when petroleum is distilled to separate the components by boiling point, the component with the highest boiling point is called

The molecular formula of the ionic compound is AlCl3, with aluminum and chloride in a molar ratio of 1:3.

To determine the molecular formula of the ionic compound, we need to know the molar mass of the compound. We can find the molar mass by dividing the mass of the sample by the number of moles present in the sample:

Molar mass = Mass of the sample / Number of moles

Molar mass = 3.70 g / 0.0141 mol

Molar mass = 262.41 g/mol

Once we know the molar mass, we can determine the molecular formula of the compound. Let's assume that the compound has the formula MX, where M is the cation and X is the anion.

The molar mass of MX can be expressed as:

Molar mass of MX = Molar mass of M + Molar mass of X

We can rearrange this equation to solve for the ratio of the cation and anion in the compound:

Molar mass of M / Molar mass of X = (Molar mass of MX - Molar mass of X) / Molar mass of X

Substituting the values, we get:

Molar mass of M / Molar mass of X = (262.41 g/mol - Molar mass of X) / Molar mass of X

Let's assume that the anion X is chloride (Cl-), which has a molar mass of 35.45 g/mol. Substituting this value, we get:

Molar mass of M / 35.45 g/mol = (262.41 g/mol - 35.45 g/mol) / 35.45 g/mol

Simplifying this equation, we get:

Molar mass of M / 35.45 = 6.41

Molar mass of M = 227.5 g/mol

This means that the cation has a molar mass of 227.5 g/mol. We can now use this information to determine the molecular formula of the compound.

Let's assume that the cation M is aluminum, which has a molar mass of 26.98 g/mol. We can calculate the ratio of aluminum to chloride by dividing the molar mass of aluminum by the molar mass of chloride:

Molar ratio of Al to Cl = Molar mass of Al / Molar mass of Cl

Molar ratio of Al to Cl = 26.98 g/mol / 35.45 g/mol

Molar ratio of Al to Cl = 0.761

This means that the molecular formula of the compound is [tex]AlCl_3[/tex].

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We were only asked to calculate the moles of zinc, the answer is simply:

moles Zn = 0.0304 mol

The given experimental data can be used to calculate the number of moles of hydrogen gas produced in the reaction with zinc.

To do so, we need to use the ideal gas law: PV = nRT, where P is the pressure of the gas, V is the volume of the container, n is the number of moles of the gas, R is the gas constant, and T is the temperature.

First, we need to calculate the volume of the container using the data for zinc metal. From the given data, we know that the mass of zinc used in the reaction is 1.9887 g. Using the molar mass of zinc (65.38 g/mol), we can calculate the number of moles of zinc:

moles Zn = 1.9887 g / 65.38 g/mol = 0.0304 mol

Next, we can use the pressure of hydrogen gas and the mass of the unknown sample to calculate the number of moles of hydrogen produced. We can assume that the volume of the container is the same for both unknown samples:

n = PV/RT = (P x V)/(R x T)

For unknown sample 1, we have:

n = (3372.7 torr x V)/(62.3637 L•torr/mol•K x 298 K)

For unknown sample 2, we have:

n = (2014.7 torr x V)/(62.3637 L•torr/mol•K x 298 K)

We can solve for V by setting the two expressions equal to each other:

(3372.7 torr x V)/(62.3637 L•torr/mol•K x 298 K) = (2014.7 torr x V)/(62.3637 L•torr/mol•K x 298 K)

Solving for V, we get V = 1.995 L

Finally, we can use the moles of hydrogen and the mass of the unknown sample to calculate the molar mass of the unknown compound. However, since we were only asked to calculate the moles of zinc, the answer is simply:

moles Zn = 0.0304 mol

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In general, the solubility of ionic compounds is dependent on their respective solubility products.

The solubility product is a constant that relates to the maximum amount of a solute that can dissolve in a solvent at a given temperature. The higher the solubility product, the more soluble the compound is.

The solubility product of Cd(OH)2 is approximately 2.5 x 10^-14, while the solubility product of ZnCO3 is approximately 2.8 x 10^-10. This means that Cd(OH)2 has a lower solubility product than ZnCO3 and therefore, Cd(OH)2 is less soluble than ZnCO3.

Hence, ZnCO3 has greater solubility compared to Cd(OH)2.

