] = W

         W = [tex]100 moles \times 100 J/mol K \times (800 K - 300 K)[/tex]

              = [tex]5 \times 10^{6}[/tex] J

              = 5 MJ

Hence, this shows that a minimum of 5 MJ work needs to be done.

Since, work is very less. Hence, it will not compress the given system to 800 K and 0.02 [tex]m^{3}[/tex].      

Answer:

757 K = 484 °C

-261 °F = -163 °C

Explanation:

The formula to convert Kelvin to degrees Celsius is:

°C = K - 273.15 = 757 - 273.15 = 484 °C

The formula to convert °F to °C is:

°C = 5/9 (°F -32) = 5/9 (-261 - 32) = -163

The theoretical yield of NH3 in the Haber-Bosch process under the given conditions is 12.036 grams.

Theoretical yield refers to the maximum amount of product that can be obtained in a chemical reaction according to the balanced chemical equation. To calculate the theoretical yield of ammonia in this reaction, you need to determine the limiting reactant first. The limiting reactant is the reactant that is completely consumed in the reaction and determines the maximum amount of product that can be obtained.

In this case, you have 1.15 g of H2 and 9.93 g of N2. To determine the limiting reactant, you can compare the moles of H2 and N2 using their molar masses:

Moles of H2 = (1.15 g H2) / (2 g/mol H2) = 0.575 mol H2

Moles of N2 = (9.93 g N2) / (28 g/mol N2) = 0.354 mol N2

Since the coefficients in the balanced equation are in a 1:1 ratio for H2 and N2, it is clear that the limiting reactant is N2 because there are fewer moles of N2 available.

Now you can use the limiting reactant to calculate the theoretical yield of NH3. According to the balanced equation, the stoichiometric ratio between N2 and NH3 is 1:2. Therefore, moles of NH3 = 2 * moles of N2 = 2 * 0.354 mol = 0.708 mol NH3. Finally, you can convert moles of NH3 to grams using the molar mass of NH3:

Mass of NH3 = (0.708 mol NH3) * (17 g/mol NH3) = 12.036 g NH3

Therefore, the theoretical yield of NH3 under the given conditions is 12.036 grams.

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To prepare a standard buffer at pH 7.00 with a total phosphate concentration of 0.100 M, you will need to calculate the amount of sodium dihydrogen phosphate (NaH2PO4 · H2O) and disodium hydrogen phosphate (Na2HPO4) needed.

To prepare a standard buffer at pH 7.00 with a total phosphate concentration of 0.100 M, you will need to calculate the amount of sodium dihydrogen phosphate (NaH2PO4 · H2O) and disodium hydrogen phosphate (Na2HPO4) needed.

Step 1: Calculate the individual concentrations of NaH2PO4 and Na2HPO4.

Using the molecular weight, you can calculate the number of moles of NaH2PO4 and Na2HPO4 needed to achieve a total phosphate concentration of 0.100 M in 1 L of solution.

NaH2PO4: (0.100 M) * (1 L) = x mol

Na2HPO4: (0.100 M) * (1 L) = y mol

Step 2: Calculate the weight of NaH2PO4 and Na2HPO4.

Using the number of moles calculated in step 1, you can calculate the weight of NaH2PO4 and Na2HPO4 needed.

NaH2PO4: (x mol) * (138 g/mol) = weight in grams

Na2HPO4: (y mol) * (142 g/mol) = weight in grams

By following these steps, you will be able to determine the weight in grams of NaH2PO4 · H2O and Na2HPO4 needed to prepare 1 L of the standard buffer solution.

Answer:

a) 0,857 g of H₃PO₄ with 9,016 g of KH₂PO₄

b) 166 mL of 0,400M NaOH

c) 22 mL of 0,400M HCl

Explanation:

a) The appropriate weak acid and conjugate salt are:

H₃PO₄ ⇄ H₂PO₄⁻ + H⁺ where pka = 2,12

Henderson–Hasselbalch equation finding pH = 3:

3 = 2,12 + log₁₀ [tex]\frac{[H2PO4-]}{[H3PO4]}[/tex]

7,59 =  [tex]\frac{[H2PO4-]}{[H3PO4]}[/tex] (1)

If buffer concentration is 0,300M:

0,300 M = [H₃PO₄] + [H₂PO₄⁻] (2)

Replacing (2) in (1):

[H₃PO₄] = 0,035 M

Thus:

[H₂PO₄⁻] = 0,265 M

Thus, to prepare this buffer you need weight:

0,035 M × 0,250 L = 8,75x10⁻³ moles × [tex]\frac{97,994 g}{1mol}[/tex] = 0,857 g of H₃PO₄

