A neutral atom and an ion of the same element have the same number of _______. They have a different number of _______ and hence reactivity.

Answer:

1. Some polar compounds form hydrogen bridge bonds with water.

2. Polar compounds DO NOT interact in the same way with hydrophobic phospholipid tails.

3. YES: interactions are formed between polar compounds and hydrophobic tails.

4. False

Explanation:

Hello!

Different types of intermolecular junctions can be formed in the molecules:

-Bridge hydrogen bond: It is formed between a hydrogen attached to a very electronegative element (such as oxygen in water) and another very electronegative element (such as oxygen, fluorine). Polar molecules that contain electronegative elements in their structure may form this junction with water.

-Dipole-dipole union: it is formed between polar molecules where the zone with positive charge density of one molecule approaches the zone with negative charge density of another.

-Dipole-induced dipole union: it is formed between polar molecules where the zone with positive charge density of a molecule causes a non-polar molecule to partially polarize. It is a weak union but becomes important in long hydrophobic chains. It is the union that is established between polar compounds and hydrophobic tails of phospholipids.

For a polar compound to enter the hydrophobic space of the phospholipid tails, the cell usually uses other transport systems other than passive transport such as transport by specialized proteins. Water, despite being a polar molecule, due to its small size it can pass through the membrane at low speed.

The bonds formed with water do not break in aqueous medium. the "lack of attraction" does not exist between the molecules, there is always attraction although it can be of different intensity.

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The question concerns the behavior of polar compounds with the phospholipid bilayer of cell membranes. Polar compounds can't interact with the hydrophobic core, not because the core repels them, but because they'd need to break their existing bonds with water and wouldn't be able to form new bonds within the hydrophobic core.

The subject is about how polar compounds interact with the phospholipid bilayer, which forms the structure of cell membranes. Polar compounds form hydrogen bonds with the water molecules outside the membrane due to their polar nature. But they cannot form a similar kind of bond with the hydrophobic tails of the phospholipids, which form the core of the cell membrane.

Hydrophobic substances do not have the necessary conditions to form hydrogen bonds with polar compounds, making them unfavorable for such interactions. For a polar compound to pass through the hydrophobic layer, it would have to break the bonds it has with water and then fail to form any new bonds within the hydrophobic core. This would require a lot of energy and is not favorable, thus making it seem like the hydrophobic core 'repels' polar compounds, when in reality, it's a lack of attraction.

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Answer: [tex]7.0\times 10^2kJ[/tex]  will be released when 0.25 moles of glucose in a can of soda is metabolized

Explanation:

Heat of combustion is the amount of heat released on complete combustion of 1 mole of substance.

Given :

Amount of heat released on combustion 1 mole of glucose = 2.8 MJ = [tex]2.8\times 10^3kJ[/tex]      [tex]1MJ=10^3kJ[/tex]

Thus we can say:  

1 mole of glucose on combustion releases = [tex]2.8\times 10^6J[/tex]

Thus 0.25 moles of glucose on combustion releases =[tex]\frac{2.8\times 10^3kJ}{1}\times 0.25=7.0\times 10^2kJ[/tex]

Thus [tex]7.0\times 10^2kJ[/tex]  will be released when 0.25 moles of glucose in a can of soda is metabolized

Answer: option A. Lab.

Explanation:

The structure of Dimethyl sulfide is H3C-S-CH3. It is produced naturally by some marine algae.

Explanation:

Answer:

Explanation:

1 - Octyne is a member of the Alkyne family. Alkyne has a general formular is CnH2n-2 which is a homologous series with a least one carbon - carbon triple bond.

Where n is the number of carbon atoms needed in the structure.

The "1" prefix is an indicator for where the triple bond of the compound will be found which is at carbon -1. Because of its triple bond it is termed an unsaturated hydrocarbon. Their properties include:

1. Highly reactivity

2. Hydrophobic compound.

Octa- decribes the number of carbon atoms present in the compound which is 8. Also note that, it is not a cycloalkyne but a straight chain compound with a triple bond at one end.

So using the general formular, CnH2n-2

For n = 8,

1-Octyne is C8H14

Below in the attac is the structural formula of 1 - Octyne

Answer:

Explanation:

1-Octyne is a member of the Alkyne series, Alkyne is a homologous series with the general molecular formula;

CnH2n-2

n is a positive whole number which equal to or greater than 2.

i.e n= number of carbon given  

In the case of the compound given (1-Octyne), the prefix “Oct” means 8, which is the number of  

carbon atoms present in the compond while the suffix “yne” is

coined from the family name “Alkyne”

Therefore, C8H(2x 8) – 2 = C8H16-2 = C8H14

It is an unsaturated hydrocarbon, each alkyne molecule contains four hydrogen atoms

corresponding to alkane and two hydrogen atom less than the corresponding alkene. This is because

each alkyne molecule contains a carbon-carbon triple bond, where two  carbon atoms are  bonded  

to each other by the sharing of three pairs of electrons.

The Alkyne family show a higher degree of unsaturation than the alkenes, hence they are chemically  

reactive than alkane and alkene .

