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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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Ilia_Sergeevich [38]

Answer:

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2 years ago
g 2BrO3- + 5SnO22-+ H2O5SnO32- + Br2+ 2OH- In the above reaction, the oxidation state of tin changes from to . How many electron
Archy [21]

Answer:

In the above reaction, the oxidation state of tin changes from 2+ to 4+.

10 moles of electrons are transferred in the reaction

Explanation:

Redox reaction is:

2BrO₃⁻ + 5SnO₂²⁻ + H2O ⇄ 5SnO₃²⁻ + Br₂ + 2OH⁻

SnO₂²⁻ → SnO₃²⁻

Tin changes the oxidation state from +2 to +4. It has increased it so this is the oxidation from the redox (it released 2 e⁻). We are in basic medium, so we add water in the side of the reaction where we have the highest amount of oxygen. We have 2 O on left side and 3 O on right side so we add 1 water on the right and we complete with OH⁻ in the opposite side to balance the H.  

SnO₂²⁻ + 2OH⁻ → SnO₃²⁻ + 2e⁻ + H₂O <u>Oxidation</u>

BrO₃⁻ →  Br₂

First of all, we have unbalance the bromine, so we add 2 on the BrO₃⁻. We have 6 O in left side and there are no O on the right, so we add 6 H₂O on the left. To balance the H, we must complete with 12OH⁻. Bromate reduces to bromine at ground state, so it gained 5e⁻. We have 2 atoms of Br, so finally it gaines 10 e⁻.

6H₂O + 10 e⁻ + 2BrO₃⁻ →  Br₂ + 12OH⁻ <u>Reduction</u>

In order to balance the main reaction and balance the electrons we multiply  (x5) the oxidation and (x1) the reduciton

(SnO₂²⁻ + 2OH⁻ → SnO₃²⁻ + 2e⁻ + H₂O) . 5

(6H₂O + 10 e⁻ + 2BrO₃⁻ →  Br₂ + 12OH⁻) . 1

5SnO₂²⁻ + 10OH⁻ + 6H₂O + 10 e⁻ + 2BrO₃⁻ → Br₂ + 12OH⁻ + 5SnO₃²⁻ + 10e⁻ + 5H₂O

We can cancel the e⁻ and we substract:

12OH⁻ - 10OH⁻ = 2OH⁻ (on the right side)

6H₂O - 5H₂O = H₂O (on the left side)

2BrO₃⁻ + 5SnO₂²⁻ + H2O ⇄ 5SnO₃²⁻ + Br₂ + 2OH⁻

6 0
3 years ago
Delta H equals 98.8 KJ, and Delta S equals 141.5 J/K. Is this reaction spontaneous or nonspontaneous at high and low pressure
Softa [21]
A is the answer

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8 0
3 years ago
Draw the major product formed when the structure shown below undergoes substitution in ch3ch2oh with heat. (use a wavy bond (sin
muminat

Answer:

Explanation:

The missing image is attached below.

The objective of this question is to draw the major product formed from the diagram attached below.

From the diagram attached, we will see the reaction of a tertiary alkyl halide together with a weak nucleophile (ch3ch2oh) undergoing a nucleophilic substitution (SN₁) mechanism to yield a racemic mixture(i.e., compound that is not optically active but contains an equal amount of dextrorotatory and levorotatory stereoisomers) as a product.

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3 years ago
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mars1129 [50]
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So --- mCΔT(copper)=mCΔT(water).

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C (Cu) = ?

So --- 38(-47)C[Cu]=15(4.184)(11)
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5 0
3 years ago
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