D
Question 1
You find an old metal ball deep in the woods one day. You determine it has a radius of 2cm and a
mass of 267.4794 grams. Calculate its volume then calculate its density to determine which type of
metal it is.
O Aluminum
O Titanium
2 pts
OZinc
O Tin
O Cast Iron
O Mild Steel
O Iron
O Stainless Steel
O Brass
O Copper
O Silver
O Lead
O Mercury
O Gold
O Tungsten
O Platinum

To determine the number of grams of KHP (potassium hydrogen phthalate, C8H5KO4) needed to react with 38.56 mL of a 0.2500 M sodium hydroxide (NaOH) solution,

We can use stoichiometry and the balanced chemical equation between KHP and NaOH. The balanced equation is:

KHP + NaOH → KNaC8H4O4 + H2O

From the balanced equation, we can see that the stoichiometric ratio between KHP and NaOH is 1:1. This means that one mole of KHP reacts with one mole of NaOH.

First, we need to calculate the number of moles of NaOH:

Volume of NaOH solution = 38.56 mL = 0.03856 L (converted to liters)

Molarity of NaOH solution = 0.2500 M

Number of moles of NaOH = Volume × Molarity = 0.03856 L × 0.2500 mol/L = 0.00964 mol

Since the stoichiometric ratio between KHP and NaOH is 1:1, the number of moles of KHP needed is also 0.00964 mol.

To calculate the mass of KHP, we need to know the molar mass of KHP, which is 204.23 g/mol.

Mass of KHP = Number of moles × Molar mass = 0.00964 mol × 204.23 g/mol = 1.969 g. Therefore, approximately 1.969 grams of KHP are needed to react with 38.56 mL of a 0.2500 M NaOH solution.

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To determine the grams of Ti metal deposited from a Tit4 solution by passing a current of 200 amps for 1 hour, we need to use Faraday's law of electrolysis.

The formula for Faraday's law of electrolysis is:

Mass of substance = (Current × Time × Atomic weight) / (Number of electrons × Faraday constant)

The atomic weight of Ti is 47.867 g/mol, and it has a valency of 4, which means it requires 4 electrons to be reduced from Ti4+ to Ti metal.

The Faraday constant is 96,485 Coulombs/mol.

Substituting the values in the formula, we get:

Mass of Ti metal = (200 A × 3600 s × 47.867 g/mol) / (4 × 96485 C/mol)

Mass of Ti metal = 42.14 g

Therefore, 42.14 grams of Ti metal will be deposited from a Tit4 solution by passing a current of 200 amps for 1 hour.

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The wide diversity of minerals can be attributed to several factors, including the elements that make up minerals and the Earth processes involved in their formation.

1. Elemental Composition: Minerals are formed from various combinations of elements. The Earth's crust contains a wide range of elements, each with its unique properties. The different combinations and proportions of these elements give rise to a vast array of minerals with distinct chemical compositions.

2. Geological Processes: Minerals are formed through a variety of geological processes. These processes include crystallization from magma or lava, precipitation from aqueous solutions, and metamorphism (changes in mineral structure due to heat and pressure). Each process creates specific conditions that influence the formation and composition of minerals.

3. Environmental Factors: Factors such as temperature, pressure, and the presence of other minerals or elements in the surroundings can also influence mineral formation. Varied environmental conditions give rise to different minerals, leading to the rich diversity observed in nature.

Overall, the wide diversity of minerals results from the interplay of elemental composition, geological processes, and environmental factors, all working together to create a multitude of unique mineral species found throughout the Earth's crust.

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In the given scenario, the maximum amount of Fe produced during the experiment needs to be determined. This can be done by analyzing the collected data and identifying the limiting reactant in the reaction. The limiting reactant is the reactant that is completely consumed and determines the maximum amount of product that can be formed.

To determine the maximum amount of Fe produced, one needs to compare the stoichiometry of the reaction and the amounts of reactants used. The balanced chemical equation for the reaction provides the molar ratio between the reactants and the product.

Once the limiting reactant is identified, its amount can be used to calculate the theoretical yield of the product, which represents the maximum amount of product that can be obtained. The theoretical yield is determined by multiplying the amount of the limiting reactant by the molar ratio between the limiting reactant and the product.

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The major organic product formed in an intramolecular aldol condensation, with heat applied, is a cyclic β-hydroxyketone.

