A compound consists of 65.45% C, 5.492% H, and 29.06% O on a mass basis and has a molar mass of 110 g/mol. Determine the molecular formula of the compound.

The pressure P2 at the temperature which CO2 gets solidifies is 0.857 atm.

Explanation:

The relation between temperature T and the pressure P is that it is proportional to each other.

                                      T ∝ P

As the temperature decreases, the pressure also decreases which is given by

                             T1 / T2 = P1 / P2

At the triple point of water, the temperature equals 273 K.

Consider T1 = 273 K,  P1 = 1.20 atm

The temperature T2 of the CO2 solidifies equals 195 K

                            (273 / 195) = (1.20 / P2)

                              P2 = (195 x 1.20) / 273

                             P2 = 0.857 atm.

The pressure P2 at the temperature which CO2 gets solidifies is 0.857 atm.

Answer:

C. the use of hydrogen gas as an electron donor.

Explanation:

Hydrogenotrophy is the convertion of hydrogen gas to other compounds as part of its metabolism.

Answer:

2Ag (s) + 2HNO₃ (aq)  2AgNO₃ (aq) + H₂ (g)

Explanation:

2Ag (s) + 2HNO₃ (aq)  2AgNO₃ (aq) + H₂ (g)

Ag is below H₂ in reactivity series. Therefore Ag does not spontaneously replace H₂ from any compound.

Answer:

The answer to the question is

2Ag (s) +2HNO₃ (aq) →2AgNO₃ (aq)+ H₂ (g)

Explanation:

The position of a cell on the reduction potential table determines if a reduction reaction will be spontaneous

The higher the positive reduction potential  the higher the spontaneity of the reduction reaction. that is if the E⁰ cell is positive the cell is a spontaneous voltaic cell, however if the E⁰ cell is negative the cell is a electrolytic non-spontaneous cell

On the standard potential table silver has a low positive potential of

Ag⁺ + e⁻ → Ag = 0.799

It is an oxidizing agent tending to scarcely release electrons, therefore for the reaction to take place, there has to be some external supply of electromotive force.

20 mL of an approximately 10% aqueous solution of ethylamine, CH3CH2NH2, is titrated with 0.3000 M aqueous HCl. Which indicator would be most suitable for this titration? The pKa of CH3CH2NH3+ is 10.75.

Answer:

Bromocresol green, color change from pH = 4.0 to  5.6

Explanation:

The equation for the reaction is as follows:

[tex]C_2H_5NH_{2(aq)[/tex]     +     [tex]H^+_{(aq)[/tex]      ⇄        [tex]C_2H_5NH_{3(aq)}^+[/tex]

Given that concentration of [tex]C_2H_5NH_{2(aq)[/tex] = 10%

i.e 10 g of [tex]C_2H_5NH_{2(aq)[/tex] in 100 ml solution

molar mass = 45.08 g/mol

number of moles = [tex]\frac{10}{45.08}[/tex]

= 0.222 mol

Molarity of [tex]C_2H_5NH_{2(aq)[/tex] = 0.222 × [tex]\frac{1000}{100}mL[/tex]

= 2.22 M

However, number of moles of [tex]C_2H_5NH_{2(aq)[/tex] in 20 mL can be determined as:

number of moles of [tex]C_2H_5NH_{2(aq)[/tex] = 20 mL × 2.22 M

= [tex]44*10^{-3} mole[/tex]

Concentration of [tex]C_2H_5NH_{2(aq)[/tex] = [tex]\frac{44*10^{-3}*1000}{20}[/tex]

= 2.22 M

Similarly, The pKa Value of [tex]C_2H_5NH_{2(aq)[/tex]  is given as 10.75

pKb value will be: 14 - pKa

= 14 - 10.75

= 3.25

Finally, the pH value at equivalence point is:

pH= [tex]\frac{1}{2}pKa - \frac{1}{2}pKb-\frac{1}{2}log[C][/tex]

pH = [tex]\frac{14}{2}-\frac{3.25}{2}-\frac{1}{2}log [2.22][/tex]

pH = 5.21

∴

The indicator that is best fit for the given titration is Bromocresol Green Color change from pH between 4.0 to 5.6.

