For the equilibrium PCl5(g) PCl3(g) + Cl2(g), Kc = 2.0 × 101 at 240°C. If pure PCl5 is placed in a 1.00-L container and allowed to come to equilibrium, and the equilibrium concentration of PCl3(g) is 0.27 M, what is the equilibrium concentration of PCl5(g)?

Answer:

Option B. 327°C

Explanation:

Absolute T° = T° in K

T° in K = T° in C + 273

T° in K - 273 = T° in C

600 K - 273 = 327°C

Answer: Option (1) is the correct answer.

Explanation:

As per Le Chatelier's principle, any disturbance caused in an equilibrium reaction will tend to shift the equilibrium in a direction away from the disturbance.

For example, [tex]H_{2}O(g) + SO_{3}(g) \rightleftarrow H_{2}SO_{4}(g)[/tex]

As this given reaction is endothermic in the forward direction and exothermic in the backward direction. Thus, in order to shift the reaction on left side we need to decrease the temperature.

Also, the number of moles are more on reactant side as compared to product side. So, when we decrease the number of moles on reactant side then the equilibrium will shift on left side.

Therefore, we can conclude that for the given reaction decreasing the pressure of the system and lowering the temperature of the system would shift the reaction to the left.

Final answer:

Decreasing the pressure of the system and lowering the temperature for the reaction H₂O(g) + SO₃(g) ⇔ H₂SO₄(g) both shift the equilibrium to the left because Le Chatelier's Principle dictates that the system will counteract changes by shifting towards more gas molecules (when pressure decreases) and by absorbing heat (when temperature is lowered for an endothermic reaction).

Explanation:

To determine how changes in pressure and temperature affect the equilibrium of the reaction H₂O(g) + SO₃(g) ⇒H₂SO₄(g), Le Chatelier's Principle can be applied. This principle states that if a system at equilibrium is subjected to a change in conditions, the system will adjust to partially counteract the effect of the change.

For the reaction given, if we decrease the pressure, the equilibrium will shift towards the side with more gas molecules to increase the pressure again. Since the left side has two moles of gas and the right side has only one, decreasing the pressure will shift the equilibrium to the left.

Regarding temperature, since it's an endothermic reaction, heat can be considered a reactant. So, if we lower the temperature, the system will shift towards the side that absorbs heat to counteract the decrease in temperature, which would be shifting the equilibrium to the left. Therefore, the correct answer is 1. decreasing, lowering.

Answer:

The root mean square speed of O2 gas molecules is

519.01 m/s

Explanation:

The root mean square velocity  :

[tex]v_{rms}=\sqrt{\frac{3RT}{M}}[/tex]

[tex]K.E_{avg}=\frac{3}{2}RT[/tex]

[tex]K.E =\frac{1}{2}mv_{rms}^{2}[/tex]

Molar mass , M

For He = 4 g/mol

For O2 = 2 x 16 = 32 g/mol

O2 = 32/1000 = 0.032 Kg/mol

First calculate the temperature at which the K.E of He is 4310J/mol

K.E of He =

[tex]K.E_{avg}=\frac{3}{2}RT[/tex]

[tex]T=\frac{2(K.E)}{3(R)}[/tex]

K.E of He = 4310 J/mol

[tex]T=\frac{2(4310J/mol)}{3(8.314J/Kmol)}[/tex]

[tex]T=345.60K[/tex]

Now , Use Vrms to calculate the velocity of O2

[tex]v_{rms}=\sqrt{\frac{3(8.314J/Kmol)(345.60K)}{0.032Kg/mol}}[/tex]

[tex]v_{rms}=\sqrt{\frac{8619.9552}{0.032}}[/tex]

[tex]v_{rms}=\sqrt{26935.001}[/tex]

[tex]v_{rms}=519.01m/s[/tex]

The root mean square speed of O2 gas molecules can be calculated using the average kinetic energy of He gas and the formula Urms = sqrt(3 * kB * T / m).

The root mean square speed (Urms) of gas molecules can be calculated using the equation:

Urms = sqrt(3 * kB * T / m)

Where kB is the Boltzmann constant (1.38 x 10^-23 J/K), T is the temperature in Kelvin, and m is the molar mass of the gas.

