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aksik [14]
3 years ago
5

Propanoic acid has a boiling point of 141 °C, propanamide has a boiling point of 213 °C, and propanal has a boiling point of 48

°C. Rationalize the differences in boiling point between these three compounds.
Chemistry
1 answer:
deff fn [24]3 years ago
4 0

Explanation:

  • As it is given that boiling point of propanamide is very high. So, reason for this is that easy formation of hydrogen bonds which are strong enough that we have to provide large amount of heat to break it.

As in -NH_{2}, the hydrogen atoms which are present are positive in nature. Due to this they are able to form hydrogen bonds with the neighboring oxygen atom.

Hence, these bonds are so strong that high heat needs to given to break them.

  • A propanoic acid contain carboxylic group as the functional group. So, this group is also able to form hydrogen bonding as it forms a hydrogen bond between an acid group and hydroxyl group of neighboring molecule.

Hence, it will also require high heat to break the bond due to which there will be increase in boiling point.

  • In propanal, there is presence of aldehyde functional group and three carbon atoms chain which will not form strong bonding with the hydrogen atom of CHO. Due to this there will exist weak Vander waal's force that is not at all strong enough.

As a result, less energy will be needed to break the bonds in propanal. Hence, it has very low boiling point.

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andrezito [222]

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3 years ago
How many H2O molecules are in 183.2 grams of H20 gas?
jek_recluse [69]

Answer: There are 61.24 \times 10^{23} molecules present in 183.2 grams of H_{2}O gas.

Explanation:

Given: Mass = 183.2 g

Number of moles is the mass of substance divided by its molar mass.

As molar mass of water is 18 g/mol. Therefore, moles of H_{2}O are calculated as follows.

Moles = \frac{mass}{molar mass}\\= \frac{183.2 g}{18 g/mol}\\= 10.17 mol

According to the mole concept, there are 6.022 \times 10^{22} molecules present in one mole of a substance.

Hence, molecules present in 10.17 moles are calculated as follows.

10.17 mol \times 6.022 \times 10^{23}\\= 61.24 \times 10^{23}

Thus, we can conclude that there are 61.24 \times 10^{23} molecules present in 183.2 grams of H_{2}O gas.

6 0
3 years ago
What is the volume of a sample of CO2 at STP that has a volume of 75.0mL at 30.0°C and 91kPa
Nezavi [6.7K]

60.7 ml is the volume of a sample of CO2 at STP that has a volume of 75.0mL at 30.0°C and 91kPa.

Explanation:

Data given:

V1 = 75 ml

T1 = 30 Degrees or 273.15 + 30 = 303.15 K

P1 = 91 KPa

V2  =?

P2 = 1 atm or 101.3 KPa

T2 = 273.15 K

At STP the pressure is 1 atm and the temperature is 273.15 K

applying Gas Law:

\frac{P1VI}{T1}= \frac{P2V2}{T2}

putting the values in the equation of Gas Law:

V2 = \frac{P1V1T2}{P2T1}

V2 = \frac{91 X 75 X 273.15}{303.15 X 101.3}

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at STP the volume of carbon dioxide gas is 60.7 ml.

4 0
4 years ago
Radon (Rn) is the heaviest, and only radioactive, member of Group 8A(18) (noble gases). It is a product of the disintegration of
Serggg [28]

Given data                Atomic mass of Ra= 226g/mol

no. of moles =1.0/226g/mol           =0.04424moles

no. of atoms in 0.044moles

no. of atoms =no. of moles x avogadro's number

= 0.044x 6.022 x10^23                  = 0.264968 x 10^22

 If 10^15  atoms of Ra produce 1,373*10^4  atoms of<u> Rn per second</u> then 2,66 *10^21  forms 3,658*10^10 atoms of Rn per second.

Day has 246060=86400 s

That means that 2,66x10^21  atoms of Ra produces 3,16 x10^15  atoms of Rn in a day.

N(Rn)=3.16* 10 ^15                           n(Rn)=N/NA

n(Rn)=5,25*10−9                              pV=nR*T

T=273.15K                                        R=8,314

p=101325Pa                                      V=n∗R∗T/p

V=5.25∗10^−9 ∗ 8.314 ∗ 273.15  /  101325

V=1.1810^−10 m^3 =   118 x10^-7 liters of Rn, measured at STP, are produced per day by 1.0 g of Ra

To know more about Ra here

brainly.com/question/9112754

#SPJ4

6 0
2 years ago
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5 0
3 years ago
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