For titration of the weak base ethylamine with a strong acid like HCl, the pH at the equivalence point will be less than 7, so an indicator that changes color in an acidic pH range such as methyl orange would be most suitable.

The appropriate indicator in this case would depend on the equivalence point of the titration. Ethylamine, CH3CH2NH2, is a weak base, and it's being titrated with HCl, a strong acid. The reaction between them generates water, and ethylammonium chloride which functions as a weak acid. This setup indicates that the pH at the equivalence point will be less than 7.

Because the pH at the equivalence point will be in the acidic range (below 7), we should select an indicator which changes color around this pH. A good choice would be methyl orange, which changes color from red to yellow across a pH range of approximately 3.1 to 4.4.

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Answer:

The limiting reactant can be described as the substance that is depleted first and stops

Explanation:

Imagine you have hydrogen and oxygen to produce water.

The reaction is:  2H₂(g) + O₂(g) → 2H₂O(g)

You have 1 mol of each reactant. As you see ratio is 2:1, so the limiting reactant is the hydrogen.

You know by stoichiometry, that 2 moles of H₂ need 1 mol of O₂ to react

If I have 1 mol of H₂, I will need the half of moles of O₂, so 0.5 moles. It is ok  because I have 1 mole, as I need the half, then half a mole will remain unreacted. This is what is called excess reagent,

If I make to react 1 mol of oxgen I need 2 moles of H₂. As I have 1 mol, of course I will need 2 moles but the thing is I have 1 mol.

This is the limiting reactant. I do not have enough of reactant so the reaction will happen until I complete to use it, that's why we can say that is depleted first and stops.

In a chemical reaction, when you have data of both reactants you can determine the limiting. Otherwise the excersise must tell that one ractant is in excess, to work with the limiting. Limiting reactant is the first step to work with the reaction, all the operations must be done by it. You do not use the reagent in excess

Answer:

secrete cytotoxic substance which triggers apoptosis of target cell.

Explanation:

Cytotoxic T cells have cell surface receptor which recognizes the antigen present on the receptor of target cell. This interaction initiates the process of killing of target cell.

After interaction cytotoxic t cell release cytotoxic substance called granzyme and perforin. Granzyme triggers apoptosis through the activation of caspases or by making the release of cytochrome c and activation of the apoptosome.

Perforin make pores in the cell and its action is similar to complement membrane attack complex. Therefore cytotoxic substances are released by Tc cells which trigger apoptosis of target cell.

Answer:

ground water / leaking storage tanks

Answer:

42 liters

Explanation:

Let the volume of 2% sulfuric acid required be X while that of 12% sulfuric acid required be Y.

According to the concentration: 0.02X + 0.12Y = 0.05(X + Y).......equation 1

According to the volume: X + Y = 60...............equation 2

Substitute equation 2 into equation 1.

0.02X + 0.12(Y) = 0.05(60)

2X + 12Y = 300

X + 6Y = 150..............................equation 3

From equation 2, Y = 60 -X...................equation 4

Substitute equation 4 into equation 3.

X + 6(60-X) = 150

X + 360 - 6X = 150

-5X = -210

X = 42

Hence, the volume of 2% sulfuric acid required is 42 liters.

Answer:

We can for 5.93 grams potassium fluoride

Explanation:

Step 1: Data given

Mass of potassium = 4.00 grams

Mass of fluorine = 3.00 grams

Molar mass potassium = 39.10 g/mol

Molar mass fluorine gas =38.00 g/mol

Step 2: The balanced equation

2K (s) + F2 (g) → 2KF (s)

Step 3: Calculate moles potassium

Moles potassium = 4.00 grams / 39.10 g/mol

Moles potassium = 0.102 moles

Step 4: Calculate moles F2

Moles F2 = 3.00 grams / 38.00 g/mol

Moles F2 = 0.0789 moles

Step 5: Calculate limiting reactant

Potassium is the limiting reactant. There will react 0.102 moles

Fluorine gas is in excess. There will react 0.102/ 2 = 0.051 moles

There will remain 0.0789 - 0.051 = 0.0279 moles

Step 6: Calculate moles potassium fluoride

For 2 moles potassium we need 1 mol fluorine to produce 2 moles potassium fluoride

For 0.102 moles K we need 0.102 moles KF

Step 7: Calculate mass KF

Mass KF = moles KF * molar mass KF

Mass KF = 0.102 moles * 58.10 g/mol

Mass KF = 5.93 grams

Final answer:

By calculating the moles of potassium and fluorine used in the reaction from their given masses, and considering the stoichiometry of the balanced equation, it's determined that 4.59 grams of potassium fluoride can be produced.