Since the average kinetic energy (KEavg) of helium gas (He) is given as 4310 J/mol, we can assume that the temperature is the same for He and oxygen gas (O2). We know that the root mean square speed of He gas is close to 500 m/s. By using the equation and the given data, we can calculate the root mean square speed of O2 gas molecules.

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Answer:

Because of less weight kick ball has more acceleration.

Explanation:

Acceleration of kick ball:

5 m/s

Acceleration of soccer ball:

3 m/s

Newton's second law:

According to newton's second law the acceleration of object depends upon two variables.

1) Mass of object

2) Force acting on it

Mathematical expression:

a = f/m

a = acceleration

f = force

m = mass

The force on balls are same thus the acceleration is depend upon the masses of balls.

The soccer ball has more weight that's why its acceleration is less while kick ball is lighter and thus its acceleration will more.

Answer:

Based on Newton's second law, if the balls are kicked with the same force, the one with less mass will have a greater acceleration. Since the kickball accelerates more than the soccer ball, it has less mass.

Explanation:

CH3CH2CH2CH3 < CH3CH2CHO < CH3CHOHCH3

Explanation:

Boiling point trend of Butane, Propan-1-ol and Propanal.

Butane is a member of the CnH2n+2 homologous series is an alkane. Alkanes have C-H and C-C bonds which have Van der waals dispersion forces which are temporary dipole-dipole forces (forces caused by the electron movement in a corner of the atom). This bond is weak but increases as the carbon chain/molecule increases.

In Propan-1-ol(Primaryalcohol), there is a hydrogen bond present in the -OH group. Hydrogen bond is caused by the attraction of hydrogen to a highly electronegative element like Cl-, O- etc. This bond is stronger than dispersion forces because of the relative energy required to break the hydrogen bond. Alcohols (CnH2n+1OH) also experience van der waals dispersion forces on its C-C chain and C-H so as the Carbon chain increases the boiling point increases in the homologous series.

Propanal which is an Aldehyde (Alkanal) with the general formula CnH2n+1CHO. This molecule has a C-O, C-C and C-H bonds only. If you notice, the Oxygen is not bonded to the Hydrogen so there is no hydrogen bond but the C-O bond has a permanent dipole-dipole force caused by the electronegativity of oxygen which is bonded to carbon. It also has van der waals dispersion forces caused by the C-C and C-H as the carbon chain increases down the homologous series. The permanent dipole-dipole forces are not as easy to break as van der waals forces.

In conclusion, the hydrogen bonds present in alcohols are stronger than the permanent dipole-dipole bonds in the aldehyde and the van der waals forces in alkanes (irrespective of the carbon chain in Butane). So Butane < Propanal < Propan-1-ol

The order of intermolecular interaction is; CH3CH2CH2OH > CH3CH2CHO > CH3CH2CH2CH3CH3CH2CH2CH3

Intermolecular forces are forces of attraction that hold a molecule together in a particular state of matter. The nature of intermolecular forces in a molecule depends on the kind of bonds between the atoms in the compounds. Compounds that contain nonpolar bonds often have weaker intermolecular interaction between molecules while molecules that have polar bonds experience a greater magnitude of intermolecular forces.

The order of intermolecular interaction is; CH3CH2CH2OH > CH3CH2CHO > CH3CH2CH2CH3CH3CH2CH2CH3

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Answer:

The concentration of CaCl2 in this solution is 0.564 molar (option A)

Explanation:

Step 1: Data given

Mass of CaCl2 = 23.7 grams

Mass of water = 375 grams

Density of solution is 1.05 g/mL

Step 2: Calculate total mass

Total mass = mass of CaCl2 + mass of water

Total mass = 23.7 grams + 375 grams = 398.7 grams

Step 3: Calculate volume of the solution

Density = mass / volume

Volume = mass / density

Volume = 398.7 grams / 1.05 g/mL

Volume = 379.7 mL = 0.3797 L

Step 4: Calculate moles CaCl2

Moles CaCl2 = mass CaCl2 / molar mass CaCl2

Moles CaCl2 = 23.7 grams / 110.98 g/mol

Moles CaCl2 = 0.214 moles

Step 5: Calculate concentration

Concentration of CaCl2 = moles / volume

Concentration of CaCl2 = 0.214 moles / 0.3797 L

Concentration of CaCl2 = 0.564 mol / L = 0.564 molar

The concentration of CaCl2 in this solution is 0.564 molar (option A)

