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ziro4ka [17]
3 years ago
12

In the figure below, two nonpolar molecules are interacting.Which interaction would most likely cause these molecules to repel e

ach other? hydrogen bonding dipole-dipole interaction dipole-induced dipole interaction induced dipole-induced dipole interaction
Chemistry
2 answers:
Fed [463]3 years ago
6 0
Two non-polar molecules are most likely to interact by
induced dipole-induced dipole interaction.

Non-polar substances do not have a permanently established charge distribution due to similar electron affinities of the atoms that are present. Moreover, due to the absence of a polar hydrogen, they cannot exhibit hydrogen bonding. They interact with one another by induced dipole-induced dipole interactions which arise from the molecules of the substances coming into close vicinity of one another.
Eva8 [605]3 years ago
3 0

The answer is,

<u>D. induced dipole-induced dipole interaction</u>

(:

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To test the accuracy of a hypothesis or any findings that have already been made  

Explanation:

by performing experiments scientists get to test out any questions they have

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Identify the group number of following oxides/hydrides in Mendeleev
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Answer:

See explanation

Explanation:

The oxides or hydrides are formed by exchange of valency between the two atoms involved. The group of the atom bonded to oxygen or hydrogen in the binary compound can be deduced by considering the subscript attached to the oxygen or hydrogen atom.

Now let us take the journey;

R2O3- refers to an oxide of a group 13 element, eg Al2O3

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3 years ago
How do you figure out sub shells please i need help
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What would happen to a sealed bag of chips left in the sun?
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A bomb calorimeter has a heat capacity of 675 J/°C and contains 925 g of water. If the combustion of 0.500 mole of a hydrocarbon
ikadub [295]

<u>Answer:</u> The enthalpy of the reaction is 269.4 kJ/mol

<u>Explanation:</u>

To calculate the heat absorbed by the calorimeter, we use the equation:

q_1=c\Delta T

where,

q = heat absorbed

c = heat capacity of calorimeter = 675 J/°C

\Delta T = change in temperature = T_2-T_1=(53.88-24.26)^oC=29.62^oC

Putting values in above equation, we get:

q_1=675J/^oC\times 29.62^oC=19993.5J

To calculate the heat absorbed by water, we use the equation:

q_2=mc\Delta T

where,

q = heat absorbed

m = mass of water = 925 g

c = heat capacity of water = 4.186 J/g°C

\Delta T = change in temperature = T_2-T_1=(53.88-24.26)^oC=29.62^oC

Putting values in above equation, we get:

q_2=925g\times 4.186J/g^oC\times 29.62^oC=114690.12J

Total heat absorbed = q_1+q_2

Total heat absorbed = [19993.5+114690.12]J=134683.62J=134.7kJ

To calculate the enthalpy change of the reaction, we use the equation:

\Delta H_{rxn}=\frac{q}{n}

where,

q = amount of heat absorbed = 134.7 kJ

n = number of moles of hydrocarbon = 0.500 moles

\Delta H_{rxn} = enthalpy change of the reaction

Putting values in above equation, we get:

\Delta H_{rxn}=\frac{134.7kJ}{0.500mol}=269.4kJ/mol

Hence, the enthalpy of the reaction is 269.4 kJ/mol

6 0
3 years ago
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