Explanation:

The question asks how many grams of potassium fluoride can form when 4.00 grams of potassium react with 3.00 grams of fluorine gas according to the balanced chemical equation 2K (s) + F2 (g) → 2KF (s). To solve this, we first find the molar mass of potassium (K) and fluorine (F2), which are approximately 39.10 g/mol and 38.00 g/mol, respectively. Then, we calculate the moles of K and F2 available by dividing the given masses by their respective molar masses. With potassium: 4.00 g / 39.10 g/mol = 0.102 moles of K; with fluorine: 3.00 g / 38.00 g/mol = 0.079 moles of F2. As the equation shows a 2:1 ratio, fluorine limits the reaction. Therefore, we can form 0.079 moles of KF. The molar mass of KF is approximately 58.10 g/mol (39.10 g/mol for K + 19.00 g/mol for F). Thus, the mass of KF that can form is 0.079 moles × 58.10 g/mol = 4.59 grams.

Answer:

B

Explanation:

The interaction between magnetism and electricity is called electromagnetism. The movement of a magnet can generate electricity. The flow of electricity can generate a magnetic field.

Answer: B: electromagnetism

Explanation:

A changing magnetic field produces an electric current in a wire or conductor.

Answer:

12 molecules of carbon dioxide

24 molecules of water

Explanation:

Given parameters:

Number of molecules of methane = 12 molecules

Unknown:

Number of molecules of carbon dioxide = ?

Number of molecules of water = ?

Solution:

The given amount of methane is the limiting reagent in this reaction. By this, we can ascertain the amount of products that will be formed and the extent of the reaction.

We first write the balanced chemical equation for this reaction;

             CH₄     +    2O₂    →     CO₂   +     2H₂O

From balanced equation;

                 1 mole of methane produced 1 mole of carbon dioxide

  12 molecules of methane will produce 12 molecules of carbon dioxide

Also;

             1 mole of methane produced 2 moles of water;

12 molecules of methane will produce (12 x 2)molecules = 24 molecules of water

When 12 molecules of methane react with excess oxygen, 12 molecules of carbon dioxide and 24 molecules of water are produced, following a 1:2:1:2 ratio of methane to oxygen to carbon dioxide to water.

If 12 molecules of methane (CH4) react with plenty of oxygen (O2), the balanced chemical equation for the combustion of methane shows a 1:2:1:2 ratio. This means for every one methane molecule, two oxygen molecules are consumed, and one carbon dioxide (CO2) molecule and two water (H2O) molecules are produced. Therefore, 12 molecules of methane reacting with excess oxygen will produce 12 molecules of carbon dioxide and 24 molecules of water.

This equation states that for every 1 molecule of methane and 2 molecules of oxygen that react, we will produce 1 molecule of carbon dioxide and 2 molecules of water.

Answer:Covalent

Explanation:

Covalent bonds are formed by sharing of electrons between two atoms. The shared electrons are normally situated between the nuclei of the both atoms. However, atoms of some elements posses an usual ability to attract the shared electrons of a bond towards itself. We say that such atoms have a high electronegativity. High electronegativity leads to the existence of polar covalent bonds since the shared electron pair is now closer to one of the bonding atoms than the other. The atom to which the electron pair is closer becomes partially negative while the other becomes partially positive. This is only possible in a covalent bond where electrons are shared between bonding atoms. The charge separation is known as a dipole.

Note that ionic bonds involve a complete transfer of electrons leading to the formation of ions and is not applicable here.

