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3 years ago

5

Hydrogen bonds form between neighboring water molecules because of:

1 answer:

12345 [234]3 years ago

4 0

*A & B*

Answers A & B are not possible, as Hydrogen “bonds” are intermolecular forces and do not actually involve transfer or sharing of electrons.

*C & D*

Viscosity and surface tension are not the answer as they are not specific enough to the question.

*E*

Polarity of water molecules is the correct answer, as water molecules are highly polar. The partial positive of the Hydrogen on one water molecule is highly attracted to the partial negative of the Oxygen (due to its lone pairs) on another water molecule.

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4 a joule is the answer

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Which term describes this molecular shape?

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Given the formulas for the ionic compounds, draw the correct ratio of cations to anions (a) BaSO4, (b) CaF2, (c) Mg3N2, (d) K2O.

Karolina [17]

Answer:

A. It formed by barium(Ba+2) ion and sulfate ( SO42- )

B. It is formed by calcium ion (Ca+2) and two fluoride ions (2F-)

C. It is formed by magnesium ion (Mg+2) and nitride ion (N3-)

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3 years ago

If 45.0 mL of ethanol (density=0.789 g/mL) initially at 9.0 C is mixed with 45.0 mL of water (density=1.0 g/mL) initially at 28.

Klio2033 [76]

Answer : The final temperature of the mixture is $22.7^oC$

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

And as we know that,

Mass = Density × Volume

Thus, the formula becomes,

$(\rho_1\times V_1)\times c_1\times (T_f-T_1)=-(\rho_2\times V_2)\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of ethanol = $2.3J/g^oC$

$c_2$ = specific heat of water = $4.18J/g^oC$

$m_1$ = mass of ethanol

$m_2$ = mass of water

$\rho_1$ = density of ethanol = 0.789 g/mL

$\rho_2$ = density of water = 1.0 g/mL

$V_1$ = volume of ethanol = 45.0 mL

$V_2$ = volume of water = 45.0 mL

$T_f$ = final temperature of mixture = ?

$T_1$ = initial temperature of ethanol = $9.0^oC$

$T_2$ = initial temperature of water = $28.6^oC$

Now put all the given values in the above formula, we get

$(0.789g/mL\times 45.0mL)\times (2.3J/g^oC)\times (T_f-9.0)^oC=-(1.0g/mL\times 45.0mL)\times 4.18J/g^oC\times (T_f-28.6)^oC$

$T_f=22.7^oC$

Therefore, the final temperature of the mixture is $22.7^oC$

4 0

3 years ago

PLZ HELP PLZ PLZ ILL MARK AS BRAINLIESTT!!!!

LuckyWell [14K]

Q.1-

Given,

mass - 10grams

volume - 24 cm³

density = mass/volume

density = 10/24

density = 0.416 g/cm³

Q.2-

Given,

mass - 700grams

volume - 1100cm³

density = mass/volume

density = 700/1100

density = 0.6363 g/cm³

5 0

3 years ago