Answer:

[OH⁻] → 1×10⁻¹⁰

Explanation:

pH = - log [H⁺]

pOH = - log [OH⁻]

pH + pOH = 14

4 + pOH = 14

14 - 4 = 10 → pOH

10^-pOH = [OH⁻]

10⁻¹⁰ = [OH⁻] → 1×10⁻¹⁰

The pOH of the lake is 10 (calculated by subtracting the pH from 14), and the hydroxide ion concentration of the lake is 10^(-10) moles per liter, obtained by raising 10 to the power of the negative pOH.

The pH of a solution is a measure of the hydrogen ion concentration, and the pOH is a measure of the hydroxide ion concentration. In water, pH and pOH are related to each other, and have a sum of 14 at room temperature. When the pH of the solution is given, you can find the pOH by subtracting the pH from 14. In this case, the pH of the lake is 4.0, so the pOH would be 14 - 4 = 10. The concentration of hydroxide ions in a solution can be calculated using the formula 10^(-pOH). So, the hydroxide ion concentration of the lake is 10^(-10) moles per liter.

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Answer:

Option b, Phosphodiester bond

Explanation:

Proteins are sequence of amino acids. Two amino acids are joined by peptide  bond. Therefore, proteins are also known as sequence of polypeptide chains.  These polypeptides chains have four level of structures:

Primary structure

secondary structure

Tertiary structure

Quaternary structure

Primary structure is simply a sequence of amino acids. In secondary, tertiary and Quaternary structure, various interactions are present.

Disulfide bond, hydrogen bond and salt bridge stabilizes tertiary structure of the protein.

Phosphodiester bond is present as link between two nucleotides and thus, present in the backbone of nucleic acid (RNA and DNA)

Therefore, the correct option is option b

Among disulfide bonds, hydrogen bonds, salt bridges, and phosphodiester bonds, the latter does not contribute to the stabilization of the tertiary structure of a protein. Instead, phosphodiester bonds are crucial in the formation of nucleic acids.

In regards to protein structure, the interaction types that contribute to the stabilization of the tertiary structure of a protein include disulfide bonds, hydrogen bonds, and salt bridges.

Disulfide bonds are covalent bonds between two sulfur atoms that stabilize the protein structure. Hydrogen bonds are weak forces of attraction between the hydrogen atom in one molecule and an electronegative atom in another. Salt bridges are ionic bonds between amino acid side chains that also stabilize the protein structure.

However, the phosphodiester bond does not play a role in tertiary protein structure. This type of bond is important in the formation of nucleic acids, such as DNA and RNA, where it links the 3' carbon of one nucleotide to the 5' carbon of another. It serves a different purpose in molecular biology and does not contribute to the stabilization of the tertiary structure of a protein.

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Answer : The mass of [tex]CO_2[/tex] produced form the combustion is, 16.43 kg

Explanation :

Density : It is defined as the mass contained per unit volume.

Formula used for density :

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

First we have to calculate the mass of octane.

Given :

Density of octane = 0.703 g/mL

Volume = 2.00 gallons = 7570 mL

conversion used : 1 gallon = 3785 mL

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

[tex]0.703g/mL=\frac{Mass}{7570mL}[/tex]

[tex]Mass=5321.71g[/tex]

Now we have to calculate the moles of octane.

[tex]\text{Moles of octane}=\frac{\text{Mass of octane}}{\text{Molar mass of octane}}[/tex]

Molar mass of octane = 114 g/mole

[tex]\text{Moles of octane}=\frac{5321.71g}{114g/mole}=46.68mole[/tex]

Now we have to calculate the moles of [tex]CO_2[/tex].

The balanced chemical combustion reaction of octane will be:

[tex]2C_8H_{18}+25O_2\rightarrow 16CO_2+18H_2O[/tex]

From the balanced chemical reaction we conclude that:

As, 2 moles of [tex]C_8H_{18}[/tex] react to give 16 moles of [tex]CO_2[/tex]

So, 46.68 moles of [tex]C_8H_{18}[/tex] react to give [tex]\frac{46.68}{2}\times 16=373.44[/tex] moles of [tex]CO_2[/tex]

Now we have to calculate the mass of [tex]CO_2[/tex].