Answer:

the value of Kp at 25°C is 2.37 × 10⁻⁴atm³

Explanation:

Given that Kp = 0.116atm

NH₄(NH₂CO₂)(s) ⇄ 2NH₃(g) + CO₂(g)

                               2x atm         xatm

Pt                     =      PNH₃         PCO₂

0.116                =      2x          +        x

0.116                =      3x

x = 0.116/3

x = 0.039 atm

PNH₃ = 2x =2(0.039) = 0.078 atm

PCO₂ = x = 0.039 atm

Now,

KP = PNH₃² × PCO₂

     = 0.078² × 0.039 atm³

     = 2.37  × 10⁻⁴ atm³

the value of Kp at 25°C is 2.37 × 10⁻⁴atm³

Answer : The value of [tex]K_p[/tex] is, [tex]2.32\times 10^{-4}[/tex]

Explanation :

For the given chemical reaction:

[tex](NH_4)(NH_2CO_2)(s)\rightleftharpoons 2NH_3(g)+CO_2(g)[/tex]

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

[tex]K_p=(P_{NH_3})^2\times P_{CO_2}[/tex]        ........(1)

                      [tex](NH_4)(NH_2CO_2)(s)\rightleftharpoons 2NH_3(g)+CO_2(g)[/tex]

Initial:                                                      0               0

At eqm:                                                   2x              x

As we are given that:

Total pressure of gas at equilibrium = 0.116 atm

2x + x = 0.116 atm

3x = 0.116 atm

x = 0.0387 atm

Putting values in expression 1, we get:

[tex]K_p=(2x)^2\times (x)[/tex]

[tex]K_p=(2\times 0.0387)^2\times (0.0387)[/tex]

[tex]K_p=2.32\times 10^{-4}[/tex]

Thus, the value of [tex]K_p[/tex] is, [tex]2.32\times 10^{-4}[/tex]

Answer : The value of new volume is, 50.0 mL

Explanation :

Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

[tex]P\propto \frac{1}{V}[/tex]

or,

[tex]P_1V_1=P_2V_2[/tex]

where,

[tex]P_1[/tex] = initial pressure at STP = 1 atm

[tex]P_2[/tex] = final pressure =  4.00  atm

[tex]V_1[/tex] = initial volume at STP = 200.0 mL

[tex]V_2[/tex] = final volume = ?

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

[tex]1atm\times 200.0mL=4.00atm\times V_2[/tex]

[tex]V_2=50.0mL[/tex]

Therefore, the value of new volume is, 50.0 mL

Answer:

There are 1.29 moles of toluene in the solution.

Explanation:

m = ∆Tb/Kb

m is molality of the solution

∆Tb is the change in boiling point of cyclohexane = 90.3 ° C - 80.9 °C = 9.4 °C

Kb is the boiling point elevation constant of cyclohexane = 2.92 °C/m

m = 9.4/2.92 = 3.22 mol/kg

Number of moles of toluene = molality × mass of cyclohexane in kilogram = 3.22 × 401/1000 = 1.29 moles

Answer: The percent change in volume will be 25 %

Explanation:

To calculate the final temperature of the system, we use the equation given by Charles' Law. This law states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

[tex]\frac{V_1}{T_1}=\frac{V_2}{T_2}[/tex]

where,

[tex]V_1\text{ and }T_1[/tex] are the initial volume and temperature of the gas.

[tex]V_2\text{ and }T_2[/tex] are the final volume and temperature of the gas.

We are given:

[tex]V_1=2L\\T_1=T_1\\V_2=?\\T_2=75\% \text{ of }T_1=0.75\times T_1[/tex]

Putting values in above equation, we get:

[tex]\frac{2L}{T_1}=\frac{V_2}{0.75\times T_1}\\\\V_2=\frac{2\times 0.75\times T_1}{T_1}=1.5L[/tex]

Percent change of volume = [tex]\frac{\text{Change in volume}}{\text{Initial volume}}\times 100[/tex]

Percent change of volume = [tex]\frac{(2-1.5)}{2}\times 100=25\%[/tex]

Hence, the percent change in volume will be 25 %

Answer: Volume of the balloon at this altitude is 46.9 L

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]

where,

[tex]P_1[/tex] = initial pressure of gas = 737 mm Hg

[tex]P_2[/tex] = final pressure of gas =  385 mm Hg

[tex]V_1[/tex] = initial volume of gas = 28.6 L

[tex]V_2[/tex] = final volume of gas = ?

