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2 years ago

Name the following amide: O CH3 CH2CH2CH2CN CH3

Chemistry

1 answer:

juin [17]2 years ago

4 0

Answer: Pentane

Explanation: According to the IUPAC(International Union of Pure and Applied Chemistry) the name of this compound will be pentane. This compound belongs to the alkane category, which has single bond between the carbon atoms. This naming is done on the basis of the number of carbon atoms and the nature of the chemical bond between the carbon atoms.

Word roots are selected based on the number of carbon atoms

1 C atom = Meth

2 C atom = Eth

3 C atom = Prop

4 C atom = But

5 C atom = Pent

6 C atom = Hex

7 C atom = Hept

8 C atom = Oct

9 C atom = Non

10 C atom = Dec

Suffixes are added based on the nature of chemical bond between the carbon atoms.

Word root + suffix (here: ane)

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2 years ago

The vapor pressure of ethanol is 1.00 × 102 mmHg at 34.90°C. What is its vapor pressure at 60.21°C? (ΔHvap for ethanol is 39.3 k

Delicious77 [7]

Answer:

The vapor pressure at 60.21°C is 327 mmHg.

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Given the vapor pressure of ethanol at 34.90°C is 102 mmHg.

We need to find vapor pressure at 60.21°C.

The Clausius-Clapeyron equation is often used to find the vapor pressure of pure liquid.

$ln(\frac{P_2}{P_1})=\frac{\Delta_{vap}H}{R}(\frac{1}{T_1}-\frac{1}{T_2})$

We have given in the question

$P_1=102\ mmHg$

$T_1=34.90\°\ C=34.90+273.15=308.05\ K\\T_2=60.21\°\ C=60.21+273.15=333.36\ K\\\Delta{vap}H=39.3 kJ/mol$

And $R$ is the Universal Gas Constant.

$R=0.008 314 kJ/Kmol$

$ln(\frac{P_2}{102})=\frac{39.3}{0.008314}(\frac{1}{308.05}-\frac{1}{333.36})\\\\ln(\frac{P_2}{102})=4726.967(\frac{333.36-308.05}{333.36\times308.05})\\\\ln(\frac{P_2}{102})=4726.967(\frac{25.31}{333.36\times308.05})\\\\ln(\frac{P_2}{102})=4726.967(\frac{25.31}{102691.548})\\\\ln(\frac{P_2}{102})=1.165$

Taking inverse log both side we get,

$\frac{P_2}{102}=e^{1.165}\\\\P_2=102\times 3.20\ mmHg\\P_2=327\ mmHg$

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3 years ago