[tex]\text{ Mass of }CO_2=\text{ Moles of }CO_2\times \text{ Molar mass of }CO_2[/tex]

Molar mass of [tex]CO_2[/tex] = 44 g/mol

[tex]\text{ Mass of }CO_2=(373.44moles)\times (44g/mole)=16431.36g=16.43kg[/tex]

Thus, the mass of [tex]CO_2[/tex] produced form the combustion is, 16.43 kg

Answer:B

Explanation:

They have a full valence shell of 8

To calculate the enthalpy change for dissolving potassium bromide, use the equation q = mcΔT. Then divide the heat gained or lost by the mass of the substance to get the enthalpy change in J/g. To convert to kJ/mol, divide by the molar mass of potassium bromide.

To calculate the enthalpy change for dissolving the salt, we need to use the equation q = mcΔT, where q is the heat, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. First, we calculate the heat gained by the water using m = 125g, c = 4.18J/goC, and ΔT = 24.2oC - 21.1oC. Next, we calculate the heat lost by the potassium bromide using m = 10.5g, c = 4.18J/goC, and ΔT = 24.2oC - 21.1oC. Finally, we can calculate the enthalpy change in J/g by dividing the heat gained or lost by the mass of the substance. To convert to kJ/mol, we need to use the molar mass of potassium bromide and divide the enthalpy change in J/g by the molar mass.

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Answer:

Phosphide ion

Explanation:

To achieve an octet, the phosphorus atom forms an ion. The name of this ion is Phosphide ion

Phosphorous have 15 electrons in shell. This means it has 5 electrons in its outer most shell well this means it requires 3 more electrons to complete its octate. Thus Phosphorus changes to phosphide ion with negative 3 sign  on it.

Answer:

Kp and Kc are 0.01266 and 145.17, respectively.

Explanation:

Please check document attached.

Answer:

b. respiration

Explanation:

Food is converted into energy that can be used by the cells of the body by the process of cellular respiration. In cellular respiration, glucose and oxygen are combined and converted into water and carbon dioxide and energy. The produced energy is transferred to ATP.

In the presence of oxygen, cells convert glucose into energy through a process called aerobic metabolism, which is also referred to as cellular respiration. This process involves several stages and results in the formation of ATP, carbon dioxide, and water.

Cellular Respiration and Energy Conversion

In the presence of oxygen, cells convert glucose into energy via a process known as aerobic metabolism (option d), which is a type of cellular respiration. The term 'respiration' is often used to describe this process as well (option b). Aerobic metabolism involves the conversion of glucose and oxygen into carbon dioxide and water, releasing energy that is stored in molecules of ATP. This energy conversion process comprises several stages, including Glycolysis, Transformation of Pyruvate, the Krebs Cycle, and Oxidative phosphorylation. In contrast, anaerobic metabolism (a) occurs in the absence of oxygen and involves different pathways such as anaerobic glycolysis or fermentation.

Cellular respiration is a crucial part of energy metabolism for all aerobic organisms, allowing them to harness energy from food. It is a complex yet beautifully orchestrated series of chemical reactions that take place within the cells, more specifically within the mitochondria. The process ultimately results in the production of ATP, which is the primary energy carrier within the cell.

Answer:  the calculation of the number of atoms conserved in the desired product rather than in waste.

Explanation:

Atom economy gives how much desired product is obtained compared to amount of starting materials.

[tex]\text {Atom economy}=\frac{\text {molecular weight of desired product}}{\text {molecular weight of all products}}\times 100%[/tex]

For example:

[tex]CH_4+2O_2\rightarrow CO_2+2H_2O[/tex]

[tex]\text {atom economy of water}=\frac{\text {molecular weight of desired product}}{\text {molecular weight of all products}}\times 100%[/tex]

[tex]\text {atom economy of water}=\frac{36}{44+36}\times 100%=45\%[/tex]

Thus atom economy for water is 45%

Answer:

a. b + s = 26

b. b = 26 — s

Explanation:

The mass of the magnesium that is produced is  [tex]2.1 * 10^{-8[/tex] g.

The ideal gas equation can also be expressed in different forms to solve for different parameters. For example, it can be rearranged to find the molar volume (V/n) or the number of moles (n) when other variables are known.

