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11111nata11111 [884]
4 years ago
9

When is a covalent bond described as polar? Choose one: when electrons are transferred from one atom to another if covalently bo

nded atoms are electrically charged if electrons are shared unequally between bonded atoms when the bonded atoms are of dif
Chemistry
1 answer:
melomori [17]4 years ago
6 0

Answer:

if electrons are shared unequally between bonded atoms

Explanation:

A polar covalent bond is a bond that is formed due to the unequal distribution of electrons between two partially charged atoms. This is observed when the difference in electronegativity between the bond atoms is between 0.5 and 1.7.

A polar bond is a covalent bond between two atoms where the electrons that form the bond are unevenly distributed. This causes the molecule to have a slight electric dipole moment where one end is slightly positive and the other is slightly negative.

The charge of the electric dipoles is less than a full unit charge, so they are considered partial charges and are called delta plus (δ +) and delta minus (δ-).

Because positive and negative charges are separated at the bond, molecules with polar covalent bonds interact with the dipoles of other molecules. This produces intermolecular dipole-dipole forces between the molecules.

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An electrical current only moves through a ____ circuit.
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Answer:

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The heat of combustion (∆H) for an unknown hydrocarbon is -8.21 kJ/mol. If 0.424 mol of the hydrocarbon is burned in a bomb calo
klio [65]

When 0.424 moles of an unknown hydrocarbon (∆Hc = -8.21 kJ/mol) is burned in a bomb calorimeter (C = 1.12 kJ/°C), the change in the temperature is 3.10 °C.

The heat of combustion (∆Hc) for an unknown hydrocarbon is -8.21 kJ/mol. The heat released by the combustion of 0.424 moles of the hydrocarbon is:

Qc = 0.424 mol \times \frac{(-8.21kJ)}{mol} = -3.48 kJ

According to the law of conservation of energy, the sum of the heat released by the combustion (Qc) and the heat absorbed by the bomb calorimeter (Qb) is zero.

Qc + Qb = 0\\\\Qb = -Qc = 3.48 kJ

Given the heat absorbed by the bomb calorimeter (Qb) and the heat capacity of the bomb calorimeter (C), we can calculate the temperature change (ΔT) using the following expression.

Qb = C \times \Delta T\\\\\Delta T = \frac{Qb}{C} = \frac{3.48 kJ}{1.12 kJ/\° C } = 3.10 \° C

When 0.424 moles of an unknown hydrocarbon (∆Hc = -8.21 kJ/mol) is burned in a bomb calorimeter (C = 1.12 kJ/°C), the change in the temperature is 3.10 °C.

Learn more: brainly.com/question/24245395

8 0
3 years ago
 
Katyanochek1 [597]

Answer:

The answer to your question is     V2 = 4.97 l

Explanation:

Data

Volume 1 = V1 = 4.40 L                    Volume 2 =

Temperature 1 = T1 = 19°C               Temperature 2 = T2 = 37°C

Pressure 1 = P1 = 783 mmHg           Pressure 2 = 735 mmHg

Process

1.- Convert temperature to °K

T1 = 19 + 273 = 292°K

T2 = 37 + 273 = 310°K

2.- Use the combined gas law to solve this problem

                  P1V1/T1  = P2V2/T2

-Solve for V2

                  V2 = P1V1T2 / T1P2

-Substitution

                  V2 = (783 x 4.40 x 310) / (292 x 735)

-Simplification

                 V2 = 1068012 / 214620

-Result

                 V2 = 4.97 l

6 0
4 years ago
Sodium (Na) and potassium (K) are in the same group in the periodic table. Sodium is in the 11th position. How many valence elec
Dafna1 [17]

potassium is just below the sodium in periodic table in s group !

so the valence electron of sodium and potassium is same and that is 1

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