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The correct coefficients to balance the equation are:

[tex]2C_2H_6 + 7O_2[/tex] → [tex]4CO_2 + 6H_2O[/tex]

We can determine the coefficients by making sure that the number of atoms of each element is the same on both sides of the equation.

On the left side of the equation, there are 2 carbon atoms and 6 hydrogen atoms. On the right side, there are 4 carbon atoms and 6 hydrogen atoms. To balance the carbons, we need a coefficient of 2 in front of [tex]CO_2[/tex]. To balance the hydrogens, we need a coefficient of 3 in front of [tex]H_2O[/tex].

Now we have:

[tex]C_2H_6 + 7O_2 = 2CO_2 + 3H_2O[/tex]

To balance the oxygen atoms, we need a coefficient of 7/2 (which can be simplified to 3.5) in front of [tex]O_2[/tex]. However, we can't have a fractional coefficient in a balanced chemical equation. So, we can multiply the entire equation by 2 to get rid of the fraction:

[tex]2C_2H_6 + 7O_2 = 4CO_2 + 6H_2O[/tex]

Now the equation is balanced, with 2 carbon atoms, 6 hydrogen atoms, and 14 oxygen atoms on both sides.

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It depends on the specific Fe(III) complexes in question and the reaction conditions. In general, the reduction of Fe(III) to Fe(II) is a reduction reaction, which involves gaining electrons and decreasing the oxidation state of the iron ion.

This type of reaction is usually exothermic and thermodynamically favorable, meaning that it releases energy and tends to proceed spontaneously in the direction of the reduced form. However, the specific thermodynamics of the reduction will depend on the nature of the Fe(III) complex and the reducing agent used, as well as the reaction conditions such as temperature, pressure, and pH.

Therefore, without more information about the specific Fe(III) complexes and reaction conditions in question, it is not possible to definitively answer whether it is thermodynamically favorable to reduce them to their Fe(II) analogs.

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To determine the mass of each substance present after the complete reaction between aluminum nitrite and ammonium chloride, we need to consider the balanced chemical equation and use stoichiometry.

The balanced chemical equation for the reaction is:

2Al(NO2)3 + 6NH4Cl → 2AlCl3 + 6NH3 + 3N2 + 6H2O

Given that 79.9 g of aluminum nitrite and 54.4 g of ammonium chloride react completely, we can calculate the moles of each reactant using their molar masses. Then, based on the stoichiometry of the balanced equation, we can determine the limiting reactant and calculate the moles and masses of the products formed.

In the second paragraph, we would perform the necessary calculations to determine the moles of aluminum nitrite and ammonium chloride, identify the limiting reactant, and calculate the moles and masses of the products (aluminum chloride, nitrogen, and water) formed.

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The rate of change of pressure of [tex]H_{2}[/tex] for the given reaction at the same time is –2.5 × 10–2 atm/s.

The given reaction is [tex]C_{6} H_{14}(g)[/tex]→ [tex]C_{6} H_{6}(g) + 4H_{2} (g)[/tex], and the value of [tex]\frac{ΔP(C_{6} 6H_{14} )}{Δt}[/tex] is –6.2 ×[tex]10^{-3}[/tex] atm/s. We need to determine [tex]\frac{ΔP(H_{2} )}{Δt}[/tex] for this reaction at the same time.

The balanced chemical equation shows that for every 1 mole of C6H14 that reacts, 4 moles of [tex]H_{2}[/tex] are produced. Therefore, we can use the stoichiometry of the reaction to relate the rate of change of pressure of [tex]H_{2}[/tex] to the rate of change of pressure of [tex]C_{6} H_{14}[/tex].

[tex]\frac{ΔP(H_{2} )}{Δt} =\frac{4}{1}×\frac{C_{6}H_{14}  }{Δt}[/tex]

After substituting we get:

= –2.5 ×  [tex]10^{-2}[/tex]  atm/s

Therefore, the answer is –2.5 × [tex]10^{-2}[/tex] atm/s.

In conclusion, the rate of change of pressure of [tex]H_{2}[/tex] for the given reaction at the same time is –2.5 × [tex]10^{-2}[/tex] atm/s.

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The correct option is c. 4.26 M is the concentration of NH3 in solution,  which was calculated from the moles of HCl used in the titration.

The balanced chemical equation for the reaction between ammonia and hydrochloric acid is:

NH3 (aq) + HCl (aq) → NH4Cl (aq)

From the balanced equation, it can be seen that 1 mole of NH3 reacts with 1 mole of HCl. Therefore, the moles of HCl used in the titration can be used to calculate the moles of NH3 present in the solution.

