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Kaylis [27]
3 years ago
9

18. Suggest an explanation for the observations that ethanol, C2H5OH, is completely miscible with water and that ethanethiol, C2

H5SH, is soluble only to the extent of 1.5 g per 100 mL of water.
Chemistry
1 answer:
nydimaria [60]3 years ago
6 0

Answer:

              Ethanol is completely miscible due to <u><em>presence</em></u> of Hydrogen bonding.

              Ethanethiol is partially miscible due to <u><em>absence</em></u> of Hydrogen Bonding.

Explanation:

                     The miscibility of liquids depend upon the intermolecular interactions between the two liquids. The stronger the intermolecular interactions the more miscible will be the liquids.

Among the two given examples, Ethanol is more miscible in water because it exhibits hydrogen bonding which is considered the strongest intermolecular interaction. Hydrogen bonding occurs when the hydrogen atom is bonded to more electronegative atoms like Fluorine, Oxygen and Nitrogen. In this way the hydrogen atom gets partial positive charge and the electronegative atom gets partial negative charge. Hence, these partial charges results in attracting the opposite charges on other surrounding atoms.

While, in case of Ethanethiol the hydrogen atom is not bonded to any high electronegative atom hence, there will be no hydrogen bonding and therefore, there will be less interactions between the neighbour atoms.

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What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from an energy level with n =
Lady bird [3.3K]

Answer:

4.86\times10^{-7}\ \text{m}

Explanation:

R = Rydberg constant = 1.09677583\times 10^7\ \text{m}^{-1}

n_1 = Principal quantum number of an energy level = 2

n_2 = Principal quantum number of an energy level for the atomic electron transition = 4

Wavelength is given by the Rydberg formula

\lambda^{-1}=R\left(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2}\right)\\\Rightarrow \lambda^{-1}=1.09677583\times 10^7\left(\dfrac{1}{2^2}-\dfrac{1}{4^2}\right)\\\Rightarrow \lambda=\left(1.09677583\times 10^7\left(\dfrac{1}{2^2}-\dfrac{1}{4^2}\right)\right)^{-1}\\\Rightarrow \lambda=4.86\times10^{-7}\ \text{m}

The wavelength of the light emitted is 4.86\times10^{-7}\ \text{m}.

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2 years ago
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kipiarov [429]

Density does not change with the amount of matter.

The density of water is 1 g/mL whether you have 1 mL or 1000 mL of water. Density is an i<em>ntensive </em>property.

Mass, volume, and weight change with the quantity of matter. For example, the mass of 1000 mL of water is greater than the mass of 1 mL of water. Mass, volume, and weight are <em>extensiv</em>e properties.


5 0
3 years ago
In the laboratory a student determines the specific heat of a metal. He heats 19.5 grams of copper to 98.27 °C and then drops it
siniylev [52]

Answer:

The specific heat of copper is 0.37 J/g°C

Explanation:

<u>Step 1: </u>Data given

Mass of copper = 19.5 grams

Initial temperature of copper = 98.27 °C

Mass of water = 76.3 grams

Initial temperature of water = 24.05 °C

Final temperature of water and copper = 25.69 °C

<u>Step 2:</u> Calculate specific heat of copper

Qgained = -Qlost

Q = m*c*ΔT

Qwater = -Qcopper

m(water) * c(water) * ΔT(water) = - m(copper) * c(copper) *ΔT(copper)

⇒ with m(water) = 76.3 grams

⇒ with c(water) = 4.184 J/g°C

⇒ with ΔT(water) = T2-T1 = 25.69 - 24.05 = 1.64

⇒ with m(copper) = 19.5 grams

⇒ with c(copper) = TO BE DETERMINED

⇒ with ΔT(copper) = T2-T1 = 25.69 - 98.27 = -72.58

76.3 * 4.184 * 1.64 = - 19.5 * c(copper) * -72.58

523.552 = 1415.31 * c(copper)

c(copper) = 0.37 J/g°C

The specific heat of copper is 0.37 J/g°C

3 0
3 years ago
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