This product is obtained by the self-condensation of a single molecule that contains both an aldehyde and a ketone functional group. The reaction involves the formation of a carbon-carbon bond between the α-carbon of the ketone and the carbonyl carbon of the aldehyde, followed by dehydration to give the cyclic product. For example, let's consider the molecule 3-hydroxy-2-pentanone. Under the influence of heat, the aldehyde and ketone groups in the same molecule can undergo intramolecular aldol condensation. The α-carbon of the ketone attacks the carbonyl carbon of the aldehyde, forming a new carbon-carbon bond. The resulting intermediate undergoes dehydration, eliminating a water molecule and forming a cyclic β-hydroxyketone. The specific product formed will depend on the starting compound and the reaction conditions. However, in general, intramolecular aldol condensations with heat favor the formation of cyclic products. These reactions are valuable in organic synthesis as they enable the construction of complex cyclic structures in a single step.

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The wavelength of light emitted by a hydrogen atom with the electron in the n=3 principle quantum number, when it jumps to the n=1 principle quantum number, is 121.6 nanometers.

This is because the energy difference between the two principle quantum numbers can be calculated using the formula ΔE = E2 - E1 = Rh(1/n1^2 - 1/n2^2), where Rh is the Rydberg constant and n1 and n2 are the initial and final principle quantum numbers respectively. Plugging in the values, we get ΔE = -2.18 x 10^-18 J.

This energy difference corresponds to the energy of a photon, which can be calculated using the formula E = hc/λ, where h is Planck's constant, c is the speed of light and λ is the wavelength of the light emitted. Rearranging this formula, we get λ = hc/ΔE, which gives us a wavelength of 121.6 nanometers for the light emitted.

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To determine the mass of chlorine required to react completely with 12.15 g of magnesium, we need to use the balanced chemical equation for the reaction:

Mg + Cl2 → MgCl2

From this equation, we can see that 1 mole of magnesium reacts with 1 mole of chlorine to produce 1 mole of magnesium chloride. The molar mass of magnesium is 24.31 g/mol, and the molar mass of chlorine is 35.45 g/mol.

We can use the given mass of magnesium and its molar mass to calculate the number of moles present:

moles of Mg = mass of Mg / molar mass of Mg
moles of Mg = 12.15 g / 24.31 g/mol
moles of Mg = 0.500 mol

Since the stoichiometry of the reaction is 1:1, we know that 0.500 moles of chlorine are required to react completely with the given amount of magnesium. We can convert this to grams of chlorine using its molar mass:

mass of Cl2 = moles of Cl2 x molar mass of Cl2
mass of Cl2 = 0.500 mol x 35.45 g/mol
mass of Cl2 = 17.72 g

Therefore, 17.72 g of chlorine would be required to react completely with 12.15 g of magnesium.

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a) The melting point of mercury at 10 bar is -118.8°C.

b) The melting point of mercury at 3540 bar is -49.5°C

The melting point of mercury at different pressures can be calculated using the Clausius-Clapeyron equation:

ln(P2/P1) = -ΔHfus/R (1/T2 - 1/T1)

where P1 and T1 are the pressure and temperature at which the heat of fusion is known (1 bar and -38.87°C, respectively), P2 is the new pressure, T2 is the new melting point temperature, ΔHfus is the heat of fusion, R is the gas constant, and ln is the natural logarithm.

We can rearrange this equation to solve for T2:

T2 = (ΔHfus/R) * (ln(P2/P1)/(-1/T1)) + 1/T1

Substituting the given values, we get:

(a) For P2 = 10 bar:

T2 = (9.75 J/g / (8.314 J/(mol*K))) * (ln(10 bar/1 bar) / (-1 / ( -38.87°C + 273.15))) + (1 / (-38.87°C + 273.15))

T2 = 155.3 K = -118.8°C

Therefore, the melting point of mercury at 10 bar is -118.8°C.

(b) For P2 = 3540 bar:

T2 = (9.75 J/g / (8.314 J/(mol*K))) * (ln(3540 bar/1 bar) / (-1 / ( -38.87°C + 273.15))) + (1 / (-38.87°C + 273.15))