Moles of HCl = 0.5000 M x 0.0426 L

= 0.0213 moles

Since 1 mole of NH3 reacts with 1 mole of HCl, there must be 0.0213 moles of NH3 in the 5.00 mL solution of ammonia cleaner.

Concentration of NH3 = 0.0213 moles / 0.00500 L

= 4.26 M

Therefore, the concentration of NH3 in solution is 4.26 M

The concentration of NH3 in solution is 4.26 M, which was calculated from the moles of HCl used in the titration and the balanced chemical equation for the reaction between NH3 and HCl.

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If we have the mentioned equilibrium constant, The corresponding ΔG∘ is -20.7 kJ/mol, and the E∘cel is 0.16 V under standard conditions.

To calculate ΔG∘, we can use the equation

where R is the gas constant (8.314 J/(mol·K)), T is the temperature in Kelvin (298 K), and K is the equilibrium constant (2.0×10⁻⁴).

Plugging in the values, we get

ΔG∘ = -(-8.314 J/(mol·K) × 298 K × ln(2.0×10⁻⁴))

≈ -20.7 kJ/mol.

To find E∘cel, we can use the relationship ΔG∘ = -nF E∘cel, where n is the number of electrons transferred (in this case, 2), and F is Faraday's constant (96,485 C/mol). Rearranging the equation, we have

E∘cel = -ΔG∘ / (nF)

= -(-20.7 kJ/mol) / (2 × 96,485 C/mol)

≈ 0.16 V.

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The percent yield of the reaction, when a reaction vessel initially containing 66.5 kg of CH4 and excess steam yields 14.9 kg of H2, is approximately 44.48%.

To determine the percent yield, we need to compare the actual yield of the desired product (H2) to the theoretical yield that could be obtained based on the stoichiometry of the reaction.

The balanced equation for the reaction between CH4 (methane) and steam (H2O) to produce H2 (hydrogen) is:

CH4 + 2H2O -> CO2 + 4H2

From the balanced equation, we can see that one mole of CH4 reacts with two moles of H2O to produce four moles of H2. Let's calculate the theoretical yield of H2 based on the given amount of CH4.

Convert the mass of CH4 to moles:

molar mass of CH4 = 12.01 g/mol (C) + 1.01 g/mol (H) × 4 = 16.05 g/mol

moles of CH4 = mass of CH4 / molar mass of CH4

moles of CH4 = 66500 g / 16.05 g/mol = 4145.17 mol

Calculate the moles of H2 using the stoichiometry of the reaction:

moles of H2 = (moles of CH4) × (4 moles of H2 / 1 mole of CH4)

moles of H2 = 4145.17 mol × (4/1) = 16580.68 mol

Convert the moles of H2 to mass:

molar mass of H2 = 1.01 g/mol (H) × 2 = 2.02 g/mol

mass of H2 = (moles of H2) × (molar mass of H2)

mass of H2 = 16580.68 mol × 2.02 g/mol = 33496.84 g = 33.5 kg

The theoretical yield of H2, based on the given amount of CH4, is 33.5 kg.

Now let's calculate the percent yield using the actual yield provided:

percent yield = (actual yield / theoretical yield) × 100

percent yield = (14.9 kg / 33.5 kg) × 100

percent yield ≈ 44.48%

The percent yield of the reaction, when a reaction vessel initially containing 66.5 kg of CH4 and excess steam yields 14.9 kg of H2, is approximately 44.48%.

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If the solution contains 120.0 g of naoh and has a volume of 6000 ml, the molarity of the solution is 0.50 mol/L.

Molarity is defined as the number of moles of solute per liter of solution. To determine the molarity of a solution, we need to first find the number of moles of the solute, which can be calculated using the formula:

moles = mass/molar mass

For sodium hydroxide (NaOH), the molar mass is 40.00 g/mol (22.99 g/mol for Na, 15.99 g/mol for O, and 1.01 g/mol for H).

