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uranmaximum [27]
2 years ago
10

Arrange the following in order of increasing boiling point: RbCl, CH3Cl, CH3OH, CH4. A. CH3OH < CH4 < CH3Cl < RbCl B. R

bCl < CH3Cl < CH3OH < CH4 C. CH4 < CH3OH < CH3Cl < RbCl D. CH3OH < CH3Cl < RbCl < CH4 E. CH4 < CH3Cl < CH3OH < RbCl
Chemistry
1 answer:
rosijanka [135]2 years ago
4 0

Answer:

E. CH₄ < CH₃Cl < CH₃OH < RbCl

Explanation:

The molecule with the stronger intermolecular forces will have the higher boiling point.

The order of strength of intermolecular forces (strongest first) is

  • Ion-Ion
  • Hydrogen bonding
  • Dipole-dipole
  • London dispersion

RbCl is a compound of a metal and a nonmetal. It is an ionic compound, so it has the highest boiling point.

CH₃Cl has a C-Cl polar covalent bond. It has dipole-dipole forces, so it has the second lowest boiling point.

CH₃OH has an O-H bond. It has hydrogen bonding, so it has the second highest boiling point.

CH₄ has nonpolar covalent C-H bonds. It has only nonpolar bonds, so the only attractive forces are London dispersion forces. It has the lowest boiling point.

Thus, the order of increasing boiling points is

CH₄ < CH₃Cl < CH₃OH < RbCl

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Using the following standard reduction potentials, Fe3+(aq) + e- → Fe2+(aq) E° = +0.77 V Ni2+(aq) + 2 e- → Ni(s) E° = -0.23 V ca
lina2011 [118]

<u>Answer:</u> The above reaction is non-spontaneous.

<u>Explanation:</u>

For the given chemical reaction:

Ni^{2+}(aq.)+2Fe^{2+}(aq.)\rightarrow 2Fe^{3+}(aq.)+Ni(s)

Here, nickel is getting reduced because it is gaining electrons and iron is getting oxidized because it is loosing electrons.

We know that:

E^o_{(Fe^{3+}/Fe^{2+})}=0.77V\\E^o_{(Ni^{2+}/Ni)}=-0.23V

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

E^o_{cell}=-0.23-0.77=-1.0V

Relationship between standard Gibbs free energy and standard electrode potential follows:

\Delta G^o=-nFE^o_{cell}

As, the standard electrode potential of the cell is coming out to be negative for the above cell. Thus, the standard Gibbs free energy change of the reaction will become positive making the reaction non-spontaneous.

Hence, the above reaction is non-spontaneous.

3 0
2 years ago
If a gas has a volume of 750 mL at 25oC, what would the volume of the gas be at 55oC?
blagie [28]

Answer:

The answer to your question is     V2 = 825.5 ml

Explanation:

Data

Volume 1 = 750 ml

Temperature 1 = 25°C

Volume 2= ?

Temperature 2 = 55°C

Process

Use the Charles' law to solve this problem

                V1/T1 = V2/T2

-Solve for V2

                V2 = V1T2 / T1

-Convert temperature to °K

T1 = 25 + 273 = 298°K

T2 = 55 + 273 = 328°K

-Substitution

                V2 = (750 x 328) / 298

-Simplification

                V2 = 246000 / 298

-Result

                V2 = 825.5 ml

             

7 0
2 years ago
Which of the following statements is true about energy quantization at the atomic level? Electrons in the outermost orbits are t
MAVERICK [17]
<h2>Answer:</h2>

The correct answer is option C which is, "Electrons in the orbit closest to the nucleus have the least amount of energy".

<h3>Explanation:</h3>
  • There are different orbitals around the nucleus on which the electrons moves around the nucleus.
  • These orbitals have a specific energy, due to which they are known as energy levels.
  • The energy level near to the nucleus has least amount of the energy and the energy of the orbitals increase as the distance of the orbitals increase to the nucleus.
8 0
2 years ago
As part of an investigation, students combined substances in a beaker to observe chemical reactions. They performed two procedur
asambeis [7]

Answer:

D.

Explanation:

8 0
2 years ago
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Consider a beaker of water sitting on the pan of an electronic scale that has been tared. A metal weight hanging from a string i
natka813 [3]

Answer:

See explanation below for answers

Explanation:

We know that the balance is tared, so the innitial weight would be zero. Now, let's answer this by parts.

a) mass of displaced water.

In this case all we need to do is to substract the 0.70 with the 0.13 g. so:

mW = 0.70 - 0.13

mW = 0.57 g of water

b) Volume of water.

In this case, we have the density of water, so we use the formula for density and solve for volume:

d = m/V

V = m/d

Replacing:

Vw = 0.57/0.9982

Vw = 0.5710 mL of water

c) volume of the metal weight

In this case the volume would be the volume displaced of water, which would be 0.5710 mL

d) the mass of the metal weight.

In this case, it would be the mass when the metal weight hits the bottom which is 0.70 g

e) density.

using the above formula of density we calculate the density of the metal

d = 0.70 / 0.5710

d = 1.2259 g/mL

4 0
2 years ago
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