T2 = 223.6 K = -49.5°C

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The IUPAC name for the given compound is 3-methylhexanoic acid. To arrive at this name, we need to follow a few rules laid down by the IUPAC. Firstly, we need to identify the longest carbon chain in the compound, which contains the functional group (-COOH) and number the carbons in the chain accordingly. Here, we can see that the longest chain has six carbons, so it is a hexanoic acid. Next, we need to identify and name any substituents attached to the main chain. In this compound, we have a methyl group attached to the third carbon, so it becomes 3-methylhexanoic acid. Therefore, the correct IUPAC name for the given compound is 3-methylhexanoic acid. It is important to use correct IUPAC names for compounds to avoid confusion and ensure that everyone is referring to the same molecule.
The IUPAC name for the given compound is 3-methylhexanoic acid. In this compound, the methyl group is attached to the third carbon in the hexanoic acid chain, which consists of six carbon atoms. When numbering the carbon atoms, start from the carboxyl group (COOH) as carbon 1, and count along the chain. The methyl group is attached to the third carbon, resulting in the name 3-methylhexanoic acid.

the pH of the buffer solution formed by mixing 0.50 mole of H3PO4 with 0.25 mole of NaOH and diluting to 1.00 L is approximately 1.06.

ToTo determine the pH of the buffer solution formed when 0.50 mole of H3PO4 is mixed with 0.25 mole of NaOH and diluted to 1.00 L, we need to consider the dissociation of H3PO4 and the subsequent reaction with NaOH.

Given:
Moles of H3PO4 = 0.50 mole
Moles of NaOH = 0.25 mole
Total volume of solution = 1.00 L

First, we need to determine which components of the H3PO4 dissociate and react with NaOH. H3PO4 is a triprotic acid, meaning it has three acidic hydrogen atoms (H+). NaOH is a strong base that will react with the acidic hydrogen ions.

Based on the given dissociation constants, the acidic hydrogen atoms with the highest Ka value (Ka1 = 7.5 x 10^-3) will react with NaOH. The other two hydrogen atoms (with Ka2 = 6.2 x 10^-8 and Ka3 = 3.6 x 10^-13) will remain as H+ ions.

Since H3PO4 is a triprotic acid, we can calculate the concentration of H+ ions from the dissociation of the first acidic hydrogen using the equation:

[H+] = √(Ka1 × (moles of H3PO4 / total To)

[H+] = √(7.5 x 10^-3 × (0.50 mole / 1.00 L))

[H+] ≈ 0.0866 M

Taking the negative logarithm (pH = -log[H+]), we can calculate the pH:

pH = -log(0.0866)

pH ≈ 1.06

Therefore, the pH of the buffer solution formed by mixing 0.50 mole of H3PO4 with 0.25 mole of NaOH and diluting to 1.00 L is approximately 1.06.

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The △G∘' for the reaction fructose-6-phosphate → glucose-6-phosphate at 37 ∘C is -1708.3 J/mol.

To calculate the △G∘' for the reaction fructose-6-phosphate → glucose-6-phosphate, we need to use the equation △G∘' = -RT ln K, where R is the gas constant (8.314 J/K·mol), T is the temperature in Kelvin (37+273=310 K), and K is the equilibrium constant (1.97).

Plugging in the values, we get:
△G∘' = -8.314 J/K·mol × 310 K × ln(1.97)
△G∘' = -8.314 J/K·mol × 310 K × 0.677
△G∘' = -1708.3 J/mol

Therefore, the △G∘' for the reaction fructose-6-phosphate → glucose-6-phosphate at 37 ∘C is -1708.3 J/mol. Note that the unit for △G∘' is J/mol, which represents the change in free energy per mole of the reaction.

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The ΔG∘' for the reaction fructose-6-phosphate → glucose-6-phosphate is -1.99 kJ/mol at 37°C.

Explanation:

The standard free energy change (ΔG∘') for a reaction can be calculated using the equation:

ΔG∘' = -RTln(K),

where R is the gas constant (8.314 J/K·mol), T is the temperature in Kelvin (37°C + 273.15 = 310.15 K), and K is the equilibrium constant (1.97).

Plugging in these values, we get:

ΔG∘' = -8.314 J/K·mol x 310.15 K x ln(1.97)

ΔG∘' = -1.99 kJ/mol

The negative sign indicates that the reaction is exergonic, meaning it releases energy. The units of ΔG∘' are in kJ/mol, which represents the amount of free energy released per mole of reactant converted to product under standard conditions.