Using the given mass of NaOH, we can calculate the number of moles:

moles = 120.0 g / 40.00 g/mol = 3.00 mol

Next, we need to convert the volume of the solution from milliliters to liters:

volume = 6000 ml / 1000 ml/L = 6.00 L

Finally, we can use the equation for molarity:

Molarity = moles / volume

Molarity = 3.00 mol / 6.00 L = 0.50 mol/L

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he value of the rate constant at 338.9 °c is 0.0523 s^-1. To calculate the value of the rate constant at 338.9 °c, we can use the Arrhenius equation which relates the rate constant (k) to the activation energy (Ea), temperature (T), and the gas constant (R):

k = Ae^(-Ea/RT)
Where A is the pre-exponential factor.
First, we need to calculate the pre-exponential factor (A). We can do this by using the rate constant value at 237.2 °c:
0.00379 = A * e^(-21.54/(8.314 * 510.35))
Here, we have converted the temperature to Kelvin (T = 237.2 + 273.15 = 510.35 K) and used the gas constant value (R = 8.314 J/K·mol).

Solving for A, we get:

A = 6.878 x 10^9 s^-1
Now, we can use this value of A and the activation energy to calculate the rate constant at 338.9 °c (T = 338.9 + 273.15 = 612.05 K):
k = 6.878 x 10^9 * e^(-21.54/(8.314 * 612.05))
k = 0.0523 s^-1 (rounded to four significant figures)

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To determine the number of particles of salt in 0.5 moles of salt, we need to use Avogadro's number, which represents the number of particles (atoms, molecules, or ions) per mole.

Avogadro's number is approximately 6.022 x 10^23 particles/mol.

Given that we have 0.5 moles of salt, we can calculate the number of particles using the following equation:

Number of particles = moles of salt * Avogadro's number

Number of particles = 0.5 moles * 6.022 x 10^23 particles/mol

Number of particles = 3.011 x 10^23 particles

Therefore, there are approximately 3.011 x 10^23 particles of salt in 0.5 moles of salt.

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Indigo is a natural dye that has been used for centuries to color textiles. Its chemical formula is C16H10N2O2, and it is derived from plants of the genus Indigofera. To calculate the degree of unsaturation of this molecule, we need to determine the number of pi bonds and rings in its structure.

Looking at the chemical formula of indigo, we can see that it contains 16 carbon atoms, 10 hydrogen atoms, 2 nitrogen atoms, and 2 oxygen atoms. To calculate the number of pi bonds, we need to subtract the number of hydrogen atoms and add one. This is because each carbon atom in the molecule is bonded to two other atoms (either carbon, nitrogen, or oxygen), and each of these bonds counts as a Sigma bond. Therefore, the number of pi bonds in indigo is: Number of pi bonds = 16 - 10 + 1 = 7
Next, we need to count the number of rings in the molecule. Indigo contains two aromatic rings, which are formed by a series of alternating double bonds and single bonds. Therefore, the degree of unsaturation of indigo is:
Degree of unsaturation = number of pi bonds + number of rings
Degree of unsaturation = 7 + 2 = 9

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The silver spoon was exposed to 0.0015 moles of H₂S.

The balanced chemical equation for the reaction of silver tarnishing is:

4Ag (s) + 2H₂S (g) + O₂ → 2Ag₂S (s) + 2H₂O (l)

From the equation, we can see that the stoichiometric ratio between Ag and H₂S is 4:2, which simplifies to 2:1. This means that for every 2 moles of H₂S, 4 moles of Ag₂S (tarnish) are produced.

Given that the silver spoon has 0.0030 moles of Ag₂S (tarnish), we can determine the moles of H₂S by dividing the moles of Ag₂S by the stoichiometric ratio.

0.0030 moles Ag₂S * (2 moles H₂S / 4 moles Ag₂S) = 0.0015 moles H₂S

Since H₂S is almost always the limiting reactant in this reaction, this quantity of H₂S was sufficient to produce 0.0030 moles of Ag₂S, which represents the tarnish on the silver spoon.

It's worth noting that the reaction assumes complete conversion of reactants to products, which might not always be the case in a real-world scenario. Additionally, the amount of tarnish on the silver spoon might not directly reflect the amount of H₂S exposure, as other factors such as surface area, time of exposure, and environmental conditions can also influence the tarnishing process.

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The most prevalent attractions between molecules of HF (hydrogen fluoride) in the liquid phase are hydrogen bonding.

Hydrogen bonding occurs when a hydrogen atom bonded to a highly electronegative atom, such as fluorine in the case of HF, interacts with a lone pair of electrons on a neighboring molecule. In HF, the electronegativity difference between hydrogen and fluorine creates a highly polar covalent bond, resulting in a partially positive hydrogen atom and a partially negative fluorine atom.