Answer:

Rate = k * [C₄H₆]²

Explanation:

It is possible to write the reaction as:

The differential rate law for a simple second order reaction of the type 2A → B is:

With the above information in mind, the rate law for the reaction of butadiene would be:

Answer: (c) CHCl3

Explanation:

From the rule : a polar solvent will dissolve a polar compound and a non polar solvent will also dissolves a non polar compound.

(a) CH4 is non- polar and water is a polar solvent. Therefore CH4 is not soluble in water

(b) CCl4 is non polar and water is a polar solvent. Therefore CCl4 is not soluble in wtaer

(c) CHCl3 is polar molecule and water is also a polar solvent.

Therefore CHCl3 is expected to be most soluble in water

Among CH4, CCl4, and CHCl3, CHCl3 (chloroform) is expected to be the most soluble in water due to its somewhat polar nature which corresponds with the polarity of water.

The gas expected to be most soluble in water among CH4, CCl4, and CHCl3 is CHCl3. This is based on the principle of 'like dissolves like'. In terms of polarity, water is polar and CHCl3 (chloroform) is also somewhat polar due to the presence of chlorine atoms, which have higher electronegativity. Since polar molecules are more likely to dissolve in another polar substance, CHCl3 would be the most soluble of the given gases in water.

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The question is incomplete, here is the complete question:

A chemistry student needs 15.00 g of 2-bromobutane for an experiment. He has available 220. g of a 30.0 % w/w solution of 2-bromobutane in ethanol. Calculate the mass of solution the student should use. If there's not enough solution, press the "No solution" button. Round your answer to 3 significant digits.

Answer: The mass of solution, the student should use is 50.0 grams

Explanation:

We are given:

30.0 % (w/w) of 2-bromobutane

This means that 30 grams of 2-bromobutane is present in 100 g of solution

Mass of solution given = 220. g

Mass of 2-bromobutane, the student needs = 15.00 g

Calculating the mass of 2-bromobutane in given amount of solution:

[tex]\Rightarrow 220\times \frac{30}{100}=66g[/tex]

To calculate the mass of solution, we use unitary method:

If 66 grams of 2-bromobutane is present in 220 grams of solution

So, 15 grams of 2-bromonutane will be present in [tex]\frac{220}{66}\times 15=50.0g[/tex] of solution

Hence, the mass of solution, the student should use is 50.0 grams

Final answer:

To calculate the mass of solution, convert the mass of -bromobutane into moles and then use the concentration of the solution to determine the amount of solution needed.

Explanation:

To calculate the mass of solution the student should use, we need to first convert the mass of -bromobutane needed into moles. Then, we can determine the amount of solution needed by using the concentration of the w/w solution of -bromobutane in ethanol. Let's go step by step:

Convert the grams of -bromobutane into liters of solution using the concentration: liters of solution = moles / concentration.

Remember to round your answer to the appropriate number of significant digits.

The correct answer is "0.000035 M/s". The rate of disappearance of NO₂ in the reaction is 0.000035 M/s, found by dividing the change in concentration of NO₂ (-0.00350 M) by the time period (100 s).

The rate of disappearance of NO₂ in the reaction 2NO₂
ightarrow 2NO₂ + NO₂ can be calculated using the change in concentration of NO₂ over time. We are given that the concentration of NO₂ drops from 0.0100 to 0.00650 M in 100 seconds.

First, we calculate the change in concentration (also known as the concentration difference):
Concentration difference = Final concentration - Initial concentration = 0.00650 M - 0.0100 M = -0.00350 M (a negative sign indicates disappearance).

To find the rate of disappearance, we divide the concentration difference by the time period:
Rate of disappearance = Concentration difference / Time period
Rate of disappearance = -0.00350 M / 100 s = -0.000035 M/s

Since we are interested in the rate of disappearance, we take the absolute value of the result, making it 0.000035 M/s.

The rate of disappearance of NO2 for the given period is [tex]\( 3.5 \times 10^{-5} \) M/s.[/tex]

To find the rate of disappearance of NO2, we need to calculate the change in concentration of NO2 over time. The reaction given is:

[tex]\[ 2NO_2 \rightarrow 2NO + O_2 \][/tex]

From the problem, we have the initial concentration of NO2 as 0.0100 M and the final concentration as 0.00650 M. The time taken for this change in concentration is 100 seconds.

The rate of disappearance of NO2 can be calculated using the formula:

[tex]\[ \text{Rate} = -\frac{\Delta [NO_2]}{\Delta t} \][/tex]

where [tex]\( \Delta [NO_2] \)[/tex] is the change in concentration of NO2, and [tex]\( \Delta t \)[/tex] is the change in time.