And:

0,265 M × 0,250 L = 6,63x10⁻² moles × [tex]\frac{136,086 g}{1mol}[/tex] = 9,016 g of KH₂PO₄

b) Using 0,400 M NaOH the equilibrium is:

H₃PO₄ + NaOH ⇄ H₂PO₄⁻ + H₂O

Knowing the equilibrium concentrations are:

[H₃PO₄] = 0,035 M = 0,300 M - x -Because in the first all 0,300 M must be of H₃PO₄-

[H₂PO₄⁻] = 0,265 M

Thus, x = 0,265 M are NaOH needed to obtain the desire pH. Those are obtained thus:

0,265 mol/ L × 0,250L × [tex]\frac{1L}{0,400mol}[/tex] = 0,166 L ≡ 166 mL of 0,400M NaOH

c) Using 0,400 M HCl the equilibrium is:

H₃PO₄ ⇄ H₂PO₄⁻ + HCl

Knowing the equilibrium concentrations are:

[H₃PO₄] = 0,035 M

[H₂PO₄⁻] = 0,265 M = 0,300 M - x -Because in the first all 0,300 M must be of H₂PO₄-

Thus, x = 0,035 M are HCl needed to obtain the desire pH. Those are obtained thus:

0,035 mol/ L × 0,250L × [tex]\frac{1L}{0,400mol}[/tex] = 0,022 L ≡ 22 mL of 0,400M HCl

I hope it helps!

Explanation:

Melting point is defined as the point at which a solid substance starts to change into liquid state.

Whereas entropy is the degree of randomness of molecules present in a substance.

Heat of fusion is defined as the amount of heat energy necessary to melt a solid substance at its melting point.

Relation between entropy and heat of fusion is as follows.

                  [tex]\Delta S = \frac{\Delta H}{T}[/tex]

where,          [tex]\Delta S[/tex] = 38.98 J/mol K

                     [tex]\Delta H[/tex] = 17.02 kJ/mol

                                    = [tex]17.02 kJ/mol \times \frac{1000 J}{1 kJ}[/tex]

                                    = 17020 J/mol

Therefore, calculate the melting point as follows.

                   [tex]\Delta S = \frac{\Delta H}{T}[/tex]

                           38.98 J/mol K = [tex]\frac{17020 J/mol}{T}[/tex]

                              T = 436.63 K

Change the temperature into degree celsius as follows.

                             [tex](436.63 - 273)^{o}C[/tex]

                                 = [tex]163.63^{o}C[/tex]

Thus, we can conclude that the melting point in [tex]^{o}C[/tex] is [tex]163.63^{o}C[/tex].

Answer:

Equilibrium constant =  [tex]2.23 \times 10^{83}[/tex]

Explanation:

[tex]Zr(s) + O_2(g) \rightarrow ZrO_2(s)[/tex]

[tex]E^0_{cell}[/tex] = 2.463 V

Equilibrium constant is related with [tex]E^0_{cell}[/tex] as

[tex]E_{cell}=E^0_{cell} - \frac{2.303 RT}{nF} ln k_{eq}[/tex]

In standard condition,

T = 25 °C = 25 + 273 = 298 K

F = 96500 C mol^-1

R = 8.314 [tex]J\ K^{-1}mol^{-1}[/tex]

On substituting values, the above expression becomes:

[tex]E_{cell}=E^0_{cell} - \frac{0.059}{n} log k_{eq}[/tex]

n = 2

At equilibrium, [tex]E_{cell}= 0[/tex]

[tex]0=E^0_{cell} - \frac{0.059}{2} log k_{eq}[/tex]

[tex]log k_{eq}=\frac{2 \times 2.463}{0.059}[/tex]

= 83.35

[tex]K_{eq} = antilog 83.35 = 2.23 \times 10^{83}[/tex]

Answer:

1. kmol of methanol= 3.12 Kmol

2. kmol of water= 5.55 Kmol

3. Liters of methanol=  126.4  L

4. L of water= 100  L  

Explanation:

1. kmol of methanol?

32.04 kg methanol ______________ 1 kmol of methanol

100 kg of methanol_______________ X=  3.12 kmol ofmethanol

2. kmol of water?

18.01 kg water ______________ 1 kmol of wáter

100 kg of wáter_______________ X=  5.55 kmol of water

3. Liters of methanol?

0.791 kg  methanol _______________________1.00  L of methanol

100kg  methanol  _______________________x= 126.4  L of methanol

4. L of water?

1kg  water _______________________1.00  L of water

100kg  water _______________________x= 100  L of water