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3 years ago

For molecules of comparable mass, why are dipole-dipole forces stronger than dispersion forces?

1 answer:

7 0

Dispersion forces, otherwise known as induced dipole-induced dipole interactions, polarize a molecule slightly into weak negative and weak positive charges. Then coulombs law dictates their electrostatic attractions. However, in dipole-dipole, these molecules are already polarized permanently and typically of a much larger magnitude than an induced dipole for a molecule of the same molar mass. This means that the electrostatic attractions are significantly stronger than the London Dispersion force's.

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Which process would require the least amount of energy?

Aleksandr [31]

Answer: a) the process for the first ionization energy

Every time you take a electron you're requiring more and more energy. Expelling the first one will require less energy than expelling the second and the second will require less than the third, and so on.
When you take the first one, the atom becomes positive and with that the negative forces of the electron will be more attracted to the positive charge. The more electrons that are lost, the more positive this ion will become, causing it to be more difficult to separate the electrons from the atom.

8 0

3 years ago

identify 2 changes of state that require energy to be used, and 2 changes of state that require the release of energy

solniwko [45]

Answer:

used=Evaporation and sublimation

released=condensation and freezing

Explanation:

8 0

3 years ago

Can someone help me write chemical equations of what went wrong in the Chernobyl nucleur accident? Eg. graphite in the rods spee

zimovet [89]

Answer:

It might be e=FCA

Explanation:

4 0

4 years ago

An automobile tire was inflated to a pressure of 24 lb in-2 (1.00 atm = 14.7 lb in-2 ) on a winter’s day when the temperature wa

const2013 [10]

Explanation:

Initial Pressure = 24 lb in-2

Initial Temperature = –5 o C = 268 K (Converting to kelvin temperature)

Final Pressure = ?

Final Temperature = 35 o C = 308 K (Converting to kelvin temperature)

No Change in Volume.

From Gay Lusaac's law; pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature.

P1T1 = P2T2

P2 = P1T1 / T2

P2 = 24 * 268 / 308 = 20.88 lb in-2

There would be a drop in pressure as the temperature increases. Appropriate measures should b taken by regularly gauging the pressure of the tire.

6 0

3 years ago

How many moles of sodium carbonate are contained by 57.3g of sodium carbonate

Lady_Fox [76]

Answer:

$\boxed {\boxed {\sf 0.541 \ mol \ Na_2CO_3}}$

Explanation:

We are asked to find how many moles of sodium carbonate are in 57.3 grams of the substance.

Carbonate is CO₃ and has an oxidation number of -2. Sodium is Na and has an oxidation number of +1. There must be 2 moles of sodium so the charge of the sodium balances the charge of the carbonate. The formula is Na₂CO₃.

We will convert grams to moles using the molar mass or the mass of 1 mole of a substance. They are found on the Periodic Table as the atomic masses, but the units are grams per mole instead of atomic mass units. Look up the molar masses of the individual elements.

Na: 22.9897693 g/mol
C: 12.011 g/mol
O: 15.999 g/mol

Remember the formula contains subscripts. There are multiple moles of some elements in 1 mole of the compound. We multiply the element's molar mass by the subscript after it, then add everything together.

Na₂ = 22.9897693 * 2= 45.9795386 g/mol
O₃ = 15.999 * 3= 47.997 g/mol
Na₂CO₃= 45.9795386 + 12.011 + 47.997 =105.9875386 g/mol

We will convert using dimensional analysis. Set up a ratio using the molar mass.

$\frac {105.9875386 \ g \ Na_2CO_3}{1 \ mol \ Na_2CO_3}$

We are converting 57.3 grams to moles, so we multiply by this value.

$57.3 \ g \ Na_2CO_3} *\frac {105.9875386 \ g \ Na_2CO_3}{1 \ mol \ Na_2CO_3}$

Flip the ratio so the units of grams of sodium carbonate cancel.

$57.3 \ g \ Na_2CO_3} *\frac {1 \ mol \ Na_2CO_3}{105.9875386 \ g \ Na_2CO_3}$

$57.3 } *\frac {1 \ mol \ Na_2CO_3}{105.9875386 }$

$\frac {57.3 }{105.9875386 } \ mol \ Na_2CO_3$

$0.5406295944 \ mol \ Na_2CO_3$

The original measurement of moles has 3 significant figures, so our answer must have the same. For the number we found that is the thousandth place. The 6 in the ten-thousandth place to the right tells us to round the 0 up to a 1.

$0.541 \ mol \ Na_2CO_3$

There are approximately 0.541 moles of sodium carbonate in 57.3 grams.

6 0

3 years ago