The change in concentration of NO2 is:

[tex]\[ \Delta [NO_2] = [NO_2]_{\text{initial}} - [NO_2]_{\text{final}} \][/tex]

[tex]\[ \Delta [NO_2] = 0.0100 \, \text{M} - 0.00650 \, \text{M} = 0.00350 \, \text{M} \][/tex]

The change in time is:

[tex]\[ \Delta t = t_{\text{final}} - t_{\text{initial}} \][/tex]

[tex]\[ \Delta t = 100 \, \text{s} - 0 \, \text{s} = 100 \, \text{s} \][/tex]

Now, we can calculate the rate:

[tex]\[ \text{Rate} = -\frac{0.00350 \, \text{M}}{100 \, \text{s}} \][/tex]

[tex]\[ \text{Rate} = -3.5 \times 10^{-5} \, \text{M/s} \][/tex]

The negative sign indicates that the concentration of NO2 is decreasing over time. However, the rate of disappearance is typically reported as a positive value, so we take the absolute value:

[tex]\[ \text{Rate} = 3.5 \times 10^{-5} \, \text{M/s} \][/tex]

Answer:

The work done is 123.5 J

Explanation:

Given that:-

Temperature = 260 K

The expression for the work done is:

[tex]W=-nRT \ln \left( \dfrac{V_2}{V_1} \right)[/tex]

Where,  

n is the number of moles = 52.0 mmol = [tex]52.0\times 10^{-3}\ moles[/tex]

W is the amount of work done by the gas

R is Gas constant having value = 8.314 J / K mol  

T is the temperature  

V₁ is the initial volume  = 300 cm³

V₂ is the final volume  = 100 cm³

Applying in the equation as:

[tex]W=-52.0\times 10^{-3}\ moles\times 8.314\ J/Kmol\times 260\ K \ln \left( \dfrac{100\ cm^3}{300\ cm^3} \right)[/tex]

[tex]W=-52.0\times 10^{-3}\times 8.314\times 260 \ln \left( \dfrac{100}{300} \right)\ J=123.5\ J[/tex]

The work done is 123.5 J

The work done in the isothermal reversible compression of an ideal gas, where the volume of gas was reduced from 300 cm3 to 100 cm3 is about 284 J.

In the given problem, we're dealing with isothermal reversible compression of an ideal gas. Under these conditions, the work done (w) can be calculated using the formula: w = -nRT ln(Vf/Vi), where n is the number of moles, R is the gas constant, T is the temperature, Vi is the initial volume, and Vf is the final volume.

Here, n = 52.0 mmol = 0.052 mol, R = 8.314 J / (mol·K) (in these units to match those of the problem), T = 260 K, Vi = 300 cm3 = 0.3 L, and Vf = 100 cm3 = 0.1 L.

Plugging these values into the formula, we get:

w = -0.052 mol * 8.314 J / (mol·K) * 260 K * ln(0.1/0.3)

Therefore, the work done in this isothermal reversible compression where the volume of gas was reduced from 300 cm3 to 100 cm3 is approximately 284 J.

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Explanation:

According to the Einstein law, it is known that

            [tex]h \times \nu = \phi + \frac{1}{2} mv^{2}[/tex]

where,   h = energy of light

           [tex]\phi[/tex] = work function

           [tex]m v^{2}[/tex] = kinetic energy of electron

It is given that the value of [tex]h \nu[/tex] is [tex]7.2 \times 10^{-19} J[/tex]. And,

               1 eV = [tex]1.6 \times 10^{-19} J[/tex]

Here,  [tex]\phi[/tex] for titanium is 4.33 eV

          = [tex](4.33 \times 1.6 \times 10^{-19})[/tex] J  

          = [tex]6.928 \times 10^{-19}[/tex] J

(a)   First of all, kinetic energy will be calculated as follows.

             [tex]\frac{1}{2}mv^{2} = h \nu - \phi[/tex]

                         = [tex](7.2 \times 10^{-19} - 6.92 \times 10^{-19})[/tex] J

                         = [tex]0.272 \times 10^{-19}[/tex] J

It is known that mass of electrons is equal to [tex]9.109 \times 10^{-31} kg[/tex].

Therefore, [tex]mv^{2} = 0.544 \times 10^{-19} J[/tex]

and,     [tex](mv)^{2} = 9.109 \times 0.544 \times 10^{-19} \times 10^{-31}[/tex]

                       = [tex]4.955 \times 10^{-50}[/tex]

                mv = [tex]2.225 \times 10^{-25}[/tex]

Now, the relation between wavelength and mv is as follows.

      [tex]\lambda = \frac{6.626 \times 10^{-34}}{2.225 \times 10^{-25}}[/tex]

                   = [tex]2.98 \times 10^{-9} m[/tex]

Therefore, the wavelength of the ejected electrons is [tex]2.98 \times 10^{-9} m[/tex].

(b)   It is known that relation between energy and wavelength is as follows.

               E = [tex]h \nu = \frac{hc}{\lambda}[/tex]

                  [tex]\lambda = \frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{7.2 \times 10^{-19}}[/tex]

                     = [tex]\frac{6.626 \times 3 \times 10^{-26}}{7.2 \times 10^{-19}}[/tex]

                    = [tex]2.76 \times 10^{-7} m[/tex]

Hence, the wavelength of the ejected electrons is [tex]2.76 \times 10^{-7} m[/tex].

