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2 years ago

Classify the compound below as an alkane, alkene, alkyne, alcohol, aldehyde, amine, and so forth

Chemistry

1 answer:

Gnoma [55]2 years ago

6 0

Answer:

Organic compounds are those which are derivatives of hydrocarbons. They are classified into following functional groups.

Alkane:

Alkane are simplest hydrocarbons.
They have general formula CnH2n+2. These hydrocarbons contain single bond.
For example ethane , H3C----CH3.

Alkene:

Alkene are most reactive.
They have general formula CnH2n.
These contain double bond in their structure.
For example , ethene, H2C=CH2

Alkyne:

These are less less reactive as compare to alkenes.
They have general formula CnH2n-2.
They contain triple bonds in their structure.
For example Acetylene HC≡CH

Alcohol:

Alcohol have functional group OH. They have general formual R---OH, R may be alkyl group.For example Ethanol H3CH2C---OH

Amine:

Amine contain NH2 F.G.
They have general formula R---NH2.
There are three types of amine like primary, secondary and tertiary amine.
For example H3CH2C---NH2

Aldehyde:

Aldehydes have CHO F.G .
They have general formula R--CHO.
For example H3CH2C---CHO

Ketone:

Ketones have R--CO--R functional group.
For example acetone H3C---CO---CH3

Carboxylic acid:

They functional group COOH.
Their general formula is R---COOH.
For example Acetic acid H3C---COOH

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Estimate the increase in the molar entropy of O2(g) when the temperature is increased at constant pressure from 298 K to 348 K,

Tamiku [17]

Explanation:

It is known that relation between entropy, heat energy and temperature is as follows.

dS = $\frac{Q}{T}$

$\int dS = \int \frac{Q}{T}$

Also we know that at constant pressure, Q = $\Delta H = C_{p} - dT$

$\int_{S_{1}}^{S_{2}} dS = \int_{T_{1}}^{T_{2}} C_{p} \frac{dT}{T}$

$\Delta S = C_{p} \int_{T_{1}}^{T_{2}} \frac{dT}{T}$

As the given data is as follows.

$T_{1}$ = 298 K, $T_{2}$ = 348 K

$C_{p}$ = 29.355 J/K mol

Now, putting the given values into the above formula as follows.

$\Delta S = C_{p} ln {T_{1}}^{T_{2}} \frac{dT}{T}$

= $29.355 [ln (348) - ln (298)]$

= $29.355 [5.85 - 5.69]$

= 4.48 J/k mol

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2 years ago

(a) The mass density of a gaseous compound was found to be 1.23 kg m^−3 at 330 K and 20 kPa. What is the molar mass of the compo

Luden [163]

Answer:

The molar mass of the compound is:- 168.82 g/mol

The molar mass of the gas is:- 16.38 g/mol

Explanation:

(a)

Using ideal gas equation as:

$PV=nRT$

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L.atm/K.mol

Also,

Moles = mass (m) / Molar mass (M)

Density (d) = Mass (m) / Volume (V)

So, the ideal gas equation can be written as:

$PM=dRt$

Given that:-

Pressure = 20 kPa = 20000 Pa

The expression for the conversion of pressure in Pascal to pressure in atm is shown below:

P (Pa) = $\frac {1}{101325}$ P (atm)

20000 Pa = $\frac {20000}{101325}$ atm

Pressure = 0.1974 atm

Temperature = 330 K

d = 1.23 kg/m³ = 1.23 g/L

Molar mass = ?

Applying the equation as:

0.1974 atm × M = 1.23 g/L × 0.0821 L.atm/K.mol × 330 K

⇒M = 168.82 g/mol

The molar mass of the compound is:- 168.82 g/mol

(b)

Given that:

Pressure = 152 Torr

Temperature = 298 K

Volume = 250 cm³ = 0.25 L

Using ideal gas equation as:

$PV=nRT$

R = $62.3637\text{torr}mol^{-1}K^{-1}$

Applying the equation as:

152 Torr × 0.25 L = n × 62.3637 L.torr/K.mol × 298 K

⇒n = 0.002045 moles

Given that :

Mass of the gas = 33.5 mg = 0.0335 g

Molar mass = ?

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$0.002045\ moles
= \frac{0.0335\ g}{Molar\ mass}$

The molar mass of the gas is:- 16.38 g/mol

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