These partially positive hydrogen atoms in one HF molecule are attracted to the partially negative fluorine atoms in neighboring HF molecules. This strong electrostatic attraction between the positive and negative charges is known as hydrogen bonding. Hydrogen bonding is stronger than other intermolecular forces such as dipole-dipole interactions or London dispersion forces, making it the dominant attractive force between HF molecules in the liquid phase.

The presence of hydrogen bonding in HF contributes to its unique physical properties, such as its relatively high boiling point and strong intermolecular interactions.

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The heat capacity of CH3CH2OH is 2.44 J/g°C.

what is The formula for specific heat?

CH3CH2OH has a heat capacity of 2.44 J/g°C. After absorbing 1.45 kJ of heat, 1 mole of CH3CH2OH will reach a final temperature of 22°C using the formula:

q = m * c * ΔT

where q is the amount of heat that CH3CH2OH absorbs, m is the substance's mass, c is its specific heat capacity, and T is the temperature change.

We are aware that one mole of CH3CH2OH weighs 46.07 g1. Solving for T by substituting these values into the formula:

1.45 kJ is equal to 46.07 g * 2.44 J/g°C * T.

ΔT = 13.9°C

As a result, after absorbing 1.45 kJ of heat at 22°C, the ultimate temperature of CH3CH2OH would be:

22°C + 13.9°C = 35.9°C.

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A chemical equation known as an "ionic equation" depicts the species involved in a reaction as ions rather than neutral molecules. It is used to describe processes like acid-base reactions and precipitation reactions that entail the exchange of electrons between species in solution.

The net ionic equations for the reaction of all hydroxide precipitates that formed complex ions upon the addition of 6 M NH3 are:

1. Aluminum hydroxide (Al(OH)3):
Al(OH)3(s) + 3 NH3(aq) + 3 H2O(l) → [Al(NH3)6]3+(aq) + 3 OH-(aq)

2. Iron (III) hydroxide (Fe(OH)3):
Fe(OH)3(s) + 3 NH3(aq) + 3 H2O(l) → [Fe(NH3)6]3+(aq) + 3 OH-(aq)

3. Zinc hydroxide (Zn(OH)2):
Zn(OH)2(s) + 4 NH3(aq) → [Zn(NH3)4]2+(aq) + 2 OH-(aq)

4. Copper (II) hydroxide (Cu(OH)2):
Cu(OH)2(s) + 4 NH3(aq) → [Cu(NH3)4]2+(aq) + 2 OH-(aq)

The example on page 6 of the introduction shows the reaction of silver nitrate with sodium chloride to form a silver chloride precipitate. The net ionic equation for that reaction is:

Ag+(aq) + Cl-(aq) → AgCl(s)

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The choice that shows the transition state for the given sn2 reaction is c) iii

When bonds participating in a chemical reaction are in a state of change, the transition state in the reaction route is referred to as a molecular configuration. This configuration specifies the point of highest energy in the reaction path. The reaction rates and processes that take place in the gas phase are thoroughly explained by transition state theory.

The hydroxyl ion attacks the alkyl halide in the aforementioned chemical process, which results in the formation of an intermediate complex. A negatively charged ion transfer creates the complex in a transition state, where a new bond will eventually form with the entering nucleophile.

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Complete Question:

which choice shows the transition state for the given sn2 reaction?

CI + NaSH ---> SH + NaCI

a) i

b) ii

c) iii

d) iv

The heat capacity of an object is given by the following equation: 65 J/K is the change in the entropy of the object associated with raising its temperature from 290 k to 380 k

To calculate the change in entropy (ΔS) of the object, we can use the equation:
ΔS = ∫(dQ/T)
where dQ is the infinitesimal amount of heat transferred to the object, and T is the temperature at which the transfer occurs.

Where T1 and T2 are the initial and final temperatures, V1 and V2 are the initial and final volumes, R is the gas constant, and ΔS is the change in entropy. Cp is the molar heat capacity at constant pressure.
Given that the heat capacity of the object is given by the equation:
[tex]C = dQ/dT[/tex]
We can express dQ in terms of dT, and substitute it into the ΔS equation, as follows:
ΔS = ∫(C/T)dT
Integrating this expression between the initial temperature (290 K) and the final temperature (380 K), we get:
ΔS = ∫(C/T)dT = ln(T2/T1)  C
where ln is the natural logarithm, T1 is the initial temperature (290 K), T2 is the final temperature (380 K), and C is the heat capacity of the object.
Substituting the values given, we get:
ΔS = [tex]ln(380/290)[/tex] °C
      = 65 J/K

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In the external circuit, electrons migrate from the Zn|Zn2+ electrode to the Cu|Cu2+ electrode. In the salt bridge, anions migrate from the Cu|Cu2+ compartment to the Zn|Zn2+ compartment.