(c)   For iron surface, [tex]\phi = 4.7 eV[/tex]

                                       = [tex](4.7 \times 1.6 \times 10^{-19})[/tex] J

                                       = [tex]7.52 \times 10^{-19}[/tex] J

Here, the value of [tex]\phi[/tex] is more than the value of UV light source. Hence, we need a shorter wavelength light as we know that,

                  [tex]E \propto \frac{1}{\lambda}[/tex]

Therefore, lesser will be the wavelength higher will be the energy.

Final answer:

In a photoelectric effect experiment, the wavelength of the ejected electrons can be calculated using the equation E = h / λ, where E is the energy of the electron, h is Planck's constant, and λ is the wavelength. The wavelength of the ejected electrons is found to be 13.8 femtometers. The wavelength of the incident UV light can be calculated using the same equation and is found to be 9.2 picometers. For an iron surface with a higher work function, a longer wavelength of light would be required to eject electrons with the same wavelength.

Explanation:

(a) To calculate the wavelength of the ejected electrons, we can use the equation E = h / λ, where E is the energy of the electron, h is Planck's constant (6.626 x 10^-34 J*s), and λ is the wavelength.

We can rearrange the equation to solve for λ: λ = h / E.

When we substitute the energy of the ejected electrons as given (3 eV = 4.8 x 10^-19 J), we get λ = (6.626 x 10^-34 J*s) / (4.8 x 10^-19 J) = 13.8 x 10^-15 m, or 13.8 femtometers.

(b) To calculate the wavelength of the incident UV light, we can use the same equation E = h / λ, but this time substitute the energy of the incident light (7.2 x 10^-19 J) and solve for λ. We get λ = (6.626 x 10^-34 J*s) / (7.2 x 10^-19 J) = 9.2 x 10^-16 m, or 9.2 picometers.

(c) The work function of an iron surface is provided (4.7 eV). Since the work function represents the minimum energy required to eject an electron, a longer wavelength of light would be required to eject electrons with the same calculated wavelength in part (a). This is because longer wavelength light corresponds to lower energy photons.

Answer:

1.472 x 10^-17 J energy of the photon is released (exothermic).

Explanation:

E = –kZ^2 / n^2

k = 2.18 x10^-18 J

Z = atomic number

Li2+ atomic number = 3

Ei = –(2.18 x 10^-18 J) x 3^2 / 2^2 = –4.905 x 10^-18 J

Ef = –(2.18 x 10^-18 J) x 3^2 / 1^2 = –1.962 x 10^-17 J

ΔE = |Ef – Ei| = |(–1.962 x 10^-17 J) – (–4.905 x 10^-18 J)| = 1.472x10^-17  

1.472 x 10-17 J energy of the photon is released (exothermic).

Using the Bohr model, [tex]\rm 1.472\times 10^{-17 } J[/tex] is the energy in joules of the photon produced when an electron in a Li2+ ion moves from the orbit with n = 2 to the orbit with n = 1.

A photon is the basic unit of light and electromagnetic radiation. It is a quantum of energy and a carrier of electromagnetic force. Light was assumed to be a continuous wave in classical physics, but scientists found in the early twentieth century that light and other kinds of electromagnetic radiation behave as both waves and particles. This dual nature is a key concept in quantum mechanics.

E =[tex]\rm -kZ^2 / n^2[/tex]

k =[tex]\rm 2.18 \times10^{-18 }[/tex]J

Z = atomic number

[tex]\rm Li^{2+}[/tex] atomic number = 3

Ei =[tex]\rm -(2.18 \times 10^{-18} J) \times 3^2 / 2^2[/tex]

  = [tex]\rm -4.905 \times 10^{-18}[/tex] J

Ef = [tex]\rm -(2.18 \times 10^{-18} J) \times 3^2 / 1^2[/tex]

   =[tex]\rm -1.962 \times 10^{-17[/tex] J

ΔE = |Ef – Ei|

    = |([tex]\rm -1.962 \times 10^{-17}[/tex] J) – ([tex]\rm -4.905 \times 10^{-18 } J[/tex])|

    = [tex]\rm 1.472\times 10^{-17 }[/tex] J

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The percent by mass of acetyl bromide in the given solution can be calculated by dividing its mass by the total mass of the solution and then multiplying by 100%, leading to a result of 43.5%.

To calculate the percent by mass of each component in the solution, we first need to find the total mass of the solution. The total mass is the sum of the mass of chloroform, acetone, and acetyl bromide, which is 96.2 g + 31.2 g + 98.1 g = 225.5 g.

Next, we calculate the percent by mass for each substance by dividing the mass of each substance by the total mass and then multiplying it by 100%.

For acetyl bromide, it's (98.1 g / 225.5 g) * 100% = 43.5%.

Please note that the number of significant digits in the answer should match the lowest number of significant digits in the input, that's why it's rounded off to three significant digits.

