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2 years ago

Which of the following has the highest induced dipole interactions in its molecule?

Chemistry

1 answer:

irinina [24]2 years ago

6 0

Answer:

Option A

Explanation:

In Dipole -induced dipole forces/interactions a polar and a non polar molecule group together to form a bond.

The higher is the electronegativity, the higher is the Dipole-Induce dipole.

As electronegativity defines the ability of an atom to attract a pair of electron in any chemical bond.

The smaller the atom, the higher is the electronegativity and hence the higher is the Dipole -induced dipole forces/interactions

Option A

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The force that one massive object exerts to attract another object is called .

Ghella [55]

Answer:

Gravity

Explanation:

The force that one massive object exerts to attract another object is called gravity.

4 0

2 years ago

Read 2 more answers

What is the symbol for Lithium, Iron, and Helium

jonny [76]

Lithium=li

Iron=Fe

Helium=He

I don't say u must have to mark my ans as brainliest but if u think it has really helped u plz don't forget to thank me....

8 0

3 years ago

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A closed, frictionless piston-cylinder contains a gas mixture with the following composition on a mass basis: 40% carbon dioxide

Hitman42 [59]

Answer:

W=-37.6kJ, therefore, work is done on the system.

Explanation:

Hello,

In this case, the first step is to compute the moles of each gas present in the given mixture, by using the total mixture weight the mass compositions and their molar masses:

$n_{CO_2}=0.8kg*0.4*\frac{1kmolCO_2}{44kgCO_2}= 0.00727kmolCO_2\\\\n_{O_2}=0.8kg*0.25*\frac{1kmolO_2}{32kgO_2}=0.00625kmolO_2\\ \\n_{Ne}=0.8kg*0.35*\frac{1kmolNe}{20.2kgNe}=0.0139kmolNe$

Next, the total moles:

$n_T=0.00727kmol+0.00625kmol+0.0139kmol=0.02742kmol$

After that, since the process is isobaric, we can compute the work as:

$W=P(V_2-V_1)$

Therefore, we need to compute both the initial and final volumes which are at 260 °C and 95 °C respectively for the same moles and pressure (isobaric closed system)

$V_1=\frac{n_TRT_1}{P}= \frac{0.02742kmol*8.314\frac{kPa*m^3}{kmol\times K}*(260+273)K}{450kPa}=0.27m^3\\ \\V_2=\frac{n_TRT_2}{P}= \frac{0.02742kmol*8.314\frac{kPa*m^3}{kmol\times K}*(95+273)K}{450kPa}=0.19m^3$

Thereby, the magnitude and direction of work turn out:

$W=450kPa(0.19m^3-0.27m^3)\\\\W=-37.6kJ$

Thus, we conclude that since it is negative, work is done on the system (first law of thermodynamics).

Regards.

7 0

3 years ago

A 453 g piece of glass at 25.7∘C is left outside on a sunny day. How much heat must the glass absorb from the sun in order to re

Tju [1.3M]

Answer:

Q = 5555.6J

Explanation:

Mass of glass piece, m = 453g

initial temperature = 25.7°C

temperature to be attained = 40.3°C

⇒change in temperature, Δt = 40.3 - 25.7 = 14.6°C

specific heat of glass, s = 0.840J/g°C

Heat absorbed, Q = msΔt

⇒Q = 453×0.840×14.6 = 5555.592J

⇒<u>Q = 5555.6J</u> (rounded to the nearest tenth)

7 0

3 years ago

The specific heat of aluminum is 0.897j/(g•°C). If a 22.6 g sample of aluminum is heated from 25°C to 250°C, then how much heat

Oxana [17]

Q=mcat

=(22.6)(.897)(250-25)

4,561.245 or 4.5 * 10^-3

3 0

3 years ago