[tex]T_1[/tex] = initial temperature of gas = [tex]26.8^oC=273+26.8=299.8K[/tex]

[tex]T_2[/tex] = final temperature of gas = [tex]-16.3^oC=273-16.3=256.7K[/tex]

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

[tex]\frac{737\times 28.6}{299.8}=\frac{385\times V_2}{256.7}[/tex]

[tex]V_2=46.9L[/tex]

Thus the volume of the balloon at this altitude is 46.9 L

Answer:

nitrogen monoxide: NO

nitrogen dioxide: NO₂

Explanation:

Nitrogen monoxide is composed by 1 atom of O (prefix "mono-") and 1 atom of N. Nitrogen dioxide is composed by 2 atoms of O (prefix "di-") and 1 atom of N. As the oxigen atom in oxides has the valency -2 (it shares 2 electrons), the nitrogen has valency +2 in NO and +4 in NO₂.

2.61 kilojoules of heat would be liberated by the condensation of 5.00 g of acetone

Explanation:

To convert grams into moles

[tex]moles = grams \times \frac{1 mole}{grams}[/tex]

We have 5.00 g acetone

[tex]moles = 5 \times \frac{1}{58.1}[/tex]

[tex]moles = 0.0861[/tex]

Heat liberated = moles [tex]\times[/tex] heat of vapourization

                         =0.0861 mol x 30.3 kJ/mol

                        = 2.61 kJ

Therefore, 2.61 kilojoules of heat would be liberated by the condensation of 5.00 g of acetone

Answer:

= 238.6J

Explanation:

According to the law of conservation of energy, energy can neither be created nor be destroyed. It can only be transformed from one form to another.

Endothermic reactions are those in which heat is absorbed by the system and thus the energy of products is higher than the energy of reactants.

For the given reaction:

A + B ⇄ C + D

Energy of A = 85.1 J

Energy of B = 87.9 J

Product C contains 38.7 J

Energy balance:  

∑ enthalpy of the reactants + energy added = ∑ enthalpy of the products + energy released.

∑ enthalpy of the reactants = 85.1 J + 87.9 J = 173 J

energy added = 104.3 J

∑ enthalpy of the products = 38.7 J + D

energy released = 0

Equation:

173J + 104.3J = 38.7 + D + 0  

⇒ D = 173J + 104.3J - 38.7J

= 238.6J

which is the chemical energy of the product D

Product D must contain 238.6 J of chemical energy based on the principles of energy conservation in a chemical reaction.

In a chemical reaction, the law of conservation of energy states that energy cannot be created or destroyed; it can only change forms. This principle applies to chemical energy as well. Therefore, the total energy in the reactants must equal the total energy in the products.

Let's analyze the given information and calculate the chemical energy in product D:

Reactant A contains 85.1 J of chemical energy.

Reactant B contains 87.9 J of chemical energy.

Product C contains 38.7 J of chemical energy.

The total initial chemical energy in the reactants (A and B) is:

Initial Energy (A + B) = 85.1 J + 87.9 J = 173.0 J

The reaction absorbs 104.3 J of chemical energy as it proceeds. This means that the total energy in the products (C + D) must be equal to the initial energy plus the absorbed energy:

Total Energy in Products (C + D) = Initial Energy (A + B) + Absorbed Energy

Total Energy in Products (C + D) = 173.0 J + 104.3 J = 277.3 J

Now, we know the total energy in the products (C + D) is 277.3 J, and we already know the energy in product C is 38.7 J. To find the energy in product D, we can subtract the energy in C from the total energy in the products:

Energy in Product D = Total Energy in Products - Energy in Product C

Energy in Product D = 277.3 J - 38.7 J = 238.6 J

Therefore, product D must contain 238.6 J of chemical energy.

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1.72 atm of hydrogen gas is created when 2.29 atm of phosphine breaks down at 953 K for 70.5 seconds.

Explanation:

The first step in this calculation is to use the first-order decay equation to find the remaining moles of phosphine. For a first-order reaction with a half-life of 35.0 s, after 70.5 s (or two half-lives), 25% of the phosphine would remain. To find the remaining moles of phosphine, we multiply 2.29 atm by 25%, yielding 0.57 atm.

From the balanced equation for the decomposition of phosphine (4PH3 -> P4 + 6H2), we see that 1 mole of phosphine yields 3/2 moles of H2. Thus, the number of moles of H2 produced is 75% of the original moles of PH3 (1.72 atm).