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Answer:

1.05 × 10²⁵ atoms H₂

General Formulas and Concepts:

Atomic Structure

Stoichiometry

Explanation:

Step 1: Define

[Given] 35.0 g H₂

[Solve] atoms H₂

Step 2: Identify Conversions

Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

[PT] Molar Mass of H - 1.01 g/mol

Molar Mass of H₂: 2(1.01) = 2.02 g/mol

Step 3: Convert

Step 4: Check

Follow sig fig rules and round. We are given 3 sig figs.

1.04342 × 10²⁵ atoms H₂ ≈ 1.04 × 10²⁵ atoms H₂

Topic: AP Chemistry

Unit: Atomic Structure

Final answer:

To find the number of hydrogen atoms in 35.0 grams of hydrogen gas, we calculated the moles of hydrogen gas and used Avogadro's number. We determined that there are 2.09 × 1025 hydrogen atoms in 35.0 grams of hydrogen gas.

Explanation:

To determine how many hydrogen atoms are in 35.0 grams of hydrogen gas, we use Avogadro's number and the concept of molar mass. The molar mass of hydrogen gas (H2) is 2.0158 grams per mole. First, we calculate the number of moles in 35.0 grams of hydrogen gas:
(35.0 grams H2) ÷ (2.0158 grams/mol) = 17.37 moles H2

Since each mole of hydrogen gas contains two hydrogen atoms, we need to multiply the number of moles by Avogadro's number, which is 6.022 × 1023 atoms/mol:
(17.37 moles H2) × (6.022 × 1023 molecules H2/mol) × (2 atoms H/molecule H2)

After calculating, we find that there are 2.09 × 1025 hydrogen atoms in 35.0 grams of hydrogen gas. Therefore, the correct answer is 2.09 × 1025.

Final answer:

A pound-mole of a substance contains approximately 6.022 × 10²³ × 453.592 atoms, as Avogadro's number is used in the definition of a mole, which is based on grams, and needs to be converted using the pound-to-gram conversion factor.

Explanation:

The concept of mole is fundamental in chemistry for measuring quantities of substances. When referring to a pound-mole is equivalent to the number of atoms in a substance weighing one pound. However, Avogadro's number, which is 6.022 × 10²³, is defined based on the international standard mole unit, which is mass in grams. To convert between the pound-mole and the gram-mole, one must use the conversion factor between pounds and grams. Since 1 pound is equal to approximately 453.592 grams, a pound-mole will contain Avogadro's number multiplied by the ratio between the pound and grams (453.592). Therefore, there are approximately 6.022 × 10²³ × 453.592 atoms in a pound-mole of a substance.

Answer:

The molarity of the solution is 9.73 M

Explanation:

Step 1: Data given

Mass of isopropanol = 585 grams

Molar mass of isopropanol = 60.1 g/mol

Volume = 1000 mL = 1 L

Step 2: Calculate moles of isopropanol

Moles isopropanol = mass isopropanol / molar mass isopropanol

Moles isopropanol = 585 grams / 60.1 g/mol

Moles isopropanol = 9.73 moles

Step 3: Calculate molarity of the solution

Molarity = moles / volume

Molarity = 9.73 moles / 1L

Molarity = 9.73 M

The molarity of the solution is 9.73 M

Answer:

Polar/Hydrophilic

Explanation:

Fluorine, Nitrogen and Oxygen are strong electronegative atoms and by definition, Electronegativity is the amount of pull or the high affinity of an atom to electrons.

Polar bond occurs when there is a high difference between the electronegativity value of both atoms that take part in the bond.

A polar molecule has a net dipole from the distribution of its positive and negayive charges. Hydrophobic and Hydrophilic (in chemistry, Polar) are terms dependent on the overall distribution of charge in its molecule.

Therefore, bonds between C-N, C-O and C-Cl are polar covalent bonds a d this is because of the jigh electronegativity possessed by Nitrogen, Oxygen and Chlorine.

Answer:

206 mL

Explanation:

In the annexed picture you can see your same question, just in another format.

First we calculate the total moles of CuF₂ that are required in the working solution:

1967 μM ⇒ 1967 / 10⁶ = 1.967 x10⁻³M

1.967 x10⁻³M * 0.275 L = 5.409x10⁻⁴ mol

Now we divide those moles by the concentration of the stock solution, to calculate the volume:

5.409x10⁻⁴ mol ⇒ 5.409x10⁻⁴ * 1000 = 0.5409 mmol

0.5409 mmol ÷ (2.63 mmol/L) = 0.206 L

0.206 L ⇒ 0.206 * 1000 = 206 mL

Answer:

1.31x10¹¹ g/cm³

Explanation:

The mass of the proton is equal to the mass of the neutron, which is 1.67x10⁻²⁴ g, so the mass of the alpha particle is 4*1.67x10⁻²⁴ = 6.68x10⁻²⁴ g.