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Answer:We are given: • P1P1 = 1200 torr. • T1T1 = 155 oCoC = 428 K

Explanation:)

The net ionic equation for the reaction is [tex]$\mathrm{CsH_5NH^+ + OH^- \rightarrow CsH_5NH_2^+ + H_2O}$[/tex]. The quantity in moles of [tex]$\mathrm{CsH_5NH^+}$[/tex] present at the start of the titration is 0.00440 mol. The quantity in moles of [tex]OH^-[/tex] present if 12.0 mL of [tex]OH^-[/tex] were added is 0.00289 mol.

a) The net ionic equation for the reaction is:

[tex]$\mathrm{CsH_5NH^+ + OH^- \rightarrow CsH_5NH_2^+ + H_2O}$[/tex]

b) The quantity in moles of [tex]CsH_5NH^+[/tex] present at the start of the titration can be calculated using the formula:

moles = concentration x volume

moles of [tex]CsH_5NH^+[/tex] = 0.220 mol/L x 0.0200 L = 0.00440 mol

c) The quantity in moles of [tex]OH^-[/tex] that would be present if 12.0 mL of OH- were added can be calculated using the formula:

moles = concentration x volume

moles of [tex]OH^-[/tex] = 0.241 mol/L x 0.0120 L = 0.00289 mol

d) After the reaction goes to completion, [tex]CsH_5NH^+[/tex] would be converted to [tex]CsH_5NH^+[/tex] and there would be no [tex]OH^-[/tex] left in the solution.

e) The quantity in moles of [tex]CsH_5NH^+[/tex] that would be left in the beaker after the reaction goes to completion can be calculated using the formula:

moles = initial moles - moles reacted

moles of [tex]CsH_5NH^+[/tex] = 0.00440 mol - 0.00289 mol = 0.00151 mol

f) The quantity in moles of CHEN that are produced after the reaction goes to completion is equal to the moles of [tex]OH^-[/tex] that reacted since the reaction is a 1:1 stoichiometric ratio. Therefore, the quantity in moles of CHEN produced is 0.00289 mol.

g) To determine the pH of the solution after the reaction goes to completion and the system reaches equilibrium, we need to calculate the concentration of [tex]H^+[/tex] ions in the solution. This can be done using the formula for the acid dissociation constant (Ka):

[tex]$\mathrm{K_a = \frac{[H^+][CsH_5NH^+]}{[CsH_5NH]}}$[/tex]

We know the values of Ka and the initial concentrations of [tex]CsH_5NH^+[/tex] and [tex]CsH_5NH[/tex], so we can rearrange the equation and solve for [[tex]H^+[/tex]]:

[tex]$\mathrm{[H^+] = \sqrt{\frac{K_a \times [CsH_5NH]}{[CsH_5NH^+]}}}$[/tex]

[tex]$\mathrm{[H^+] = \sqrt{\frac{5.9 \times 10^{-6} \times 0.220}{0.00440-0.00289}}}$[/tex]

[tex][H^+] = 0.000826 M[/tex]

[tex]$\mathrm{pH = -\log_{10}[H^+]}$[/tex]

[tex]$\mathrm{pH = -\log_{10}(0.000826)}$[/tex]

pH = 3.08

Therefore, the pH of the solution after the reaction goes to completion and the system reaches equilibrium is 3.08.

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Answer:

0.64A

Explanation:

There is a well-known relationship between the bond length of a diatomic molecule and the atomic radius of its constituent atoms, known as the covalent radius. Specifically, the covalent radius of an atom is half the bond length between two identical atoms in a diatomic molecule.

Therefore, to determine the atomic radius of chlorine (Cl), we can use the bond length of fluorine (F2) and the fact that the two atoms in F2 are identical.

Since the bond length of F2 is given as 1.28 A, the covalent radius of fluorine is 1.28/2 = 0.64 A.

Since both fluorine and chlorine are halogens, they have similar electronic configurations and form similar covalent bonds. Therefore, we can use the covalent radius of fluorine as an estimate for the covalent radius of chlorine.

Thus, the atomic radius of chlorine is approximately 0.64 A

The Lewis structure for the sulfate polyatomic ion (SO4)2- is:

      O
      ||
-O - S - O-
      ||
      O

     O    
      ||    
O = S - O-
      ||      
    -O  

There are a total of 6 equivalent resonance structures that can be drawn for the sulfate ion. These structures differ only in the placement of the double bonds between sulfur and oxygen atoms. One structure has two double bonds between sulfur and oxygen atoms, while the other has one double bond and one single bond between sulfur and oxygen atoms.

The Lewis structure for the sulfate polyatomic ion (SO₄²⁻) consists of a central sulfur atom surrounded by four oxygen atoms, with each oxygen atom forming a double bond with the sulfur atom.

There are a total of 32 valence electrons in this structure. Due to the nature of the double bonds and the overall charge, there are 6 equivalent resonance structures that can be drawn for the sulfate ion. This resonance stabilization contributes to the stability of the ion.