The sum of the partial pressures of phosphine and hydrogen is 0.57 + 1.72 = 2.29 atm, which is the final pressure in the 2.20 L vessel after 70.5 s.

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The question is incomplete, here is the complete question:

An aqeous solution of oxalic acid was prepared by dissolving a 0.5842 g of solute in enough water to make a 100 ml solution a 10 ml aliquot of this solution was then transferred to a volumetric flask and diluted to a final volume of 250 ml. How many grams of oxalic acid are in 100. mL of the final solution?

Answer: The mass of oxalic acid in final solution is 0.0234 grams

Explanation:

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

[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}[/tex]     ......(1)

Given mass of oxalic acid = 0.5842 g

Molar mass of oxalic acid = 90 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

[tex]\text{Molarity of oxalic acid solution}=\frac{0.5842\times 1000}{90\times 100}\\\\\text{Molarity of oxalic acid solution}=0.0649M[/tex]

To calculate the molarity of the diluted solution, we use the equation:

[tex]M_1V_1=M_2V_2[/tex]

where,

[tex]M_1\text{ and }V_1[/tex] are the molarity and volume of the concentrated oxalic acid solution

[tex]M_2\text{ and }V_2[/tex] are the molarity and volume of diluted oxalic acid solution

We are given:

[tex]M_1=0.0649M\\V_1=10mL\\M_2=?M\\V_2=250.0mL[/tex]

Putting values in above equation, we get:

[tex]0.0649\times 10=M_2\times 250.0\\\\M_2=\frac{0.0649\times 10}{250}=0.0026M[/tex]

Now, calculating the mass of glucose by using equation 1, we get:

Molarity of oxalic acid solution = 0.0026 M

Molar mass of oxalic acid = 90 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

[tex]0.0026=\frac{\text{Mass of oxalic acid solution}\times 1000}{90\times 100}\\\\\text{Mass of oxalic acid solution}=\frac{0.0026\times 90\times 100}{1000}=0.0234g[/tex]

Hence, the mass of oxalic acid in final solution is 0.0234 grams

The question involves dilution of an oxalic acid solution and subsequent calculation of the molarity for titration, which is a typical procedure in a college-level chemistry course.

The student's question pertains to the dilution of an aqueous solution of oxalic acid and the calculation of molarity after dilution and titration. Starting with a mass of 0.5842g of oxalic acid dissolved to make a 100 ml solution, a 10 ml aliquot is further diluted to 250 ml. The goal is to understand the change in concentration as a result of the dilution process and to apply this understanding to various problems presented, such as titration against potassium permanganate (KMnO4).

To find the amount of NaCl needed to increase the boiling temperature of water by 1.500°C, one must understand molal boiling point elevation and van't Hoff factor. Using the formula and given constants, we calculate the grams of NaCl required through the steps of determining molality, calculating the moles of NaCl, and then converting to grams.

To calculate the amount of NaCl needed to raise the boiling temperature of water by 1.500°C, we first understand the concept of molal boiling point elevation and van't Hoff factor. Given that the Kb for water is 0.5100°C/m, we use the formula ΔTb = i*Kb*m, where ΔTb is the boiling point elevation, i is the van't Hoff factor for NaCl (which is 2 because NaCl disassociates into Na+ and Cl- ions), Kb is the ebullioscopic constant for the solvent (water), and m is the molality of the solution.

To find the molality (m), we rearrange the equation as m = ΔTb / (i*Kb). Substituting the known values, we get m = 1.500 / (2*0.5100) = 1.4706 mol/kg. Knowing the molality and the mass of the solvent (water), we can then calculate the moles of NaCl required, which is molality * mass of solvent in kg. Finally, converting moles of NaCl to grams using its molar mass (58.44 g/mol), gives us the total grams of NaCl needed. This step-by-step process provides a clear link between the theoretical concepts and their practical application.

Approximately 85.78 grams of [tex]NaCl[/tex] would need to be added to 1001 grams of water to increase the boiling temperature of the solution by [tex]1.500 \°C[/tex].