1 fm = 1.0x10⁻²³ cm, thus the radius of the alpha particle is 2.3x10⁻¹² cm. If the particle is a sphere, the volume of it is:

V = (4/3)*π*r³, where r is the radius, so:

V = (4/3)*π*(2.3x10⁻¹²)³

V = 5.1x10⁻³⁵ cm³

The density of the particle is the how mass exists per unit of volume, so, it's the mass divided by the volume:

d = 6.68x10⁻²⁴/5.1x10⁻³⁵

d = 1.31x10¹¹ g/cm³

An alpha particle is a helium-4 nucleus consisting of two protons and two neutrons. The density of an alpha particle can be calculated by dividing its mass by its volume. For an alpha particle with a mass of approximately 4 atomic mass units and a radius of 2.6 femtometers, the density is approximately 4.90 x 10⁶ grams per cubic centimeter.

An alpha particle is a helium-4 nucleus, consisting of two protons and two neutrons. The density of an alpha particle can be calculated by dividing its mass by its volume. Since the mass of an alpha particle is approximately 4 atomic mass units (amu) and its radius is given as 2.6 fm (femtometers), we can use the formula for the volume of a sphere to find the volume of the alpha particle.

V = (4/3)πr³ = (4/3)π(2.6 fm)³ = (4/3)π(2.6 x 10⁻¹⁵ m)³ = (4/3)π(2.6 x 10⁻¹⁵ x 10⁻¹⁵ x 10⁻¹⁵) m³ = (4/3)π(1.94 x 10⁻⁴) m³ ≈ 8.15 x 10⁻¹³ m³

To convert this volume from cubic meters to cubic centimeters, we can multiply by the conversion factor 1 m³ = 1 x 10⁶ cm³.

Volume in cm³ = (8.15 x 10⁻¹³ m³) x (1 x 10⁶ cm³/m³) ≈ 8.15 x 10⁻⁷ cm³

To find the density, we divide the mass of the alpha particle (4 amu) by its volume (8.15 x 10⁻⁷ cm³).

Density = mass/volume = (4 amu) / (8.15 x 10⁻⁷ cm³) ≈ 4.90 x 10⁶ g/cm³

This question does not contain the structures of the molecules. The structures in Daylight SMILES format are:

I. C1=CC=CC=C1C(=O)C

II. C1=CC=CC=C1CC=O

III. C1=CC(C)=CC=C1C=O

IV. C1=CC=CC=C1CCC

V. C1=CC=CC=C1C(C)C

The structures are also attached

Answer:

The structure of compound IV is consistent with the information obtained analysis

Proposed structures for the ions with m/z values of 120, 105,77 and 43 are (also attached):

C1=CC=CC=C1C(=[OH0+])C |^1:7|

C1C([CH0+]=O)=CC=CC=1

C1[CH0+]=CC=CC=1

C(#[OH0+])C

respectively

Explanation:

The IR peak at 1687 cm⁻¹ is indicative of an α unsaturated carbonyl carbon. While the 1H NMR singlet is of the methyl group next to carbonyl and the multiplet near 7.1 ppm is a characteristic peak of benzene. This data shows points towards structure I.

Mass spectrum peak at 120 m/z is of molecular ion peak. In the case of carbonyl-containing molecule, this peak is observable. The signal at 105 shows the loss of a methyl group next to the carbonyl. m/z value of 77 is the characteristic cationic peak of benzene, while the peak at 43 infers the formation of acylium ion (RCO+) due to α-cleavage. All this data agrees with the structure of acetophenone (Structure 1)

Answer: The pH of the solution is 4.57

Explanation:

To calculate the number of moles for given molarity, we use the equation:

[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}[/tex]

Molarity of benzoic acid = 50 mM = 0.05 M      (Conversion factor:  1 M = 1000 mM)

Volume of solution = 100 mL

Putting values in above equation, we get:

[tex]0.05M=\frac{\text{Moles of benzoic acid}\times 1000}{100mL}\\\\\text{Moles of benozic acid}=\frac{(0.05\times 100)}{1000}=0.005mol[/tex]

Molarity of sodium hydroxide = 50 mM  = 0.05 M

Volume of solution = 70 mL

Putting values in above equation, we get:

[tex]0.05M=\frac{\text{Moles of sodium hydroxide}\times 1000}{50mL}\\\\\text{Moles of sodium hydroxide}=\frac{0.05\times 70}{1000}=0.0035mol[/tex]

The chemical reaction for sodium hydroxide and benzoic acid follows the equation:

                [tex]C_6H_5COOH+NaOH\rightarrow C_6H_5COONa+H_2O[/tex]  

Initial:        0.005            0.0035    

Final:         0.0015        -                   0.0035

Volume of solution = 100 + 70 = 170 mL = 0.170 L    (Conversion factor:  1 L = 1000 mL)

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

[tex]pH=pK_a+\log(\frac{[salt]}{[acid]})[/tex]

[tex]pH=pK_a+\log(\frac{[C_6H_5COONa]}{[C_6H_5COOH]})[/tex]

We are given:

[tex]pK_a[/tex] = negative logarithm of acid dissociation constant of benzoic acid = 4.2

[tex][C_6H_5COONa]=\frac{0.0035}{0.170}[/tex]

[tex][C_6H_5COOH]=\frac{0.0015}{0.170}[/tex]

pH = ?