Sulfur has 6 valence electrons, and each oxygen has 6 valence electrons, giving a total of 32 valence electrons for the sulfate ion (6 from sulfur + 4 x 6 from oxygen). To complete the Lewis structure, we add formal charges to each atom to make sure the overall charge of the ion is -2. The sulfur atom has a formal charge of 0, while each oxygen atom has a formal charge of -1.

These structures have the same overall charge and the same number of valence electrons, but the distribution of electrons is different.

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The Lewis structure for the sulfate polyatomic ion can be drawn by following a few steps. There are equivalent resonance structures that can be drawn for the ion.

The Lewis structure for the sulfate polyatomic ion (SO42-) can be drawn by following these steps:

There are equivalent resonance structures that can be drawn for the sulfate polyatomic ion because the double bond can be moved around among the oxygen atoms while maintaining the same overall structure.

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The difference in boiling points between neon and HF can be explained by the intermolecular forces present in each substance, with HF exhibiting stronger intermolecular forces due to hydrogen bonding.

The boiling points of substances are determined by the strength of intermolecular forces between their molecules. Neon (Ne) is a noble gas that exists as individual atoms, and its boiling point is very low (-246.1°C). The weak van der Waals forces between neon atoms are easily overcome, requiring minimal energy to transition from a liquid to a gas state.

On the other hand, hydrogen fluoride (HF) exhibits higher boiling point (19.5°C) due to the presence of hydrogen bonding. HF molecules form strong dipole-dipole interactions through the electronegativity difference between hydrogen and fluorine. Hydrogen bonding is a particularly strong type of dipole-dipole interaction that occurs when hydrogen is bonded to highly electronegative atoms such as fluorine, oxygen, or nitrogen.

The hydrogen bonding in HF requires a significant amount of energy to break the strong intermolecular forces, resulting in a higher boiling point compared to neon.

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It can be concluded that Solid A has a lower melting point than Solid B and Solid C. Solid B has a higher melting point than both Solid A and Solid C. Solid C has the highest melting point among the three solids.

The melting point of a substance is the temperature at which it changes from a solid to a liquid state. From the information provided, we can deduce the following:

Solid A and Solid B:

When a 50/50 mixture of Solid A and Solid B is formed, it has a lower melting point of 140-147 C. This suggests that Solid A has a lower melting point than Solid B since the mixture's melting point is below the individual melting points of both A and B.

Solid B and Solid C:

When a 70/30 mixture of Solid B and Solid C is formed, it has the same melting point as Solid C, which is 170-171 C. This indicates that Solid B has a higher melting point than Solid C since the mixture's melting point is equal to Solid C's melting point.

Combining these conclusions, we can summarize that Solid A has the lowest melting point, Solid B has a higher melting point than Solid A but lower than Solid C, and Solid C has the highest melting point among the three solids.

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The statement "collision frequency per square centimeter of surface made by O2 molecules" is false because it is not clear what surface is being referred to.

In a gas-phase reaction, the rate of reaction is determined by the frequency of collisions between the reactant molecules. The collision frequency is dependent on the concentration of the reactants, their velocities, and the surface area available for collisions.

The rate of collision of O2 molecules with a surface can be expressed as the collision frequency per unit area of the surface, also known as the flux. The flux of O2 molecules is dependent on the concentration of O2 and the velocity of the molecules, as well as the surface area available for collisions.

However, we can say that the collision frequency of O2 molecules with a surface is dependent on the concentration of O2, the velocity of the molecules, and the surface area available for collisions.

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The acetamido group (-NHCOCH3) is an ortho, para-directing group because it can donate electron density to the aromatic ring via resonance. The acetamido group is less effective in activating the aromatic ring towards further substitution compared to an amino group (-NH2) due to the presence of the carbonyl group (C=O) in the acetamido group.

1. The acetamido group (-NHCOCH3) is an ortho, para-directing group because it has a lone pair of electrons on the nitrogen atom that can participate in resonance with the aromatic ring. This resonance effect stabilizes the positive charge developed during the electrophilic aromatic substitution reaction on the ortho and para positions relative to the acetamido group.

2. The acetamido group is less effective in activating the aromatic ring towards further substitution compared to an amino group (-NH2) due to the presence of the carbonyl group (C=O) in the acetamido group. The carbonyl group has a higher electron-withdrawing inductive effect, which weakens the electron-donating capability of the nitrogen atom. Consequently, the overall activating effect of the acetamido group is reduced compared to the amino group, which does not have an electron-withdrawing group attached to it.