To solve this problem, we will use the concept of boiling point elevation, which states that the boiling point of a solvent is increased by an amount proportional to the molal concentration of the solute. The proportionality constant is known as the ebullioscopic constant (Kb). The formula to calculate the boiling point elevation is:

[tex]\[ \Delta T_b = i \cdot K_b \cdot m \][/tex]

 where:

- [tex]\( \Delta T_b \)[/tex] is the increase in boiling point temperature,

- [tex]\( i \)[/tex] is the van 't Hoff factor (the number of moles of particles in solution per mole of solute; for [tex]NaCl[/tex], [tex]\( i = 2 \)[/tex] because it dissociates into two ions, [tex]Na^+ and Cl^-)[/tex],

- [tex]\( K_b \)[/tex] is the ebullioscopic constant for the solvent (given as [tex]0.5100 \°C[/tex]/m for water),

- [tex]\( m \)[/tex] is the molality of the solution (moles of solute per kilogram of solvent).

Given:

- [tex]\( \Delta T_b = 1.5000 °C \)[/tex]

- [tex]\( K_b = 0.5100 °C/m \)[/tex]

- [tex]\( i = 2 \)[/tex] for [tex]NaCl[/tex]

- Mass of water (solvent) = 1001 g (which is approximately 1 kg, since 1000 g = 1 kg)

First, we will rearrange the formula to solve for the molality (m):

[tex]\[ m = \frac{\Delta T_b}{i \cdot K_b} \][/tex]

Substitute the given values:

[tex]\[ m = \frac{1.5000 °C}{2 \cdot 0.5100 °C/m} \][/tex]

[tex]\[ m = \frac{1.5000 °C}{1.0200 °C/m} \][/tex]

[tex]\[ m \approx 1.4695 m \][/tex]

Now that we have the molality, we can calculate the number of moles of [tex]NaCl[/tex] needed to achieve this molality in 1 kg of water:

[tex]\[ \text{moles of NaCl} = m \cdot \text{mass of solvent in kg} \][/tex]

[tex]\[ \text{moles of NaCl} = 1.4695 \text{ m} \cdot 1 \text{ kg} \][/tex]

[tex]\[ \text{moles of NaCl} \approx 1.4695 \][/tex]

The molar mass of [tex]NaCl[/tex] is approximately 58.44 g/mol. To find the mass of [tex]NaCl[/tex] needed, we multiply the number of moles by the molar mass:

[tex]\[ \text{mass of NaCl} = \text{moles of NaCl} \cdot \text{molar mass of NaCl} \][/tex]

[tex]\[ \text{mass of NaCl} \approx 1.4695 \text{ mol} \cdot 58.44 \text{ g/mol} \][/tex]

[tex]\[ \text{mass of NaCl} \approx 85.7797 \text{ g} \][/tex]

The number of liters of oxygen formed at STP when 20.8 g of potassium chlorate decomposes is 5.70 L.

Potassium chlorate decomposes according to the following balanced equation:

2KClO3 → 2KCl + 3O2

The molar mass of KClO3 is 122.55 g/mol.

To calculate the number of moles of KClO3, we divide the given mass (20.8 g) by the molar mass:

20.8 g / 122.55 g/mol = 0.1697 mol

According to the balanced equation, for every 2 moles of KClO3, 3 moles of O2 are produced.

Therefore, the number of moles of O2 produced is:

0.1697 mol × (3/2) = 0.2546 mol

At Standard Temperature and Pressure (STP), 1 mole of any gas occupies approximately 22.4 liters.

Therefore, the number of liters of O2 produced at STP is:

0.2546 mol × 22.4 L/mol = 5.70 L

Answer:

a) pPCl5 = 0.856 atm

b)pPCl5 = 0.0557 atm

pCl2 = pCl3 = 0.800 atm

c)  Ptotal = 1.66 atm

d) 93.5

Explanation:

Step 1: Data given

Temperature = 600 K

Kp = 11.5

Mass of PCl5 = 2.010 grams

Volume of the bulb = 555 mL = 0.555 L

The bulb is heated to 600 K

Step 2: The balanced equation

PCl5(g) ⇄ PCl3(g) + Cl2(g)

Step 3:

a) pv = nrt

⇒with p = the pressure = TO BE DETERMINED

⇒with V = the volume = 0.555 L

⇒ with n =the number of moles PCl5 = 2.010 grams / 208.24 g/mol = 0.00965 moles

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

⇒ with T = the temperature = 600K

p = (0.00965 *0.08206*600)/0.555

pPCl5 = 0.856 atm

b)

