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Basile [38]
3 years ago
15

1. 1-propanol can form __________ with _______, but butane cannot.2. 1-propanol is more soluble than ethyl methyl ether because

it can form more _______ with _______.3. Ethanol is more soluble in water than 1-hexanol because it has ______.
Chemistry
1 answer:
12345 [234]3 years ago
8 0

Answe:

1. 1-propanol can form Hydrogen Bonds with other alcohol molecules but butane cannot.

2. 1-propanol is more soluble than ethyl methyl ether because it can form more hydrogen bonds with water.

3. Ethanol is more soluble in water than 1-hexanol because it has ethanol has shorter carbon chain whereas 1-hexanol has longer carbon chains.

Explanation:

Propanol can react with other alcohol molecules through hydrogen bonding to form ethers because alcohols do not dehydrate to form alkenes. They do this through that Williamson Ether Synthesis. In other words, alcohols can serve as an acid and a base in reactions. Butane do not follow this process.

Propanol can form hydrogen bonds with water making it more soluble than ethyl methyl ether. Propanol can donate and accept electrons which makes them interact with water better than ethyl methyl ether which can only receive proton from water.

Ethanol is mode soluble than hexanol beacuse it's non polar chain is shorter unlike in hexanol with a longer chain. Both alcohols has hydoxyl group which makes them soluble in water but their solubility is greatly affected by the remaining non polar side of the alcohol. Long non polar chain reduces solubility and vice versa.

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2.01atm

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List the following molecules in order of increasing boiling point: Br2, Cl2, F2, I2.
Andrej [43]

Answer:

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3 years ago
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The degree to which a weak base dissociates is given by the base-ionization constant, Kb. For the generic weak base, B B(aq)+H2O
Colt1911 [192]

Answer:

pH of a 0,245 M ammonia solution is 11,3 and percent ionization is 0,86%

Explanation:

For the equilibrium buffer of NH₃:

NH₃ + H₂O ⇄ NH₄⁺ + OH⁻; kb = 1,8x10⁻⁵

kb = [NH₄⁺] [OH⁻] / [NH₃] <em>(1)</em>

When 0,245 M of NH₃ is added, the equilibrium concentrations are:

[NH₃] = 0,245 - x

[NH₄⁺] = x

[OH⁻] = x.

Replacing this values in (1)

1,8x10^{-5} = \frac{x^2}{0,245 - x}

x² + 1,8x10⁻⁵x - 4,41x10⁻⁶ = 0

Solving for x:

x = -0,00211 No physical sense. There are not negative concentrations.

x = 0,00211 Real answer

Thus [OH⁻] in equilibrium is 0,00211 M.

As pOH = -log [OH⁻] and 14 = pH + pOH

pH of 0,00211 M is <em>11,3</em>

It is possible to calculate the percent ionization thus:

Percent ionization = [OH−] equilibrium / [B] initial×100%

Replacing:

0,00211 / 0,245 × 100 = <em>0,86%</em>

I hope it helps!

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Answer:

Explanation:

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