In a voltaic cell, chemical reactions take place in two half-cells which are connected by a salt bridge. The reaction that occurs in the cell can be written as:

Cu2+(aq) + Zn(s) --> Cu(s) + Zn2+(aq)

This reaction involves the transfer of electrons from the zinc electrode to the copper electrode. The zinc electrode loses electrons and is therefore the anode, while the copper electrode gains electrons and is the cathode.

The anode reaction is: Zn(s) --> Zn2+(aq) + 2e-

The cathode reaction is: Cu2+(aq) + 2e- --> Cu(s)

In the external circuit, the electrons migrate from the Zn|Zn2+ electrode to the Cu|Cu2+ electrode. This flow of electrons generates an electric current which can be used to power a device or perform work.

In the salt bridge, anions migrate from the Cu|Cu2+ compartment to the Zn|Zn2+ compartment to balance out the charge and maintain electrical neutrality.

I hope this helps! Let me know if you have any further questions or if you need me to elaborate on anything. Also, I apologize for the long answer but I wanted to make sure I covered all the necessary information.
In a voltaic cell, the anode and cathode reactions are:

Anode reaction (oxidation): Zn(s) → Zn2+(aq) + 2e-

Cathode reaction (reduction): Cu2+(aq) + 2e- → Cu(s)

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The pressure of H2 gas in atm is 0.951 atm. To find the pressure of H2 gas in atm, we need to use the total pressure of the gas mixture and subtract the partial pressure of water vapor to get the partial pressure of H2 gas.

First, we need to convert the barometric pressure and water vapor pressure from torr to atm:

Barometric pressure = 740.8 torr = 0.974 atm (using 1 atm = 760 torr)

Water vapor pressure = 17.5 torr = 0.023 atm

The total pressure of the gas mixture is the sum of the barometric pressure and the partial pressure of the gases:

Total pressure = barometric pressure + partial pressure of gases

Assuming that the H2 gas is the only other gas present in the mixture, the partial pressure of H2 gas is:

Partial pressure of H2 gas = total pressure - water vapor pressure

= (0.974 atm) - (0.023 atm)

= 0.951 atm

Therefore, the pressure of H2 gas in atm is 0.951 atm.

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The acids are ranked from most acidic to least acidic as follows: HI > HBr > HCl > HF.

This ranking is based on the stability of the conjugate bases and the strength of the corresponding acids. The trend can be explained by the increasing electronegativity of the halogen atoms, which leads to stronger acid strength due to increased polarity and more efficient stabilization of the conjugate base.

The ranking of acids is determined by the stability of their conjugate bases. In this case, we are comparing hydrohalic acids: HI, HBr, HCl, and HF.

HI is the most acidic because iodine is the least electronegative halogen. The resulting conjugate base, I-, is the most stable among the conjugate bases of these acids. The larger size and lower electronegativity of iodine allows for better dispersion of the negative charge, leading to greater stability.

As we move across the periodic table, the electronegativity of the halogen atoms increases. This results in a stronger pull on the shared electron pair in the H-X bond, making it easier to dissociate the hydrogen ion. Therefore, the acid strength increases from HBr to HCl to HF.

In the case of HF, fluorine is the most electronegative halogen. The small size and high electronegativity of fluorine result in strong hydrogen bonding interactions and a relatively unstable conjugate base, F-. The strong hydrogen bonding in HF makes it less likely to dissociate, leading to a weaker acid compared to the hydrohalic acids with larger halogens.

In summary, the ranking of these acids from most acidic to least acidic is HI > HBr > HCl > HF, based on the stability of the conjugate bases resulting from the increasing electronegativity of the halogen atoms.

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Answer:

The balanced single replacement reaction for the given chemical equation "Ag + CoBr₂" is:

2Ag + CoBr₂ → 2AgBr + Co

In this reaction, silver (Ag) replaces cobalt (Co) in the compound CoBr₂ (cobalt(II) bromide) to form silver bromide (AgBr) and solid cobalt (Co). The reaction is balanced because the number of atoms of each element is equal on both the reactant and product sides of the equation.