We have that;

PV = nRT

n = PV/RT

n = [tex]3.4 * 10^{-6[/tex] * 0.432/62.4 * 55

n = [tex]4.3 * 10^{-10[/tex] moles

If 2 moles of Mg produces 1 mole of oxygen

x moles of oxygen produces  [tex]4.3 * 10^{-10[/tex] moles of oxygen

x =[tex]8.6 * 10^{-10[/tex] moles

Mass of magnesium = [tex]8.6 * 10^{-10[/tex] moles * 24 g/mol

= [tex]2.1 * 10^{-8[/tex] g

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Answer:

The answer to your question is number 1. Pb

Explanation:

Data

10 grams of Na, C, Pb, and Ne

Process

Calculate the moles of each element

                        23 g of Na ----------------- 1 mol

                        10 g of Na ------------------- x

                               x = 0.43 moles of Na

                        20 g of Ne ---------------- 1 mol

                        10 g of Ne ----------------- x

                               x = 0.5 moles of Ne

                         64 g of Cu -------------- 1 mol

                          10 g of Cu -------------- x

                              x = 0.16 moles of Cu

                         207 g of Pb ------------- 1 mol

                            10 g of Pb -------------- x

                               x = 0.048 moles of Pb

Pb has the smallest number of moles

Answer:

Pb contains the smallest no of moles

Mole = mass/atomic mass

For lead no of mole = 10g/207.2g/mol

= 0.04826mol

For Cu no of mole = 10g/63.546g/mol = 0.1574mol

For Ne no of mole = 10g/20.1797g/mol = 0.4955mol

For Na no of mole = 10g/22.9898g/mol = 0.5350mol

This is an incomplete question, here is a complete question.

For each trial, calculate the number moles of 6.0 M HCl used in the reaction?

Trial 1 : Volume of HCl = 15.0ml

Trial 2 : Volume of HCl = 14.9ml

Trial 3 : Volume of HCl = 15.2ml

Answer :

The number moles of HCl for trial 1, 2 and 3 is, 0.090 mol, 0.089 mol and 0.091 mol

Explanation :

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

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

In this question, the solute is HCl.

Now we have to calculate the number of moles of HCl for trial 1.

Volume of HCl = 15.0 mL = 0.015 L

[tex]6.0M=\frac{\text{Moles of HCl}}{0.015L}[/tex]

[tex]\text{Moles of HCl}=0.090mol[/tex]

Now we have to calculate the number of moles of HCl for trial 2.

Volume of HCl = 14.9 mL = 0.0149 L

[tex]6.0M=\frac{\text{Moles of HCl}}{0.0149L}[/tex]

[tex]\text{Moles of HCl}=0.089mol[/tex]

Now we have to calculate the number of moles of HCl for trial 3.

Volume of HCl = 15.2 mL = 0.0152 L

[tex]6.0M=\frac{\text{Moles of HCl}}{0.0152L}[/tex]

[tex]\text{Moles of HCl}=0.091mol[/tex]

Answer: Arginine.

Explanation: Arginine is an amino acid that is used for the biosynthesis of protein. It is made up of an α-amino group, an α-carboxylic acid group, & a side chain consisting of a 3-carbon aliphatic straight chain ending in a guanidino group. At physiological pH, the carboxylic acid is deprotonated (−COO−), the amino group is protonated (−NH3+), and the guanidino group is also protonated to give the guanidinium form (-C-(NH2)2+), making arginine a charged, aliphatic amino acid.

The amino acid when incorporated into a polypeptide chain (not at the N or C terminus), make the charge of the polypeptide more positive is Arginine.

Thus, the amino acid is Arginine.

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Answer : The new volume of was, 5.17 L

Explanation :

Charles' Law : It states that volume of the gas is directly proportional to the temperature of the gas at constant pressure and number of moles of gas.