The initial pressures

pPCl5 = 0.856 atm

pCl2 = pCl3 = 0 atm

For 1 mol PCl5 we'll have PCl3 and 1 mol Cl2

The pressure at the equilibrium

pPCl5 = (0.856 -x) atm

pCl2 = pCl3 = x atm

Kp = pPCl3 * pCl2/pPCl5  

11.5 = x*x / (0.856 - x)

11.5 = x²/(0.856- x)

x = 0.8003

pPCl5 = (0.856 -x) atm = 0.0557 atm

pCl2 = pCl3 = x atm = 0.800 atm

c) Since x = 0.8003 and PCl3 and PCl2 are x  

Ptotal = 0.8003 + 0.8003 +0.0557 = 1.66 atm

d)

The degree of dissociation = (x / initial pressure PCl5)

(0.8003/0.856) * 100 = 93.5

Answer : The final temperature of the aluminum is, [tex]155.9^oC[/tex]

Explanation :

Formula used :

[tex]q=m\times c\times (T_{final}-T_{initial})[/tex]

where,

q = heat = 3600 cal = 15062.4 J    (1 cal = 4.184 J)

m = mass of aluminum = 125 g

c = specific heat of aluminum = [tex]0.90J/g^oC[/tex]

[tex]T_{final}[/tex] = final temperature = ?

[tex]T_{initial}[/tex] = initial temperature = [tex]22^oC[/tex]

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

[tex]15062.4J=125g\times 0.90J/g^oC\times (T_{final}-22)^oC[/tex]

[tex]T_{final}=155.9^oC[/tex]

Thus, the final temperature of the aluminum is, [tex]155.9^oC[/tex]

Answer:

0.3 Coulomb charge flows through the toy car.

0.9 Joules of energy is lost by the battery

Explanation:

[tex]Current(I)=\frac{charge(Q)}{Time (T)}[/tex]

a)

Current drawn from the battery = 0.50 mA = (0.50 × 0.001 A)

milli Ampere = 0.001 Ampere

Duration of time current drawn = T = 10 min = 10 × 60 s = 600 s

1 min = 60 seconds

Charge flows through the toy car be Q

[tex]0.50\times 0.001 A=\frac{Q}{600 s}[/tex]

[tex]Q=0.50 \times 0.001 A\times 600 s=0.3 C[/tex]

0.3 Coulomb charge flows through the toy car.

b)

[tex]Heat(H)=Voltage(V)\times Current(I)\times Time(T)[/tex]

Voltage of the battery = V = 3.0 V

Current drawn from the battery = 0.50 mA = (0.50 × 0.001 A)

Duration of time current drawn = T = 10 min = 10 × 60 s = 600 s

[tex]H=V\times I\time T=3.0 V\times 0.50 \times 0.001 A\times 600 s[/tex]

H = 0.9 Joules

0.9 Joules of energy is lost by the battery

Answer:

The answer to the question is;

The inter-molecular forces of water are stronger than those of hydrogen.

Explanation:

Ammonia is a compressible gas at room temperature with molecules free to move about and so fill up the volume of the container in which it is placed due to the weaker inter-molecular Van der Waals forces such as Keesom, Debye and London dispersion forces holding the particles of ammonia together in a given volume of the compound.

The inter-molecular forces between water molecules is hydrogen binding and dipole moments due to the strongly electronegative oxygen and hydrogen which tends to move the electrons towards the oxygen creating a charge imbalance that causes the hydrogen surrounding the water molecule to aggregate to neutralize the the charge imbalance  forming the bases for the strong hydrogen bonds.

Therefore water is a liquid at room temperature while ammonia is a gas due to the difference in strength of their inter-molecular forces.

The intermolecular forces in water are stronger than the intermolecular forces in ammonia.

The intermolecular forces hold substances together in a particular state of matter. There are three states of matter;

The strongest degree of intermolecular interaction occurs between matter in the solid state. weaker intermolecular interactions occur in the liquid state and the weakest intermolecular interaction occurs in the gaseous state.

Since ammonia is a gas at room temperature, it has weaker intermolecular interaction between its molecules compared to molecules of water at room temperature:

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