Note that the coefficients are 2 in front of Ag and AgBr, indicating that two molecules of Ag and two molecules of AgBr are required to balance the reaction.

Metal X, with its weaker atomic bonding, typically exhibits lower melting points, reduced mechanical strength, higher electrical conductivity, increased malleability and ductility, and reduced hardness compared to metal Y.

If atomic bonding in metal X is weaker than metal Y, then metal X generally has:

1. Lower melting point: Weaker atomic bonds require less energy to break, so metal X would have a lower melting point compared to metal Y.

2. Reduced mechanical strength: Weaker bonds result in a lower tensile and compressive strength, making metal X less durable and more prone to deformation or breakage under stress compared to metal Y.

3. Higher electrical conductivity: Weaker atomic bonding often allows electrons to move more freely, resulting in metal X having higher electrical conductivity compared to metal Y.

4. Increased malleability and ductility: Metal X, with its weaker atomic bonds, is more likely to be malleable (able to be hammered into thin sheets) and ductile (able to be drawn into wires) compared to the stronger-bonded metal Y.

5. Reduced hardness: Metal X would have a lower hardness compared to metal Y, meaning it would be easier to scratch or dent due to the weaker atomic bonds.

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Q is less than Ksp, the solution is unsaturated, and no precipitate will form, where  KClO₄is the only species in liquid water

Based on the given information, we can determine if a precipitate will form in the equilibrium by comparing the ion product (Q) with the Ksp value.
The equilibrium expression for KClO₄ is:
Ksp = [K⁺][ClO₄⁻]
Since KClO4 is the only species in liquid water, the concentrations of K⁺ and ClO₄⁻ ions are equal. In this case, the potassium concentration is given as 3.9 × 10⁻⁷ M, which is also the concentration of ClO₄⁻ ions.
Now, let's calculate the ion product (Q):
Q = [K⁺][ClO₄⁻] = (3.9 × 10⁻⁷ M)(3.9 × 10⁻⁷ M) = 1.521 × 10⁻¹³
Now compare Q with Ksp:
Q (1.521 × 10⁻¹³) < Ksp (5.2 × 10⁻¹¹)
The solution is unsaturated and will not precipitate if Q < Ksp.

The solution is supersaturated and will precipitate if Q > Ksp.

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When the amino acid leucine is treated with C₆H₅CH₂OH and H⁺,  dipeptide is formed.

When the amino acid leucine is treated with C₆H₅CH₂OH and H⁺, it undergoes esterification reaction to form a dipeptide. Specifically, the carboxylic acid group (-COOH) of leucine reacts with the hydroxyl group (-OH) of benzyl alcohol (C₆H₅CH₂OH) in the presence of an acid catalyst (H⁺) to form an ester bond (-COO-). The resulting product is benzyl leucinate, which is a dipeptide composed of benzyl alcohol and leucine.

The stereochemistry of the product depends on the stereochemistry of the starting material, leucine. Leucine has one chiral center, so there are two possible stereoisomers: L-leucine and D-leucine. The reaction will produce the dipeptide with the same stereochemistry as the starting material.

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In this case, the starting material is an isotope of phosphorus, denoted as 32/15P. The missing product is the alpha particle that is emitted during the decay process. Therefore, the complete equation would be:
32/15P → 28/13Al + 4/2He

Based on the given information, it appears that the reaction is a type of nuclear decay reaction. Specifically, it seems to be an example of alpha decay, which occurs when an atomic nucleus emits an alpha particle, consisting of two protons and two neutrons.
In this case, the starting material is an isotope of phosphorus, denoted as 32/15P. The missing product is the alpha particle that is emitted during the decay process. Therefore, the complete equation would be:
32/15P → 28/13Al + 4/2He
This indicates that the isotope of phosphorus decays into an isotope of aluminum, with the emission of an alpha particle. The resulting product nucleus has a mass number of 28 and an atomic number of 13, while the alpha particle has a mass number of 4 and an atomic number of 2.
Overall, alpha decay is an important type of nuclear decay that occurs in many isotopes, particularly those that are heavy and unstable. By understanding the products and mechanisms of nuclear decay reactions, scientists can gain insights into the behavior of matter at the atomic and subatomic level.

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