Putting values in above equation, we get:

[tex]pH=4.2+\log(\frac{0.0035/0.170}{0.0015/0.170})\\\\pH=4.57[/tex]

Hence, the pH of the solution is 4.57

Answer:

Fe³⁺ and Mn⁺²

B⁻ and C

Explanation:

Isoelectronic are species that have the same number of electrons. A neutral species has the same number of protons and electrons, and the number of protons in the atomic number (Z) found in the periodic table.

A cation (positive ions) lost the numbe of electrons indicate in its charge, and an anion (negative ions) gain the number of electrons indicate in its charge. So, let's identify the number of electrons (e-) in each one the atoms:

First column:

Zn: Z = 30, e- = 30

Fe³⁺: Z = 26, e- = 26 - 3 = 23

Mn⁺²: Z = 25, e- = 25 - 2 = 23

Ar: Z = 18

Isoelectronic: Fe³⁺ and Mn⁺²

Second column:

Ge⁺²: Z = 32, e- = 32-2 = 30

Cl⁻: Z = 17, e- = 17 + 1 = 18

B⁻: Z = 5, e- = 5 + 1 = 6

C: Z = 6, e- = 6

Isoelectronic B⁻ and C

Isoelectronic species are species that contain the same number of electrons.

Each column as we can see is made up of four chemical species which may be ions or neutral atoms. The species that are isoelectronic are those that contain the same number of electrons. We shall now examine the species in each column to determine which ones are isoelectronic.

In the first column, Fe³⁺ and Mn²⁺ are isoelectronic species, they both contain 23 electrons. In the second column, B⁻ and C are isoelectronic species, they both contain six electrons.

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Answer:

18.75 mL

Explanation:

We must apply the dilute factor formula to solve this:

Diluted volume . Diluted M = Concentrated volume . concentrated M

250 mL . 0.15 M = Concentrated volume . 2 M

(250 mL . 0.15 M) / 2M = Concentrated volume

18.75 mL = Concentrated volume

Answer:

A

Explanation:

The molecule would not have a net dipole moment because the dipole moments cancel out. Being a vector quantity, dipole moment is affected by the direction of the dipole. If the dipoles are such that they are oriented opposite each other, theory simply cancel out. Polarity of individual bonds in a molecule does not automatically imply that the molecule must poses a net dipolev moment. The polarity of the Si-H and Si-Cl bonds cancel out cause they are oriented opposite each other hence the molecule is non polar.

The molecular shape of the SiH2Cl2 molecule is approximately tetrahedral and its dipole moment is zero.

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Answer:

The temperature of water will be 60 degree Celsius.

Explanation:

For every chemical there present a particular solubility temperature. The phenomenon occurs as the positive charge is attracted towards the negative and hence gives rise to a cohesive structure. When the polar compounds as well as the ions gets added to the water, they start getting into smaller parts and hence dissolve and becomes solution. The partial charge of water starts attracting different parts of the compound and makes them soluble in water.

Final answer:

To dissolve 91.6 g of KCl in 200 g of water, a temperature higher than 25°C will likely be necessary, but an exact temperature cannot be provided without specific solubility data for KCl.

Explanation:

To determine the temperature at which 91.6 g of KCl (potassium chloride) will dissolve in 200 g of water, we need to refer to the solubility data for KCl. Solubility tables or graphs provide this information and show the amount of a solute that can dissolve in a solvent at various temperatures. Generally, the solubility of ionic compounds like KCl increases with temperature. Without specific solubility data for KCl at different temperatures, an exact answer cannot be provided. However, it is known that at 25°C (77°F), approximately 34 g of KCl will dissolve in 100 mL of water. Therefore, for 91.6 g of KCl, a temperature higher than 25°C will likely be necessary to dissolve the entire amount in 200 g of water. More accurate determination requires a solubility chart or experimental data for KCl.

Answer:

19.91 J/K

Explanation:

The entropy is a measure of the randomness of the system, and it intends to increase in nature, thus for a spontaneous reaction ΔS > 0.

The entropy variation can be found by:

ΔS = ∑n*S° products - ∑n*S° reactants

Where n is the coefficient of the substance. The value of S° (standard molar entropy) can be found at a thermodynamic table.

S°, Cl(g) = 165.20 J/mol.K

S°, O3(g) = 238.93 J/mol.K

S°, O2(g) = 205.138 J/mol.K

So:

ΔS = (1*205.138 + 1*218.9) - (1*165.20 + 1*238.93)

ΔS = 19.91 J/K

Answer:

a.Carbon :proton number =6

             electron number =6

b. Fluorine: proton number=9

                  electron number =9

c. tin:  proton number=50

           electron number =50

d. nickel:proton number=28

           electron number =28

Explanation:

In the neutral state of an atom the number of protons is always equal to the number of electrons and that whats makes it electrically neutral as the positive charges of protons balances the negative charges of electrons.

Answer: some other substance must be reduced.