In summary, the acetamido group is an ortho, para-directing group due to resonance involving the lone pair on the nitrogen atom, but it is less effective in activating the aromatic ring than an amino group because of the electron-withdrawing effect of the carbonyl group present in the acetamido group.

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The acetamido group is an ortho, para-directing group because it contains a lone pair of electrons that can interact with the pi-electron system of the aromatic ring through resonance.

This interaction results in a partial positive charge on the ortho and para positions, making these positions more attractive to electrophilic attack. However, the acetamido group is less effective in activating the aromatic ring towards further substitution than an amino group because the lone pair of electrons on the nitrogen of the acetamido group is partially delocalized into the carbonyl group, reducing its availability for resonance with the aromatic ring.

To obtain a pure sample of o-nitroaniline from a mixture with p-nitroaniline using ethanol as the solvent, one possible flow scheme is:

1. Dissolve the mixture of o-nitroaniline and p-nitroaniline in ethanol.

2. Add a strong base, such as sodium hydroxide, to the solution to convert the nitro groups to their corresponding sodium salts, which are more soluble in ethanol.

3. Acidify the solution with hydrochloric acid to protonate the amino groups, which will precipitate out the nitroanilines as their hydrochloride salts.

4. Collect the precipitate by filtration and wash with cold ethanol to remove any impurities.

5. Recrystallize the o-nitroaniline hydrochloride from hot ethanol, which will selectively dissolve the o-nitroaniline hydrochloride due to its higher solubility, leaving the p-nitroaniline hydrochloride behind as a solid.

6. Treat the o-nitroaniline hydrochloride with a base, such as sodium hydroxide, to regenerate o-nitroaniline in its free base form.

7. Finally, purify the o-nitroaniline by recrystallization from a suitable solvent, such as ethanol or acetone.

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One molecule of polyethylene with a molar mass of 17,500 g contains approximately 623 H2C-CH2 monomeric units.

To determine the number of H2C-CH2 monomeric units in one molecule of polyethylene with a molar mass of 17,500 g, we first need to understand the molecular formula of polyethylene. Polyethylene is a polymer made up of repeating monomeric units of ethylene, which has the chemical formula H2C=CH2.

The molar mass of polyethylene is given as 17,500 g. To calculate the number of monomeric units in one molecule of polyethylene, we need to divide the molar mass of polyethylene by the molar mass of one monomeric unit of ethylene.

The molar mass of one monomeric unit of ethylene can be calculated by adding the atomic masses of each element in the molecule. The atomic mass of hydrogen is 1.01 g/mol and the atomic mass of carbon is 12.01 g/mol. Therefore, the molar mass of one monomeric unit of ethylene is 2*(1.01 g/mol) + 2*(12.01 g/mol) = 28.05 g/mol.

Dividing the molar mass of polyethylene (17,500 g/mol) by the molar mass of one monomeric unit of ethylene (28.05 g/mol) gives us the number of monomeric units in one molecule of polyethylene.

17,500 g/mol ÷ 28.05 g/mol ≈ 623.08

Therefore, one molecule of polyethylene with a molar mass of 17,500 g contains approximately 623 H2C-CH2 monomeric units.

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The molar solubility of copper(I) bromide is 7.9 × 10^-5 mol/L, which is the concentration of Cu+ and Br- ions in the solution when the solution is saturated with CuBr at equilibrium.

The solubility product constant (Ksp) expression for copper(I) bromide (CuBr) is:

CuBr(s) ⇌ Cu+(aq) + Br-(aq)

Ksp = [Cu+][Br-]

Since the concentration of CuBr is assumed to be very small compared to the concentration of Cu+ and Br- ions in the solution, the concentrations of the ions can be approximated as equal to the molar solubility of CuBr (x) in the solution. Therefore, the Ksp expression can be simplified as follows:

Ksp = x^2

Substituting the given value of Ksp into the equation, we get:

6.3 × 10^-9 = x^2

Taking the square root of both sides, we get:

x = √(6.3 × 10^-9) = 7.9 × 10^-5 mol/L

Therefore, the molar solubility of copper(I) bromide is 7.9 × 10^-5 mol/L, which is the concentration of Cu+ and Br- ions in the solution when the solution is saturated with CuBr at equilibrium.

Note that the molar solubility is the maximum amount of solute that can dissolve in a given solvent to form a saturated solution at a particular temperature and pressure. Any further addition of the solute will lead to the formation of a precipitate of the solute.

