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VLD [36.1K]
3 years ago
15

Which compound in the following pair would you expect to have the greater dipole moment μ? Why? Select the single best answer. N

aF or HF Hydrogen fluoride has the larger dipole moment because sodium has a greater atomic radius than hydrogen. Sodium fluoride has the larger dipole moment because hydrogen has a greater atomic radius than sodium. Hydrogen fluoride has the larger dipole moment, because sodium is more electronegative than hydrogen. Sodium fluoride has the larger dipole moment, because hydrogen is more electronegative than sodium.
Chemistry
1 answer:
Luba_88 [7]3 years ago
7 0

Sodium fluoride has the larger dipole moment, because hydrogen is more electronegative than sodium.

Option D.

8.156

1.86

<h3><u>Explanation:</u></h3>

Dipole moment of a molecule is defined as the product of the distance between two atoms of a molecule and their partial charges. This value of dipole moment of a molecule gives the idea of how polar a molecule will be. The molecules with lower dipole moment are apolar, and the molecules with higher dipole moment are polar.

The dipole moment becomes higher in those molecules where one of the participating atoms are highly electropositive and the other one is highly electronegative. This leads to the electropositive atom donating the electron pull towards the electronegative atom, thereby increasing the partial charge difference.

In case of HF and NaF, the sodium atom is more electropositive than hydrogen atom. This leads to more difference in partial charges between the corresponding atoms, hence more dipole moment.

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This can be emitted during the combustion of fossil fuels and traps radiant energy from the:a Greenhouse Effect
lisov135 [29]

Answer:

D. Greenhouse Gas

Explanation:

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8 0
3 years ago
if i have 4 moles of a gas at a pressure of 5.6 atm and a volume of 12 liters, what is the temperature? (205 K)
Thepotemich [5.8K]

Hello!

If i have three moles of a gas at a pressure of 5.6 atm and a volume of 3 liters, what is the temperature ?

We have the following data:

n (number of moles) = 3 moles

P (pressure) = 5.6 atm

V (volume) = 12 L

T (temperature) = ? (in Kelvin)

R (gas constant) = 0.082 atm.L / mol.K

We apply the data above to the Clapeyron equation (gas equation), let's see:

P*V = n*R*T

5.6*12 = 4*0.082*T

67.2 = 0.328\:T

0.328\:T = 67.2

T = \dfrac{67.2}{0.328}

T = 204.8780488...

\boxed{\boxed{T \approx 205\:K}}\end{array}}\qquad\checkmark

Answer:

The temperature is approximately 205 Kelvin

_______________________________  

I Hope this helps, greetings ... Dexteright02! =)

3 0
3 years ago
What is the organelle that packages and releases materials other parts of the cell?
VARVARA [1.3K]
The vacuoles are the answer
7 0
4 years ago
If all of the energy from burning 281.0 g of propane (ΔHcomb,C3H8 = –2220 kJ/mol) is used to heat water, how many liters of wate
lapo4ka [179]

This problem is providing us with the mass of propane, its enthalpy of combustion, and the initial and final temperature of water that can be heated from the burning of this fuel. At the end, the result turns out to be 42.27 L.

<h3>Combustion:</h3>

In chemistry, combustion reactions are based on the burning of fuels by using oxygen and producing both carbon dioxide and water. For propane, we will have:

C_3H_8+5O_2\rightarrow 3CO_2+4H_2O

Hence, we can calculate the heat released from this reaction by using the mass, which has to be converted to moles, and the given enthalpy of combustion:

Q=281.0g*\frac{1mol}{44.09g}*-2220\frac{kJ}{mol}*\frac{1000J}{1kJ}\\ \\ Q=-1.415x10^7 J

<h3>Calorimetry:</h3>

In chemistry, we can analyze the mass-specific heat-temperature-heat relationship via the most general heat equation:

Q=mC\Delta T

Thus, since Q was obtained from the previous problem, but the sign change because the released heat is now absorbed by the water, one can calculate the mass of water that rises from 20.0°C to 100.0°C with this heat:

m=\frac{Q}{C\Delta T} =\frac{1.415x10^7J}{4.184\frac{J}{g\°C}(100.0\°C-20.0\°C)}\\ \\m=4.227x10^4g

Finally, we convert it to liters as required:

V=4.227 x10^4g*\frac{1mL}{1.00g}*\frac{1L}{1000mL}  \\\\V=42.27L

Learn more about calorimetry: brainly.com/question/1407669

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Answer:1. Energy 2. Medium 3. Original position 4. Waves 5. Matter

Explanation:

3 0
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