Explanation:

Oxidation-reduction reaction or redox reaction is defined as the reaction in which oxidation and reduction reactions occur simultaneously.

Oxidation reaction is defined as the reaction in which a substance looses its electrons. The oxidation state of the substance increases.

[tex]M\rightarrow M^{n+}+ne^-[/tex]

The substance which gets oxidized itself reduces others and thus is called as reducing agent.

Reduction reaction is defined as the reaction in which a substance gains electrons. The oxidation state of the substance gets reduced.

[tex]M^{n+}+ne^-\rightarrow M[/tex]

The substance which gets itself reduced , oxidise others and thus is called as oxidising agent.

Oxidation is the loss of electrons and is usually coupled with the reduction of another substance, known as the oxidizing agent. The production of hydronium ions is not a core characteristic of all oxidation processes.

Whenever a substance is oxidized, it means that it has lost electrons. This process is usually coupled with the reduction of another substance, which gains the electrons. Thus, statement 1 is incorrect, and statement 2 is correct: when a substance is oxidized, another substance must be reduced.

Statement 3 is also correct: a substance that is oxidized and causes the reduction of another substance is called the oxidizing agent. Statement 4 is incorrect: the production of hydronium ions is not a general characteristic of oxidation reactions, it’s specific to certain reactions that involve acids.

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Answer:1. False, 2. False, 3. True, 4. True

Explanation:

1. FALSE. Pure crystalline solids have sharp melting points, while the present of impurity lower the melting point. This is True, but the exception is a eutectic mixture. A properly mixed eutectic mixture have a sharp melting point also.

2. FALSE, although the present of impurity lower the melting point of a organic compound, but if a molecular size impurity with a significantly higher melting point and is present in large quantities the melting point of the organic compound will not be lowered.

3. TRUE, two compounds with same melting point, when mixed together will not change the melting point of either compounds. But if A has higher MP than B and the addition of A to B creates mixture with same melting point as pure A then the statement will be false.

4. TRUE, a properly mixed eutectic mixture will have a sharp melting point.

Melting point and freezing point of a pure compound will be identical because the liquid form of a pure substance when cooled it forms a stable arrangement that supercooling is prevented, freezing occur at the same temperature at which the pure solid phase melts

Answer:

For 1: The value of [tex]K_c[/tex] is 10.24

For 2: The value of [tex]Q_c[/tex] is 9.07

For 3: The new equilibrium concentration of A is 0.220 M

Explanation:

We are given:

Volume of the container = 2.00 L

Equilibrium moles of A = 0.50 moles

Equilibrium moles of B = 0.50 moles

Equilibrium moles of C = 1.60 moles

Equilibrium moles of D = 1.60 moles

We know that:

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

For the given chemical reaction:

[tex]A(g)+B(g)\rightleftharpoons C(g)+D(g)[/tex]

The expression of [tex]K_c[/tex] for above equation follows:

[tex]K_c=\frac{[C][D]}{[A][B]}[/tex]

We are given:

[tex][A]_{eq}=\frac{0.50}{2.00}=0.25[/tex]

[tex][B]_{eq}=\frac{0.50}{2.00}=0.25[/tex]

[tex][C]_{eq}=\frac{1.60}{2.00}=0.8[/tex]

[tex][D]_{eq}=\frac{1.60}{2.00}=0.8[/tex]

Putting values in above equation, we get:

[tex]K_c=\frac{0.8\times 0.8}{0.25\times 0.25}\\\\K_c=10.24[/tex]

Hence, the value of [tex]K_c[/tex] is 10.24

Added moles of B = 0.10 moles

Added moles of C = 0.10 moles

[tex]Q_c[/tex] is the quotient of activities of products and reactants at any stage other than equilibrium of a reaction.

[tex]Q_c=\frac{[C][D]}{[A][B]}[/tex]

Now,

[tex][A]=\frac{0.50}{2.00}=0.25[/tex]

[tex][B]=\frac{0.60}{2.00}=0.3[/tex]

[tex][C]=\frac{1.70}{2.00}=0.85[/tex]

[tex][D]=\frac{1.60}{2.00}=0.8[/tex]

Putting values in above equation, we get:

[tex]Q_c=\frac{0.85\times 0.8}{0.25\times 0.3}\\\\Q_c=9.07[/tex]

Hence, the value of [tex]Q_c[/tex] is 9.07

Taking equilibrium constant as 10.24 for calculating the equilibrium concentration of A.

[tex]K_c=10.24[/tex]

[tex][B]_{eq}=\frac{0.60}{2.00}=0.3[/tex]

[tex][C]_{eq}=\frac{1.70}{2.00}=0.85[/tex]

[tex][D]_{eq}=\frac{1.60}{2.00}=0.8[/tex]

Putting values in expression 1, we get:

[tex]10.24=\frac{0.85\times 0.8}{[A]\times 0.3}[/tex]

[tex][A]_{eq}=\frac{0.85\times 0.8}{0.3\times 10.24}=0.220[/tex]

Hence, the new equilibrium concentration of A is 0.220 M