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=5.00L\\T_1=25^oC=(25+273)K=298K\\V_2=?\\T_2=35^oC=(35+273)K=308K[/tex]

Putting values in above equation, we get:

[tex]\frac{5.00L}{298K}=\frac{V_2}{308K}\\\\V_2=5.17L[/tex]

Therefore, the new volume of was, 5.17 L

Answer:

Sum of coefficients in balanced equation is 18

Explanation:

Unbalanced equation: [tex]Mg_{3}(PO_{4})_{2}+C\rightarrow Mg_{3}P_{2}+CO[/tex]

Balance O: [tex]Mg_{3}(PO_{4})_{2}+C\rightarrow Mg_{3}P_{2}+8CO[/tex]

Balance C: [tex]Mg_{3}(PO_{4})_{2}+8C\rightarrow Mg_{3}P_{2}+8CO[/tex]

Balanced equation:[tex]Mg_{3}(PO_{4})_{2}+8C\rightarrow Mg_{3}P_{2}+8CO[/tex]

Sum of coefficients in balanced equation = (1+8+1+8) = 18

So, option (4) is correct.

Answer:

The coefficient of LiH is 1 (non written)

Explanation:

The equation is this:

LiH (s)  +  H₂O (l) → LiOH (aq)  +  H₂ (g)

Ratio is 1:1 between reactants and products.

The coefficient of LiH is 1

Li2O(s)+H2O(l)→2LiOH(aq)

Answer:

They should obtain the same Rf for the same compounds.

Explanation:

The Rf is defined as A/B. Where A is the displacement of the substance of interest, and B is the solvent front.

By dividing the substance's displacement by B, we make it so that the Rf factor is equal for identical compounds in the same mobile phase, no matter what the solvent front is.

Answer:

The answer to your question is 7.77 x 10²² molecules of Glucose

Explanation:

Data

Number of molecules = ?

Mass = 23.3 g

Molecular mass of Glucose = 180 g

Avogadro's number = 6.023 x 10²³

Process

1.- Calculate the moles of Glucose in 23.3    

                      180 g ------------------ 1 mol

                       23.3 g ---------------- x

                       x = (23.3 x 1) / 180

                       x = 0.129 moles

2.- Use Avogadro's number to calculate the number of molecules

                        1 mol of Glucose ------------- 6.023 x 10²³ molecules

                      0.129 moles          --------------   x

                         x = (0.129 x 6.023 x 10²³) / 1

                         x = 7.77 x 10²² molecules                      

The number of molecules of glucose in 23.3 g of the substance is  7.77 x 10^23

To determine the number of molecules of glucose (C6H12O6) in 23.3 g of the substance, you need to use the concept of moles. First, calculate the molar mass of glucose by adding up the atomic masses of its constituent elements. This gives you a molar mass of 180.16 g/mol. Next, use the formula:

moles = mass (g) / molar mass (g/mol)

Substituting the values, you get:

moles = 23.3 g / 180.16 g/mol = 0.129 moles of glucose

Since one mole of any substance contains Avogadro's number of molecules (6.022 x 10^23), multiply the number of moles by Avogadro's number to get the number of molecules:

0.129 moles x 6.022 x 10^23 molecules/mol = 7.77 x 10^23 molecules of glucose

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Answer:

9.1L

Explanation:

Firstly, to solve the problem, we need a balanced chemical equation.

2H2 + O2 ———> 2H2O

2 moles of hydrogen reacted one mole of oxygen. Now, we know that at STP, one mole of a gas occupies a volume of 22.4L, now let us get the number of moles in 4.55L of oxygen.

This means 4.55/22.4 = 0.203125 mole

Since in theory we have 2 moles of hydrogen reacting one mole of oxygen. Hence the number of moles of hydrogen we have is 0.203125 * 2 = 0.40625 mole

We now proceed to get the volume of hydrogen gas. Since 1 mole is 22.4L, then

0.40625 Is 0.40625 * 22.4 = 9.1L

To react with 4.55 L of oxygen gas completely, you need 9.10 L of hydrogen gas. This follows the balanced chemical equation 2H₂ + O₂ → 2H₂O, which indicates a 2:1 ratio between hydrogen and oxygen volumes.

To determine the volume of hydrogen gas needed to react completely with 4.55 L of oxygen gas to produce water vapor, we begin with the balanced chemical equation:

2H₂(g) + O₂(g) → 2H₂O(g)

This tells us that 2 volumes of hydrogen gas react with 1 volume of oxygen gas. Therefore, to find the necessary volume of hydrogen gas:

Recognize the stoichiometric ratio from the equation: 2 volumes of hydrogen gas (H₂) react with 1 volume of oxygen gas (O₂).  