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The reaction between a weak acid and a strong base results in the formation of a salt and water.

When a weak acid reacts with a strong base, they undergo a neutralization reaction. The acid donates a proton (H+) to the base, forming water and a salt. Since the acid is weak, it does not completely dissociate in water, resulting in a partial reaction. The strong base, on the other hand, completely dissociates into ions. The formation of water and a salt in the reaction leads to a decrease in the concentration of H+ ions in the solution. As a result, the pH of the solution increases and becomes more basic compared to the initial pH of the weak acid.

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The equilibrium constant for the given reaction can be expressed as Kc = ([CH2Cl2]^16 [H2]^8)/([CH3Cl]^16 [Cl2]^8), where [ ] represents the molar concentration of the respective species at equilibrium.

To express the equilibrium constant for the reaction 16CH3Cl(g) + 8Cl2(g) ⇌ 16CH2Cl2(g) + 8H2(g), we will use the terms equilibrium constant (K) and equilibrium expression.

The equilibrium constant (K) is a value that describes the ratio of the concentrations of products to reactants when a chemical reaction is at equilibrium. The equilibrium expression is written as:

K = [Products]^coefficients / [Reactants]^coefficients

For the given reaction:

16CH3Cl(g) + 8Cl2(g) ⇌ 16CH2Cl2(g) + 8H2(g)

The equilibrium expression will be:

K = [CH2Cl2]¹⁶ * [H2]⁸ / [CH3Cl]¹⁶ * [Cl2]⁸

This is the equilibrium constant expression for the given reaction, with the concentrations of each species raised to the power of their respective stoichiometric coefficients.

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When, a buffer will be prepared by mixing 86.4 ml of 1.05 m hbr and 274 ml of 0.833 M ethylamine. Then, the pH of the buffer after 0.068 mol NaOH is added is 5.72.

To solve this problem, we use the Henderson-Hasselbalch equation;

pH = pKa + log([base]/[acid])

First, we need to find the concentrations of the acid and base in the buffer solution;

[acid] = 1.05 M (HBr)

[base] = 0.833 M (ethylamine)

The pKa of HBr is -9, so we can assume that the concentration of H⁺ions is equal to the concentration of HBr. Therefore, the pH of the buffer before adding NaOH is;

pH = -log[H⁺] = -log(1.05) = 0.978

To calculate pH after adding 0.068 mol NaOH, we need to determine the new concentrations of the acid and base. We know that 0.068 mol NaOH will react with some of the HBr in the buffer, so we calculate how much HBr will be left.

1 mol HBr reacts with 1 mol NaOH, so 0.068 mol NaOH will react with 0.068 mol HBr. The amount of HBr remaining in the buffer is;

0.068 mol HBr - 0.068 mol NaOH = 0.054 mol HBr

The concentration of HBr is now;

[acid] = 0.054 mol / 0.3604 L = 0.1499 M

To calculate the concentration of the conjugate base, we need to determine how much of the ethylamine will react with the remaining H⁺ ions. Since ethylamine is a weak base, we need to use the [tex]K_{b}[/tex] equation;

[tex]K_{b}[/tex] = [BH⁺][OH⁻] / [B]

We can assume that all of the remaining H⁺ ions will react with the ethylamine to form the conjugate acid. The amount of ethylamine that reacts can be calculated using the stoichiometry of the reaction;

C₂H₅NH₂ + H⁺ → C₂H₅NH₃⁺

1 mol C₂H₅NH₂reacts with 1 mol H⁺, so 0.054 mol H⁺ will react with 0.054 molC₂H₅NH₂. The amount of C₂H₅NH₂ remaining in the buffer is;

.833 mol - 0.054 mol = 0.779 mol

The concentration of the conjugate base is;

[base] = 0.779 mol / 0.3604 L = 2.160 M

Now we use the Henderson-Hasselbalch equation to calculate the pH;

pH = pKa + log([base]/[acid])

pH = 9 - log(2.160/0.1499)

pH = 5.72

Therefore, the pH of the buffer after 0.068 mol NaOH is added is 5.72.

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The Gibbs free-energy change at 298 K for 2 KClO₃(s) → 2 KCl(s) + 3 O₂(g) is -2.38 kJ/mol and would be negative, so the reaction is spontaneous at all temperatures.

The Gibbs free-energy change can be calculated using the equation:

ΔG = ΔH - TΔS

where ΔH is the enthalpy change, ΔS is the entropy change, and T is the temperature in Kelvin.

