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fredd [130]
3 years ago
15

Molecules that possess only London dispersion forces generally have higher melting points than those with hydrogen bonding. True

or False?
Chemistry
2 answers:
Alex73 [517]3 years ago
7 0
It is True not false ok
Drupady [299]3 years ago
7 0

Answer:

The correct option is False

Explanation:

London dispersion forces are a part of Van Der Waal's forces acting between atoms and molecules. They are a temporary attractive force that are formed when two atoms close to each other form a temporary dipole and are hence the weakest intermolecular force of attraction. They are usually the force of attraction found between non-polar molecules.

While a hydrogen bonding a strong intermolecular force (dipole-dipole) of attraction formed as a result of the interaction between an electronegative atom (like oxygen or chlorine) and a hydrogen atom bonded to another electronegative atom (like fluorine). Since, molecules that posses hydrogen bonding have stronger intermolecular forces, they thus possess higher melting point than molecules that possess only London dispersion forces.

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A student dissolves 10.7 g of lithium chloride (LiCl) in 300. g of water in a well-insulated open cup. He then observes the temp
Novosadov [1.4K]

Answer:

1) Exothermic.

2) Q_{rxn}=-8580J

3) \Delta _rH=-121.0kJ/mol

Explanation:

Hello there!

1) In this case, for these calorimetry problems, we can realize that since the temperature increases the reaction is exothermic because it is releasing heat to solution, that is why the temperature goes from 22.0 °C to 28.6 °C.

2) Now, for the total heat released by the reaction, we first need to assume that all of it is absorbed by the solution since it is possible to assume that the calorimeter is perfectly isolated. In such a way, it is also valid to assume that the specific heat of the solution is 4.184 J/(g°C) as it is mostly water, therefore, the heat released by the reaction is:

Q_{rxn}=-m_{Total}C(T_2-T_1)\\\\Q_{rxn}=-(300g+10.7g)*4.184 \frac{J}{g\°C} (28.6\°C-22.0\°C)\\\\Q_{rxn}=-8580J

3) Finally, since the enthalpy of reaction is calculated by dividing the heat released by the reaction over the moles of the solute, in this case LiCl, we proceed as follows:

\Delta _rH=\frac{Q_{rxn}}{n_{LiCl}} \\\\\Delta _rH=\frac{-8580J}{10.7g*\frac{1mol}{150.91g} }*\frac{1kJ}{1000J}  \\\\\Delta _rH=-121.0kJ/mol

Best regards!

8 0
3 years ago
What are the hypotheses on which Dalton's atomic theory is based?
worty [1.4K]

Answer:

 Dalton's theory are based on the two laws  that are: Law of conservation of mass and law of constant composition. This theory basically described their properties of atoms.

This theory state that all the atoms are made up of matter which are invisible and in the elements all the atoms are identical in mass and properties.

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3 years ago
in a titration 20.0 mL of 0.150 M NaOH(aq) Solution exactly neutralize is 24.0 mL of an HCl solution what is the concentration o
dusya [7]

Answer:

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Explanation:

i took a test and got it right choosing this answer hopes this help

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3 years ago
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3 years ago
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The naturally occurring radioactive decay series that begins with 23592U stops with formation of the stable 20782Pb nucleus. The
dsp73

Answer: There are 7 alpha-particle emissions and 4 beta-particle emissions involved in this series

Explanation:

Alpha Decay: In this process, a heavier nuclei decays into lighter nuclei by releasing alpha particle. The mass number is reduced by 4 units and atomic number is reduced by 2 units.

Beta Decay : It is a type of decay process, in which a proton gets converted to neutron and an electron. This is also known as -decay. In this the mass number remains same but the atomic number is increased by 1.

In radioactive decay the sum of atomic number or mass number of reactants must be equal to the sum of atomic number or mass number of products .

_{92}^{235}\textrm{U}\rightarrow _{82}^{207}\textrm{Pb}+X_2^4\alpha+Y_{-1}^0e

Thus for mass number : 235 = 207+4X

4X= 28

X = 7

Thus for atomic number : 92 = 82+2X-Y

2X- Y = 10

2(7) - Y= 10

14-10 = Y

Y= 4

_{92}^{235}\textrm{U}\rightarrow _{82}^{207}\textrm{Pb}+7_2^4\alpha+4_{-1}^0e

Thus there are 7 alpha-particle emissions and 4 beta-particle emissions involved in this series

3 0
3 years ago
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