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galina1969 [7]
3 years ago
6

What is the expected product when l-propanol is oxidized?

Chemistry
2 answers:
bearhunter [10]3 years ago
6 0
Answer is: <span>C. propanal.
1-propanol is primary alcohol. With weak oxidizing agent primary alcohol gives aldehyde and with strong oxidizing agent primary alcohol gives carbonic acid.
</span>Oxidation of 1-propanol with <span>Na</span>₂<span>Cr</span>₂<span>O</span>₇<span> and sulfuric acid</span><span> gives propanal and o</span><span>xidation with </span>chromic acid<span> gives </span>propionic acid<span>.</span>
mafiozo [28]3 years ago
3 0

Answer:

C. propanal

Explanation:

Organic compounds have different functional groups. It is possible to convert from one functional group to another by using proper oxidizing and reducing agents.

Alcohols are oxidized to aldehydes . Here porpanal is the aldehyde and propanol is the alcohol.

Propanone is a ketone and propene is an alkene.

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If 801 J of heat is available, what is the mass in grams of iron (specific heat = 0.45 J/g・°C) that can be heated from 22.5°C to
inysia [295]

Answer:

The correct answer will be "18.25 g".

Explanation:

The given values are:

Specific heat,

C = 0.45 J/g・°C

Heat involved,

q =  801 J

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ΔT = 120.0°C-22.5°C

     = 97.5°C

As we know,

⇒  C = \frac{q}{m \Delta T}

On substituting the given values, we get

⇒  0.45=\frac{801}{m(97.5)}

⇒  m = 18.25 \ g

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<u>Explanation:</u>

To calculate the molar solubility, we use the equation given by Henry's law, which is:

C_{O_2}=K_H\times p_{O_2}

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K_H = Henry's constant = ?

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p_{O_2 = partial pressure of oxygen gas = 2.1 atm

Putting values in above equation, we get:

0.0043g/120.3mL=K_H\times 2.1atm\\\\K_H=\frac{0.0043g}{120.3mL\times 2.1atm}=1.702\times 10^{-5}g/mL.atm

Hence, the Henry's law constant for oxygen gas in water is 1.702\times 10^{-5}g/mL.atm

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2 years ago
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