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Assoli18 [71]
3 years ago
6

Identify the major force between molecules of pentane. dispersion dipole-dipole hydrogen bonding ion-dipole ion-ion

Chemistry
2 answers:
Alex Ar [27]3 years ago
8 0

Answer:

The correct answer is option A, Dispersion Force

Explanation:

The molecules and atoms of n-pentane are non-polar and thus there is no chance of any dipole –dipole interaction between the molecules of n –pentane. There exists repulsion between the electrons moving in near to each other in respective orbits. The molecule of n-pentane processes instantaneous dipole moment  due to instantaneous distribution of electrons and since the molecules are very close to each other instantaneous dipole moment   in one molecule induces instantaneous dipole moment   in the other molecule and thus there exists an attractive force between the non polar molecules of n –pentane  known as Dispersion Forces

Lyrx [107]3 years ago
5 0
It wouldn’t be hydrogen bonding because hydrogen bonding takes place with highly electronegative elements like N,O & F being the most electronegative. It’s not ion - dipole because there is no ion present. So I’m sure it is dispersion
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5 0
3 years ago
What is the temperature of 0.460 mol of gas at a pressure of 1.2 atm and a volume of 12.5 L ? Express your answer using two sign
Taya2010 [7]

The temperature is 370K.

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5 0
2 years ago
A gas sample occupies 3.25 liters at 297.5K and 2.4 atm. Determine the temperature at which the gas will occupy 4.25 L at 1.50 a
lorasvet [3.4K]

For equal moles of  gas, temperature can be calculated from ideal gas equation as follows:

P×V=n×R×T ...... (1)

Initial volume, temperature and pressure of gas is 3.25 L, 297.5 K and 2.4 atm respectively.

2.4 atm ×3.25 L=n×R×297.5 K

Rearranging,

n\times R=0.0262 atm L/K

Similarly at final pressure and volume from equation (1),

1.5 atm ×4.25 L=n×R×T

Putting the value of n×R in above equation,

1.5 atm ×4.25 L=0.0262 (atm L/K)×T

Thus, T=243.32 K


7 0
3 years ago
SOMEONE PLEASE HELPPP​
xxTIMURxx [149]
A is the correct awnser Beacuse it ether right kne
5 0
3 years ago
The rate constant k for a certain reaction is measured at two different temperatures:
bogdanovich [222]

Answer:

Ea=5.5 Kcal/mole

Explanation:

Let rate constant are K_1  and K_2  at temperature T_1  and T_2

By using Arrhenius equation at two different two different temperature,

Log K_1/K_2 =E_a/2.303R*(1/T_2 -1/T_2 );T_1=273+376=649K  ;K_1=4.8*10^8;T_2=273+280=553K  ;K_2=2.3*10^8;R=2 cal/(mole.K);Log (4.8*10^8)/(2.3*10^8 )=E_a/2.303R*(1/553K-1/649); Log 4.8/2.3=E_a/2.303R*96/358897 ;0.32=E_a/2.303R*96/358897;E_a=(0.32*2.303R*358897)/96;  

By putting value of R=2 cal/mole.K

E_a=5510.265cal/mole;

By rounding off upto 2 significant figure;

E_a=5.5Kcal/mole;

8 0
4 years ago
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