Given that there are 4.55 L of oxygen gas, apply the ratio:

Volume of H₂ needed = 2 times the volume of O₂ = 2 x 4.55 L = 9.10 L

Thus, 9.10 L of hydrogen gas is required to react completely with 4.55 L of oxygen gas under the same conditions of temperature and pressure.

Approximately 1764.21 grams of sodium bicarbonate are needed to neutralize a spill of 1.75 L of concentrated HCl, assuming a concentration of 12 M for the HCl.

To determine the mass of sodium bicarbonate (NaHCO₃) needed to neutralize a spill of 1.75 L of concentrated HCl, one must first know the concentration of the hydrochloric acid (HCl). However, if we assume that 'concentrated' HCl is approximately 12 M (molar), we could proceed with the calculation. The balanced chemical equation for the reaction between sodium bicarbonate and hydrochloric acid is:

From this equation, there is a one-to-one mole ratio between HCl and NaHCO₃. Thus, the moles of HCl that spilled are:

Since the mole ratio is 1:1, 21 moles of sodium bicarbonate are also required. The molar mass of NaHCO₃ is 84.01 g/mol, so you would need:

Therefore, to neutralize the acid spill, you would need approximately 1764.21 grams of sodium bicarbonate.

Answer:

Diamond is burnt

Diamond (Carbon)+ Oxygen → CO2

Water electrolysed

2H2O → 2H2 + O2

Explanation:

Dalton atomic theory states that in a chemical reaction, the atoms of the elements join together or combine in simple whole number ratios

Hence when, at very high temperatures, diamond which consists of carbon is burnt we have

Diamond (Carbon)+ Oxygen → CO2

Also when water is electrolysed it decomposes as follows

2H2O → 2H2 + O2

Answer:

The new total body concentration would be 300 mOsM

Explanation:

In order to do this, we need to convert all concentrations to moles.

First, with the total body concentration, we have the initial volume of 3 liters and the concentration of 300 M (I will omit til the end the part of mOs)

The moles of the body concentration in this volume is:

moles = M * V

moles = 300 * 3 = 900 moles

To this moles, we add 150 moles of NaCL so, the total moles now is:

moles = 900 + 150 = 1050 moles

Finally, we can calculate the concentration with the new volume of 3.5 L (the sum of 3 and 0.5 liters added):

M = 1050 / 3.5

M = 300 mOsM

So the concentration remains the same as initial

Answer: Option (4) is the correct answer.

Explanation:

Heat of vaporization is defined as the heat energy which is necessarily added to a liquid substance in order to transform the quantity of the substance into a gas.

For example, in [tex]H_{2}O[/tex] there will be presence of strong hydrogen bonding and in order to break this bond high amount of heat energy is required.

Whereas [tex]H_{2}[/tex], [tex]F_{2}[/tex] and [tex]SiF_{4}[/tex] are all covalent compounds which are bonded together by Vander waal forces. As these forces are weak in nature hence, they require less amount of heat energy to convert into vapor state.

Hence, they have low value of [tex]\Delta H_{vap}[/tex]. Also, Ar is a noble gas and it has only Vander waal forces. So, it will also have low value of [tex]\Delta H_{vap}[/tex].

Therefore, we can conclude that out of the given options [tex]H_{2}O[/tex] have the highest [tex]\Delta H_{vap}[/tex].

4. H₂O

Enthalpy of vaporization

The enthalpy of vaporization (symbol ∆Hvap), also known as the (latent) heat of vaporization or heat of evaporation, is the amount of energy (enthalpy) that must be added to a liquid substance to transform a quantity of that substance into a gas.

In terms of formula it can be written as:

[tex]\Delta H_\mathrm{vap}=\Delta U_\mathrm{vap}+p \Delta V[/tex]

Lets look at all the options one by one:

1. In case of H₂O molecule, there is a strong hydrogen bonding thus greatest energy is required to break this bonding.

2. While in case of H₂, F₂ and SiF₄ molecules are all covalent compounds that are bonded via weak vanderwaal forces thus it needs lesser heat energy to convert into vapor state.

3. Noble gases usually have weak vanderwaal forces thus Ar has lower ∆Hvap.

So, we can conclude that H₂O has the the highest ΔHvap.

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