ΔH for the reaction is the sum of the enthalpies of formation of the products minus the sum of the enthalpies of formation of the reactants:

ΔH = [2 mol KCl(g) + 3 mol O₂(g)] - [2 mol KClO₃(s)]

ΔH = (-869.6 kJ/mol) - (-924.4 kJ/mol)

ΔH = 54.8 kJ/mol

ΔS for the reaction is the sum of the entropies of the products minus the sum of the entropies of the reactants:

ΔS = [2 mol KCl(g) + 3 mol O₂(g)] - [2 mol KClO₃(s)]

ΔS = (205.2 J/K mol) + (231.0 J/K mol) - (238.7 J/K mol)

ΔS = 197.5 J/K mol

Substituting these values into the equation for ΔG:

ΔG = 54.8 kJ/mol - (298 K)(197.5 J/K mol)

ΔG = -2.38 kJ/mol

Since the ΔG value is negative, the reaction is spontaneous at all temperatures.

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The compound H2N-C-COOH, with two hydrogen atoms attached to the central carbon, is an amino acid.

The compound H2N-C-COOH represents an amino acid. Amino acids are organic compounds that serve as the building blocks of proteins. They contain an amino group (H2N) and a carboxyl group (COOH) attached to a central carbon atom. The presence of the amino and carboxyl groups gives amino acids their characteristic properties and reactivity. In proteins, amino acids are linked together through peptide bonds to form polypeptide chains. These chains then fold and interact to create the complex three-dimensional structures of proteins, which play crucial roles in biological processes.

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The predominant type of intermolecular force in each of the following compounds are:
- Hydrogen bonding
- London dispersion forces
- Dipole-dipole interactions

Hydrogen bonding is the predominant type of intermolecular force in compounds that contain hydrogen bonded directly to a highly electronegative atom, such as nitrogen, oxygen, or fluorine. This type of bonding is stronger than other intermolecular forces and can result in high boiling points and surface tensions. In the given compounds, ethanol contains a hydrogen bonded directly to an oxygen atom, which allows for hydrogen bonding to occur.

London dispersion forces are the predominant type of intermolecular force in nonpolar compounds, such as hydrocarbons. This type of force results from the temporary dipole that occurs when electrons are unevenly distributed around a molecule. London dispersion forces are the weakest intermolecular force and result in low boiling points and surface tensions. In the given compounds, pentane is a nonpolar hydrocarbon, which allows for London dispersion forces to occur.
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The mass of platinum plated on the electrode is 53.6 mg, which corresponds to answer choice (c).

To calculate the mass of platinum plated on the electrode, we need to use Faraday's law of electrolysis, which relates the amount of substance produced at an electrode to the quantity of electricity passed through an electrolytic cell. The formula is:

mass of substance = (current x time x atomic weight) / (Faraday constant x valence)

Where:

current is the electric current (in amperes)

time is the duration of the electrolysis (in seconds)

atomic weight is the atomic weight of the substance being plated (in grams per mole)

Faraday constant is the charge on one mole of electrons (96485 C/mol)

valence is the number of electrons transferred per mole of substance

For [tex]Pt(NO_3)_2[/tex], the atomic weight of platinum is 195.08 g/mol, and the valence is 2 (since each platinum ion accepts 2 electrons to form neutral platinum atoms). Plugging in the values:

mass of Pt = (0.500 A x 55.0 s x 195.08 g/mol) / (96485 C/mol x 2) = 0.0536 g = 53.6 mg

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The ionic strength of the 0.0020 M MgCl2 solution at 298 K is 0.0060 mol/L.

The ionic strength of a solution is a measure of the concentration of ions in the solution. It is calculated using the following formula:

I = 1/2 * ∑(Ci * zi^2)

where I is the ionic strength, Ci is the molar concentration of each ion in the solution, and zi is the charge of the ion.

For MgCl2, the compound dissociates into Mg2+ and 2 Cl- ions in solution. Therefore, the concentration of Mg2+ and Cl- in the solution are both 0.0020 mol/L.

Using the formula above, we can calculate the ionic strength of the solution:

I = 1/2 * [(0.0020 mol/L * 2^2) + (0.0020 mol/L * (-1)^2 * 2)]

I = 1/2 * (0.0080 + 0.0040)

I = 0.0060 mol/L

Therefore, the ionic strength of the 0.0020 M MgCl2 solution at 298 K is 0.0060 mol/L.

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