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fiasKO [112]
3 years ago
13

Explain the difference in the boiling point of 2-methylpropane and 2-iodo-2-methylpropane in terms of both molecular polarity an

d imf
Chemistry
1 answer:
Len [333]3 years ago
6 0
The Boiling Point of 2-methylpropane is approximately -11.7 °C, while, Boiling Point of <span>2-iodo-2-methylpropane is approximately 100 </span>°C.

As both compounds are Non-polar in nature, So there will be no dipole-dipole interactions between the molecules of said compounds.

The Interactions found in these compounds are London Dispersion Forces.

And among several factors at which London Dispersion Forces depends, one is the size of molecule.

Size of Molecule:
                          There is direct relation between size of molecule and London Dispersion forces. So, 2-iodo-2-methylpropane containing large atom (i.e. Iodine) experience greater interactions. So, due to greater interactions 2-iodo-2-methylpropane need more energy to separate from its partner molecules, Hence, high temperature is required to boil them.
You might be interested in
a 108ml sampl of gas has a mass of 77.96mg at a pressure of 1140mmhg and temperature of 183°c .what is the molar mass of the gas
madam [21]

Answer:

146.85 g/mol

Explanation:

PV=nRT

n=mass/molar mass

covert from mmhg to atm = 0.184 atm

convert from ml to L= 0.108 L

convert from degree C to K= 456.15 K

convert from mg to g= 0.07796g

then rearrange the formula:

n=PV/RT

=(0.184)(0.108)/(0.08206)(456.15)

n= 5.308*10^(-4)

rearrange the n formula interms of molar mass:

Molar mass= mass/n

=0.07796/(5.308*10^-4)

molar mass= 146.85g/mol

3 0
2 years ago
Read 2 more answers
Pls HELPPP
Yuri [45]

maybe c

Explanation:

I'm not sure but maybe

8 0
2 years ago
Read 2 more answers
Dissolution of KOH, ΔHsoln:
swat32

Using Hess's law we found:

1) By <em>adding </em>reaction 10.2 with the <em>reverse </em>of reaction 10.1 we get reaction 10.3:

KOH(aq) + HCl(aq)  → H₂O(l) + KCl(aq)   ΔH  (10.3)

2) The ΔHsoln must be subtracted from ΔHneut to get the <em>total </em>change in enthalpy (ΔH).    

The reactions of dissolution (10.1) and neutralization (10.2) are:

KOH(s) → KOH(aq)   ΔHsoln    (10.1)

KOH(s) + HCl(aq) → H₂O(l) + KCl(aq)     ΔHneut     (10.2)

1) According to Hess's law, the total change in enthalpy of a reaction resulting from <u>differents changes</u> in various <em>reactions </em>can be calculated as the <u>sum</u> of all the <em>enthalpies</em> of all those <em>reactions</em>.      

Hence, to get reaction 10.3:

KOH(aq) + HCl(aq) → H₂O(l) + KCl(aq)    (10.3)

We need to <em>add </em>reaction 10.2 to the <u>reverse</u> of reaction 10.1

KOH(s) + HCl(aq) + KOH(aq) → H₂O(l) + KCl(aq) + KOH(s)

<u>Canceling</u> the KOH(s) from both sides, we get <em>reaction 10.3</em>:

KOH(aq) + HCl(aq)  → H₂O(l) + KCl(aq)    (10.3)

2) The change in enthalpy for <em>reaction 10.3</em> can be calculated as the sum of the enthalpies ΔHsoln and ΔHneut:

\Delta H = \Delta H_{soln} + \Delta H_{neut}

The enthalpy of <em>reaction 10.1 </em>(ΔHsoln) changed its sign when we reversed reaction 10.1, so:

\Delta H = \Delta H_{neut} - \Delta H_{soln}

Therefore, the ΔHsoln must be <u>subtracted</u> from ΔHneut to get the total change in enthalpy ΔH.

Learn more here:

  • brainly.com/question/2082986?referrer=searchResults
  • brainly.com/question/1657608?referrer=searchResults  

I hope it helps you!

6 0
2 years ago
1. Based on the observed performance of the air bag models and the amounts of sodium bicarbonate and acetic acid (vinegar) neede
Ira Lisetskai [31]

Sodium bicarbonate and acetic acid are not good substitute for sodium azide in airbags since the require more mass and produce less gas.

<h3>Which is the better chemical for an airbag?</h3>

The chemical equation for the production of nitrogen gas from sodium azide is given below:

  • NaN₃ → 2 Na + 3 N₂

1 mole or 66 go of sodium azide produces 3 moles or 67.2 L of nitrogen gas.

The equation for the production of carbon dioxide from sodium bicarbonate and acetic acid is given below:

  • Na₂CO₃ + CH₃COOH → CH₃COONa + CO₂ + H₂O

1 mole, 106 g of Na₂CO₃ and 1 mole, 82 g of CH₃COOH are required to produce 1 mole or 22.4 L of CO₂.

The mass of sodium azide required is less than that of sodium bicarbonate and acetic acid required. Also, sodium azide produces a greater volume of gas. Therefore, sodium bicarbonate and acetic acid are not good substitute for sodium azide in airbags.

In conclusion, sodium azide is a better choice in airbags.

Learn more about airbags at: brainly.com/question/14954949

#SPJ1

7 0
2 years ago
38.2mm converted into cm
Alex
1 mm ---------- 0.1 cm
38.2 mm ------ ?

38.2 x 0.1 / 1

= 3.82 cm
8 0
3 years ago
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